tag:blogger.com,1999:blog-72041714905306054162017-06-19T12:17:12.784-07:00ENGENIEERING ECONOMIC ANALYSIS.Edin O'Briennoreply@blogger.comBlogger110125tag:blogger.com,1999:blog-7204171490530605416.post-57194117464558613662015-02-12T10:05:00.000-08:002015-02-12T10:05:02.426-08:00THE BALANCE SHEETThe balance sheet is a snapshot of a company’s financial assets, liabilities, and the value of the company to its owner—often referred to as net worth or equity—at a specific point in time. Balance sheets are commonly prepared at the end of each month and at the end of the fiscal year. A typical balance sheet for a construction company using the percentage-of-completion accounting method is shown in <b>Figure 2-2</b>.<br /><br />The balance sheet is divided into three sections: assets, liabilities, and owner’s equity. The balance sheet reports the values of each of the accounts in the balance sheet portion of the chart of accounts at the time the balance sheet is printed. For example, the amount reported as cash on the balance sheet in Figure 2-2 comes from account number 110 from the chart of accounts shown in <b>Figure 2-1</b>. To prevent the balance sheet from becoming too complicated multiple accounts may be summarized by combining two or more consecutive accounts into a single line on the balance sheet. Other items on the balance sheet may be calculated from other lines on the balance sheet. For example, the Total Current Assets is the sum of the Cash, Accounts Receivable-Trade, Accounts Receivable-Retention, Costs and Profits in Excess of Billings, Notes Receivable, Prepaid Expenses, and Other Current Assets or accounts 110 through 199 on the chart of accounts in Figure 2-1. Not all companies will use all the accounts shown in Figure 2-1. For example, the construction company in Figure 2-2 does not use the inventory account.<br /><br />On the balance sheet the relationship between assets, liabilities, and equity is as follows:<br /><br />Asset = Liabilities + Equity (2-1)<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-u1fTGQmY0Ec/VNzq_B7PGyI/AAAAAAAAOtM/Zboe4-hJZas/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img alt="Chart of Accounts" border="0" src="http://1.bp.blogspot.com/-u1fTGQmY0Ec/VNzq_B7PGyI/AAAAAAAAOtM/Zboe4-hJZas/s1600/1.gif" title="Chart of Accounts" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-vEAebFiVPj0/VNzq3_TuDeI/AAAAAAAAOtA/yIOCrXfRLVs/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img alt="Balance Sheet for Big W Construction" border="0" src="http://3.bp.blogspot.com/-vEAebFiVPj0/VNzq3_TuDeI/AAAAAAAAOtA/yIOCrXfRLVs/s1600/1.gif" title="Balance Sheet for Big W Construction" /></a></div>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com1tag:blogger.com,1999:blog-7204171490530605416.post-23955246251861332892015-01-28T18:54:00.000-08:002015-01-28T18:54:43.493-08:00COST REPORTING VERSUS COST CONTROLCost reporting is where the accounting system provides management with the accounting data after the opportunity has passed for management to respond to and correct the problems indicated by the data. When companies wait to enter the cost of their purchases until the bills are received, management does not know if they are under or over budget until the bills are entered, at which time the materials purchased have been delivered to the project and may have been consumed. The extreme case of cost reporting is where companies only look at the costs and profit for each project after the project is finished. Cost reporting is typified by the accounting reports showing where a company has been financially without giving management an opportunity to proactively respond to the data.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-VS6xxuPIC_Q/VMmGfrBubuI/AAAAAAAAOiU/s9F-HCpVHaE/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img alt="COST REPORTING VERSUS COST CONTROL" border="0" src="http://1.bp.blogspot.com/-VS6xxuPIC_Q/VMmGfrBubuI/AAAAAAAAOiU/s9F-HCpVHaE/s1600/1.gif" title="COST REPORTING VERSUS COST CONTROL" /></a></div><br />Cost control is where the accounting system provides management with the accounting data in time for management to analyze the data and make corrections in a timely manner. Companies that enter material purchase orders and subcontracts, along with their associated costs, into their accounting system as committed costs before issuing the purchase order or subcontract allow management time to address cost overruns before ordering the materials or work. Committed costs are those costs that the company has committed to pay and can be identified before a bill is received for the costs. For example, when a contractor signs a fixed-price subcontract he or she has committed to pay the subcontractor a fixed price once the work has been completed and, short of any change orders, knows what the work is going to cost. Accounting systems that track committed costs give management time to identify the cause of the overrun early on, identify possible solutions, and take corrective action. Cost control is typified by identifying problems early and giving management a chance to proactively address the problem. A lot of money can be saved by addressing pervasive problems—such as excessive waste—early in the project.<br /><br />If a company’s accounting system is going to allow management to control costs rather than just report costs, the accounting system must have the following key components:<br /><br /><b>First,</b> the accounting system must have a strong job cost and equipment tracking system. The accounting system should update and report costs, including committed costs and estimated cost at completion on a weekly basis. Having timely, up-to-date costs for the project and the equipment is a must if management is going to manage costs and identify problems early. <br /><br /><b>Second,</b> the accounting system must utilize the principle of management by exception. It can be easy for managers to get lost in the volumes of data generated by the accounting system. The accounting system should provide reports that allow management to quickly identify problem areas and address the problems. For example, as soon as bills are entered into the accounting system, management should get a report detailing all bills that exceed the amount of their purchase order or subcontract. Problems that are buried in volumes of accounting data are often never addressed because management seldom has time to pour through all of the data to find the problems or if they are found they are often found too late for management to address the problem. Providing reports that flag transactions that fall outside the acceptable limits is a necessity if management is going to control costs. By having reports that flag items that fall outside acceptable limits, management can make addressing these items a priority.<br /><br /><b>Third,</b> accounting procedures need to be established to ensure that things do not fall through the cracks. These procedures should include things such as who can issue purchase orders and what to do when a bill is received for a purchase order that has not been issued. The procedures should also identify the acceptable limits for different types of transactions. Procedures ensure that the accounting is handled in a<br />consistent manner and give management confidence in the data that it is using to manage the company.<br /><br /><b>Finally,</b> the data must be easily and quickly available to management and other employees who are directly responsible for controlling costs. It does little good to collect cost data for use in controlling costs if the data cannot be accessed. Where possible the reports should be automatically prepared by the accounting software. This eliminates the time and effort needed to prepare the reports manually. Additionally, frontline supervisors who are responsible for control costs should readily have access to their costs. Holding supervisors responsible for costs at the end of a job while not giving them access to their costs throughout the project denies them the opportunity to proactively control costs.<br /><br />The accounting system for many construction companies consists of three different ledgers: the general ledger, the job cost ledger, and the equipment ledger. The general ledger tracks financial data for the entire company and is used to prepare the company’s financial statements and income taxes. The job cost ledger is used to track the financial data for each of the construction projects. The equipment ledger is used to track financial data for heavy equipment and vehicles. All construction companies should have a general ledger and a job cost ledger. Companies with lots of heavy equipment or vehicles should have an equipment ledger.Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-38093213900703499982015-01-22T20:24:00.000-08:002015-01-22T20:24:08.465-08:00Managing Cash FlowsFinancial managers are responsible for managing the cash flows for the company. Many profitable companies fail because they simply run out of cash and are unable to pay their bills. The duties of a financial manager include the following:<br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-45NUhEZxHNg/VMHMN5r-WXI/AAAAAAAAOc0/Iln3h5b7eD4/s1600/Fotolia_12268633_XS-179x300.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img alt="Managing Cash Flows" border="0" src="http://1.bp.blogspot.com/-45NUhEZxHNg/VMHMN5r-WXI/AAAAAAAAOc0/Iln3h5b7eD4/s1600/Fotolia_12268633_XS-179x300.jpg" title="Managing Cash Flows" /></a></div><br /><span class="sign">■</span> Matching the use of in-house labor and subcontractors to the cash available for use on a project.<br /><span class="sign">■</span> Ensuring that the company has sufficient cash to take on an additional project.<br /><span class="sign">■</span> Preparing an income tax projection for the company.<br /><span class="sign">■</span> Preparing and updating annual cash flow projections for the company.<br /><span class="sign">■</span> Arranging for financing to cover the needs of the construction company.Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-8655306703193895202015-01-22T20:14:00.001-08:002015-01-22T20:14:22.680-08:00Managing Costs and ProfitsFinancial managers are responsible for managing the company’s costs and earning a profit for the company’s owners. Financial managers rely heavily on the reports from the accounting system in their management of costs. Managing the company’s costs and profits includes the following duties:<br /><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-G9tUXBMucU4/VMHKkbLVZ1I/AAAAAAAAOcs/W1OJXlvXMps/s1600/analisis-costo-beneficio.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img alt="Managing Costs and Profits" border="0" src="http://4.bp.blogspot.com/-G9tUXBMucU4/VMHKkbLVZ1I/AAAAAAAAOcs/W1OJXlvXMps/s1600/analisis-costo-beneficio.jpg" title="Managing Costs and Profits" /></a></div><br /><span class="sign">■</span> Controlling project costs.<br /><span class="sign">■</span> Monitoring project and company profitability.<br /><span class="sign">■</span> Setting labor burden markups.<br /><span class="sign">■</span> Developing and tracking general overhead budgets.<br /><span class="sign">■</span> Setting the minimum profit margin for use in bidding.<br /><span class="sign">■</span> Analyzing the profitability of different parts of the company and making the necessary changes to improve profitability.<br /><span class="sign">■</span> Monitoring the profitability of different customers and making the necessary marketing changes to improve profitability.<br /><br /><br /><br /><br />Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-42203394592902851832014-07-02T09:46:00.002-07:002014-07-02T09:46:43.076-07:00Nominal and Effective Interest Rate StatementsHere we discuss nominal and effective interest rates, which have the same basic relationship. The difference here is that the concepts of nominal and effective must be used when interest is compounded more than once each year. For example, if an interest rate is expressed as 1% per month, the terms nominal and effective interest rates must be considered. <br /><br />To understand and correctly handle effective interest rates is important in engineering practice as well as for individual finances. The interest amounts for loans, mortgages, bonds, and stocks are commonly based upon interest rates compounded more frequently than annually. The engineering economy study must account for these effects. In our own personal ﬁ nances, we manage most cash disbursements and receipts on a nonannual time basis. Again, the effect of compounding more frequently than once per year is present. First, consider a nominal interest rate.<br /><br />A nominal interest rate r is an interest rate that does not account for compounding. By definition,<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-WCCMDNUHv7w/U7Q1ydxeKAI/AAAAAAAAN4A/Y_AsQ2fAAmY/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-WCCMDNUHv7w/U7Q1ydxeKAI/AAAAAAAAN4A/Y_AsQ2fAAmY/s1600/1.gif" /></a></div><br />A nominal rate may be calculated for any time period longer than the time period stated by using Equation [4.1]. For example, the interest rate of 1.5% per month is the same as each of the following nominal rates. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-2esWHMXR9fc/U7Q17qtrChI/AAAAAAAAN4I/MEeQC1wUeBU/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-2esWHMXR9fc/U7Q17qtrChI/AAAAAAAAN4I/MEeQC1wUeBU/s1600/1.gif" /></a></div><br />Note that none of these rates mention anything about compounding of interest; they are all of the form “ r % per time period.” These nominal rates are calculated in the same way that simple rates are calculated using Equation [1.7], that is, interest rate times number of periods.<br /><br />After the nominal rate has been calculated, the compounding period (CP) must be in- cluded in the interest rate statement. As an illustration, again consider the nominal rate of 1.5% per month. If we deﬁ ne the CP as 1 month, the nominal rate statement is 18% per year, compounded monthly, or 4.5% per quarter, compounded monthly . Now we can consider an effective interest rate. <br /><br />An effective interest rate i is a rate wherein the compounding of interest is taken into account. Effective rates are commonly expressed on an annual basis as an effective annual rate; however, any time basis may be used.<br /><br />The most common form of interest rate statement when compounding occurs over time periods shorter than 1 year is “% per time period, compounded CP-ly,” for example, 10% per year, compounded monthly, or 12% per year, compounded weekly. An effective rate may not always include the compounding period in the statement. If the CP is not mentioned, it is understood to be the same as the time period mentioned with the interest rate. For example, an interest rate of “1.5% per month” means that interest is compounded each month; that is, CP is 1 month. An equivalent effective rate statement, therefore, is 1.5% per month, compounded monthly.<br /><br />All of the following are effective interest rate statements because either they state they are effective or the compounding period is not mentioned. In the latter case, the CP is the same as the time period of the interest rate.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-UdPbOV6s3Ak/U7Q2RxqxbYI/AAAAAAAAN4Q/w3TTD3AoyH4/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-UdPbOV6s3Ak/U7Q2RxqxbYI/AAAAAAAAN4Q/w3TTD3AoyH4/s1600/1.gif" /></a></div><br />All nominal interest rates can be converted to effective rates. The formula to do this is discussed in the next section. <br /><br />All interest formulas, factors, tabulated values, and spreadsheet functions must use an effective interest rate to properly account for the time value of money. <br /><br />The term APR (Annual Percentage Rate) is often stated as the annual interest rate for credit cards, loans, and house mortgages. This is the same as the nominal rate . An APR of 15% is the same as a nominal 15% per year or a nominal 1.25% on a monthly basis. Also the term APY (Annual Percentage Yield) is a commonly stated annual rate of return for investments, certiﬁcates of deposit, and saving accounts. This is the same as an effective rate . The names are different, but the interpretations are identical. As we will learn in the following sections, the effective rate is always greater than or equal to the nominal rate, and similarly APY APR.<br /><br />Based on these descriptions, there are always three time-based units associated with an interest rate statement. <br /><br /><b>Interest period ( t ) —</b>The period of time over which the interest is expressed. This is the t in <br />the statement of r % per time period t , for example, 1% per month. The time unit of 1 year is by far the most common. It is assumed when not stated otherwise. <br /><br /><b>Compounding period (CP) —</b>The shortest time unit over which interest is charged or earned. This is deﬁ ned by the compounding term in the interest rate statement, for example, 8% per year, compounded monthly. If CP is not stated, it is assumed to be the same as the interest period. <br /><br /><b>Compounding frequency (m) —</b>The number of times that compounding occurs within the interest period t. If the compounding period CP and the time period t are the same, the com-pounding frequency is 1, for example, 1% per month, compounded monthly. <br /><br />Consider the (nominal) rate of 8% per year, compounded monthly. It has an interest period t of 1 year, a compounding period CP of 1 month, and a compounding frequency m of 12 times per year. A rate of 6% per year, compounded weekly, has t = 1 year, CP = 1 week, and m = 52, based on the standard of 52 weeks per year. <br /><br />In previous chapters, all interest rates had t and CP values of 1 year, so the compounding frequency was always m = 1. This made them all effective rates, because the interest period and compounding period were the same. Now, it will be necessary to express a nominal rate as an effective rate on the same time base as the compounding period. <br /><br />An effective rate can be determined from a nominal rate by using the relation<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-f-hmbQwEtOc/U7Q3CqQEWvI/AAAAAAAAN4Y/0FRWBERDmhI/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-f-hmbQwEtOc/U7Q3CqQEWvI/AAAAAAAAN4Y/0FRWBERDmhI/s1600/1.gif" /></a></div><br />As an illustration, assume r = 9% per year, compounded monthly; then m = 12. Equation [4.2] is used to obtain the effective rate of 9% / 12 = 0.75% per month, compounded monthly. <br /><br />Note that changing the interest period t does not alter the compounding period, which is 1 month in this illustration. Therefore, r = 9% per year, compounded monthly, and r = 4.5% per 6 months, compounded monthly, are two expression of the same interest rate. Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-54440948039292344352014-06-17T09:16:00.002-07:002014-06-17T09:16:31.498-07:00Calculations For Shifted GradientsPreviously, we derived the relation P/G ( P/G , i , n ) to determine the present worth of the arithmetic gradient series. The P/G factor, Equation [2.25], was derived for a present worth in year 0 with the gradient first appearing in year 2.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-iaykETG1GSI/U6BnBFD6mhI/AAAAAAAANyA/iEjoGe1wBsM/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-iaykETG1GSI/U6BnBFD6mhI/AAAAAAAANyA/iEjoGe1wBsM/s1600/1.gif" /></a></div><br />The present worth of an arithmetic gradient will always be located two periods before the gradient starts. <br /><br />Refer to<b> Figure 2–14</b> as a refresher for the cash flow diagrams.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-VIP5zQzv-js/U6Bmol1229I/AAAAAAAANx4/jzylZWf22Zs/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-VIP5zQzv-js/U6Bmol1229I/AAAAAAAANx4/jzylZWf22Zs/s1600/1.gif" /></a></div><br />The relation A/G ( A/G , i , n ) was also derived previously. The A/G factor in Equation [2.27] performs the equivalence transformation of a gradient only into an A series from years 1 through n (Figure 2–15). Recall that the base amount must be treated separately. Then the equivalent P or A values can be summed to obtain the equivalent total present worth P<span style="font-size: xx-small;">T</span> and total annual series A/T.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-mogBOp6y1M8/U6BnK6QCx_I/AAAAAAAANyI/i1OKcbxqels/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-mogBOp6y1M8/U6BnK6QCx_I/AAAAAAAANyI/i1OKcbxqels/s1600/1.gif" /></a></div><br />A conventional gradient series starts between periods 1 and 2 of the cash flow sequence. A gradient starting at any other time is called a shifted gradient . The n value in the P/G and A/G factors for a shifted gradient is determined by renumbering the time scale. The period in which the gradient first appears is labeled period 2. The n value for the gradient factor is determined by the renumbered period where the last gradient increase occurs. <br /><br />Partitioning a cash flow series into the arithmetic gradient series and the remainder of the cash flows can make very clear what the gradient n value should be.<br /><br />It is important to note that the A/G factor cannot be used to find an equivalent A value in periods 1 through n for cash flows involving a shifted gradient. Consider the cash flow diagram of <b>Figure 3–11</b>. To find the equivalent annual series in years 1 through 10 for the gradient series only, first find the present worth P<span style="font-size: xx-small;">G</span> of the gradient in actual year 5, take this present worth back to year 0, and annualize the present worth for 10 years with the A/P factor. If you apply the annual series gradient factor ( A/G , i ,5) directly, the gradient is converted into an equivalent annual series over years 6 through 10 only, not years 1 through 10, as requested. <br /><br />To find the equivalent A series of a shifted gradient through all the n periods, first find the present worth of the gradient at actual time 0, then apply the (A /P, i, n) factor<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-qxA5BbFVDQk/U6Bnzur7oII/AAAAAAAANyU/8oXQ2EoTsDY/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-qxA5BbFVDQk/U6Bnzur7oII/AAAAAAAANyU/8oXQ2EoTsDY/s1600/1.gif" /></a></div>Previously, we derived the relation Pg = A<span style="font-size: xx-small;">1</span> ( P /A,g,i,n ) to determine the present worth of a geometric gradient series, including the initial amount A<span style="font-size: xx-small;">1</span>. The factor was derived to find the present worth in year 0, with A<span style="font-size: xx-small;">1</span> in year 1 and the first gradient appearing in year 2. The present worth of a geometric gradient series will always be located two periods before the gradient stars, and the initial amount is included in the resulting present worth. Refer to<b> Figure 2–21</b> as a refresher for the cash flows.<br /><br />Equation [2.35] is the formula used for the factor. It is not tabulated. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-gRAAdECFwa8/U6Boisfqr9I/AAAAAAAANyg/RnJ-o_w9vus/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-gRAAdECFwa8/U6Boisfqr9I/AAAAAAAANyg/RnJ-o_w9vus/s1600/1.gif" height="276" width="640" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-8f6vx_XR24M/U6BozYWaF9I/AAAAAAAANyo/SX8h0FmNH20/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-8f6vx_XR24M/U6BozYWaF9I/AAAAAAAANyo/SX8h0FmNH20/s1600/1.gif" height="79" width="320" /></a></div><br />Decreasing arithmetic and geometric gradients are common, and they are often shifted gradient series . That is, the constant gradient is – G or the percentage change is – g from one period to the next, and the first appearance of the gradient is at some time period (year) other than year 2 of the series. Equivalence computations for present worth P and annual worth A are basically the same as discussed previously, except for the following. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-FKNbpq-jf-A/U6BpNTTSovI/AAAAAAAANyw/TpGiYoNNpXI/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-FKNbpq-jf-A/U6BpNTTSovI/AAAAAAAANyw/TpGiYoNNpXI/s1600/1.gif" height="548" width="640" /></a></div><br />For shifted, decreasing gradients:<br /><br />• The base amount A (arithmetic) or initial amount A<span style="font-size: xx-small;">1</span> (geometric) is the largest amount in the first year of the series.<br />• The gradient amount is subtracted from the previous year’s amount, not added to it.<br />• The amount used in the factors is –G for arithmetic and –g for geometric gradient series.<br />• The present worth P<span style="font-size: xx-small;">G</span> or P<span style="font-size: xx-small;">g</span> is located 2 years prior to the appearance of the first gradient; however, a P F factor is necessary to find the present worth in year 0<br /><br /><b>Figure 3–15</b> partitions a decreasing arithmetic gradient series with G = $−100 that is shifted 1 year forward in time. P/G occurs in actual year 1, and P<span style="font-size: xx-small;">T</span> is the sum of three components.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-9dcqMid4BXA/U6BppsvfzTI/AAAAAAAANy4/x0IxtST75Z0/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-9dcqMid4BXA/U6BppsvfzTI/AAAAAAAANy4/x0IxtST75Z0/s1600/1.gif" /></a></div>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-31771209932491452522014-05-29T12:50:00.001-07:002014-05-29T12:50:10.929-07:00Calculations Involving Uniform Series and Randomly Placed Single Amounts When a cash flow includes both a uniform series and randomly placed single amounts, the procedures are applied to the uniform series and the single-amount formulas are applied to the one-time cash flows. This approach, illustrated in Examples 3.3, is merely a combination of previous ones. For spreadsheet solutions, it is necessary to enter the net cash flows before using the NPV and other functions.<br /><br /><b>EXAMPLE 3.3</b><br /><br />An engineering company in Wyoming that owns 50 hectares of valuable land has decided to lease the mineral rights to a mining company. The primary objective is to obtain long-term income to finance ongoing projects 6 and 16 years from the present time. The engineering company makes a proposal to the mining company that it pay $20,000 per year for 20 years beginning 1 year from now, plus $10,000 six years from now and $15,000 sixteen years from now. If the mining company wants to pay off its lease immediately, how much should it pay now if the investment is to make 16% per year? <br /><br /><b>Solution</b><br /><br />The cash flow diagram is shown in Figure 3–6 from the owner’s perspective. Find the present worth of the 20-year uniform series and add it to the present worth of the two one-time amounts to determine P<span style="font-size: xx-small;">T</span>,<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-SSIYZjp9aTQ/U4ePBpkwfRI/AAAAAAAANoQ/WDTSGv-eGjw/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-SSIYZjp9aTQ/U4ePBpkwfRI/AAAAAAAANoQ/WDTSGv-eGjw/s1600/1.gif" /></a></div><br /><br />Note that the $20,000 uniform series starts at the end of year 1, so the P/A factor determines the present worth at year 0. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-ztiP8gdIBZ0/U4ePMULnN_I/AAAAAAAANoY/pZ_7TNtDc4s/s1600/2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-ztiP8gdIBZ0/U4ePMULnN_I/AAAAAAAANoY/pZ_7TNtDc4s/s1600/2.gif" /></a></div><br /><br />When you calculate the A value for a cash ﬂ ow series that includes randomly placed single amounts and uniform series, ﬁ rst convert everything to a present worth or a future worth. Then you obtain the A value by multiplying P or F by the appropriate A/P or A/F factor. <br /><br /><br />Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-62564257827364124932014-05-19T12:25:00.003-07:002014-05-19T12:25:52.734-07:00EXAMPLES - Uniform Series That Are Shifted<span style="font-size: large;"><b>EXAMPLE 1</b></span><br /><br />The offshore design group at Bechtel just purchased upgraded CAD software for $5000 now and annual payments of $500 per year for 6 years starting 3 years from now for annual upgrades. What is the present worth in year 0 of the payments if the interest rate is 8% per year?<br /><br /><span style="font-size: small;"><b>Solution </b></span><br /><br />The cash flow diagram is shown in Figure 3–4. The symbol P<span style="font-size: xx-small;">A</span> is used throughout this chapter to represent the present worth of a uniform annual series A , and P'<span style="font-size: xx-small;">A</span> represents the present worth at a time other than period 0. Similarly, P T represents the total present worth at time 0. The correct placement of P'<span style="font-size: xx-small;">A</span> and the diagram renumbering to obtain n are also indicated. Note that P'<span style="font-size: xx-small;">A</span> is located in actual year 2, not year 3. Also, n = 6, not 8, for the P/A factor. First find the value of P'<span style="font-size: xx-small;">A </span>of the shifted series.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-cdLNBmn5uAY/U3pZfcrOwWI/AAAAAAAANkU/ahnQJMElnqo/s1600/99.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-cdLNBmn5uAY/U3pZfcrOwWI/AAAAAAAANkU/ahnQJMElnqo/s1600/99.gif" /></a></div><br /><span style="font-size: large;"><b>EXAMPLE 2</b></span><br /><br />Recalibration of sensitive measuring devices costs $8000 per year. If the machine will be reca-<br />librated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent <br />uniform series at 16% per year. Show hand and spreadsheet solutions.<br /><br /><b>Solution by Hand </b><br /><br />Figure 3–5 a and b shows the original cash ﬂ ows and the desired equivalent diagram. To convert the $8000 shifted series to an equivalent uniform series over all periods, first convert the uniform series into a present worth or future worth amount. Then either the A/P factor or the A/F factor can be used. Both methods are illustrated here. <br /><br /><b>Present worth method.</b> (Refer to Figure 3–5 a .) Calculate P'<span style="font-size: xx-small;">A</span> for the shifted series in year 2, followed by P<span style="font-size: xx-small;">T</span> in year 0. There are 6 years in the A series.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-yJXUc7LfDoo/U3paRlJldRI/AAAAAAAANkc/Cu-6qE5imV8/s1600/98.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-yJXUc7LfDoo/U3paRlJldRI/AAAAAAAANkc/Cu-6qE5imV8/s1600/98.gif" /></a></div><br /><b>Future worth method. </b> (Refer to Figure 3–5 a .) First calculate the future worth F in year 8.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-LUP6sTtJ3GY/U3paexYvhGI/AAAAAAAANkk/Y_6GuMViuu8/s1600/97.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-LUP6sTtJ3GY/U3paexYvhGI/AAAAAAAANkk/Y_6GuMViuu8/s1600/97.gif" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-dBqFzEA-kgc/U3paojj3LzI/AAAAAAAANks/Nw9BlGQBB1A/s1600/96.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-dBqFzEA-kgc/U3paojj3LzI/AAAAAAAANks/Nw9BlGQBB1A/s1600/96.gif" /></a></div>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-5836395262837180232014-05-07T11:33:00.003-07:002014-05-07T11:33:56.116-07:00Calculations for Uniform Series That Are Shifted When a uniform series begins at a time other than at the end of period 1, it is called a <b>shifted series</b>. In this case several methods can be used to ? nd the equivalent present worth P . For example, P of the uniform series shown in <b>Figure 3–1</b> could be determined by any of the following methods: <br /><br />• Use the P/F factor to find the present worth of each disbursement at year 0 and add them.<br />• Use the F/P factor to find the future worth of each disbursement in year 13, add them, and then find the present worth of the total, using P/F ( P/F , i ,13).<br />• Use the F/A factor to ? nd the future amount F/A ( F/A , i ,10), and then compute the present<br />worth, using P/F ( P/F , i ,13).<br />• Use the P/A factor to compute the “present worth” P<span style="font-size: xx-small;">3</span> = A ( P/A , i ,10) (which will be located in year 3, not year 0), and then ? nd the present worth in year 0 by using the ( P/F , i ,3) factor.<br /><br />Typically the last method is used for calculating the present worth of a uniform series that does not begin at the end of period 1. For <b>Figure 3–1,</b> the “present worth” obtained using the P/A factor is located in year 3. This is shown as P<span style="font-size: xx-small;">3</span> in F<b>igure 3–2</b>. Note that a P value is always located m1 year or period prior to the beginning of the first series amount. Why? Because the P/A factor was derived with P in time period 0 and A beginning at the end of period 1. The most common mistake made in working problems of this type is improper placement of P . Therefore, it is extremely important to remember:<br /><br />The present worth is always located one period prior to the ﬁ rst uniform series amount when using the P/A factor. <br /><br />To determine a future worth or F value, recall that the F/A factor derived in Section 2.3 had the F located in the same period as the last uniform series amount. <b>Figure 3–3</b> shows the location of the future worth when F/A is used for<b> Figure 3–1 </b>cash flows.<br /><br />The future worth is always located in the same period as the last uniform series amount when using the F/A factor. <br /><br />It is also important to remember that the number of periods n in the P/A or F/A factor is equal <br />to the number of uniform series values. It may be helpful to renumber the cash ﬂ ow diagram to <br />avoid errors in counting. <b>Figures 3–2 and 3–3 show Figure 3–1</b> renumbered to determine n = 10.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-56qjSEyVihA/U2p734i7mvI/AAAAAAAANio/W8DXqVyeGpM/s1600/30.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-56qjSEyVihA/U2p734i7mvI/AAAAAAAANio/W8DXqVyeGpM/s1600/30.gif" /></a></div><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/--aM5Q-AbPgE/U2p8GGpSwGI/AAAAAAAANiw/ekXk2HE9aFM/s1600/31.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/--aM5Q-AbPgE/U2p8GGpSwGI/AAAAAAAANiw/ekXk2HE9aFM/s1600/31.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-GwO6VaJozd8/U2p8SnFjrdI/AAAAAAAANi4/p3ZoilIQbHI/s1600/3.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-GwO6VaJozd8/U2p8SnFjrdI/AAAAAAAANi4/p3ZoilIQbHI/s1600/3.gif" height="161" width="320" /></a></div>As stated above, several methods can be used to solve problems containing a uniform series that is shifted. However, it is generally more convenient to use the uniform series factors than the single-amount factors. Specific steps should be followed to avoid errors: <br /><br /> 1. Draw a diagram of the positive and negative cash flows. <br /> 2. Locate the present worth or future worth of each series on the cash flow diagram. <br /> 3. Determine n for each series by renumbering the cash flow diagram. <br /> 4. Draw another cash flow diagram representing the desired equivalent cash flow. <br /> 5. Set up and solve the equations. Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-5104698871246756302014-04-19T19:33:00.002-07:002014-04-19T19:33:50.119-07:00Examples - Determining i or n for Known Cash Flow Values<span style="font-size: large;"><b>EXAMPLE</b></span><br /><br /> If Laurel made a $30,000 investment in a friend’s business and received $50,000 5 years later, determine the rate of return.<br /><br /><span style="font-size: large;"><b>Solution </b></span><br /><br />Since only single amounts are involved, i can be determined directly from the P/F factor.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-qzi3DH63ars/U1Mw9-Ivy8I/AAAAAAAANbA/WVAWhCdE0OI/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-qzi3DH63ars/U1Mw9-Ivy8I/AAAAAAAANbA/WVAWhCdE0OI/s1600/1.gif" /></a></div><br />Alternatively, the interest rate can be found by setting up the standard P/F relation, solving for the factor value, and interpolating in the tables.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-Kq6HejqFSy4/U1MxF5cEiAI/AAAAAAAANbI/7rieGd4V8p8/s1600/2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-Kq6HejqFSy4/U1MxF5cEiAI/AAAAAAAANbI/7rieGd4V8p8/s1600/2.gif" /></a></div><br />From the interest tables, a P/F factor of 0.6000 for n = 5 lies between 10% and 11%. Interpolate between these two values to obtain i = 10.76%.<br /><br /><span style="font-size: large;"><b>EXAMPLE</b></span><br /><br />Pyramid Energy requires that for each of its offshore wind power generators $5000 per year be placed into a capital reserve fund to cover unexpected major rework on field equipment. In one case, $5000 was deposited for 15 years and covered a rework costing $100,000 in year 15. What rate of return did this practice provide to the company?<br /><br /><span style="font-size: large;"><b>Solution </b></span><br /><br />The cash flow diagram is shown in Figure 2–24 . Either the A/F or F/A factor can be used. Using A/F, <br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-neUtkiyaDz8/U1Mxp5JUmQI/AAAAAAAANbQ/8XpCK4nrfOg/s1600/3.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-neUtkiyaDz8/U1Mxp5JUmQI/AAAAAAAANbQ/8XpCK4nrfOg/s1600/3.gif" /></a></div><br />From the A /F interest tables for 15 years, the value 0.0500 lies between 3% and 4%. By interpolation, i = 3.98%. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-epbuXIYq9Dw/U1Mx4-nbN4I/AAAAAAAANbY/Yy_RCJ726cE/s1600/5.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-epbuXIYq9Dw/U1Mx4-nbN4I/AAAAAAAANbY/Yy_RCJ726cE/s1600/5.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–24 Diagram to determine the rate <br />of return</td></tr></tbody></table>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-73832588356861322212014-03-17T17:52:00.001-07:002014-03-17T17:52:18.138-07:00Determining i or n for Known Cash Flow ValuesWhen all the cash flow values are known or have been estimated, the i value (interest rate or rate of return) or n value (number of years) is often the unknown. An example for which i is sought may be stated as follows: A company invested money to develop a new product. After the net annual income series is known following several years on the market, determine the rate of return i on the investment. There are several ways to find an unknown i or n value, depending upon the nature of the cash flow series and the method chosen to find the unknown. The simplest case involves only single amounts ( P and F ) and solution utilizing a spreadsheet function. The most difficult and complex involves finding i or n for irregular cash flows mixed with uniform and gradient series utilizing solution by hand and calculator. The solution approaches are summa-rized below, followed by examples.<br /><br /><b>Single Amounts — P and F Only </b><br /><span style="color: #0b5394;"><b>Hand or Calculator Solution</b></span> Set up the equivalence relation and (1) solve for the variable <br />using the factor formula, or (2) find the factor value and interpolate in the tables. Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find n . (See below and Appendix A for details.) <br /><br /><b>Uniform Series — A Series </b><br /><span style="color: #0b5394;"><b>Hand or Calculator Solution</b></span> Set up the equivalence relation using the appropriate factor <br />( P/A , A/P , F/A , or A/F ), and use the second method mentioned above. Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find n . <br /><br /><b>Mixed A Series, Gradients, and/or Isolated Values </b><br /><span style="color: #0b5394;"><b>Hand or Calculator Solution</b></span> Set up the equivalence relation and use (1) trial and error or (2) the calculator functions. <br /><br /><b>Spreadsheet Solution</b> Use the IRR or RATE function to find i or the NPER function to find n . (This is the recommended approach.)<br /><br />Besides the PV, FV, and NPV functions, other spreadsheet functions useful in determining i are IRR (internal rate of return) and RATE, and NPER (number of periods) to find n . The formats are shown here and the inside front cover with a detailed explanation in Appendix A. In all three of these functions, at least one cash flow entry must have a sign opposite that of others in order to find a solution.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-4GvAQEUuYcI/UyeXaad4jMI/AAAAAAAANSQ/lGe3RDhf6bs/s1600/100.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-4GvAQEUuYcI/UyeXaad4jMI/AAAAAAAANSQ/lGe3RDhf6bs/s1600/100.gif" /></a></div><br />To use IRR to find i , enter all cash flows into contiguous cells, including zero values. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-pE_M57tLGr0/UyeXe8CzV8I/AAAAAAAANSY/RsMp6k-5wsk/s1600/101.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-pE_M57tLGr0/UyeXe8CzV8I/AAAAAAAANSY/RsMp6k-5wsk/s1600/101.gif" /></a></div><br />The single-cell RATE function finds i when an A series and single P and/or F values are involved. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-VXHv9Mc35Pw/UyeXqUFpvsI/AAAAAAAANSg/vcHBVSeKDj8/s1600/101.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-VXHv9Mc35Pw/UyeXqUFpvsI/AAAAAAAANSg/vcHBVSeKDj8/s1600/101.gif" /></a></div><br /> NPER is a single-cell function to find n for single P and F values, or with an A series. Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-61820360574356766682014-03-07T12:08:00.003-08:002014-03-07T12:08:43.307-08:00Examples - Geometric Gradient Series Factors<span style="font-size: large;"><b>EXAMPLE 2.11</b></span><br /><br />A coal-fired power plant has upgraded an emission control valve. The modification costs only $8000 and is expected to last 6 years with a $200 salvage value. The maintenance cost is expected to be high at $1700 the first year, increasing by 11% per year thereafter. Determine the equivalent present worth of the modification and maintenance cost by hand and by spreadsheet at 8% per year.<br /><br /><span style="font-size: large;"><b>Solution by Hand </b></span><br /><br />The cash flow diagram (Figure 2–22) shows the salvage value as a positive cash flow and all <br />costs as negative. Use Equation [2.35] for g <span class="st">≠</span> i to calculate P<span style="font-size: xx-small;">g</span>. Total P<span style="font-size: xx-small;">T</span> is the sum of three <br />present worth components.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-LjjbM5uE6ls/UxolYYa0-hI/AAAAAAAANMc/rpy04_OYPOc/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-LjjbM5uE6ls/UxolYYa0-hI/AAAAAAAANMc/rpy04_OYPOc/s1600/1.gif" /></a></div><br /><br /><span style="font-size: large;"><b>Solution by Spreadsheet </b></span><br /><br />Figure 2–23 details the spreadsheet operations to find the geometric gradient present worth P<span style="font-size: xx-small;">g</span> and total present worth P<span style="font-size: xx-small;">T</span>. To obtain P<span style="font-size: xx-small;">T</span> = $- 17,999, three components are summed—first cost, present worth of estimated salvage in year 6, and Pg . Cell tags detail the relations for the second and third components; the first cost occurs at time 0.<br /><br /><b>Comment </b><br /><br />The relation that calculates the ( P /A,g,i%,n ) factor is rather complex, as shown in the cell tag and formula bar for C9. If this factor is used repeatedly, it is worthwhile using cell reference formatting so that A<span style="font-size: xx-small;">1</span> , i , g, and n values can be changed and the correct value is always obtained. Try to write the relation for cell C9 in this format.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-LQw_i7mYzeU/UxomAqr7LVI/AAAAAAAANMk/mHUFpcow5t8/s1600/2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-LQw_i7mYzeU/UxomAqr7LVI/AAAAAAAANMk/mHUFpcow5t8/s1600/2.gif" /></a></div><br /><span style="font-size: large;"><b>EXAMPLE 2.12 The Cement Factory Case</b></span><br /><br />Now let’s go back to the proposed Houston American Cement plant in Georgia. The revenue series estimate of $50 million annually is quite optimistic, especially since there are many other cement product plants operating in Florida and Georgia on the same limestone deposit. (The website for the HAC plant shows where they are located currently; it is clear that keen competition will be present.) Therefore, it is important to be sensitive in our analysis to possibly declining and increasing revenue series, depending upon the longer-term success of the plant’s marketing, quality, and reputation. Assume that revenue may start at $50 million by the end of the first year, but then decreases geometrically by 12% per year through year 5. Deter- mine the present worth and future worth equivalents of all revenues during this 5-year time frame at the same rate used previously, that is, 10% per year. <br /><br /><span style="font-size: large;"><b>Solution </b></span><br /><br />The cash flow diagram appears much like Figure 2–21 b , except that the arrows go up for revenues. In year 1, A<span style="font-size: xx-small;">1</span> $50 M and revenues decrease in year 5 to <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-oQIsXgEvhUw/UxomfztYVvI/AAAAAAAANM0/uaP2Uvs8YpI/s1600/4.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-oQIsXgEvhUw/UxomfztYVvI/AAAAAAAANM0/uaP2Uvs8YpI/s1600/4.gif" /></a></div><br />First, we determine Pg in year 0 using Eq. [2.35] with i 0.10 and g 0.12, then we calculate F in year 5. In $1 million units, <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-OWLensknkWM/Uxoms0xyLMI/AAAAAAAANM8/j8VxLCe9ORg/s1600/8.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-OWLensknkWM/Uxoms0xyLMI/AAAAAAAANM8/j8VxLCe9ORg/s1600/8.gif" /></a></div>This means that the decreasing revenue stream has a 5-year future equivalent worth of $246.080 M. If you look back to Example 2.6, we determined that the F in year 5 for the uniform revenue series of $50 M annually is $305.255 M. In conclusion, the 12% declining geometric gradient has lowered the future worth of revenue by $59.175 M, which is a sizable amount from the perspective of the owners of Votorantim Cimentos North America, Inc. Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-30749647885882933232014-02-19T09:39:00.001-08:002014-02-19T09:39:01.275-08:00Geometric Gradient Series Factors It is common for annual revenues and annual costs such as maintenance, operations, and labor to go up or down by a constant percentage, for example, +5% or -3% per year. This change occurs every year on top of a starting amount in the ﬁ rst year of the project. A deﬁ nition and description of new terms follow. <br /><br />A geometric gradient series is a cash flow series that either increases or decreases by a constant percentage each period. The uniform change is called the rate of change<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-OnErW-CbqCs/UwTqJgNW5GI/AAAAAAAAND4/_B3Mz3mrOHs/s1600/2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-OnErW-CbqCs/UwTqJgNW5GI/AAAAAAAAND4/_B3Mz3mrOHs/s1600/2.jpg" /></a></div>Note that the initial cash flow A<span style="font-size: xx-small;">1</span> is not considered separately when working with geometric gradients. <br /><br /><b>Figure 2–21 </b>shows increasing and decreasing geometric gradients starting at an amount A<span style="font-size: xx-small;">1</span> in time period 1 with present worth Pg located at time 0. The relation to determine the total present worth Pg for the entire cash flow series may be derived by multiplying each cash flow in <b>Figure 2–21</b> a by the <br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-B85wmV8Ky1k/UwTqt6M4A2I/AAAAAAAANEA/orBg4ncIDQ8/s1600/10.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-B85wmV8Ky1k/UwTqt6M4A2I/AAAAAAAANEA/orBg4ncIDQ8/s1600/10.jpg" /></a></div><br /><br /><div class="separator" style="clear: both; text-align: center;"></div><br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-OAvYzX4XloU/UwTq4WM-K_I/AAAAAAAANEI/AgBbZkJjHbs/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-OAvYzX4XloU/UwTq4WM-K_I/AAAAAAAANEI/AgBbZkJjHbs/s1600/1.gif" /></a></div>The term in brackets in Equation [2.32] is the ( P/A , g , i , n ) or geometric gradient series present worth factor for values of g not equal to the interest rate i. When g = i , substitute i for g in Equation [2.31] and observe that the term 1/(1 + i) appears n times.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-m5QFxS2ZSH4/UwTrWLS1thI/AAAAAAAANEY/Zd4_kFA3uZI/s1600/3.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://1.bp.blogspot.com/-m5QFxS2ZSH4/UwTrWLS1thI/AAAAAAAANEY/Zd4_kFA3uZI/s1600/3.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–21<br />Cash flow diagram of (a) increasing and (b) decreasing geometric gradient series and present worth Pg.</td></tr></tbody></table><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-ZIP4_DrhKZw/UwTrfJGbARI/AAAAAAAANEg/oufHDZobqKQ/s1600/4.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-ZIP4_DrhKZw/UwTrfJGbARI/AAAAAAAANEg/oufHDZobqKQ/s1600/4.gif" /> </a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">The (P A,g,i,n) factor calculates Pg in period t = 0 for a geometric gradient series starting in period 1 in the amount A<span style="font-size: xx-small;">1</span> and increasing by a constant rate of g each period. </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"> The equation for Pg and the ( P/A , g , i , n ) factor formula are</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-Qzsj9JSQxm0/UwTr1xlBITI/AAAAAAAANEo/WpJ6fSJ1kYo/s1600/8.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-Qzsj9JSQxm0/UwTr1xlBITI/AAAAAAAANEo/WpJ6fSJ1kYo/s1600/8.gif" /> </a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"> It is possible to derive factors for the equivalent A and F values; however, it is easier to determine <br />the Pg amount and then multiply by the A/P or F/P factor. </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">As with the arithmetic gradient series, there are no direct spreadsheet functions for geometric gradient series. Once the cash flows are entered, P and A are determined using the NPV and PMT functions, respectively. </div><div class="separator" style="clear: both; text-align: left;"><br /></div>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-13510725470852378202014-02-16T09:09:00.003-08:002014-02-16T09:11:41.364-08:00Examples Arithmetic Gradient Factors (P/ G and A/ G)EXAMPLE 2.8<br /><br />A local university has initiated a logo-licensing program with the clothier Holister, Inc. Esti-<br />mated fees (revenues) are $80,000 for the ﬁ rst year with uniform increases to a total of $200,000 <br />by the end of year 9. Determine the gradient and construct a cash ﬂ ow diagram that identiﬁ es <br />the base amount and the gradient series. <br /><br />Solution <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-nx0v2w0uUyQ/UwDsHEgcOxI/AAAAAAAAM_s/BWpCyTdOpSc/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-nx0v2w0uUyQ/UwDsHEgcOxI/AAAAAAAAM_s/BWpCyTdOpSc/s1600/1.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-ywQDWoon-jg/UwDwCs_CCjI/AAAAAAAANBc/dMD2OSHql-E/s1600/31.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-ywQDWoon-jg/UwDwCs_CCjI/AAAAAAAANBc/dMD2OSHql-E/s1600/31.gif" /></a></div><br /> The cash ﬂ ow diagram (Figure 2–13) shows the base amount of $80,000 in years 1 through 9 and the $15,000 gradient starting in year 2 and continuing through year 9. <br /><br />EXAMPLE 2.9<br /><br />Neighboring parishes in Louisiana have agreed to pool road tax resources already designated for bridge refurbishment. At a recent meeting, the engineers estimated that a total of $500,000 will be deposited at the end of next year into an account for the repair of old and safety-questionable bridges throughout the area. Further, they estimate that the deposits will increase by $100,000 per year for only 9 years thereafter, then cease. Determine the equivalent (a) present worth and (b) annual series amounts, if public funds earn at a rate of 5% per year.<br /><br />Solution<br /><br />(a) The cash ﬂ ow diagram of this conventional arithmetic gradient series from the perspective of the parishes is shown in Figure 2–16. According to Equation [2.19], two computations must be made and added: the ﬁrst for the present worth of the base amount PA and the second for the present worth of the gradient PG. The total present worth PT occurs in year 0. This is illustrated by the partitioned cash ﬂ ow diagram in Figure 2–17. In $1000 units, the total present worth is<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-nTCj4zyPyjE/UwDwUYr5IlI/AAAAAAAANBk/UpHupleOVds/s1600/32.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-nTCj4zyPyjE/UwDwUYr5IlI/AAAAAAAANBk/UpHupleOVds/s1600/32.gif" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-14yLBQ_9dZI/UwDwewQ-oSI/AAAAAAAANBs/YkQ9R4G5lWI/s1600/35.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-14yLBQ_9dZI/UwDwewQ-oSI/AAAAAAAANBs/YkQ9R4G5lWI/s1600/35.gif" /></a></div><br /><br /><br /><br />(b) Here, too, it is necessary to consider the gradient and the base amount separately. The total annual series AT is found by Equation [2.20] and occurs in years 1 through 10.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-ye0RnlJ6GPc/UwDwolgxWFI/AAAAAAAANB0/nDQGcnFR88E/s1600/36.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-ye0RnlJ6GPc/UwDwolgxWFI/AAAAAAAANB0/nDQGcnFR88E/s1600/36.gif" /></a></div><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-LccGeh5IbDE/UwDwy27yugI/AAAAAAAANB8/UFtFmf2he3c/s1600/38.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-LccGeh5IbDE/UwDwy27yugI/AAAAAAAANB8/UFtFmf2he3c/s1600/38.gif" /></a></div><br />Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-83341399234425069322014-02-12T17:30:00.000-08:002014-02-16T09:10:16.468-08:00Arithmetic Gradient Factors (P/ G and A/ G) Assume a manufacturing engineer predicts that the cost of maintaining a robot will increase by $5000 per year until the machine is retired. The cash flow series of maintenance costs involves a constant gradient, which is $5000 per year.<br /><br />An arithmetic gradient series is a cash flow series that either increases or decreases by a constant amount each period. The amount of change is called the gradient.<br /><br />Formulas previously developed for an A series have year-end amounts of equal value. In the case of a gradient, each year-end cash flow is different, so new formulas must be derived. First, assume that the cash flow at the end of year 1 is the base amount of the cash flow series and, therefore, not part of the gradient series. This is convenient because in actual applications, the base amount is usually significantly different in size compared to the gradient. For example, if you purchase a used car with a 1-year warranty, you might expect to pay the gasoline and insurance costs during the first year of operation. Assume these cost $2500; that is, $2500 is the base amount. After the first year, you absorb the cost of repairs, which can be expected to increase each year. If you estimate that total costs will increase by $200 each year, the amount the second year is $2700, the third $2900, and so on to year n , when the total cost is 2500 ( n - 1)200. The cash flow diagram is shown in <b>Figure 2–11.</b> Note that the gradient ($200) is first observed between year 1 and year 2, and the base amount ($2500 in year 1) is not equal to the gradient. Define the symbols G for gradient and CF<span style="font-size: xx-small;">n</span> for cash flow in year n as follows.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-8s43OkaQT4Y/UvwfvnAmH5I/AAAAAAAAM9o/zjrvQGX4EAA/s1600/1.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-8s43OkaQT4Y/UvwfvnAmH5I/AAAAAAAAM9o/zjrvQGX4EAA/s1600/1.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–11 Cash flow diagram of an <br />arithmetic gradient series.</td></tr></tbody></table><br />G constant arithmetic change in cash flows from one time period to the next; G may be positive or negative.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-UanfzYXg_qk/UvwfXPb7VdI/AAAAAAAAM9g/67o0usPbUPo/s1600/5.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-UanfzYXg_qk/UvwfXPb7VdI/AAAAAAAAM9g/67o0usPbUPo/s1600/5.gif" /></a></div><br />It is important to realize that the base amount defines a uniform cash flow series of the size A that occurs eash time period. We will use this fact when calculating equivalent amounts that involve arithmetic gradients. If the base amount is ignored, a generalized arithmetic (increasing) gradient cash flow diagram is as shown in <b>Figure 2–12</b>. Note that the gradient begins between years 1 and 2. This is called a conventional gradient . <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-F1_G7avSJnQ/Uvwf7eO2bAI/AAAAAAAAM9w/6E8lXlyp4Cc/s1600/2.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-F1_G7avSJnQ/Uvwf7eO2bAI/AAAAAAAAM9w/6E8lXlyp4Cc/s1600/2.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b>Figure 2–12</b> Conventional arithmetic <br />gradient series without <br />the base amount.</td></tr></tbody></table>The total present worth P<span style="font-size: xx-small;">T</span> for a series that includes a base amount A and conventional arithmetic gradient must consider the present worth of both the uniform series deﬁ ned by A and the rithmetic gradient series. The addition of the two results in P<span style="font-size: xx-small;">T</span>. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-NDdMRkpnAmM/UwDs34acBdI/AAAAAAAAM_0/daDtMZNJpYo/s1600/2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-NDdMRkpnAmM/UwDs34acBdI/AAAAAAAAM_0/daDtMZNJpYo/s1600/2.gif" height="30" width="320" /></a></div><br />where P<span style="font-size: xx-small;">A</span> is the present worth of the uniform series only, P<span style="font-size: xx-small;">G</span> is the present worth of the gradient series only, and the + or - sign is used for an increasing ( +G ) or decreasing ( -G ) gradient, respectively. <br /> <br />The corresponding equivalent annual worth A<span style="font-size: xx-small;">T</span> is the sum of the base amount series annual worth A<span style="font-size: xx-small;">A</span> and gradient series annual worth A<span style="font-size: xx-small;">G</span> , that is,<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-B7ofi47sF5E/UwDtNHUSjCI/AAAAAAAAM_8/asIW-bUpcv0/s1600/5.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-B7ofi47sF5E/UwDtNHUSjCI/AAAAAAAAM_8/asIW-bUpcv0/s1600/5.gif" height="28" width="320" /></a></div><br />Three factors are derived for arithmetic gradients: the P/G factor for present worth, the A/G <br />factor for annual series, and the F/G factor for future worth. There are several ways to derive <br />them. We use the single-payment present worth factor ( P/F , i , n ), but the same result can be ob-<br />tained by using the F/P , F/A , or P/A factor.<br /><br />In Figure 2–12, the present worth at year 0 of only the gradient is equal to the sum of the present worths of the individual cash ﬂ ows, where each value is considered a future amount. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-bso5T9pYNQo/UwDtnhC3NFI/AAAAAAAANAE/En6SSuDn5D4/s1600/11.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-bso5T9pYNQo/UwDtnhC3NFI/AAAAAAAANAE/En6SSuDn5D4/s1600/11.gif" /></a></div><br />The left bracketed expression is the same as that contained in Equation [2.6], where the P/A factor was derived. Substitute the closed-end form of the P/A factor from Equation [2.8] <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-OEZxF5UhU0o/UwDuBIAQYnI/AAAAAAAANAM/edaciOXa7xY/s1600/12.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-OEZxF5UhU0o/UwDuBIAQYnI/AAAAAAAANAM/edaciOXa7xY/s1600/12.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–14<br />Conversion diagram from an arithmetic gradient to a present worth.</td></tr></tbody></table><br /><br />into Equation [2.23] and simplify to solve for P/G, the present worth of the gradient series only. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-HBXw4U0Yhis/UwDuOYH9MEI/AAAAAAAANAU/LGOYJmAtAls/s1600/14.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-HBXw4U0Yhis/UwDuOYH9MEI/AAAAAAAANAU/LGOYJmAtAls/s1600/14.gif" /></a></div><br />Equation [2.24] is the general relation to convert an arithmetic gradient G (not including the base amount) for n years into a present worth at year 0 . Figure 2–14 a is converted into the equivalent cash flow in Figure 2–14 b . The arithmetic gradient present worth factor, or P/G factor, may be expressed in two forms:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-cFS7h7sSlZs/UwDuqs3D5lI/AAAAAAAANAk/kgtaMxM0rfU/s1600/17.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-cFS7h7sSlZs/UwDuqs3D5lI/AAAAAAAANAk/kgtaMxM0rfU/s1600/17.gif" /></a></div><br /> Equation [2.24] expressed as an engineering economy relation is<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-9VDJLYVVWJ4/UwDuzV52NXI/AAAAAAAANAw/NWBMsMKYDGU/s1600/18.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-9VDJLYVVWJ4/UwDuzV52NXI/AAAAAAAANAw/NWBMsMKYDGU/s1600/18.gif" height="29" width="320" /></a></div><br />which is the rightmost term in Equation [2.19] to calculate total present worth. The G carries a <br />minus sign for decreasing gradients.<br /><br />The equivalent uniform annual series A/G for an arithmetic gradient G is found by multiplying the present worth in Equation [2.26] by the ( A/P , i , n ) formula. In standard notation form, the equivalent of algebraic cancellation of P can be used.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-tzHK1ItVMEk/UwDu-vfO4pI/AAAAAAAANA0/O-Yau35kAIY/s1600/19.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-tzHK1ItVMEk/UwDu-vfO4pI/AAAAAAAANA0/O-Yau35kAIY/s1600/19.gif" /></a></div> In equation form,<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-fKK2fhjoSz8/UwDvINPmZiI/AAAAAAAANA8/L1MKT1xYT0Y/s1600/20.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-fKK2fhjoSz8/UwDvINPmZiI/AAAAAAAANA8/L1MKT1xYT0Y/s1600/20.gif" height="80" width="320" /></a></div><br />which is the rightmost term in Equation [2.20]. The expression in brackets in Equation [2.27] is called the arithmetic gradient uniform series factor and is identiﬁ ed by ( A/G , i , n ). This factor converts Figure 2–15 a into Figure 2–15 b . <br /><br />The P/G and A/G factors and relations are summarized inside the front cover. Factor values <br />are tabulated in the two rightmost columns of factor values at the rear of this text. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-R6YceYqqR2I/UwDvT9Dj08I/AAAAAAAANBE/U_hSB4bLv2E/s1600/29.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://3.bp.blogspot.com/-R6YceYqqR2I/UwDvT9Dj08I/AAAAAAAANBE/U_hSB4bLv2E/s1600/29.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–15<br />Conversion diagram of an arithmetic gradient series to an equivalent uniform annual series.</td></tr></tbody></table><br /><br /><br />There is no direct, single-cell spreadsheet function to calculate P<span style="font-size: xx-small;">G</span> or A<span style="font-size: xx-small;">G</span> for an arithmetic gradient. Use the NPV function to display P<span style="font-size: xx-small;">G </span> and the PMT function to display A<span style="font-size: xx-small;">G</span> after entering all cash ﬂ ows (base and gradient amounts) into contiguous cells. General formats for these functions are<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-uRArkQ4D2aQ/UwDvmN0ixYI/AAAAAAAANBQ/P1Zr5LOVJdU/s1600/27.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-uRArkQ4D2aQ/UwDvmN0ixYI/AAAAAAAANBQ/P1Zr5LOVJdU/s1600/27.gif" /></a></div><span id="goog_1584900291"></span><span id="goog_1584900292"></span><br /> The word entries in italic are cell references, not the actual numerical values. (See Appendix A, <br />Section A.2, for a description of cell reference formatting.) <br /><br />An F/ G factor ( arithmetic gradient future worth factor ) to calculate the future worth F/G of a gradient series can be derived by multiplying the P/G and F/P factors. The resulting factor, ( F/G , i , n ), in brackets, and engineering economy relation is<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-o9Z6DGjEcU4/UwDv1rkt9JI/AAAAAAAANBU/BrfNTN5X-Xw/s1600/30.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-o9Z6DGjEcU4/UwDv1rkt9JI/AAAAAAAANBU/BrfNTN5X-Xw/s1600/30.gif" /></a></div><br />Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-48105356393204257322014-01-30T14:11:00.000-08:002014-01-30T14:14:18.210-08:00Factor Values for Untabulated i or n Values Often it is necessary to know the correct numerical value of a factor with an i or n value that is <br />not listed in the compound interest tables in the rear of the book. Given specific values of i and n , there are several ways to obtain any factor value. <br /><br /> • Use the formula listed in this chapter or the front cover of the book, <br /> • Use an Excel function with the corresponding P , F , or A value set to 1. <br /> • Use linear interpolation in the interest tables. <br /><br />When the formula is applied, the factor value is accurate since the specific i and n values are input. However, it is possible to make mistakes since the formulas are similar to each other, es-<br />pecially when uniform series are involved. Additionally, the formulas become more complex <br />when gradients are introduced, as you will see in the following sections.<br /><br />A spreadsheet function determines the factor value if the corresponding P , A , or F argument in the function is set to 1 and the other parameters are omitted or set to zero. For example, the P F factor is determined using the PV function with A omitted (or set to 0) and F = 1, that is, PV( i %, n ,0,1) or PV( i %, n ,0,1). A minus sign preceding the function identifier causes the factor to have a positive value. Functions to determine the six common factors are as follows. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-w8iUp0bcSVQ/UurMZCy4t5I/AAAAAAAAM58/VyDY1oDYMk8/s1600/10.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-w8iUp0bcSVQ/UurMZCy4t5I/AAAAAAAAM58/VyDY1oDYMk8/s1600/10.jpg" /></a></div><br />Figure 2–9 shows a spreadsheet developed explicitly to determine these factor values. When it is made live in Excel, entering any combination of i and n displays the exact value for all six factors. The values for i = 3.25% and n = 25 years are shown here. As we already know, these same functions will determine a ﬁ nal P , A , or F value when actual or estimated cash ﬂ ow amounts are entered. <br /><br /><b>Linear interpolation </b> for an untabulated interest rate i or number of years n takes more time to complete than using the formula or spreadsheet function. Also interpolation introduces some level of inaccuracy, depending upon the distance between the two boundary values selected for i or n , as the formulas themselves are nonlinear functions. Interpolation is included here for indi- viduals who wish to utilize it in solving problems. Refer to Figure 2–10 for a graphical description of the following explanation. First, select two tabulated values ( x<span style="font-size: xx-small;">1</span> and x<span style="font-size: xx-small;">2</span> ) of the parameter for which the factor is requested, that is, i or n , ensuring that the two values surround and are not too distant from the required value x . Second, ﬁnd the corresponding tabulated factor values ( f<span style="font-size: xx-small;">1</span> and f<span style="font-size: xx-small;">2</span> ). Third, solve for the unknown, linearly interpolated value f using the formulas below, where the differences in parentheses are indicated in <b>Figure 2–10</b> as a through c. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-s6eILfsiqGs/UurM8ph8m2I/AAAAAAAAM6E/QiTfJr52ivI/s1600/13.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-s6eILfsiqGs/UurM8ph8m2I/AAAAAAAAM6E/QiTfJr52ivI/s1600/13.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-ozuYBOgvnqc/UurNGNEHesI/AAAAAAAAM6M/K5E2gEjqAi8/s1600/14.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-ozuYBOgvnqc/UurNGNEHesI/AAAAAAAAM6M/K5E2gEjqAi8/s1600/14.jpg" /></a></div><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-4Fce39JoGkQ/UurNNNn_V2I/AAAAAAAAM6U/tx33x7nlIEY/s1600/16.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-4Fce39JoGkQ/UurNNNn_V2I/AAAAAAAAM6U/tx33x7nlIEY/s1600/16.jpg" /></a></div>The value of d will be positive or negative if the factor is increasing or decreasing, respectively, in value between x<span style="font-size: xx-small;">1</span> and x<span style="font-size: xx-small;">2</span>. <br /><br />Determine the P /A factor value for i = 7.75% and n 10 years, using the three methods described previously. <br /><br /><span style="font-size: large;"><b>Solution </b></span><br /><br />Factor formula: Apply the formula from inside the front cover of the book for the P A factor. Showing 5-decimal accuracy,<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-mYHYlANElfY/UurNmAn6KrI/AAAAAAAAM6c/4dv-u4makUU/s1600/5.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-mYHYlANElfY/UurNmAn6KrI/AAAAAAAAM6c/4dv-u4makUU/s1600/5.jpg" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-y0Bd1yHipE8/UurNweE5JPI/AAAAAAAAM6k/_5ZtH6LTAlc/s1600/50.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-y0Bd1yHipE8/UurNweE5JPI/AAAAAAAAM6k/_5ZtH6LTAlc/s1600/50.jpg" /></a></div><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><b>Comment </b><br /><br />Note that since the P/A factor value decreases as i increases, the linear adjustment is negative at 0.2351. As is apparent, linear interpolation provides an approximation to the correct factor value for 7.75% and 10 years, plus it takes more calculations than using the formula or spread- sheet function. It is possible to perform two-way linear interpolation for untabulated i and n values; however, the use of a spreadsheet or factor formula is recommended. <u><br /></u>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-6239415429389580062014-01-10T17:30:00.002-08:002014-01-10T17:30:45.379-08:00 Sinking Fund Factor and Uniform Series Compound Amount Factor ( A/F and F/A )The simplest way to derive the A/F factor is to substitute into factors already developed. If P from Equation [2.3] is substituted into Equation [2.9], the following formula results. <br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-lzj1JJUcEe8/UtCchSVUxeI/AAAAAAAAMy4/V_Uilme8Q3c/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-lzj1JJUcEe8/UtCchSVUxeI/AAAAAAAAMy4/V_Uilme8Q3c/s1600/1.gif" /></a></div><br />The expression in brackets in Equation [2.12] is the A/F or sinking fund factor. It determines the uniform annual series. A that is equivalent to a given future amount F . This is shown graphically in Figure 2–6 a , where. A is a uniform annual investment. <br /><br />The uniform series A begins at the end of year (period) 1 and continues through the year of the given F. The last. A value and F occur at the same time. <br /><br />Equation [2.12] can be rearranged to ﬁ nd F for a stated A series in periods 1 through n (Figure 2–6 b ).<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-z-CFlmRbz1M/UtCc_F0d9mI/AAAAAAAAMzA/eq2rlqXMSPA/s1600/4.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-z-CFlmRbz1M/UtCc_F0d9mI/AAAAAAAAMzA/eq2rlqXMSPA/s1600/4.gif" /></a></div><br />The term in brackets is called the uniform series compound amount factor (USCAF), or F/A factor. <br />When multiplied by the given uniform annual amount A , it yields the future worth of the uniform <br />series. It is important to remember that the future amount F occurs in the same period as the last A . <br /><br />Standard notation follows the same form as that of other factors. They are ( F /A , i , n ) and <br />( A/F , i , n ). Table 2–3 summarizes the notations and equations, as does the inside front cover. <br /><br />As a matter of interest, the uniform series factors can be symbolically determined by using an abbreviated factor form. For example, F/A = ( F/P )( P/A ), where cancellation of the P is correct. <br />Using the factor formulas, we have<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-To3SCTcA6is/UtCdXZMYqwI/AAAAAAAAMzI/UiFPXWsJxc0/s1600/5.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-To3SCTcA6is/UtCdXZMYqwI/AAAAAAAAMzI/UiFPXWsJxc0/s1600/5.gif" /></a></div><br />For solution by spreadsheet, the FV function calculates F for a stated A series over n years. The format is<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-5rXkPJxf02Y/UtCdg38XlRI/AAAAAAAAMzQ/nwthox3v6YA/s1600/6.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-5rXkPJxf02Y/UtCdg38XlRI/AAAAAAAAMzQ/nwthox3v6YA/s1600/6.gif" /></a></div><br />The P may be omitted when no separate present worth value is given. The PMT function deter-<br />mines the A value for n years, given F in year n and possibly a separate P value in year 0. The <br />format is<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-0z2o9fPF77Y/UtCdp7pSg5I/AAAAAAAAMzY/cwpQDZVfem4/s1600/7.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-0z2o9fPF77Y/UtCdp7pSg5I/AAAAAAAAMzY/cwpQDZVfem4/s1600/7.gif" /></a></div><br />If P is omitted, the comma must be entered so the function knows the last entry is an F value. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-9NSbX35WNUE/UtCeAiFNToI/AAAAAAAAMzg/aq2y8hwoTyk/s1600/8.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-9NSbX35WNUE/UtCeAiFNToI/AAAAAAAAMzg/aq2y8hwoTyk/s1600/8.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–6 Cash flow diagrams to ( a ) find A, given F , and ( b ) find F, given A . </td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-tqFQiHCj8eo/UtCeN8OxSqI/AAAAAAAAMzo/J4XsweWdWfo/s1600/10.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-tqFQiHCj8eo/UtCeN8OxSqI/AAAAAAAAMzo/J4XsweWdWfo/s1600/10.gif" /></a></div><br /><span style="font-size: large;"><b>EXAMPLE 2.5</b></span><br /><br />The president of Ford Motor Company wants to know the equivalent future worth of a $1 million capital investment each year for 8 years, starting 1 year from now. Ford capital earns at a rate of 14% per year. <br /><br /><b>Solution </b><br /><br />The cash flow diagram (Figure 2–7) shows the annual investments starting at the end of year 1 and ending in the year the future worth is desired. In $1000 units, the F value in year 8 is found by using the F A factor. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-USEX6IngCM8/UtCeh2wCsmI/AAAAAAAAMzw/SQqF-gW5tO4/s1600/11.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-USEX6IngCM8/UtCeh2wCsmI/AAAAAAAAMzw/SQqF-gW5tO4/s1600/11.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–7 Diagram to find F for a uniform series, Example 2.5. </td></tr></tbody></table><br />Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-29497398271406627422014-01-07T11:24:00.003-08:002014-01-07T11:24:26.798-08:00Uniform Series Present Worth Factor and Capital Recovery Factor ( P/A and A/P )The equivalent present worth P of a uniform series. A of end-of-period cash flows (investments) is shown in Figure 2–4 a . An expression for the present worth can be determined by considering each A value as a future worth F , calculating its present worth with the P/F factor, Equation [2.3], and summing the results.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-fIoSgsVBHEE/UsxSC62ewiI/AAAAAAAAMw0/tRhmjGDDW8Y/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-fIoSgsVBHEE/UsxSC62ewiI/AAAAAAAAMw0/tRhmjGDDW8Y/s1600/1.gif" /></a></div><br />The terms in brackets are the P /F factors for years 1 through n , respectively. Factor out A <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-7FHI84V1EIQ/UsxSJLAnQEI/AAAAAAAAMw8/WOxpKW28Flk/s1600/2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-7FHI84V1EIQ/UsxSJLAnQEI/AAAAAAAAMw8/WOxpKW28Flk/s1600/2.gif" /></a></div>To simplify Equation [2.6] and obtain the P A factor, multiply the n -term geometric progression <br />in brackets by the ( P/F , i %,1) factor, which is 1 (1 + i ). This results in Equation [2.7]. Now subtract the two equations, [2.6] from [2.7], and simplify to obtain the expression for P when i <span class="st">≠</span> 0 (Equation [2.8])<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-EsTkGbTEgP4/UsxSe6wPHJI/AAAAAAAAMxE/6L-eznU05Pg/s1600/5.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-EsTkGbTEgP4/UsxSe6wPHJI/AAAAAAAAMxE/6L-eznU05Pg/s1600/5.gif" /></a></div><br />The term in brackets in Equation [2.8] is the conversion factor referred to as the uniform series <br />present worth factor (USPWF). It is the P /A factor used to calculate the equivalent P value in <br />year 0 for a uniform end-of-period series of A values beginning at the end of period 1 and extend-<br />ing for n periods. The cash flow diagram is Figure 2–4 a .<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-sYm8E479pbY/UsxS3tSUaWI/AAAAAAAAMxM/5GY9nrW-5_M/s1600/10.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-sYm8E479pbY/UsxS3tSUaWI/AAAAAAAAMxM/5GY9nrW-5_M/s1600/10.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–4 Cash flow diagrams used to determine (a) P, given a uniform series A, and (b) A, given a present worth P.</td></tr></tbody></table><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-KMF1LZjRHyE/UsxTBpOVzzI/AAAAAAAAMxU/x__yub_Zs3c/s1600/11.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-KMF1LZjRHyE/UsxTBpOVzzI/AAAAAAAAMxU/x__yub_Zs3c/s1600/11.gif" /></a></div><br />To reverse the situation, the present worth P is known and the equivalent uniform series amount A is sought (Figure 2–4 b ). The first A value occurs at the end of period 1, that is, one period after P occurs. Solve Equation [2.8] for A to obtain<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-ut_6juttt-w/UsxTJQ-BPoI/AAAAAAAAMxc/NSxWcTsitKk/s1600/15.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-ut_6juttt-w/UsxTJQ-BPoI/AAAAAAAAMxc/NSxWcTsitKk/s1600/15.gif" /></a></div>The term in brackets is called the capital recovery factor (CRF), or A /P factor. It calculates the equivalent uniform annual worth. A over n years for a given P in year 0, when the interest rate is i. <br /><br />The P/A and A/ P factors are derived with the present worth P and the first uniform annual amount A one year (period) apart. That is, the present worth P must always be located one period prior to the first A. <br /><br />The factors and their use to find P and A are summarized in Table 2–2 and inside the front cover. <br />The standard notations for these two factors are ( P/A , i %, n ) and ( A /P , i %, n ). Tables at the end of <br />the text include the factor values. As an example, if i 15% and n 25 years, the P/A factor <br />value from Table 19 is ( P /A ,15%,25) 6.4641. This will find the equivalent present worth at 15% per year for any amount A that occurs uniformly from years 1 through 25. <br /> <br />Spreadsheet functions can determine both P and A values in lieu of applying the P/A and A/P factors. The PV function calculates the P value for a given A over n years and a separate F value in year n , if it is given. The format, is<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-VvqN4BQ7eiE/UsxTnlgYTeI/AAAAAAAAMxk/JgC0efMLngg/s1600/23.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-VvqN4BQ7eiE/UsxTnlgYTeI/AAAAAAAAMxk/JgC0efMLngg/s1600/23.gif" /></a></div>Similarly, the A value is determined by using the PMT function for a given P value in year 0 and a separate F , if given. The format is<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-2j21KVhu48I/UsxTvL0gOdI/AAAAAAAAMxs/vC99Lt0ENKA/s1600/25.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-2j21KVhu48I/UsxTvL0gOdI/AAAAAAAAMxs/vC99Lt0ENKA/s1600/25.gif" /></a></div><br />Table 2–2 includes the PV and PMT functions.<br /><br /><span style="font-size: large;"><b>EXAMPLE 2.3</b></span><br /><br />How much money should you be willing to pay now for a guaranteed $600 per year for 9 years starting next year, at a rate of return of 16% per year? <br /><br /><b>Solution </b><br /><br />The cash flows follow the pattern of Figure 2–4 a , with A $600, i 16%, and n 9. The present worth is<br /><br /> P = 600( P/A ,16%,9) = 600(4.6065) = $2763.90 <br /><br />The PV function PV(16%,9,600) entered into a single spreadsheet cell will display the answer P = ($2763.93). Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-90663126832247872062014-01-04T09:07:00.001-08:002014-01-04T09:07:44.234-08:00Single-Amount Factors ( F/P and P/F ) The most fundamental factor in engineering economy is the one that determines the amount <br />of money F accumulated after n years (or periods) from a single present worth P, with interest <br />compounded one time per year (or period). Recall that compound interest refers to interest paid <br />on top of interest. Therefore, if an amount P is invested at time t = 0, the amount F<span style="font-size: xx-small;">1</span> accumulated <br />1 year hence at an interest rate of i percent per year will be<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-selZ9wz4PsY/Usg7W1UEyTI/AAAAAAAAMsU/VpNCXX1QuRA/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-selZ9wz4PsY/Usg7W1UEyTI/AAAAAAAAMsU/VpNCXX1QuRA/s1600/1.gif" /></a></div><br />where the interest rate is expressed in decimal form. At the end of the second year, the amount <br />accumulated F<span style="font-size: xx-small;">2</span> is the amount after year 1 plus the interest from the end of year 1 to the end of <br />year 2 on the entire F<span style="font-size: xx-small;">1</span> <br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/--uNrmP8TBx4/Usg7Z1TT5EI/AAAAAAAAMsg/7cxvYCGiIss/s1600/2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/--uNrmP8TBx4/Usg7Z1TT5EI/AAAAAAAAMsg/7cxvYCGiIss/s1600/2.gif" /></a></div><br />The amount F<span style="font-size: xx-small;">2</span> can be expressed as<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-Q8Uav6M9TMA/Usg7p8qLttI/AAAAAAAAMsk/iEXqC26Nj6o/s1600/3.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-Q8Uav6M9TMA/Usg7p8qLttI/AAAAAAAAMsk/iEXqC26Nj6o/s1600/3.gif" /></a></div><br />Similarly, the amount of money accumulated at the end of year 3, using Equation [2.1], will be<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-ZkLIyY6Ql-Y/Usg7xi9nEaI/AAAAAAAAMss/5dJzb0Bec1c/s1600/5.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-ZkLIyY6Ql-Y/Usg7xi9nEaI/AAAAAAAAMss/5dJzb0Bec1c/s1600/5.gif" /></a></div><span id="goog_956142840"></span><span id="goog_956142841"></span><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-2RmkI_VpJ9Q/Usg764GeFWI/AAAAAAAAMs0/fNbo9WAudIU/s1600/9.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="196" src="http://3.bp.blogspot.com/-2RmkI_VpJ9Q/Usg764GeFWI/AAAAAAAAMs0/fNbo9WAudIU/s640/9.gif" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2–1 Cash flow diagrams for single-payment factors: (a) ﬁnd F, given P, and (b) ﬁ nd P, given F.</td></tr></tbody></table><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-TCSJo53tbjY/Usg8ZwtUj_I/AAAAAAAAMs8/QezIZ4KuX2g/s1600/10.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-TCSJo53tbjY/Usg8ZwtUj_I/AAAAAAAAMs8/QezIZ4KuX2g/s1600/10.gif" /></a></div><span id="goog_956142840"></span><span id="goog_956142841"></span><br /><span id="goog_956142840">A standard notation has been adopted for all factors. The notation includes two cash ﬂ ow sym- bols, the interest rate, and the number of periods. It is always in the general form ( X Y , i , n ). The letter X represents what is sought, while the letter Y represents what is given. For example, F /P means ﬁ nd F when given P. The i is the interest rate in percent, and n represents the number of periods involved. </span><br /><br /><span id="goog_956142840">Using this notation, ( F P ,6%,20) represents the factor that is used to calculate the future amount F accumulated in 20 periods if the interest rate is 6% per period. The P is given. The standard notation, simpler to use than formulas and factor names, will be used hereafter. Table 2–1 summarizes the standard notation and equations for the F/P and P/F factors. This information is also included inside the front cover. </span><span id="goog_956142841"></span><br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-_9mg0zp0dv0/Usg8y1IrxII/AAAAAAAAMtE/ZHnQxSngJtw/s1600/11.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="132" src="http://4.bp.blogspot.com/-_9mg0zp0dv0/Usg8y1IrxII/AAAAAAAAMtE/ZHnQxSngJtw/s640/11.gif" width="640" /></a></div><br />To simplify routine engineering economy calculations, tables of factor values have been pre- pared for interest rates from 0.25% to 50% and time periods from 1 to large n values, depending on the i value. These tables, found at the rear of the book, have a colored edge for easy identiﬁ cation. They are arranged with factors across the top and the number of periods n down the left side. The word discrete in the title of each table emphasizes that these tables utilize the end-of-period convention and that interest is compounded once each interest period. For a given factor, interest rate, and time, the correct factor value is found at the intersection of the factor name and n . For example, the value of the factor ( P/F ,5%,10) is found in the P/F column of Table 10 at period 10 as 0.6139. This value is determined by using Equation [2.3].<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-xIbyOTDrVAc/Usg9GcRsobI/AAAAAAAAMtM/EyhdTI7Y_N8/s1600/13.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-xIbyOTDrVAc/Usg9GcRsobI/AAAAAAAAMtM/EyhdTI7Y_N8/s1600/13.gif" /></a></div><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-lyauv1cXb4Y/Usg9NxkxYHI/AAAAAAAAMtU/yvmfKJtP1us/s1600/14.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-lyauv1cXb4Y/Usg9NxkxYHI/AAAAAAAAMtU/yvmfKJtP1us/s1600/14.gif" /></a></div><br /><span style="font-size: large;"><b>EXAMPLE 2.1</b></span><br /><br />Sandy, a manufacturing engineer, just received a year-end bonus of $10,000 that will be invested immediately. With the expectation of earning at the rate of 8% per year, Sandy hopes to take the entire amount out in exactly 20 years to pay for a family vacation when the oldest daughter is due to graduate from college. Find the amount of funds that will be available in 20 years by using <b>(a)</b> hand solution by applying the factor formula and tabulated value and <b>(b)</b> a spreadsheet function.<br /><br /><span style="font-size: large;">SOLUTION</span><br /><br />The cash ﬂow diagram is the same as Figure 2–1a. The symbols and values are<br /> P = $10,000 F = ? i = 8% per year n = 20 years<br /><br /><b>(a)</b> Factor formula: Apply Equation [2.2] to ﬁ nd the future value F. Rounding to four decimals, we have<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-p6z1yUcnw00/Usg9_Y0ypwI/AAAAAAAAMtc/ckCzHOEOhGo/s1600/21.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-p6z1yUcnw00/Usg9_Y0ypwI/AAAAAAAAMtc/ckCzHOEOhGo/s1600/21.gif" /></a></div><br />Standard notation and tabulated value: Notation for the F/P factor is (F/ P,i%,n).<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-y_FqjvAmF5M/Usg-FW-RD9I/AAAAAAAAMtk/b6BsP7dHujs/s1600/22.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-y_FqjvAmF5M/Usg-FW-RD9I/AAAAAAAAMtk/b6BsP7dHujs/s1600/22.gif" /></a></div>Table 13 provides the tabulated value. Round-off errors can cause a slight difference in the final answer between these two methods.<br /><br /><b>(b)</b> Spreadsheet: Use the FV function to ﬁ nd the amount 20 years in the future. The format is that shown in Equation [2.4]; the numerical entry is FV(8%,20,,10000). The spread- sheet will appear similar to that in the right side of Figure 1–13, with the answer ($46,609.57) displayed. (You should try it on your own computer now.) The FV function has performed the computation in part (a) and displayed the result. The equivalency statement is: If Sandy invests $10,000 now and earns 8% per year every year for 20 years, $46,610 will be available for the family vacation.<br /><br /><span style="font-size: large;"><b>EXAMPLE 2.2 The Cement Factory Case</b></span><br /><br />As discussed in the introduction to this chapter, the Houston American Cement factory will require an investment of $200 million to construct. Delays beyond the anticipated implementation year of 2012 will require additional money to construct the factory. Assuming that the cost of money is 10% per year, compound interest, use both tabulated factor values and spread- sheet functions to determine the following for the board of directors of the Brazilian company that plans to develop the plant.<br /><br /><b>(a)</b> The equivalent investment needed if the plant is built in 2015.<br /><b>(b)</b> The equivalent investment needed had the plant been constructed in the year 2008.<br /><br /><span style="font-size: large;">Solution</span><br /><br />Figure 2–2 is a cash ﬂ ow diagram showing the expected investment of $200 million ($200 M) in 2012, which we will identify as time t = 0. The required investments 3 years in the future and 4 years in the past are indicated by F<span style="font-size: xx-small;">3</span> = ? and P <span style="font-size: xx-small;">-4</span> = ?, respectively.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-dJVYPldBOmg/Usg-vqypTCI/AAAAAAAAMt0/Hk9XnpjGfvY/s1600/22.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="147" src="http://4.bp.blogspot.com/-dJVYPldBOmg/Usg-vqypTCI/AAAAAAAAMt0/Hk9XnpjGfvY/s400/22.gif" width="400" /></a></div><b>(a)</b> To ﬁnd the equivalent investment required in 3 years, apply the F P factor. Use $1 million units and the tabulated value for 10% interest (Table 15).<br /><br /><a href="http://2.bp.blogspot.com/-SzgLPguxWZY/Usg-8Mi2fGI/AAAAAAAAMt8/f7GIQfg2NS0/s1600/25.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-SzgLPguxWZY/Usg-8Mi2fGI/AAAAAAAAMt8/f7GIQfg2NS0/s1600/25.gif" /></a><br /><br /> Now, use the FV function on a spreadsheet to ﬁnd the same answer, F<span style="font-size: xx-small;">3</span> = $266.20 million. (Refer to Figure 2–3, left side.)<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-TtApH9ffTbQ/Usg_KOXmh9I/AAAAAAAAMuE/7rBcXqLIx24/s1600/26.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-TtApH9ffTbQ/Usg_KOXmh9I/AAAAAAAAMuE/7rBcXqLIx24/s1600/26.gif" /></a></div><b>(b)</b> The year 2008 is 4 years prior to the planned construction date of 2012. To determine the equivalent cost 4 years earlier, consider the $200 M in 2012 (t 0) as the future value F and apply the P/ F factor for n = 4 to ﬁnd P <span style="font-size: xx-small;">-4</span>. (Refer to Figure 2–2.) Table 15 supplies the tabulated value.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-jpdDABsCzNQ/Usg_oXpforI/AAAAAAAAMuM/mCXZi6Iher0/s1600/31.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-jpdDABsCzNQ/Usg_oXpforI/AAAAAAAAMuM/mCXZi6Iher0/s1600/31.gif" /></a></div><br />The PV function PV(10%,4,,200) will display the same amount as shown in Figure 2–3, right side.<br /><br />This equivalence analysis indicates that at $136.6 M in 2008, the plant would have cost about 68% as much as in 2012, and that waiting until 2015 will cause the price tag to increase about 33% to $266 M.Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-85939273495378589392014-01-02T17:04:00.000-08:002014-01-02T17:06:00.166-08:00Engineering Economics: Introduction to Spreadsheet UseThe functions on a computer spreadsheet can greatly reduce the amount of hand work for equivalency computations involving compound interest and the terms P , F , A , i , and n . The use of a calculator to solve most simple problems is preferred by many students and professors as described in Appendix D. However, as cash ﬂ ow series become more complex, the spreadsheet offers a good alternative. Microsoft Excel is used throughout this book because it is readily available and easy to use. Appendix A is a primer on using spreadsheets and Excel. The functions used in engineering economy are described there in detail, with explanations of all the parameters. Appendix A also includes a section on spreadsheet layout that is useful when the economic analysis is presented to someone else—a coworker, a boss, or a professor. <br /><br />A total of seven Excel functions can perform most of the fundamental engineering economy calculations. The functions are great supplemental tools, but they do not replace the understanding of engineering economy relations, assumptions, and techniques. Using the symbols P , F , A , i , and n deﬁ ned in the previous section, the functions most used in engineering economic analysis are formulated as follows. <br /><br />If some of the parameters don’t apply to a particular problem, they can be omitted and zero is assumed. For readability, spaces can be inserted between parameters within parentheses. If the parameter omitted is an interior one, the comma must be entered. The last two functions require that a series of numbers be entered into contiguous spreadsheet cells, but the ﬁ rst ﬁ ve can be used with no supporting data. In all cases, the function must be preceded by an equals sign ( = ) in the cell where the answer is to be displayed. <br /><br /> To understand how the spreadsheet functions work, look back at Example 1.6 a , where the <br />equivalent annual amount A is unknown, as indicated by A = ?. To find A using a spreadsheet function, simply enter the PMT function = PMT(5%,5,5000). <b>Figure 1–13</b> is a screen image of a spreadsheet with the PMT function entered into cell B4. The answer ($1154.87) is dis- played. The answer may appear in red and in parentheses, or with a minus sign on your screen to indicate a negative amount from the perspective of a reduction in the account balance. The right side of <b>Figure 1–13</b> presents the solution to Example 1.6 b. The future value F is determined by using the FV function. The FV function appears in the formula bar; and many examples throughout this text will include cell tags, as shown here, to indicate the format of important entries. <br /><br />The following example demonstrates the use of a spreadsheet to develop relations (not built-in functions) to calculate interest and cash ﬂ ows. Once set up, the spreadsheet can be used to perform sensitivity analysis for estimates that are subject to change. We will illustrate the use of spreadsheets throughout the chapters. ( Note: The spreadsheet examples may be omitted, if spreadsheets are not used in the course. A solution by hand is included in virtually all examples.)<br /><br /><a href="http://engineeringandeconomicanalysis.blogspot.com/2013/12/engineering-economics-terminology-and.html"><span style="font-size: large;"><b>EXAMPLE 1.6</b></span></a><br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-6Y3DoQIatpY/UsYLdY7jNlI/AAAAAAAAMrM/1vNK_0rjfFA/s1600/1.jpg" style="margin-left: auto; margin-right: auto;"><img border="0" height="200" src="http://1.bp.blogspot.com/-6Y3DoQIatpY/UsYLdY7jNlI/AAAAAAAAMrM/1vNK_0rjfFA/s640/1.jpg" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–13</b> Use of spreadsheet functions PMT and FV, Example 1.6. </td></tr></tbody></table><br /><br /><span style="font-size: large;"><b>EXAMPLE 1.17</b></span><br /><br />A Japan-based architectural ﬁ rm has asked a United States–based software engineering group to infuse GPS sensing capability via satellite into monitoring software for high-rise structures in order to detect greater than expected horizontal movements. This software could be very beneficial as an advance warning of serious tremors in earthquake-prone areas in Japan and the United States. The inclusion of accurate GPS data is estimated to increase annual revenue over that for the current software system by $200,000 for each of the next 2 years, and by $300,000 for each of years 3 and 4. The planning horizon is only 4 years due to the rapid advances made internationally in building-monitoring software. Develop spreadsheets to answer the questions below. <br /><br /> <b> (a)</b> Determine the total interest and total revenue after 4 years, using a compound rate of return of 8% per year. <br /> <b> (b)</b> Repeat part ( a ) if estimated revenue increases from $300,000 to $600,000 in years 3 and 4. <br /> <b> (c)</b> Repeat part ( a ) if inﬂ ation is estimated to be 4% per year. This will decrease the real rate of return from 8% to 3.85% per year (Chapter 14 shows why).<br /><br /><span style="font-size: large;">Solution by Spreadsheet</span><br /><br />Refer to Figure 1–14 a to d for the solutions. All the spreadsheets contain the same information, but some cell values are altered as required by the question. (Actually, all the questions can be answered on one spreadsheet by changing the numbers. Separate spreadsheets are shown here for explanation purposes only.) <br /><br />The Excel functions are constructed with reference to the cells, not the values them- selves, so that sensitivity analysis can be performed without function changes. This ap- proach treats the value in a cell as a global variable for the spreadsheet. For example, the 8% rate in cell B2 will be referenced in all functions as B2, not 8%. Thus, a change in the rate requires only one alteration in the cell B2 entry, not in every relation where 8% is used. See Appendix A for additional information about using cell referencing and building spreadsheet relations. <br /><br /><b>(a)</b> Figure 1–14 a shows the results, and <b>Figure 1–14</b> b presents all spreadsheet relations for estimated interest and revenue (yearly in columns C and E, cumulative in columns D and F). As an illustration, for year 3 the interest I<span style="font-size: xx-small;">3</span> and revenue plus interest R<span style="font-size: xx-small;">3</span> are<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-fVzw7GyEcco/UsYL9UF0-cI/AAAAAAAAMrU/N0akuhLRnvA/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-fVzw7GyEcco/UsYL9UF0-cI/AAAAAAAAMrU/N0akuhLRnvA/s1600/1.gif" /></a></div><br />The equivalent amount after 4 years is $1,109,022, which is comprised of $1,000,000 in total revenue and $109,022 in interest compounded at 8% per year. The shaded cells in Figure 1–14 a and b indicate that the sum of the annual values and the last entry in the cu- mulative columns must be equal. <br /><br /><b>(b)</b> To determine the effect of increasing estimated revenue for years 3 and 4 to $600,000, use the same spreadsheet and change the entries in cells B8 and B9 as shown in Figure 1–14 c . Total interest increases 22%, or $24,000, from $109,222 to $133,222. <br /><br /><b>(c)</b> Figure 1–14 d shows the effect of changing the original i value from 8% to an inﬂ ation- adjusted rate of 3.85% in cell B2 on the ﬁ rst spreadsheet. [Remember to return to the $300,000 revenue estimates for years 3 and 4 after working part ( b ).] Inﬂ ation has now reduced total interest by 53% from $109,222 to $51,247, as shown in cell C10. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-7cyUSVM40j4/UsYMVVuQE8I/AAAAAAAAMrc/khwf0U_dvCw/s1600/2.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="640" src="http://3.bp.blogspot.com/-7cyUSVM40j4/UsYMVVuQE8I/AAAAAAAAMrc/khwf0U_dvCw/s640/2.gif" width="440" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–14 </b> Spreadsheet solutions with sensitivity analysis, Example 1.17 a to c . </td></tr></tbody></table><span style="font-size: large;">Comment </span><br /><br />Later we will learn how to utilize the NPV and FV Excel ﬁ nancial functions to obtain the same answers determined in Figure 1–14 , where we developed each basic relation. When you are working with an Excel spreadsheet, it is possible to display all of the entries and functions on the screen as shown in Figure 1–14 b by simultaneously touching the <Ctrl> and < `> keys, which may be in the upper left of the keyboard on the key with <~>. Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-82972658236420182502013-12-23T07:16:00.000-08:002013-12-23T07:16:06.284-08:00Minimum Attractive Rate of ReturnFor any investment to be profitable, the investor (corporate or individual) expects to receive more money than the amount of capital invested. In other words, a fair rate of return, or return on investment, must be realizable. The definition of ROR in Equation [1.4] is used in this discussion, that is, amount earned divided by the principal.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-Ck4iIuiqBXM/UquUvsZs0UI/AAAAAAAAMcY/LsI81MRXZlw/s1600/9.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-Ck4iIuiqBXM/UquUvsZs0UI/AAAAAAAAMcY/LsI81MRXZlw/s1600/9.jpg" /></a></div><br />Engineering alternatives are evaluated upon the prognosis that a reasonable ROR can be expected. Therefore, some reasonable rate must be established for the selection criteria (step 4) of the engineering economy study ( <a href="http://engineeringandeconomicanalysis.blogspot.com/2013/12/cash-flows-estimation-and-diagramming.html"><b>Figure 1–1</b></a> ).<br /><br />The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return established for the evaluation and selection of alternatives. A project is not economically viable unless it is e<b>xpected to return at least the MARR</b>. MARR is also referred to as the hurdle rate, cutoff rate, benchmark rate, and minimum acceptable rate of return.<br /><b><br /></b><b>Figure 1–12</b> indicates the relations between different rate of return values. In the United States, the current U.S. Treasury Bill return is sometimes used as the benchmark safe rate. The MARR will always be higher than this, or a similar, safe rate. The MARR is not a rate that is calculated as a ROR. The MARR is established by (financial) managers and is used as a criterion against which an alternative’s ROR is measured, when making the accept/reject investment decision. <br /><br />To develop a foundation-level understanding of how a MARR value is established and used to make investment decisions. Although the MARR is used as a criterion to decide on investing in a project, the size of MARR is fundamentally connected to how much it costs to obtain the needed capital funds. It always costs money in the form of interest to raise capital. The interest, expressed as a percentage rate per year, is called the <b>cost of capital</b>. As an example on a personal level, if you want to purchase a new widescreen HDTV, but do not have sufficient money (capital), you could obtain a bank loan for, say, a cost of capital of 9% per year and pay for the TV in cash now. Alternatively, you might choose to use your credit card and pay off the balance on a monthly basis. This approach will probably cost you at least 15% per year. Or, you could use funds from your savings account that earns 5% per year and pay cash. This approach means that you also forgo future returns from these funds. The 9%, 15%, and 5% rates are your cost of capital estimates to raise the capital for the system by different methods of capital financing. In analogous ways, corporations estimate the <b>cost of capital</b> from different sources to raise funds for engineering projects and other types of projects. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-6nVlzV4457w/UrhRd2x0nqI/AAAAAAAAMpw/hz8TGcE8l-M/s1600/31.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-6nVlzV4457w/UrhRd2x0nqI/AAAAAAAAMpw/hz8TGcE8l-M/s1600/31.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–12 </b> Size of MAAR relative to other rate of return values. </td></tr></tbody></table>In general, capital is developed in two ways—equity financing and debt financing. A combination of these two is very common for most projects. Chapter 10 covers these in greater detail, but a snapshot description follows. <br /> <br />Equity financing. The corporation uses its own funds from cash on hand, stock sales, or retained earnings. Individuals can use their own cash, savings, or investments. In the example above, using money from the 5% savings account is equity financing. <br /><br />Debt financing. The corporation borrows from outside sources and repays the principal and interest according to some schedule, much like the plans in<b> Table 1–1</b>. Sources of debt capital may be bonds, loans, mortgages, venture capital pools, and many others. Individuals, too, can utilize debt sources, such as the credit card (15% rate) and bank options (9% rate) described above.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-f9S98BxTSJ8/UrhTnl__0SI/AAAAAAAAMqE/dZOg9aa62cQ/s1600/10.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-f9S98BxTSJ8/UrhTnl__0SI/AAAAAAAAMqE/dZOg9aa62cQ/s1600/10.gif" /></a></div><br />Combinations of debt-equity financing mean that a weighted average cost of capital (WACC) results. If the HDTV is purchased with 40% credit card money at 15% per year and 60% savings account funds earning 5% per year, the weighted average cost of capital is 0.4(15) 0.6(5) 9% per year. <br /><br />For a corporation, the established MARR used as a criterion to accept or reject an investment alternative will usually be equal to or higher than the WACC that the corporation must bear to obtain the necessary capital funds. So the inequality <br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://4.bp.blogspot.com/-qFd_DgG5MVA/UrhRy3Li5DI/AAAAAAAAMp4/dL14Ya95ZlI/s1600/32.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-qFd_DgG5MVA/UrhRy3Li5DI/AAAAAAAAMp4/dL14Ya95ZlI/s1600/32.gif" /></a></div><br /> must be correct for an accepted project. Exceptions may be government-regulated requirements (safety, security, environmental, legal, etc.), economically lucrative ventures expected to lead to other opportunities, etc. <br /><br />Often there are many alternatives that are expected to yield a ROR that exceeds the MARR as indicated in <b>Figure</b> <b>1–12</b> , but there may not be sufficient capital available for all, or the project’s risk may be estimated as too high to take the investment chance. Therefore, new projects that are undertaken usually have an expected return at least as great as the return on another alternative that is not funded. The expected rate of return on the unfunded project is called the opportunity cost.<br /><br />The opportunity cost is the rate of return of a forgone opportunity caused by the inability to pursue a project. Numerically, it is the <b>largest rate of return of all the projects not accepted (forgone) due to the lack of capital funds or other resources</b>. When no specific MARR is established, the de facto MARR is the opportunity cost, i.e., the ROR of the first project not undertaken due to unavailability of capital funds. <br /><br />As an illustration of opportunity cost, refer to <b>Figure 1–12</b> and assume a MARR of 12% per year. Further, assume that a proposal, call it A, with an expected ROR = 13% is not funded due to a lack of capital. Meanwhile, proposal B has a ROR = 14.5% and is funded from available capital. Since proposal A is not undertaken due to the lack of capital, its estimated ROR of 13% is the opportunity cost; that is, the opportunity to make an additional 13% return is forgone. Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-56344681953951822692013-12-21T18:19:00.002-08:002013-12-21T18:20:17.023-08:00Simple and Compound InterestThe terms interest, interest period, and interest rate are useful in calculating equivalent sums of money for one interest period in the past and one period in the future. However, for more than one interest period, the terms simple interest and compound interest become important. <br /><br />Simple interest is calculated using the principal only, ignoring any interest accrued in preceding interest periods. The total simple interest over several periods is computed as<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-5krQPWI3lsA/UrZE7YQd7lI/AAAAAAAAMlA/gj8soAyOwqA/s1600/1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-5krQPWI3lsA/UrZE7YQd7lI/AAAAAAAAMlA/gj8soAyOwqA/s1600/1.jpg" /></a></div><br />where I is the amount of interest earned or paid and the interest rate i is expressed in decimal form. <br /><br /><span style="font-size: large;"><b>EXAMPLE 1.14</b></span><br /><br />GreenTree Financing lent an engineering company $100,000 to retroﬁ t an environmentally unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money will the firm repay at the end of 3 years? <br /><br /><b>Solution </b><br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-y4BcXuG3dc4/UrZFgTGaXZI/AAAAAAAAMlQ/BsxxObAUyVo/s1600/6.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-y4BcXuG3dc4/UrZFgTGaXZI/AAAAAAAAMlQ/BsxxObAUyVo/s1600/6.gif" /></a></div><br />The interest accrued in the first year and in the second year does not earn interest. The interest due each year is $10,000 calculated only on the $100,000 loan principal. <br /><br />In most financial and economic analyses, we use compound interest calculations.<br /><br />For compound interest, the interest accrued for each interest period is calculated on the <b>principal plus the total amount of interest accumulated in all previous periods</b>. Thus, compound interest means interest on top of interest. <br /><br />Compound interest reflects the effect of the time value of money on the interest also. Now the interest for one period is calculated as<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-2EEJ2PDyFe8/UrZF50MbmvI/AAAAAAAAMlY/M1X4B5MZRvU/s1600/10.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-2EEJ2PDyFe8/UrZF50MbmvI/AAAAAAAAMlY/M1X4B5MZRvU/s1600/10.gif" /></a></div><br />In mathematical terms, the interest It for time period t may be calculated using the relation.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-CRsOPkf0QtM/UrZGFialWFI/AAAAAAAAMlg/zG1CEFe2n3Q/s1600/11.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-CRsOPkf0QtM/UrZGFialWFI/AAAAAAAAMlg/zG1CEFe2n3Q/s1600/11.gif" /></a></div><br /><span style="font-size: large;"><b>EXAMPLE 1.15</b></span><br /><br />Assume an engineering company borrows $100,000 at 10% per year compound interest and will pay the principal and all the interest after 3 years. Compute the annual interest and total amount due after 3 years. Graph the interest and total owed for each year, and compare with the previous example that involved simple interest. <br /><br /><b>Solution </b><br /><br />To include compounding of interest, the annual interest and total owed each year are calculated by Equation [1.8]. <br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-DzixVuHfg7I/UrZGWizytMI/AAAAAAAAMlo/P9wSRqOQYkM/s1600/13.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-DzixVuHfg7I/UrZGWizytMI/AAAAAAAAMlo/P9wSRqOQYkM/s1600/13.gif" /></a></div><br /> The repayment plan requires no payment until year 3 when all interest and the principal, a total of $133,100, are due. <b>Figure 1–11</b> uses a cash flow diagram format to compare end-of-year (a) simple and (b) compound interest and total amounts owed. The differences due to com- pounding are clear. An extra $133,100 – 130,000 = $3100 in interest is due for the compounded interest loan. <br /><br />Note that while simple interest due each year is constant, the compounded interest due grows geometrically. Due to this geometric growth of compound interest, the difference between simple and compound interest accumulation increases rapidly as the time frame increases. For example, if the loan is for 10 years, not 3, the extra paid for compounding interest may be calculated to be $59,374. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-Qtx-iTBccJA/UrZG-nUAJyI/AAAAAAAAMl0/EFbuBjqONW4/s1600/16.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-Qtx-iTBccJA/UrZG-nUAJyI/AAAAAAAAMl0/EFbuBjqONW4/s1600/16.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b>Figure 1–11</b> Interest I owed and total amount owed for ( a ) simple interest (Example 1.14) and ( b ) compound interest <br />(Example 1.15). </td></tr></tbody></table>A more efﬁ cient way to calculate the total amount due after a number of years in Example 1.15 is to utilize the fact that compound interest increases geometrically. This allows us to skip the year-by-year computation of interest. In this case, the total amount due at the end of each year is<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-RV74B3xm63U/UrZHPYC4-BI/AAAAAAAAMl8/tq8-nFzDKAg/s1600/17.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-RV74B3xm63U/UrZHPYC4-BI/AAAAAAAAMl8/tq8-nFzDKAg/s1600/17.gif" /></a></div>This allows future totals owed to be calculated directly without intermediate steps. The general form of the equation is<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-_z7ad_mCw-s/UrZHb26mtgI/AAAAAAAAMmI/0uk5LHH7dGA/s1600/20.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-_z7ad_mCw-s/UrZHb26mtgI/AAAAAAAAMmI/0uk5LHH7dGA/s1600/20.jpg" /></a></div><br />where i is expressed in decimal form. Equation [1.10] was applied above to obtain the $133,100 due after 3 years. This fundamental relation will be used many times in the upcoming chapters. <br /><br />We can combine the concepts of interest rate, compound interest, and equivalence to demon- strate that different loan repayment plans may be equivalent, but differ substantially in amounts paid from one year to another and in the total repayment amount. This also shows that there are many ways to take into account the time value of money. <br /><br /><span style="font-size: large;"><b>EXAMPLE 1.16</b></span><br /><br /><b>Table 1–1</b> details four different loan repayment plans described below. Each plan repays a $5000 loan in 5 years at 8% per year compound interest. <br /><br /><b>• Plan 1:</b> Pay all at end. No interest or principal is paid until the end of year 5. Interest accumulates each year on the total of principal and all accrued interest.<br />• <b>Plan 2:</b> Pay interest annually, principal repaid at end. The accrued interest is paid each year, and the entire principal is repaid at the end of year 5.<br />• <b> Plan 3:</b> Pay interest and portion of principal annually. The accrued interest and one-ﬁ fth of the principal (or $1000) are repaid each year. The outstanding loan balance decreases each year, so the interest (column 2) for each year decreases.<br />• <b>Plan 4:</b> Pay equal amount of interest and principal. Equal payments are made each year with a portion going toward principal repayment and the remainder covering the accrued interest. Since the loan balance decreases at a rate slower than that in plan 3 due to the equal end-of-year payments, the interest decreases, but at a slower rate.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-MY4My3ELJMs/UrZLVGRtsMI/AAAAAAAAMmU/xDUuVCV7cfs/s1600/25.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-MY4My3ELJMs/UrZLVGRtsMI/AAAAAAAAMmU/xDUuVCV7cfs/s1600/25.jpg" /></a></div><br /><b>(a)</b> Make a statement about the equivalence of each plan at 8% compound interest.<br /><b>(b)</b> Develop an 8% per year simple interest repayment plan for this loan using the same approach as plan 2. Comment on the total amounts repaid for the two plans.<br /><br /><b>Solution </b> <br /><br /><b>(a)</b> The amounts of the annual payments are different for each repayment schedule, and the total amounts repaid for most plans are different, even though each repayment plan requires exactly 5 years. The difference in the total amounts repaid can be ex- plained by the time value of money and by the partial repayment of principal prior to year 5. <br /><br />A loan of $5000 at time 0 made at 8% per year compound interest is equivalent to each of the following: <br /><br /><b>Plan 1</b> $7346.64 at the end of year 5<br /><b>Plan 2</b> $400 per year for 4 years and $5400 at the end of year 5<br /><b>Plan 3</b> Decreasing payments of interest and partial principal in years 1 ($1400) through 5 ($1080)<br /><b>Plan 4</b> $1252.28 per year for 5 years <br /><br />An engineering economy study typically uses plan 4; interest is compounded, and a constant amount is paid each period. This amount covers accrued interest and a partial amount of principal repayment. <br /><br /><b>(b)</b> The repayment schedule for 8% per year simple interest is detailed in Table 1–2. Since the annual accrued interest of $400 is paid each year and the principal of $5000 is repaid in year 5, the schedule is exactly the same as that for 8% per year compound interest, and the total amount repaid is the same at $7000. In this unusual case, simple and compound interest result in the same total repayment amount. Any deviation from this schedule will cause the two plans and amounts to differ.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-1JgMf8u9cuA/UrZL5YQaZdI/AAAAAAAAMmc/m9sPJJ2lrq0/s1600/26.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-1JgMf8u9cuA/UrZL5YQaZdI/AAAAAAAAMmc/m9sPJJ2lrq0/s1600/26.jpg" /></a></div>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-20915057051153747782013-12-18T06:24:00.001-08:002013-12-18T06:24:21.097-08:00Economic EquivalenceEconomic equivalence is a fundamental concept upon which engineering economy computations are based. Before we delve into the economic aspects, think of the many types of equivalency we may utilize daily by transferring from one scale to another. Some example transfers between scales are as follows:<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-REl53D3OHdE/UrGticbsTjI/AAAAAAAAMjU/4XOw2rLz9Nw/s1600/1.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-REl53D3OHdE/UrGticbsTjI/AAAAAAAAMjU/4XOw2rLz9Nw/s1600/1.gif" /></a></div><br />Often equivalency involves two or more scales. Consider the equivalency of a speed of 110 kilometers per hour (kph) into miles per minute using conversions between distance and time scales with three-decimal accuracy. <br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-sKcGSuDwi84/UrGtr54VUII/AAAAAAAAMjc/yaCyL81g1C4/s1600/2.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-sKcGSuDwi84/UrGtr54VUII/AAAAAAAAMjc/yaCyL81g1C4/s1600/2.gif" /></a></div><br />Four scales—time in minutes, time in hours, length in miles, and length in kilometers—are combined to develop these equivalent statements on speed. Note that throughout these statements, the fundamental relations of 1 mile = 1.609 kilometers and 1 hour = 60 minutes are applied. If a fundamental relation changes, the entire equivalency is in error. <br /><br />Now we consider economic equivalency. <br /><br /><b>Economic equivalence</b> is a combination of <b>interest rate</b> and <b>time value of money</b> to determine the different amounts of money at different points in time that are equal in economic value. <br /><br />As an illustration, if the interest rate is 6% per year, $100 today (present time) is equivalent to $106 one year from today. <br /><br /> Amount accrued = 100 + 100(0.06) = 100(1 + 0.06) = $106<br /><br />If someone offered you a gift of $100 today or $106 one year from today, it would make no difference which offer you accepted from an economic perspective. In either case you have $106 one year from today. However, the two sums of money are equivalent to each other only when the interest rate is 6% per year. At a higher or lower interest rate, $100 today is not equivalent to $106 one year from today. <br /><br />In addition to future equivalence, we can apply the same logic to determine equivalence for previous years. A total of $100 now is equivalent to $100 1.06 = $94.34 one year ago at an interest rate of 6% per year. From these illustrations, we can state the following: $94.34 last year, $100 now, and $106 one year from now are equivalent at an interest rate of 6% per year. The fact that these sums are equivalent can be veriﬁ ed by computing the two interest rates for 1-year interest periods. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-2nREAXJj9Z4/UrGuiHdTLSI/AAAAAAAAMjk/dISs85wsSXo/s1600/4.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-2nREAXJj9Z4/UrGuiHdTLSI/AAAAAAAAMjk/dISs85wsSXo/s1600/4.gif" /></a></div><br />The cash flow diagram in <b>Figure 1–10</b> indicates the amount of interest needed each year to make these three different amounts equivalent at 6% per year. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-QGLLH_-FNo0/UrGuv7MfgqI/AAAAAAAAMjs/lXaf6f3rVP8/s1600/6.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-QGLLH_-FNo0/UrGuv7MfgqI/AAAAAAAAMjs/lXaf6f3rVP8/s1600/6.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–10 </b> Equivalence of money at 6% per year interest. </td></tr></tbody></table><br /><span style="font-size: large;"><b>EXAMPLE 1.12</b></span><br /><br />Manufacturers make backup batteries for computer systems available to Batteries + dealers through privately owned distributorships. In general, batteries are stored throughout the year, and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner. Assume you own the City Center Batteries + outlet. Make the calculations necessary to show which of the following statements are true and which are false about battery costs. <br /><br />(a) The amount of $98 now is equivalent to a cost of $105.60 one year from now.<br />(b) A truck battery cost of $200 one year ago is equivalent to $205 now.<br />(c) A $38 cost now is equivalent to $39.90 one year from now.<br />(d) A $3000 cost now is equivalent to $2887.14 one year earlier.<br />(e) The carrying charge accumulated in 1 year on an investment of $20,000 worth of batteries is $1000. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-QPd16l6XQYc/UrGvZS_RfSI/AAAAAAAAMj0/nXadM7ZVc74/s1600/5.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-QPd16l6XQYc/UrGvZS_RfSI/AAAAAAAAMj0/nXadM7ZVc74/s1600/5.gif" /></a></div><br />Comparison of alternative cash flow series requires the use of equivalence to determine when the series are economically equal or if one is economically preferable to another. The keys to the analysis are the interest rate and the timing of the cash flows. Example 1.13 demonstrates how easy it is to be misled by the size and timing of cash flows. <br /><br /><span style="font-size: large;"><b>EXAMPLE 1.13</b></span><br /><br />Howard owns a small electronics repair shop. He wants to borrow $10,000 now and repay it over the next 1 or 2 years. He believes that new diagnostic test equipment will allow him to work on a wider variety of electronic items and increase his annual revenue. Howard received 2-year repayment options from banks A and B.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-TiC7aiIVl_4/UrGvqnqxeAI/AAAAAAAAMj8/hIp7x77yiKE/s1600/6.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-TiC7aiIVl_4/UrGvqnqxeAI/AAAAAAAAMj8/hIp7x77yiKE/s1600/6.gif" /></a></div><br />After reviewing these plans, Howard decided that he wants to repay the $10,000 after only 1 year based on the expected increased revenue. During a family conversation, Howard’s brother-in-law offered to lend him the $10,000 now and take $10,600 after exactly 1 year. Now Howard has three options and wonders which one to take. Which one is economically the best?<br /><br /><b>Solution</b> <br /><br />The repayment plans for both banks are economically equivalent at the interest rate of 5% per year. (This is determined by using computations that you will learn in Chapter 2.) Therefore, Howard can choose either plan even though the bank B plan requires a slightly larger sum of money over the 2 years. <br /><br />The brother-in-law repayment plan requires a total of $600 in interest 1 year later plus the principal of $10,000, which makes the interest rate 6% per year. Given the two 5% per year options from the banks, this 6% plan should not be chosen as it is not economically better than the other two. Even though the sum of money repaid is smaller, the timing of the cash flows and the interest rate make it less desirable. The point here is that cash flows themselves, or their sums, cannot be relied upon as the primary basis for an economic decision. The interest rate, timing, and economic equivalence must be considered. Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com1tag:blogger.com,1999:blog-7204171490530605416.post-76693234326285224172013-12-17T13:46:00.002-08:002013-12-17T13:46:45.037-08:00Cash Flows: Estimation and DiagrammingAs mentioned in earlier sections, cash flows are the amounts of money estimated for future projects or observed for project events that have taken place. All cash flows occur during specific time periods, such as 1 month, every 6 months, or 1 year. Annual is the most common time period. For example, a payment of $10,000 once every year in December for 5 years is a series of 5 outgoing cash flows. And an estimated receipt of $500 every month for 2 years is a series of 24 incoming cash flows. Engineering economy bases its computations on the timing, size, and direction of cash flows. <br /><br /><b>Cash inflows</b> are the receipts, revenues, incomes, and savings generated by project and business activity. A <b>plus sign</b> indicates a cash inflow. <br /><br /><b>Cash outflows</b> are costs, disbursements, expenses, and taxes caused by projects and business activity. A <b>negative or minus sign</b> indicates a cash outflow. When a project involves only costs, the minus sign may be omitted for some techniques, such as benefit/cost analysis. <br /><br />Of all the steps in <b>Figure 1–1</b> that outline the engineering economy study, estimating cash flows (step 3) is the most difficult, primarily because it is an attempt to predict the future. Some examples of cash flow estimates are shown here. As you scan these, consider how the cash inflow or outflow may be estimated most accurately.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-8_k_jE0zA-s/UqhYHFZMQjI/AAAAAAAAMZ4/S7ZVJr36VjU/s1600/1.gif" style="margin-left: auto; margin-right: auto;"><img alt="Steps in an engineering economy study. " border="0" src="http://1.bp.blogspot.com/-8_k_jE0zA-s/UqhYHFZMQjI/AAAAAAAAMZ4/S7ZVJr36VjU/s1600/1.gif" title="Steps in an engineering economy study. " /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–1</b> Steps in an engineering economy study. </td></tr></tbody></table><br /><b>Cash Inflow Estimates </b><br /><br /> Income: $150,000 per year from sales of solar-powered watches <br /> Savings: $24,500 tax savings from capital loss on equipment salvage <br /> Receipt: $750,000 received on large business loan plus accrued interest <br /> Savings: $150,000 per year saved by installing more efficient air conditioning <br /> Revenue: $50,000 to $75,000 per month in sales for extended battery life iPhones <br /><br /><b>Cash Outflow Estimates </b><br /><br /> Operating costs: $230,000 per year annual operating costs for software services <br /> First cost: $800,000 next year to purchase replacement earthmoving equipment <br /> Expense: $20,000 per year for loan interest payment to bank <br /> Initial cost: $1 to $1.2 million in capital expenditures for a water recycling unit <br /><br />All of these are point estimates , that is, single-value estimates for cash flow elements of an alternative, except for the last revenue and cost estimates listed above. They provide a range estimate, because the persons estimating the revenue and cost do not have enough knowledge or experience with the systems to be more accurate. For the initial chapters, we will utilize point estimates. The use of risk and sensitivity analysis for range estimates is covered in the later chapters of this book.<br /><br />Once all cash inflows and outflows are estimated (or determined for a completed project), the net cash flow for each time period is calculated.<br /><br /><div class="separator" style="clear: both; text-align: left;"><a href="http://2.bp.blogspot.com/-EZGJDkj5HSY/UrDBpGBGUUI/AAAAAAAAMhI/6TsJ0e35Bqg/s1600/10.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-EZGJDkj5HSY/UrDBpGBGUUI/AAAAAAAAMhI/6TsJ0e35Bqg/s1600/10.jpg" /></a></div><br /> where NCF is net cash flow, R is receipts, and D is disbursements. <br /><br />At the beginning of this section, the timing, size, and direction of cash flows were mentioned as important. Because cash flows may take place at any time during an interest period, as a matter of convention, all cash flows are assumed to occur at the end of an interest period. <br /><br />The end-of-period convention means that all cash inflows and all cash outflows are assumed to take place at the <b>end of the interest period</b> in which they actually occur. When several inflows and outflows occur within the same period, the net cash flow is assumed to occur at the end of the period. <br /><br /> In assuming end-of-period cash flows, it is important to understand that future (F) and uniform annual (A) amounts are located at the end of the interest period, which is not necessarily December 31. If in <a href="http://engineeringandeconomicanalysis.blogspot.com/2013/12/engineering-economics-terminology-and.html"><b>Example 1.7</b></a> the lump-sum deposit took place on July 1, 2011, the withdraw- als will take place on July 1 of each succeeding year for 6 years. Remember, end of the period means end of interest period, not end of calendar year. <br /><br />The cash flow diagram is a very important tool in an economic analysis, especially when the cash flow series is complex. It is a graphical representation of cash flows drawn on the y axis with a time scale on the x axis. The diagram includes what is known, what is estimated, and what is needed. That is, once the cash flow diagram is complete, another person should be able to work the problem by looking at the diagram. <br /><br />Cash flow diagram time t = 0 is the present, and t = 1 is the end of time period 1. We assume that the periods are in years for now. The time scale of <b>Figure 1–4</b> is set up for 5 years. Since the end-of-year convention places cash flows at the ends of years, the “1” marks the end of year 1.<br /><br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-JmvXbbc8XSI/UrDD_ztbxWI/AAAAAAAAMhc/UKJxVPbspyM/s1600/25.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://4.bp.blogspot.com/-JmvXbbc8XSI/UrDD_ztbxWI/AAAAAAAAMhc/UKJxVPbspyM/s1600/25.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–4</b> A typical cash flow time scale for 5 years. </td></tr></tbody></table><br />While it is not necessary to use an exact scale on the cash flow diagram, you will probably avoid errors if you make a neat diagram to approximate scale for both time and relative cash flow magnitudes. <br /><br />The direction of the arrows on the diagram is important to differentiate income from outgo. A vertical arrow pointing up indicates a positive cash flow. Conversely, a down-pointing arrow indicates a negative cash flow. We will use a bold, colored arrow to indicate what is unknown and to be determined. For example, if a future value F is to be determined in year 5, a wide, colored arrow with F = ? is shown in year 5. The interest rate is also indicated on the diagram. <b>Figure 1–5 </b> illustrates a cash inflow at the end of year 1, equal cash outflows at the end of years 2 and 3, an interest rate of 4% per year, and the unknown future value F after 5 years. The arrow for the unknown value is generally drawn in the opposite direction from the other cash flows; however, the engineering economy computations will determine the actual sign on the F value. <br /><br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-v3Q4wa0difM/UrDEJ3D1KAI/AAAAAAAAMhk/8JSC5b3XTV8/s1600/27.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://2.bp.blogspot.com/-v3Q4wa0difM/UrDEJ3D1KAI/AAAAAAAAMhk/8JSC5b3XTV8/s1600/27.gif" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–5 </b> Example of positive and negative cash flows. </td></tr></tbody></table><br />Before the diagramming of cash flows, a perspective or vantage point must be determined so that or – signs can be assigned and the economic analysis performed correctly. Assume you borrow $8500 from a bank today to purchase an $8000 used car for cash next week, and you plan to spend the remaining $500 on a new paint job for the car two weeks from now. There are several perspectives possible when developing the cash flow diagram—those of the borrower (that’s you), the banker, the car dealer, or the paint shop owner. The cash flow signs and amounts for these perspectives are as follows. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-RKFeeombzic/UrDCXuul9pI/AAAAAAAAMhQ/FGhM0iW10fw/s1600/21.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-RKFeeombzic/UrDCXuul9pI/AAAAAAAAMhQ/FGhM0iW10fw/s1600/21.gif" /></a></div><br />One, and only one, of the perspectives is selected to develop the diagram. For your perspective, all three cash flows are involved and the diagram appears as shown in <b>Figure 1–6</b> with a time scale of weeks. Applying the end-of-period convention, you have a receipt of $8500 now (time 0) and cash outflows of $8000 at the end of week 1, followed by $500 at the end of week 2. <br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-FY4tdqpqWCg/UrDEd1eTROI/AAAAAAAAMhs/TXudQ3HuSyU/s1600/29.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://4.bp.blogspot.com/-FY4tdqpqWCg/UrDEd1eTROI/AAAAAAAAMhs/TXudQ3HuSyU/s320/29.gif" width="299" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><b> Figure 1–6 </b> Cash flows from perspective of borrower for loan and purchases. </td></tr></tbody></table><span style="font-size: large;"><b>EXAMPLE 1.9</b></span><br /><br />Each year Exxon-Mobil expends large amounts of funds for mechanical safety features throughout its worldwide operations. Carla Ramos, a lead engineer for Mexico and Central American operations, plans expenditures of $1 million now and each of the next 4 years just for the improvement of field-based pressure-release valves. Construct the cash flow diagram to find the equivalent value of these expenditures at the end of year 4, using a cost of capital esti- mate for safety-related funds of 12% per year.<br /><br /><b>Solution </b><br /><br />Figure 1–7 indicates the uniform and negative cash flow series (expenditures) for five periods, and the unknown F value (positive cash flow equivalent) at exactly the same time as the fifth expenditure. Since the expenditures start immediately, the first $1 million is shown at time 0, not time 1. Therefore, the last negative cash flow occurs at the end of the fourth year, when F also occurs. To make this diagram have a full 5 years on the time scale, the addition of the year 1 completes the diagram. This addition demonstrates that year 0 is the end-of-period point for the year 1. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-NKzZ0YQRbzE/UrDFDgEb7VI/AAAAAAAAMh0/szNvuXyTq0Q/s1600/30.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-NKzZ0YQRbzE/UrDFDgEb7VI/AAAAAAAAMh0/szNvuXyTq0Q/s1600/30.gif" /></a></div><br /><span style="font-size: large;"><b>EXAMPLE 1.10</b></span><br /><br />An electrical engineer wants to deposit an amount P now such that she can withdraw an equal annual amount of A1 = $2000 per year for the first 5 years, starting 1 year after the deposit, and a different annual withdrawal of A2 = $3000 per year for the following 3 years. How would the cash flow diagram appear if i = 8.5% per year? <br /><br /><b>Solution </b><br /><br />The cash flows are shown in Figure 1–8. The negative cash outflow P occurs now. The with- drawals (positive cash inflow) for the A<span style="font-size: xx-small;">1</span> series occur at the end of years 1 through 5, and A<span style="font-size: xx-small;">2</span> occurs in years 6 through 8. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-A-_Q1ZJ81u0/UrDFfpUKUQI/AAAAAAAAMh8/L1A64F2kvq8/s1600/31.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-A-_Q1ZJ81u0/UrDFfpUKUQI/AAAAAAAAMh8/L1A64F2kvq8/s1600/31.gif" /></a></div><br /><span style="font-size: large;"><b>EXAMPLE 1.11</b></span><br /><br />A rental company spent $2500 on a new air compressor 7 years ago. The annual rental income from the compressor has been $750. The $100 spent on maintenance the first year has in- creased each year by $25. The company plans to sell the compressor at the end of next year for $150. Construct the cash flow diagram from the company’s perspective and indicate where the present worth now is located. <br /> <br /><b>Solution </b><br /><br />Let now be time t = 0. The incomes and costs for years 7 through 1 (next year) are tabulated below with net cash flow computed using Equation [1.5]. The net cash flows (one negative, eight positive) are diagrammed in Figure 1–9 . Present worth P is located at year 0. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-z20MjrSgoDw/UrDFuRLb2PI/AAAAAAAAMiE/hcqoaZsf5QA/s1600/34.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-z20MjrSgoDw/UrDFuRLb2PI/AAAAAAAAMiE/hcqoaZsf5QA/s1600/34.gif" /></a></div>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0tag:blogger.com,1999:blog-7204171490530605416.post-20761053348077614012013-12-14T05:52:00.003-08:002013-12-14T05:52:50.961-08:00Engineering Economics - Terminology and Symbols The equations and procedures of engineering economy utilize the following terms and symbols. <br />Sample units are indicated. <br /><br /> P = value or amount of money at a time designated as the present or time 0. Also P is referred to as present worth (PW), present value (PV), net present value (NPV), discounted cash ﬂ ow (DCF), and capitalized cost (CC); monetary units, such as dollars <br /> F = value or amount of money at some future time. Also F is called future worth (FW) and future value (FV); dollars <br /> A = series of consecutive, equal, end-of-period amounts of money. Also A is called the annual worth (AW) and equivalent uniform annual worth (EUAW); dollars per year, euros per month <br /><br /> n = number of interest periods; years, months, days <br /> i = interest rate per time period; percent per year, percent per month <br /> t = time, stated in periods; years, months, days<br /><br />The symbols P and F represent one-time occurrences: A occurs with the same value in each inter- est period for a speciﬁ ed number of periods. It should be clear that a present value P represents a single sum of money at some time prior to a future value F or prior to the ﬁ rst occurrence of an equivalent series amount A . <br /><br />It is important to note that the symbol A always represents a uniform amount (i.e., the same <br />amount each period) that extends through consecutive interest periods. Both conditions must exist before the series can be represented by A . <br /><br />The interest rate i is expressed in percent per interest period, for example, 12% per year. Un- less stated otherwise, assume that the rate applies throughout the entire n years or interest periods. The decimal equivalent for i is always used in formulas and equations in engineering econ- omy computations. <br /><br />All engineering economy problems involve the element of time expressed as n and interest rate i . In general, every problem will involve at least four of the symbols P , F , A , n , and i , with at least three of them estimated or known. <br /><br />Additional symbols used in engineering economy are deﬁ ned in Appendix E<br /><br /><span style="font-size: large;"><b>EXAMPLE 1.6</b></span><br /><br />Today, Julie borrowed $5000 to purchase furniture for her new house. She can repay the loan in either of the two ways described below. Determine the engineering economy symbols and their value for each option. <br /><br /> <b>(a)</b> Five equal annual installments with interest based on 5% per year. <br /> <b> (b)</b> One payment 3 years from now with interest based on 7% per year. <br /> <br /><b>Solution </b><br /><br /> <b> (a)</b> The repayment schedule requires an equivalent annual amount A , which is unknown.<br /> P = $5000 i =5% per year n = 5 years A = ? <br /> <b>(b)</b> Repayment requires a single future amount F, which is unknown.<br /> P = $5000 i = 7% per year n = 3 years F = ? <br /><br /><span style="font-size: large;"><b>EXAMPLE 1.7</b></span><br /><br />You plan to make a lump-sum deposit of $5000 now into an investment account that pays 6% per year, and you plan to withdraw an equal end-of-year amount of $1000 for 5 years, starting next year. At the end of the sixth year, you plan to close your account by withdrawing the re- maining money. Deﬁ ne the engineering economy symbols involved. <br /> <br /><b>Solution </b><br /><br />All five symbols are present, but the future value in year 6 is the unknown.<br /><br />P = $5000<br />A = $1000 per year for 5 years<br />F = ? at end of year 6<br />i = 6% per year<br />n = 5 years for the A series and 6 for the F value <br /><br /><span style="font-size: large;"><b>EXAMPLE 1.8</b></span><br /><br />Last year Jane’s grandmother offered to put enough money into a savings account to generate $5000 in interest this year to help pay Jane’s expenses at college. (a) Identify the symbols, and (b) calculate the amount that had to be deposited exactly 1 year ago to earn $5000 in interest now, if the rate of return is 6% per year. <br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-YuERL4MviWs/UqxipPDcEgI/AAAAAAAAMe4/ZVZhkpNr2so/s1600/3.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-YuERL4MviWs/UqxipPDcEgI/AAAAAAAAMe4/ZVZhkpNr2so/s1600/3.gif" /></a></div>Edin O'Brienhttps://plus.google.com/108220878186807451656noreply@blogger.com0