Example: Equivalence Calculations - Solve P.

Consider the following situation:

Solve for P assuming a 12% interest rate and using the compound interest tables. Recall that receipts have a plus sign and disbursements or payments have a negative sign. Thus, the diagram is:


It its important to understandjust what the solution, $625.16, represents.We can say that $625.16
is tbe amount of moneythat would need to be invested at 12% annual interest to allow for the withdrawal of $400 at the end of 3 years and $600 at the end of 5 years.

Let's examine the computations further.
If  $625.16 is invested for one year at 12% interest, it will increase to [625.16+0.12(625.16)] = $700.18. If for the second year the $700.18 is invested at 12%, it will increase to [700.18+0.12(700.18)] = $784.20. And if this is repeated for another year, [784.20+ 0.12(784.20)] =$878.30.

We are now at the end of Year 3. The original $625.16 has increased through the addition of interest to $878.30. It is at this point that the $400 is paid out. Deducting $400 from $878.30 leaves $478.30.

The $478.30 can be invested at 12%for the fourth year and will increase to [478.30+0.12(478.30)] =$535.70.Andifleft at interest for anotheryear, itwill increaseto [535.70+0.12(535.70)] = $600.We are now at the end of Year 5; with a $600 payout; there is money remaining in the account.

In other words, the $625.16 was just enough money,at a 12%interest rate, to exactly provide for a $400 disbursement at the end of Year 3 and also a $60 0disbursement at the end of Year 5. Weend up neither short of moneyn or with money left over: this is an illustration of equivalence.The initial $625.16 is equivalent to the combination of a $400 disbursement at the end of Year 3 and a $600 disbursementat the end of Year 5.


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