Calculations For Shifted Gradients

Previously, 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.


The present worth of an arithmetic gradient  will always be located two periods  before  the gradient starts. 

Refer to Figure 2–14 as a refresher for the cash flow diagrams.


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  PT  and total  annual series   A/T.


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.

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.

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 Figure 3–11. To find the equivalent annual series in years 1 through 10 for the gradient series only, first find the present worth PG  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.

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

Previously, we derived the relation   Pg = A1 (  P /A,g,i,n ) to determine the present worth of a geometric gradient  series, including the initial amount   A1. The factor was derived to find the present worth in year 0, with   A1  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 Figure 2–21 as a refresher for the cash flows.

Equation [2.35] is the formula used for the factor. It is not tabulated.


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. 


For shifted, decreasing gradients:

•  The base amount A (arithmetic) or initial amount A1 (geometric) is the largest amount in the first year of the series.
•  The gradient amount is subtracted from the previous year’s amount, not added to it.
•  The amount used in the factors is –G for arithmetic and –g for geometric gradient series.
•  The present worth PG or Pg  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

Figure 3–15 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  PT  is the sum of three components.

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