Some problems involve receipts or expenses that are projected to increase or decrease by a uniform amount each period, thus constituting an arithmetic sequence of cash flows. For example, because of leasing a certain type of equipment, maintenance and repair savings relative to purchasing the equipment may increase by a roughly constant amount each period. This situation can be modeled as a uniform gradient of cash flows.
Figure 4-13 is a cash-flow diagram of a sequence of end-of-period cash flows increasing by a constant amount, G, in each period. The G is known as the uniform gradient amount. Note that the timing of cash flows on which the derived formulas and tabled values are based is as follows:
End of Period Cash Flows
1 (0)G
2 (1)G
3 (2)G
· ·
· ·
· ·
N − 1 (N − 2)G
N (N − 1)G
Notice that the first uniform gradient cash flow, G, occurs at the end of period two.
(N ! 1)G (N ! 2)G
(N ! 3)G
3G 2G
G 1
0 2 3 4 N ! 2 N ! 1 N
End of Period i " Effective Interest
Rate per Period
Figure 4-13 Cash-Flow Diagram for a Uniform Gradient Increasing by G Dollars per Period
4.11.1 Finding P when Given G
The present equivalent, P, of the arithmetic sequence of cash flows shown in Figure 4-13 is
If we add in the dummy term G(0)/(1 + i)1to represent the “missing” cash flow at time one, we can rewrite the above equation as:
P = G
!N n=1
(n − 1) (1 + i)n.
Recognizing the above equation as the summation of a geometric sequence, we can make the appropriate substitutions as we did in the development of Equation (4-6).
After some algebraic manipulation, we have P = G
The term in braces in Equation (4-23) is called the gradient to present equivalent conversion factor. It can also be expressed as (1/i)[(P/A, i%, N) − N(P/F, i%, N)].
Numerical values for this factor are given in column 8 of Appendix C for a wide assortment of i and N values. We shall use the functional symbol (P/G, i%, N) for this factor. Hence,
P = G (P/G, i%, N). (4-24)
4.11.2 Finding A when Given G
From Equation (4-23), it is easy to develop an expression for A as follows:
A = P(A/P, i%, N)
The term in brackets in Equation (4-25) is called the gradient to uniform series conversion factor. Numerical values for this factor are given on the right-hand side
of Appendix C for a range of i and N values. We shall use the functional symbol (A/G, i%, N) for this factor. Thus,
A = G (A/G, i%, N). (4-26)
4.11.3 Finding F when Given G
We can develop an equation for the future equivalent, F, of an arithmetic series using Equation (4-23):
F = P(F/P, i%, N)
= G
!1 i
"
(1 + i)N − 1
i(1 + i)N − N (1 + i)N
#$
(1 + i)N
= G
!1 i
"
(1 + i)N − 1
i − N
#$
= G
i (F/A, i%, N) − NG
i . (4-27)
It is usually more practical to deal with present and annual equivalents of arithmetic series.
4.11.4 Computations Using G
Be sure to notice that the direct use of gradient conversion factors applies when there is no cash flow at the end of period one, as in Example 4-20. There may be an A amount at the end of period one, but it is treated separately, as illustrated in Examples 4-21 and 4-22. A major advantage of using gradient conversion factors (i.e., computational time savings) is realized when N becomes large.
EXAMPLE 4-20 Using the Gradient Conversion Factors to Find P and A
As an example of the straightforward use of the gradient conversion factors, suppose that certain EOY cash flows are expected to be $1,000 for the second year, $2,000 for the third year, and $3,000 for the fourth year and that, if interest is 15% per year, it is desired to find
(a) present equivalent value at the beginning of the first year,
(b) uniform annual equivalent value at the end of each of the four years.
Solution
0 1 2 3 4
$1,000 $2,000 $3,000
End of Year
Observe that this schedule of cash flows fits the model of the arithmetic gradient formulas with G = $1,000 and N = 4. Note that there is no cash flow at the end of the first period.
(a) The present equivalent can be calculated as
P0= G(P/G, 15%, 4) = $1,000(3.79) = $3,790.
(b) The annual equivalent can be calculated from Equation (4-26) as A = G(A/G, 15%, 4) = $1,000(1.3263) = $1,326.30.
Of course, once P0is known, the value of A can be calculated as A = P0(A/P, 15%, 4) = $3,790(0.3503) = $1,326.30.
EXAMPLE 4-21 Present Equivalent of an Increasing Arithmetic Gradient Series As a further example of the use of arithmetic gradient formulas, suppose that we have cash flows as follows:
End of Year Cash Flows ($)
1 5,000
2 6,000
3 7,000
4 8,000
Also, assume that we wish to calculate their present equivalent at i = 15% per year, using gradient conversion factors.
Solution
The schedule of cash flows is depicted in the left-hand diagram of Figure 4-14.
The right two diagrams of Figure 4-14 show how the original schedule can be broken into two separate sets of cash flows, an annuity series of $5,000 plus an arithmetic gradient of $1,000 that fits the general gradient model for which factors are tabled. The summed present equivalents of these two separate sets of cash flows equal the present equivalent of the original problem. Thus, using the symbols shown in Figure 4-14, we have
P0T = P0A+ P0G
= A(P/A, 15%, 4) + G(P/G, 15%, 4)
= $5,000(2.8550) + $1,000(3.79) = $14,275 + 3,790 = $18,065.
The annual equivalent of the original cash flows could be calculated with the aid of Equation (4-26) as follows:
AT = A + AG
= $5,000 + $1,000(A/G, 15%, 4) = $6,326.30.
AT is equivalent to P0T because $6,326.30(P/A, 15%, 4) = $18,061, which is the same value obtained previously (subject to round-off error).
P0T
$7,000 $8,000
0 1 2 3 4
$6,000
$5,000
End of Year
P0A
0 1 2 3 4
$5,000
End of Year
! "
P0G
0 1 2 3 4
$1,000 $2,000 $3,000
End of Year Figure 4-14 Breakdown of Cash Flows for Example 4-21
EXAMPLE 4-22 Present Equivalent of a Decreasing Arithmetic Gradient Series For another example of the use of arithmetic gradient formulas, suppose that we have cash flows that are timed in exact reverse of the situation depicted in Example 4-21. The left-hand diagram of Figure 4-15 shows the following sequence of cash flows:
End of Year Cash Flows ($)
1 8,000
2 7,000
3 6,000
4 5,000
Calculate the present equivalent at i = 15% per year, using arithmetic gradient interest factors.
Solution
The right two diagrams of Figure 4-15 show how the uniform gradient can be broken into two separate cash-flow diagrams. In this example, we are subtracting an arithmetic gradient of $1,000 from an annuity series of $8,000.
P0T
$7,000
$8,000
0 1 2 3 4
$6,000
$5,000
End of Year
P0A
0 1 2 3 4
$8,000
End of Year
! "
P0G
0 1 2 3 4
$1,000 $2,000 $3,000
End of Year Figure 4-15 Breakdown of Cash Flows for Example 4-22
So,
P0T = P0A− P0G
= A(P/A, 15%, 4) − G(P/G, 15%, 4)
= $8,000(2.8550) − $1,000(3.79)
= $22,840 − $3,790 = $19,050.
Again, the annual equivalent of the original decreasing series of cash flows can be calculated by the same rationale:
AT = A − AG
= $8,000 − $1,000(A/G, 15%, 4)
= $6,673.70.
Note from Examples 4-21 and 4-22 that the present equivalent of $18,065 for an increasing arithmetic gradient series of payments is different from the present equivalent of $19,050 for an arithmetic gradient of payments of identical amounts, but with reversed timing (decreasing series of payments). This difference would be even greater for higher interest rates and gradient amounts and exemplifies the marked effect of the timing of cash flows on equivalent values.