Arithmetic 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.

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 costs $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 Figure (2–5). 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_n for cash flow in year n as follows.

Figure (2–5): Cash flow diagram of an arithmetic gradient series.

It is important to realize that the base amount defines a uniform cash flow series of the size A that occurs each 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 Figure (2–6). Note that the gradient begins between years 1 and 2. This is called a conventional gradient.

Figure (2–6): Conventional arithmetic gradient series without the base amount.

A local university has initiated a logo-licensing program with the clothier Holister, Inc. Estimated fees (revenues) are $80,000 for the first year with uniform increases to a total of $200,000 by the end of year 9. Determine the gradient and construct a cash flow diagram that identifies the base amount and the gradient series.

The year 1 base amount is CF₁ = $80,000, and the total increase over 9 years is

Equation (2-15), solved for G, determines the arithmetic gradient.

The cash flow diagram Figure (2–7) 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.

Figure (2–7): Diagram for gradient series, Example (2-4).

The total present worth P_T for a series that includes a base amount A and conventional arithmetic gradient must consider the present worth of both the uniform series defined by A and the arithmetic gradient series. The addition of the two results in P_T.

where P_A is the present worth of the uniform series only, P_G is the present worth of the gradient series only, and the – or + sign is used for an increasing (–G) or decreasing (+G) gradient, respectively.

The corresponding equivalent annual worth A_T is the sum of the base amount series annual worth A_A and gradient series annual worth A_G, that is,

Three factors are derived for arithmetic gradients: The P/G factor for present worth, the A/G factor for annual series, and the F/G factor for future worth. There are several ways to derive them. We use the single-payment present worth factor (P/F, i, n ), but the same result can be obtained by using the F/P , F/A , or P/A factor.

In Figure (2–6), the present worth at year 0 of only the gradient is equal to the sum of the present worth’s of the individual cash flows, where each value is considered a future amount.

Factor out G and use the P/F formula.

Multiplying both sides of Equation (2-18) by yields

Subtract Equation (2-18) from Equation (2-19) and simplify.

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 (2-7).

Figure (2–8): Conversion diagram from an arithmetic gradient to a present worth.

into Equation (2-20) and simplify to solve for P_G, the present worth of the gradient series only.

Equation (2-21) 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 (2–8 a) is converted into the equivalent cash flow in Figure (2–8 b). The arithmetic gradient present worth factor, or P/G factor, may be expressed in two forms:

Equation (2-21) expressed as an engineering economy relation is

which is the rightmost term in Equation (2-16) to calculate total present worth. The G carries a minus sign for decreasing gradients.

The equivalent uniform annual series A_G for an arithmetic gradient G is found by multiplying the present worth in Equation (2-23) by the (A/P, i, n) formula. In standard notation form, the equivalent of algebraic cancellation of P can be used.

In equation form,

which is the rightmost term in Equation (2-17). The expression in brackets in Equation (2-24) is called the arithmetic gradient uniform series factor and is identified by (A/ G, i, n). This factor converts Figure (2–9 a) into Figure (2–9 b).

The P/G and A/G factors and relations are summarized inside the front cover. Factor values are tabulated in the two rightmost columns of factor values at the rear of this text.

Figure (2–9): Conversion diagram of an arithmetic gradient series to an equivalent uniform annual series

There is no direct, single-cell spreadsheet function to calculate P_G or A_G for an arithmetic gradient. Use the NPV function to display P_G and the PMT function to display A_G after entering all cash flows (base and gradient amounts) into contiguous cells. General formats for these functions are:

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

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.

1)      Determine the equivalent present worth

2)      Determine the equivalent annual series amounts, if public funds earn at a rate of 5% per year.

1)      The cash flow diagram of this conventional arithmetic gradient series from the perspective of the parishes is shown in Figure (2–10). According to Equation (2-16), two computations must be made and added: the first for the present worth of the base amount P_A and the second for the present worth of the gradient P_G. The total present worth P_T occurs in year 0. This is illustrated by the partitioned cash flow diagram in Figure (2–11). In $1000 units, the total present worth is

Figure (2–10): Cash flow series with a conventional arithmetic gradient (in $1000 units), Example (2-5).

Figure (2–11): Partitioned cash flow diagram (in $1000 units), Example (2-5).

2)      Here, too, it is necessary to consider the gradient and the base amount separately. The total annual series A_T is found by Equation (2-17) and occurs in years 1 through 10.