Calculations for Cash Flows That Are Shifted

When a uniform series begins at a time other than at the end of period 1, it is called a shifted series. In this case several methods based on factor equations or tabulated values can be used to find the equivalent present worth P.

Figure (2–15): A uniform series that is shifted.

Figure (2–16): Location of present worth for the shifted uniform series in Figure (2–15)

For example, P of the uniform series shown in Figure (2-15) could be determined by any of the following methods:

§  Use the P/F factor to find the present worth of each disbursement at year 0 and add them.

§  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).

§  Use the F/A factor to find the future amount F=A (F/A, i, 10), and then compute the present worth using P = F (P/F, i, 3).

§  Use the P/A factor to compute the “present worth” (which will be located in year 3 not year 0), and then find the present worth in year 0 by using the (P/F, i, 3) factor. (Present worth is enclosed in quotation marks here only to represent the present worth as determined by the P/A factor in year 3, and to differentiate it from the present worth in year 0.)

Typically, the last method is used. For Figure (2-15), the “present worth” obtained using the P/A factor is located in year 3. This is shown as P₃ in Figure (2-16).

To determine a future worth or F value, recall that the F/A factor has the F located in the same period as the last uniform-series amount. Figure (2-17) shows the location of the future worth when F/A is used for Figure (2-15) cash flows.

It is also important to remember that the number of periods n in the P/A or F/A factor is equal to the number of uniform-series values. It may be helpful to renumber the cash flow diagram to avoid

errors in counting. Figure (2-17) shows Figure (2-15) renumbered to determine n = 10.

Figure (2–17): Placement of F and renumbering for n for the shifted uniform series of Figure (2-15)

As stated above, there are several methods that 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. There are specific steps that should be followed in order to avoid errors:

1.      Draw a diagram of the positive and negative cash flows.

2.      Locate the present worth or future worth of each series on the cash flow diagram.

3.      Determine n for each series by renumbering the cash flow diagram.

4.      Set up and solve the equations.

An engineering technology group just purchased new 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 of the payments if the interest rate is 8% per year?

The cash flow diagram is shown in Figure (2-18). The symbol P_A is used to represent the present worth of a uniform annual series A, and  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  and the diagram renumbering to obtain n are also indicated.

Figure (2–18): Cash flow diagram with placement of P values, Example (2-6)

 Note that is located in actual year 2, not year 3. Also, n = 6 not 8, for the P/A factor. First find the value of   the shifted series.

Since  is located in year 2, now find P_A in year 0.

The total present worth is determined by adding P_A and the initial payment P₀ in year 0.

To determine the present worth for a cash flow that includes both uniform series and single amounts at specific times, use the P/F factor for the single amounts and the P/A factor for the series. To calculate A for the cash flows, first convert everything to a P value in year 0, or an F value in the last year. Then obtain the A value using the A/P or A/F factor, where n is the total number of years over which the A is desired.

Many of the considerations that apply to shifted uniform series apply to arithmetic gradient series as well. Recall that 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 the 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 factor is determined by the renumbered period where the last gradient increase occurs. The P/G factor values and placement of the gradient series present worth P_G for the shifted arithmetic gradients in Figure (2-19) are indicated.

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 (2-19 b). To find the equivalent annual series in years 1 through 10 for the arithmetic gradient series only, first find the present worth of the gradient in year 5, take this present worth back to year 0, and then 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.

If the cash flow series involves a geometric gradient and the gradient starts at a time other than between periods 1 and 2, it is a shifted gradient. The P_g is located in a manner similar to that for P_G above, and Equation (2-28) is the factor formula.

Figure (2–19): Determination of G and n values used in factors for shifted gradients

Chemical engineers at a Coleman Industries plant in the Midwest have determined that a small amount of a newly available chemical additive will increase the water repellency of Coleman’s tent fabric by 20%. The plant superintendent has arranged to purchase the additive through a 5-year contract at $7000 per year, starting 1 year from now. He expects the annual price to increase by 12% per year starting in the sixth year and thereafter through year 13. Additionally, an initial investment of $35,000 was made now to prepare a site suitable for the contractor to deliver the additive. Use i = 15% per year to determine the equivalent total present worth for all these cash flows.

Figure (2-20) presents the cash flows. The total present worth PT is found using g = 0.12 and i = 0.15. Equation (2-28) is used to determine the present worth P_g for the entire geometric series at actual year 4, which is moved to year 0 using (P/F, 15%, 4).

Figure (2–20): Cash flow diagram including a geometric gradient with g = 12%, Example (2-8).

Note that n = 4 in the (P/A,15%,4) factor because the $7000 in year 5 is the initial amount A₁ in Equation (2-20) for the geometric gradient.