1. Combining Factors

2. 1- Calculation for Uniform Serious that are shifted When a uniform series begins at a time other than at the end of period 1, it is called at shifted series. In this case several methods can be used to find the equivalent present worth P. Forexample, P of the uniform series shown in Figure below could be determined by any of the following methods:

3. Use the P/F factor to find the present worth of each disbursement at year O and add them. Use the F/P factor to find the future worth of each disbursement in year I3, add them, and then find the present worth of the total, using P = F(P/F,i,l3). Use the F/A factor to find the future amount F = A(F/A,i,l0), and then compute the present worth, using P = F(P/F,i,l3). Use the P/A factor to compute the “present worth“ P3 = A(P/A,i,10) (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. Typically the last method is used for calculating the present worth of a uniform series that does not begin at the end of period 1. The present worth is always located one period prior to the first uniform series amount when using the P/A factor.

4. To determine a future worth or F value, recall that the F/A factor the F located in the same period as the last uniform series amount. The future worth is always located in the same period as the last uniform series amount when using the F /A factor. 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.

5. Specific steps should be followed 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. Draw another cash flow diagram representing the desired equivalent cash flow. 5. Set up and solve the equations.

6. The offshore design group at Bechtel just purchased upgraded 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 in year 0 of the payments if the interest rate is 8% per year? Sol: The cash flow diagram is shown in Figure The symbol PA is used throughout this chapter to represent the present worth of a uniform annual series A, and P'A represents the present worth at a time other than period 0. Similarly, PT represents the total present worth at time 0. The correct placement of P'A and the diagram renumbering to obtain n are also indicated. Note that P'A is located in actual year 2, not year 3. Also, n = 6, not 8, for the P/A factor: First find the value of P'A of the shifted series. P; = $500(P/A,8%,6) Since P'A is located in year 2, now find PA in year 0. PA = P'A(P/F,8%,2) The total present worth is determined by adding PA and the initial payment PO in year O. Pt = P0 + PA = 5000 + 500(P/A,8%,6)(P/F,8%,2) = 5000 + 500(4.6229)(O.8573) = $6981.60

7. Ex: Recalibration of sensitive measuring devices costs $8000 per year. If the machine will be recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent uniform series at 16% per year.

8. Sol: Figure a and b shows the original cash flows and the desired equivalent diagram. To con- vert the $8000 shifted series to an equivalent uniform series over all periods, first convert the uniform series into a present worth or future worth amount. Then either the A/P factor or the A/F factor can be used. Both methods are illustrated here. Present worth method. (Refer to Figure a.) Calculate P'A for the shifted series in year 2, followed by PT in year 0. There are 6 years in the A series. P'A = 8000(P/A,l6%,6) PT = P'A(P/F,l6%,2) = 8000(P/A,l6%,6)(P/F.,l6%,2) = 8000(3.6847)(0.7432) = $21,907.75 The equivalent series A‘ for 8 years can now be determined via the A/P factor. A' = PT(A/P,l6%,8) = $5043.60 Future worth method. (Refer to Figure a.) First calculate the future worth F in year 8. F = 8000(F/A,l6%,6) = $7l,820 The A/F factor is now used to obtain A' over all 8 years. A' = F(A/F,16%,8) = $5043.20

9. 2- Calculations Involving Uniform Serious and Randomly Placed Single Amounts When a cash flow includes both a uniform series and randomly placed single amounts, the procedures of Section (1) are applied to the uniform series and the single-amount formulas are applied to the one-time cash flows. Ex: An engineering company in Wyoming that owns 50 hectares of valuable land has decided to lease the mineral rights to a mining company. The primary objective is to obtain long-term income to finance ongoing projects 6 and l6 years from the present time. The engineering company makes a proposal to the mining company that it pay $20,000 per year for 20 years beginning l year from now, plus $10,000 six years from now and $15,000 sixteen years from now. If the mining company wants to pay off its lease immediately, how much should it pay now if the investment is to make 16% per year?

10. Solution: The cash flow diagram is shown in Figure from the owner‘s perspective. Find the present worth of the 20-year uniform series and add it to the present worth of the two one-time amounts to detemine PT PT= 20,000(P/A,l6%,20) + l0,000(P/F,l6%,6) + l5,000(P/F,16%,l6) = $124,075 Note that the $20,000 uniform series starts at the end of year 1, so the P/A factor determines the present worth at year 0. Diagram including a uniform series and single amounts.

11. Ex: A design-build-operate engineering company in Texas that owns a sizable amount of land plans to lease the drilling rights (oil and gas only) to a mining and exploration company. The contract calls for the mining company to pay $20,000 per year for 20 years beginning 3 years from now (i.e., beginning at the end of year 3 and continuing through year 22) plus $10,000 six years from now and $15,000 sixteen years from now. Utilize engineering economy relations to determine the five equivalent values listed below at 16% per year. 1-Total present worth PT in year 0 2- Future worth Fin year 22 3-Annual series over all 22 years 4- Annual series over the first 10 years 5- Annual series over the last 12 years

13. 4. A over years 1 to 10: This and (5), which follows, are special cases that often occur in engineering economy studies. The equivalent A series is calculated for a number of years different from that covered by the original cash flows. This occurs when a defined study period or planning horizone is preset for the analysis.

15. 3- Calculations for Shifted Gradients 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 [ 8 ], 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. 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. Example 3.5 illustrates this partitioning.

16. Ex: Fujitsu, lnc. has tracked the average inspection cost on a robotics manufacturing line for 8 years. Cost averages were steady at $100 per completed unit for the first 4 years, but have increased consistently by $50 per unit for each of the last 4 years. Analyze the gradient increase, using the P/G factor. Where is the present worth located for the gradient? What is the general relation used to calculate total present worth in year 0 ?

18. Set up the engineering economy relations to compute the equivalent annual series in years 1 through 7 for the cash flow estimates in Figure Example Diagram of a shifted gradient,

19. Diagram used to determine A for a shifted gradient,

20. 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. Example 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. 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 thereafter for the next 8 years. 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.

22. 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, except for the following.

23. Partitioned cash flow of :1 shifted arithmetic gradient.,(a) = (b)— (c).

24. 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.

25. Morris Glass Company has decided to invest funds for the next 5 years so that development of “smart” glass is well funded in the future. This type of new-technology glass uses electro chrome coating to allow rapid adjustment to sun and dark in building glass, as well as assisting with internal heating and cooling cost reduction. The financial plan is to invest first, allow appreciation to occur, and then use the available funds in the future. All cash flow estimates are in $1000 units, and the interest rate expectation is 8% per year. Years 1 through 5: Invest $7000 in year1, decreasing by $1000 per year through year 5. Years 6 through 10: No new investment and no withdrawals. Years 11 through 15: Withdraw $20,000 in year 11, decreasing 20% per year through year 15. Determine if the anticipated withdrawals will be covered by the investment and appreciation plans. If the withdrawal series is over-or underfunded, what is the exact amount available in year 11, provided all other estimates remain the same?

26. Investment series: Decreasing, conventional arithmetic series starting in year 2 with A =$—7000, G = $— I000, and gradient n = 5 years. The present worth in year 0 is PG = —[7000(P/A,8%,5) — l000(P/G,8%,5)] = $—2O,577 Investment and withdrawal series,