Chapter 7

1. Chapter 7 Rate of Return Analysis: Single Alternative

2. Learning Objectives • Definition of ROR • ROR using PW and AW • Cautions about ROR • Multiple RORs • Composite ROR

3. Rate of return (ROR) is the rate paid on the unpaid balance of borrowed money, or the rate earned on the unrecovered balance of an investment, so that the final payment of receipt brings the balance to exactly zero with interest considered.

4. Example: Wells Fargo bank lent a newly graduated engineer $1000 at i=10% per year for 4 years to buy home office equipment. From the bank’s perspective (the lender), the investment in this young engineer is expected to produce an equivalent net cash flow of $315.47 for each of 4 years. A = $1000 (A/P,10%,4) = $315.47 This represents a 10% per year rate of return on the bank’s unrecovered balance. Compute the amount of the unrecovered investment for each of the 4 years using a) the rate of return on the unrecovered balance (the correct basis) b) the return on the initial $1000 investment. c) explain why all the initial $1000 amount is not recovered by the final payment in part b)

5. a)

6. b)

7. Rate of Return Calculation Using a PW or AW Equation • The present worth of costs or disbursements PWD is equated to the present worth of incomes or receiptsPWR. Net Present Worth = 0 • The annual worth approach utilizes the AW values in the same fashion to solve for i. Net Annual Worth = 0 • The I value that makes these equations numerically correct is called i*. To determine if the alternative’s cash flow series is viable, compare i* with the established MARR. • If i* >= MARR, accept the alternative as economically viable. • If i* < MARR, the alternative is not economically viable.

8. $1500 $500 0 5 3 $1000 • Example: • If you deposit $1000 now and are promised payments of $500 three years from now and $1500 five years from now, the rate of return using PW factors is NPV = -1000+500(P/F,i*,3)+1500(P/F,i*,5) 0 = -1000+500(P/F,i*,3)+1500(P/F,i*,5) i* = ?

9. $700,000 $10,000 0 10 $500,000 • Example: • The HVAC engineer for a company constructing one of the world’s tallest buildings (Shanghai Financial Center in Peoples’ Republic of China) has requested that $500,000 be spent now during construction on software and hardware to improve the efficiency of the environmental control systems. This is expected to save $10,000 per year for 10 years in energy costs and $700,000 at the end of 10 years in equipment refurbishment costs. Find the rate of return. i* = ?

10. Solution: NPV = -500000+100000(P/A,i*,10)+700000(P/F,i*,10) 0 = -500000+100000(P/A,i*,10)+700000(P/F,i*,10) Use 5% as the first trial: 0 = -500000+100000(P/A,5%,10)+700000(P/F,5%,10) 0 < $6946 The result is positive, indicating that the return is more than 5% Use 6% as the second trial: 0 = -500000+100000(P/A,6%,10)+700000(P/F,6%,10) 0 > $-35,519 Since the interest rate of 6% is too high, linearly interpolate between 5% and 6%. i* = 5.16%

11. Cautions When Using the ROR Method • Multiple i* values. There maybe more than one real-number root to the ROR equation, resulting in more than one i* value. • Reinvestment at i*. When i* is not close to the MARR (e.g., if i* is substantially larger than MARR), this is an unrealistic assumption. • Computational difficulty versus understanding. • Special procedure for multiple alternatives. To correctly use the ROR method to choose from two or more mutually exclusive alternatives requires an analysis procedure significantly different from that used in PW and AW.

12. Multiple Rate of Return Values • Descartes’ rule of signs: • The total number of real-number roots is always less than or equal to the number of sign changes in the series.

13. Cumulative cash flow sign test (Norstrom’s criterion) • Only one sign change in the series of cumulative cash flows which starts negatively, indicates that there is one positive root to the polynomial relation. • To perform this test, determine the series S0, S1,…, Sn. Only if S0<0 and signs change one time in the series is there a single, real-number, positive i*. • Example: The engineering design and testing group for Honda Motor Corp. does contract-based work for automobile manufacturers throughout the world. During the last 3 years, the net cash flows for contract payments have varied widely, as shown below, primarily due to a large manufacturer’s inability to pay its contract fee. • Write the PW-based ROR relation and approximate the i* value(s) by plotting PW vs. i.

14. 100 75 50 25 0 25 50 75 100 • The PW relation is PW = 2000–500(P/F,i,1)-8100(P/F,i,2)+6800(P/F,i,3) • Plot PW vs. i. i*1 = 8% i*2 = 41%

15. Composite Rate of Return: Removing Multiple i* Values • The composite rate of return will be used when: • There are multiple i* values. • The net positive cash flows from the project will earn at a stated rate that is different from any of the multiple i* values. • The composite rate of return i’ is the unique rate of return for a project that assumes that net positive cash flows, which represent money not immediately needed by the project, are reinvested at the reinvestment rate c (Net investment Procedure).

16. Net-Investment Procedure Ft = Ft-1(1+i)+Ct Where t = 1,2,…,n n = total years in project Ct = net cash flow in year t i = c if Ft-1 > 0 (net positive investment) = i’ if Ft-1 < 0 (net negative investment) Set the net-investment relation for year n equal to zero (Fn = 0) and solve for i’. The i’ value obtained is unique for a stated reinvestment rate c.

17. Example: Year Cash Flow 0 50 1 -200 2 50 3 100 C = MARR = 15%; Find the composite rate i’. The net investment for year t=0 is F0 = $50 (positive) F1 = F0(1+c)+C1 = 50(1+0.15)-200 = $-142.50 (negative) F2 = F1(1+i’)+C2 = -142.5(1+i’)+50 (negative) F3 = F2(1+i’)+C3 = [-142.5(1+i’)+50](1+i’)+100 0 = [-142.5(1+i’)+50](1+i’)+100 i’= 3.13%

18. A summary of the relations between c, i’ and i’*.