2/22/2016rd Multiple Rates of Return nCash FlowCumulativeAdjusted 10% 20% 0-$500-$500 -$500 -$ (quadratic –660)  ( ) (irr '( –660) 0.6)  10%; 20%

2/22/2016rd Multiple RoR n cf Compute RoR on internal investment with 10% external rate. New-cash flow (1.1) = $22 added to –40 get –18 and 1 sign change. (IRR ‘(-50 0 – ))  14.91%. (IRR '( ))  15.38%

2/22/2016rd Use rate of return (RoR) analysis for the following 3 mutually exclusive alternatives in reference to an unknown MARR. A B C First Cost$200$300$600 Uniform annual benefits Useful life (years) End salvage Computed RoR 15% 9% 11.7% Incremental RoR B - A => 100 = 17.4(P/A, i%, 5) => i = -4.47% C - A => 400 = 105.5(P/A, i%,5) => i = 10% C - B => 300 = 88.1(P/A,i%,5) => i = 14.3% Conclude: if MARR  9% Choose C 9%  MARR  10% Choose C Reject B 10%  MARR  11.7% Choose A Reject B 11.7%  MARR  15% Choose A

2/22/2016rd Multiple RoRs n cf PW(20%) = (1.2) -1 –5580(1.2) (1.2) -3 = – = 0 PW(40%) = (1.4) -1 –5580(1.4) (1.4) -3 = 0 PW(50%) = (1.5) -1 –5580(1.5) (1.5) -3 = 0.

2/22/2016rd 7A-17 n cf External rate at 12% (IRR ‘( ))  7.22% (IRR ‘( ))  43.96% (MIRR ‘( –394) 6 12)  9.5% At 12% the $358 in year 5 can be transformed to pay at n = * 1.12 = (IRR ‘( ))  7.63% (list-pgf '( ) )  e-4

2/22/2016rd 7A-18 n A K1K 1K (IRR ‘( ))  9.995% (Cum-add ‘( – )) returns ( ) => unique RoR (list-pgf '( ) 9.995) 

2/22/2016rd Incremental Analysis MARR = 8%ABA-B First Cost$100$50 50 UAB Life (years) RoR15%20% 9.6% => A 0 < MARR < 9.6% A is better If 9.6% < MARR < 20%, B is better. NPW A (9.6%) = $24.59 = NPW B (9.6%) A earned at B’s rate (20%) for the first $50 and at 9.6% for the next $50 (increment).

2/22/2016rd Incremental Analysis MARR = 6% ABCDE 1 st Cost UAB Life (years) RoR15%20%11%10%6% Start with D, better than Do Nothing, Challenger is B. RoR B-D (UIRR )  29.12% => B is better than D RoR A-B (UIRR )  9.63% => A is better than B RoR C-A (UIRR )  1.97% => A is better than C RoR E-A (UIRR )  -4.65% => A is best

2/22/2016rd Investment Decision Net cash flow: –1,000,000 2,300,000 –1,320,000 (2 years) MARR = 15%, quadratic roots => 10% and 20% RoRs NPW(15%) = (1.15) (1.15) -2 = $ > 0 => Invest cautiously. 2,300K -1320K(P/F, 15%, 1) = $1,152, New cash flow [–1,000,000 1,152, ] with RoR at 15.22% if MARR rate of 15% is used to transfer year-one amount to cover year-two amount.

2/22/2016rd Higher IRR Not Sufficient Mutually Exclusive Alternatives n A B IRR 100% 40% PW(10%) $818.18$ , B is better

2/22/2016rd View Point n0123 IRR A % B % B - A % (lending or investing) A – B %(borrowing) Do you see why we strive to make the first difference negative?

2/22/2016rd IRR on Incremental Investment n A B A - B IRR 18% 20% 14.71% If MARR = 12%, then A is better

2/22/2016rd Unequal Service Lives nA B BB - A MARR = 10% and can repeat service life. (IRR '( ))  44.22% B is better.

2/22/2016rd Infinite Cash Flow Find the rate of return for the following infinite cash flow: -18,976 3, , , … Ans. 17%. Perpetuity => RoR = / 18,976 (irr (cons (list-of ))) 

2/22/2016rd Find X Given RoR Find minimum X to make at least a 10% return on investment. n cf X1200ans. $ X = [2000 – 1000(1.1) -1 – 1200(1.1) -3 ] / (1.1) -2

Computing the MIRR Compute the MIRR for the following cash flow using 6% for the borrowing rate and 12% for the investing rate. n0123 cf (mirr '( ) 6 12)  11.87% (list-pgf '( ) 6)  $ (list-fgp '( ) 12)  $ (igpfn )  11.87% 2/22/2016rd

2/22/2016rd Compute the MIRR Find the MIRR for the following cash flow by using 5% for borrowing rate and 9% for the investment rate. n cf P 0 for the negatives at the borrowing rate F n for the positives at the investing rate Then find i given P. F and n Ans % 2/22/2016rd17