2/20/2016rd1 Engineering Economic Analysis Chapter 8  Incremental Analysis

2/20/2016rd2 X Y $15 $10 Year X Y Y - X 0-$10-$20 $ MARR = 6% Y is preferred at RoR 40% over X at 50% Increment earns at 30% Benefit-Cost Graph Rejection $20

2/20/2016rd3 MARR = 6% A B C 20-years First cost $2000$4000$5000 UAB PWBenefits Benefit-Cost Graph A B C i=6% NPW=0

Incremental Analysis ABCMARR = 10% First cost$18K$25K$15KLife 25 years UAB Salvage ~ 0 IRR (%) IRR A-C = (UIRR )  -7.86% C > A IRR B-C = (UIRR 10E )  10.04% B > C 2/20/2016rd4

2/20/2016rd5 Incremental Analysis MARR = 8% A B C First Cost UA Benefits Salvage value Life IRR15%11.81%9.33% Take in order of increasing first cost: A > 15% RoR B-A (IRR ‘( – ))  9.8% => B > A RoR C-B (IRR –1000 – – )) returns 6.75% => B is best.

Incremental RoR Analysis ABCD5-year life First cost Annual income Annual cost IRR (%) B – A ~ (UIRR )  30.35% C – B ~ (UIRR )  - 14% D – B ~ (UIRR )  10% If MARR > 15%Do Nothing 15% >MARR > 0% Select B 2/20/2016rd6

Problem 8-27 A  ( ) B  ( ) B – A  (mapcar #'- b a))  ( ) (sum *)  0 => 0% => a is better than b for any positive rate Check I: (cum+ '( )) ( ) => no positive rate of return for the B – A increment exists Check II: (list-pgf a 8) --> ; (list-pgf b 8) --> (mapcar #'- a c)  A – C  ( ) (cum-add *)  ( ) There is no positive rate of return for which A is better than C => Reject A Check: (list-pgf c 8)  $ > $ (mapcar #' - c d)  ( ) (cum-add *)  ( ) => UIRR (IRR '( ) 0.95)  8.97% > 8% => C is better than D and is best. Check: (list-pgf d 8) --> /20/2016rd7

2/20/2016rd8 Problem 8-2 X Y X - Y First Cost UAB Life (years) RoR9.93%12.11% 7.71% Which is better if a) MARR = 6%?X b) MARR = 9%Y c) MARR = 10%Y d) MARR = 14%Do Nothing

2/20/2016rd9 Problem 8-3 A B B-A First Cost UAB Life (years) 5 5 IRR (%) (UIRR )  15.24% for A (UIRR )  13.34% for B (UIRR )  9.43% for B – A Which is better if a) MARR = 6%? Bb) MARR = 8%? B c) MARR = 10% Ad) MARR = 16% Do Nothing

2/20/2016rd10 Example 8-6 MARR = 6%ABCDE First Cost4K2K6K1K9K UAB Life (UIRR )  9.84 > 6% D is better than MARR (UIRR )  29.12% => B > D (UIRR )  9.62% => A > B (UIRR )  1.97% => A > C (UIRR )  -4.65% => A > E Choose A

Problem 8-8 2/20/2016rd11 Mutually ExclusiveNeutralizationPrecpitation First cost$700K$500K Annual chemical cost 40K 110K Salvage value 175K 125K Life, years 5 5 (UIRR )  26.05% => Select Neutralization

2/20/2016rd12 Problem 8-? MARR = 6% A B C20-year analysis First cost$10K$15K$20K UAB Life (UIRR 10e )  9.96% A (UIRR 15e )  8.84% B (UIRR 20e )  7.01% C (UIRR 5e e3)  7.18% B – A 10-year (UIRR 5e )  0.56% C – B 20-year

2/20/2016rd13 Problem 8-14 MARR = 8%ABC no replacement First cost UAB Salvage Life (yrs) RoR 15%11.83%13.3% (IRR '( ))  9.8% B > A (IRR '( ))  6.75% for C – B Reject C; Conclude B is best.

2/20/2016rd14 Problem 8-15 Year X YY- X IRR (%) Over what range of MARR is Y preferred over X? Y is better for MARR 50%.

2/20/2016rd15 Problem 8-19 Replace B and C when needed. Use MARR = 8% A B C First cost$100$150$200 UAB Life (years) ∞ 20 5 Capitalized Costs Analysis NPW A = 10/0.08 – 100 = $25 NAW B = (A/P, 8%, 20) = $2.34 perpetuity NPW B = 2.34/0.08 = $29.28 NAW C = (A/P, 8%, 5) = $5.39 or perpetuity NPW C = 5.39/0.08 = $67.36 *** C

2/20/2016rd16 Problem 8-21 MARR = 12% n A$-20K10K 5K10K6K B -20K10K10K10K0 C -20K 5K 5K 5K 15K (IRR '( ))  21.35% A (IRR '( ))  23.38% B (IRR '( ))  14.98% C (IRR '(-5 0 6))  9.54% A – B Reject A (IRR '( ))  0.0% C – B Reject C Choose B

2/20/2016rd17 Problem 8-25 A B C D First Cost100K130K200K330K UAB26.38K38.78K47.48K 91.55K Life RoR10%15% 6%12% At a MARR of 8%, which to choose? Reject C & Do Nothing B dominates A as its return is greater for a larger investment. D – B => (UIRR )  9.84% => D is best.  RoR D-B = 9.84% > 8% MARR => Select D.

Problem 8-32 Option A $30,976 tax free retirement annuity Option B $359.60/month for test of life or 20 years Option C $513.80/month for next 10 years What to d 12 * = $ , 12 * = B – A (UIRR )  12.64% C – A (UIRR )  14.97% B – C (IRR '( ) 0.8)  8.836% MARR < 8.836% Choose B < MARR < 14.9% Choose C at 9% 14.9% < MARR < i% Choose A30976(A/P, i%, 20) 2/20/2016rd18

2/20/2016rd19 Problem 8-27 MARR = 6% A B C D First Cost, $ 2K 5K 4K 3K Ann-Benefits Salvage 2K1.5K 1.4K 3K Life RoR 40% -2.4% 1% 43.3% Reject B & C (UIRR )  40%; (UIRR )  -2.38% (UIRR )  0.98%; (UIRR )  43.33% (IRR '( ))  51.9% D - A (cum-add '( ))  ( ) => unique positive RoR 2800(P/F, 6%, 1) = $ (IRR '( ))  41.1%

2/20/2016rd20 Problem 8-29 MARR = 8%Atlas Zippy First cost$6700$16,900 AO&M cost UAB Salvage Life (years) 3 6 (UIRR )  % for Atlas (UIRR )  8.983% for Zippy Atlas cf: Zippy cf: (IRR '( ))  6.802% Select Atlas

2/20/2016rd21 Problem 8-33 A B B-A C First Cost$100K $300K 200K$500K Annual Benefit 30K 66K 33K 80K Profit Rate (%) 30% 22% 18% 16% MARR = 20% thus eliminating C. B – A cash flow is -200K 36K returns a profit rate of 18%. Thus best to choose A as the $200K difference can be making MARR money at 20%.

Problem 8-34 ABCD $30K Budget First cost$10K$18K$25K$30K MARR = 15% AB 4K 6K 7.5K 9K AOC 2K 3K 3K 4K A earns 2K * 20K = $5K / year B earns 3K * 12K = $4800 / year C earns 4.5K * 5K = $5200 / year *** D earns 5K * 0 = $5K Choose C 2/20/2016rd22

Problem month lease costing $267/month for $9400 car which can be bought for 24 equal monthly payments at 12% APR. Assume car salvage value is $4700. Lease or buy? 9400(A/P, 1%, 24) = $ = ( – 267)(F/A, i%, 24) (F/A, i%, 24) = % APR => Lease at rate above 11.28%. 2/20/2016rd23

ME; MARR = 9%; Life 10 years ABCDE First cost$4K$5K$2K$3K$6K UAB$797$885$259$447$ (UIRR )  5% Reject 2.(UIRR )  8% Reject D 3.(UIRR )  15% Accept A 4.(UIRR )  -% => Reject B 5. (UIRR 2K )  5.52% => Reject C; A is best. 2/20/2016rd24