2/20/2016rd1 engineering economic analysis chapter 8 incremental analysis
DESCRIPTION
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=0TRANSCRIPT
05/05/2305/05/23 rdrd 11
Engineering Economic AnalysisEngineering Economic Analysis
Chapter 8 Incremental Analysis
05/05/2305/05/23 rdrd 22
X
Y
$15
$10
Year X Y Y - X 0 -$10 -$20 $-10 1 15 28 13
MARR = 6% Y is preferred
at RoR 40% over X at 50%
Increment earns at 30%
Benefit-Cost Graph
Rejection
$20
05/05/2305/05/23 rdrd 33
MARR = 6% A B C 20-yearsFirst cost $2000 $4000 $5000UAB 410 639 700PWBenefits 4703 7329 8029
Benefit-Cost Graph
4703
73298029
A
BC
2000 4000 5000
i=6% NPW=0
Incremental AnalysisIncremental Analysis
A B C MARR = 10%First cost$18K $25K $15K Life 25 years
UAB 1055 2125 1020 Salvage ~ 0IRR (%) 7 9 8
IRRA-C = (UIRR 3000 35 25 0) -7.86% C > A
IRRB-C = (UIRR 10E3 1105 25 0) 10.04% B > C
05/05/2305/05/23 rdrd 44
05/05/2305/05/23 rdrd 55
Incremental AnalysisIncremental Analysis
MARR = 8% A B CFirst Cost 1000 2000 3000UA Benefits 150 150 0Salvage value 1000 2700 5600Life 5 6 7IRR 15% 11.81% 9.33%
Take in order of increasing first cost: A > 15%
RoRB-A (IRR ‘(-1000 0 0 0 0 –1000 2850)) 9.8% => B > A
RoRC-B (IRR –1000 –150 -150 -150 –150 -150 -2850 5600)) returns 6.75% => B is best.
Incremental RoR AnalysisIncremental RoR Analysis
A B C D 5-year life First cost100 130 200 330
Annual income 100 90.78 160 164.55Annual cost 73.62 52.00 112.52 73.00
IRR (%) 10 15 6 12B – A ~ (UIRR 30 12.4 5 0) 30.35% C – B ~ (UIRR 70 8.7 5 0) - 14%D – B ~ (UIRR 200 52.77 5 0) 10%If MARR > 15% Do Nothing
15% > MARR > 0% Select B
05/05/2305/05/23 rdrd 66
Problem 8-27Problem 8-27A (-1300 100 130 160 190 220 250 280 310 340 370)B (-1300 10 60 110 160 210 260 310 360 410 460)B – A (mapcar #'- b a)) (0 -90 -70 -50 -30 -10 10 30 50 70 90)(sum *) 0 => 0% => a is better than b for any positive rateCheck I: (cum+ '(0 -90 -70 -50 -30 -10 10 30 50 70 90))
(0 -90 -160 -210 -240 -250 -240 -210 -160 -90 0)=> no positive rate of return for the B – A increment exists
Check II: (list-pgf a 8) --> 150.31; (list-pgf b 8) --> 65.94(mapcar #'- a c) A – C (0 -160 -130 -100 -70 -40 -10 20 50 80 110)(cum-add *) (0 -160 -290 -390 -460 -500 -510 -490 -440 -360 -250)
There is no positive rate of return for which A is better than C => Reject ACheck: (list-pgf c 8) $444.62 > $ 150.31(mapcar #' - c d) (0 -190 -140 -90 -40 10 60 110 160 210 260)(cum-add *) (0 -190 -330 -420 -460 -450 -390 -280 -120 90 350) => UIRR
(IRR '(-190 -140 -90 -40 10 60 110 160 210 260) 0.95) 8.97% > 8% => C is better than D and is best. Check: (list-pgf d 8) --> 420.69
05/05/2305/05/23 rdrd 77
05/05/2305/05/23 rdrd 88
Problem 8-2Problem 8-2
X Y X - YFirst Cost -100 -50 -50UAB 31.5 16.5 15Life (years) 4 4 4RoR 9.93% 12.11% 7.71%Which is better if a) MARR = 6%? Xb) MARR = 9% Yc) MARR = 10% Yd) MARR = 14% Do Nothing
05/05/2305/05/23 rdrd 99
Problem 8-3Problem 8-3 A B B-A
First Cost -100 -150 -50 UAB 30 43 13Life (years) 5 5IRR (%) 15.24 13.34 9.43
(UIRR 100 30 5 0) 15.24% for A(UIRR 150 43 5 0) 13.34% for B(UIRR 50 13 5 0) 9.43% for B – AWhich is better if a) MARR = 6%? B b) MARR = 8%? B
c) MARR = 10% A d) MARR = 16% Do Nothing
05/05/2305/05/23 rdrd 1010
Example 8-6Example 8-6
MARR = 6% A B C D EFirst Cost 4K 2K 6K 1K 9KUAB 639 410 761 117 785Life 20 20 20 20 20
(UIRR 1 0.117 20) 9.84 > 6% D is better than MARR(UIRR 1 0.293 20) 29.12% => B > D(UIRR 2 0.229 20) 9.62% => A > B(UIRR 2 0.122 20) 1.97% => A > C(UIRR 5 0.146 20) -4.65% => A > E Choose A
Problem 8-8Problem 8-8
05/05/2305/05/23 rdrd 1111
Mutually Exclusive Neutralization PrecpitationFirst cost $700K $500K
Annual chemical cost 40K 110K
Salvage value 175K 125K
Life, years 5 5
(UIRR 200 70 5 50) 26.05% => Select Neutralization
05/05/2305/05/23 rdrd 1212
Problem 8-?Problem 8-?
MARR = 6% A B C 20-year analysisFirst cost $10K $15K $20KUAB 1625 1625 1890Life 10 20 20(UIRR 10e3 1625 10) 9.96% A(UIRR 15e3 1625 20) 8.84% B(UIRR 20e3 1890 20) 7.01% C(UIRR 5e3 0 10 10e3) 7.18% B – A 10-year(UIRR 5e3 265 20 0) 0.56% C – B 20-year
05/05/2305/05/23 rdrd 1313
Problem 8-14Problem 8-14
MARR = 8% A B C no replacementFirst cost 1000 2000 3000UAB 150 150 0Salvage 1000 2700 5600Life (yrs) 5 6 7RoR 15% 11.83% 13.3%
(IRR '(-1000 0 0 0 0 -1000 2850)) 9.8% B > A (IRR '(-1000 -150 -150 -150 -150 -150 -2850 5600)) 6.75% for C – B < 8% => Reject C; Conclude B is best.
05/05/2305/05/23 rdrd 1414
Problem 8-15Problem 8-15
Year X Y Y- X0 -10 -20 -101 15 28 13IRR (%) 50 40 30
Over what range of MARR is Y preferred over X?
Y is better for MARR < 30%X is better for 30% < MARR < 50%Do Nothing for MARR > 50%.
05/05/2305/05/23 rdrd 1515
Problem 8-19Problem 8-19Replace B and C when needed. Use MARR = 8%
A B CFirst cost $100 $150 $200UAB 10 17.62 55.48Life (years) ∞ 20 5
Capitalized Costs AnalysisNPWA = 10/0.08 – 100 = $25
NAWB = 17.62 -150(A/P, 8%, 20) = $2.34 perpetuity NPWB = 2.34/0.08 = $29.28
NAWC = 55.48 -200(A/P, 8%, 5) = $5.39 orperpetuity NPWC = 5.39/0.08 = $67.36 *** C
05/05/2305/05/23 rdrd 1616
Problem 8-21 MARR = 12%Problem 8-21 MARR = 12%
n 0 1 2 3 4A $-20K 10K 5K 10K 6K B -20K 10K 10K 10K 0C -20K 5K 5K 5K 15K
(IRR '(-20 10 5 10 6)) 21.35% A(IRR '(-20 10 10 10 0)) 23.38% B(IRR '(-20 5 5 5 15)) 14.98% C(IRR '(-5 0 6)) 9.54% A – B Reject A(IRR '(-5 -5 -5 15)) 0.0% C – B Reject CChoose B
05/05/2305/05/23 rdrd 1717
Problem 8-25Problem 8-25
A B C DFirst Cost 100K 130K 200K 330KUAB 26.38K 38.78K 47.48K 91.55KLife 5 5 5 5RoR 10% 15% 6% 12%At a MARR of 8%, which to choose? Reject C & Do NothingB dominates A as its return is greater for a larger investment.D – B => (UIRR 200 52.55 5) 9.84% => D is best.RoRD-B = 9.84% > 8% MARR => Select D.
Problem 8-32Problem 8-32
Option A $30,976 tax free retirement annuityOption B $359.60/month for test of life or 20 yearsOption C $513.80/month for next 10 years What to d12 * 359.6 = $4315.20, 12 * 513.80 = 6165.60 B – A (UIRR 30976 4315.20 20) 12.64%C – A (UIRR 30976 6165.60 10) 14.97%B – C (IRR '(-1850.4 -1850.4 -1850.4 -1850.4 -1850.4 -1850.4 -1850.4
-1850.4 -1850.4 -1850.4 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2 4315.2) 0.8) 8.836%
MARR < 8.836% Choose B8.836 < MARR < 14.9% Choose C at 9%14.9% < MARR < i% Choose A 30976(A/P, i%, 20)
05/05/2305/05/23 rdrd 1818
05/05/2305/05/23 rdrd 1919
Problem 8-27Problem 8-27MARR = 6% A B C DFirst Cost, $ 2K 5K 4K 3KAnn-Benefits 800 500 400 1300Salvage 2K 1.5K 1.4K 3KLife 5 6 7 4RoR 40% -2.4% 1% 43.3% Reject B & C
(UIRR 2 0.8 5 2) 40%; (UIRR 5 0.5 1.5 6 1.5) -2.38%(UIRR 4 0.4 7 1.4) 0.98%; (UIRR 3 1.3 4 3) 43.33%
(IRR '(-1000 500 500 500 3500 -2800)) 51.9% D - A
(cum-add '(-1000 500 500 500 3500 -2800)) (-1000 -500 0 500 4000 1200) => unique positive RoR2800(P/F, 6%, 1) = $2641.51
(IRR '(-1000 500 500 500 858.49 0)) 41.1%
05/05/2305/05/23 rdrd 2020
Problem 8-29Problem 8-29MARR = 8% Atlas ZippyFirst cost $6700 $16,900AO&M cost 1500 1200UAB 4000 4500Salvage 1000 3500Life (years) 3 6(UIRR 6700 2500 3 1000) 12.134% for Atlas(UIRR 16900 3300 6 3500) 8.983% for Zippy
Atlas cf: -6700 2500 2500 -3200 2500 2500 3500Zippy cf: -16900 3300 3300 3300 3300 3300 6800 (IRR '(-10200 800 800 6500 800 800 3300)) 6.802%
Select Atlas
05/05/2305/05/23 rdrd 2121
Problem 8-33Problem 8-33
A B B-A CFirst Cost $100K $300K 200K $500KAnnual Benefit 30K 66K 33K 80KProfit 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 8-34Problem 8-34
A B C D $30K BudgetFirst cost$10K $18K $25K $30K MARR = 15%
AB 4K 6K 7.5K 9KAOC 2K 3K 3K 4K
A earns 2K + 0.15 * 20K = $5K / yearB earns 3K + 0.15 * 12K = $4800 / yearC earns 4.5K + 0.15 * 5K = $5200 / year ***D earns 5K + 0.15 * 0 = $5K
Choose C
05/05/2305/05/23 rdrd 2222
Problem 8-35Problem 8-35
24-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) = $442.49 4700 = (442.49 – 267)(F/A, i%, 24)
(F/A, i%, 24) = 26.78215 < 1% 0.94% => 11.28% APR
=> Lease at rate above 11.28%.
05/05/2305/05/23 rdrd 2323
ME; MARR = 9%; Life 10 yearsME; MARR = 9%; Life 10 years
A B C D EFirst cost$4K $5K $2K $3K $6K
UAB $797 $885 $259 $447 $1063
1. (UIRR 2000 259 10 0) 5% Reject 2. (UIRR 3000 447 10 0) 8% Reject D3. (UIRR 4000 797 10 0) 15% Accept A4. (UIRR 1000 88 10 0) -% => Reject B5. (UIRR 2K 266 10 0) 5.52% => Reject C; A is best.
05/05/2305/05/23 rdrd 2424