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8/14/2019 Ch 9 Solutions
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9.? (a) @se A$cel and assume an in#inite li#e. /alculate the capitaliBed costs #or all annual amount estimates.
(b) /hange cell 5 to 8"00%000 to get 6/ 7 1.0"3.
/hapter 9 "
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9.C 1.0 7 (1")(value o# a li#e)D ("00)(90%000%000)
Ealue o# a li#e 7 81.! billion
9.9 nnual cost 7 30%000(0.0"!) 7 8?!0 per year6household
Fet $ 7 number o# households *otal annual cost% / 7 (?!0)($)
Fet y 7 8 health bene#it per household #or the 1- o# households *otal annual bene#its% 7 (0.01$)(y)
1.0 7 6/ 7 6(?!0)($) 7 (?!0)($)
ubstitute 7 (0.01$)(y)
(0.01$)(y) 7 ?!0$ y 7 8?!%000 per year
/hapter 9 3
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9.10 ll parts are solved on the spreadsheet once it is #ormatted using cell re#erences.
>ote in part (b) how much larger (81!0%000) than the median income (830%000)the required bene#it becomes as #ewer households are a##ected
9.11 (a) /ost 7 +%000%000(0.0+) 300%000 7 8+50%000 per year 6/ 7 !!0%000 , 90%000 7 1.0
+50%000
(b) /ost 7 +%000%000(0.0+) 7 8150%000 per year
/ 7 (!!0%000 , 90%000) , (150%000 300%000) 7 0.0
*he project is just economically acceptable using bene#it6cost analysis.
/hapter 9 +
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9.1" /ost 7 1!0%000(6P%3-%"0) 1"%000 7 1!0%000(0.05?"") 1"%000 7 8""%0C3 per year
ene#its 7 "+%000(")(0.!0) 7 8"+%000 per year
6/ 7 "+%006""%0C3 7 1.0C? *he project is marginally economically justi#ied.
9.13 (a) yhand solution2 Girst% set up H value relation o# the initial cost% P capitaliBed a ?-. *hen determine P #or 6/ 7 1.3.
1.3 7 500%000DDDDP(0.0?) 300%000
P 7 I(500%00061.3) , 300%000J60.0? 7 8"%30?%59"
(b) preadsheet solution2 et up the spreadsheet to calculate P 7 8"%30?%59".
/hapter 9 !
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9.1+ ame spreadsheet% e$cept change the discount rate and equations #or H and 6/. *he 6/ value is the same at 1.3% so the project is still justi#ied.
/hapter 9 5
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9.1! 1.? 7 1!0%000 , 'K4 costsD 1%000%000(6P%5-%30)
1.? 7 1!0%000 , 'K4 costs 1%000%000(0.0?"5!)
'K4 costs 7 8"5%+9! per year
9.15 /onvert all estimates to PH values.
PH disbene#its 7 +!%000(P6%5-%1!) 7 +!%000(9.?1"") 7 8+3?%0+9
PH 'K4 /ost 7 300%000(P6%5-%1!) 7 300%000(9.?1"") 7 8"%913%550
6/ 7 3%C00%000 , +3?%0+9DD "%"00%000 "%913%550
7 3%35"%9!16!%113%5507 0.55
9.1? (a) H o# /ost 7 30%000%000(0.0C) 100%000 7 8"%!00%000 per year
6/ 7 "%C00%000 7 1.1""%!00%000
/onstruct the dam.
(b) /alculate the // o# the initial cost to obtain H #or denominator.
B/C = 1.12
9.1C H 7 / 7 "%"00%000(0.1") 10%000 5!%000(6G%1"-%1!) 7 "5+%000 10%000 5!%000(0.0"5C")
7 8"?!%?+3/hapter 9 ?
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B/C = (2,800,000)/(100,000+30,000,000*(0.08))
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nnual ene#it 7 7 90%000 , 10%000 7 8C0%000
6/ 7 C0%0006 "?!%?+3 7 0."9
ince 6/ L 1.0% the dam should not be constructed.
9.19 /alculate the H o# initial cost% then the 3 6/ measures o# worth. *he roadwayshould not be built.
/hapter 9 C
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9."0 (a) H 7 / 7 1%!00%000(6P%5-%"0) "!%000 7 1%!00%000(0.0C?1C) "!%000
7 81!!%??0
nnual revenue 7 7 81?!%000
6/ 7 1?!%00061!!%??0 7 1.1"ince 6/ M 1.0% the canals should be e$tended.
(b) Gor modi#ied 6/ ratio%/ 7 1%!00%000(6P%5-%"0) 7 8130%??0 7 1?!%000 , "!%000 7 1!0%000
'odi#ied 6/ 7 1!0%0006130%??0 7 1.1!
ince modi#ied 6/ M 1.0% canals should be e$tended.9."1 (a) etermine the H o# the initial cost% annual cost and recurring dredging cost%
/hapter 9 9
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then calculate ( , )6/.
*he disbene#it o# 81!%000 per year and the dredging cost each third year havereduced the 6/ ratio to below 1.0 the canals should not be e$tended now.
(b) H 7 / 7 1%!00%000(6P%5-%"0) "!%000 50%000[(P6G%5-%3) (P6G%5-%5)
(P6G%5-%9) (P6G%5-%1") (P6G%5-%1!) (P6G%5-%1C)](6P%5-%"0)
7 1%!00%000 (0.0C?1C) "!%000 50%000 [0.C395 0.?0!0 0.!919
0.+9?0 0.+1?3 0.3!03](0.0C?1C)
7 81?3%!50
nnual disbene#it 7 7 81!%000nnual revenue 7 7 81?!%000
( , )6/ 7 (1?!%000 , 1!%000)61?3%!50 7 0.9""
s above% the disbene#it o# 81!%000 per year and the dredging cost each thirdyear have reduced the 6/ ratio to below 1.0 the canals should not bee$tended now.
/hapter 9 10
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9."" lternative has a larger total annual cost it must be incrementally justi#ied. @se PH values. ene#it is the di##erence in damage costs. Gor incrementally over 2
=ncr cost 7 (C00%000 , 500%000) (?0%000 , !0%000)(P6%C-%"0) 7 8"00%000 "0%000(9.C1C1)
7 8395%35"
=ncr bene#it 7 (9!0%000 , "!0%000)(P6G%C-%5) 7 ?00%000(0.530") 7 ++1%1+0
=ncr 6/ 7 ++1%1+06395%35" 7 1.11
elect alternative .
9."3 nnual cost o# long route 7 "1%000%000(0.05) +0%000 "1%000%000(0.10) (6G%5-%10) 7 1%"50%000 +0%000 "%100%000(0.0?!C?) 7 81%+!9%3"?
nnual cost o# short route 7 +!%000%000(0.05) 1!%000 +!%000%000(0.10)(6G%5-%10)
7 "%?00%000 1!%000 +%!00%000(0.0?!C?) 7 83%0!5%+1!
*he short route must be incrementally justi#ied.
A$tra cost #or short route 7 3%0!5%+1! , 1%+!9%3"? 7 81%!9?%0CC
=ncremental bene#its o# short route 7 +00%000(0.3!)("! , 10) 900%000 7 83%000%000
=ncr 6/short 7 3%000%000 1%!9?%0CC 7 1.CC
uild the short route.
9."+ Nusti#y e$tra cost o# downtown (*) location.A$tra cost #or * site 7 11%000%000(0.0C)
7 8CC0%000A$tra bene#its #or * site 7 3!0%000 +00%000
7 8?!0%000=ncremental 6/* 7 ?!0%0006CC0%000
7 0.C!
*he city should build on the west side site.
/hapter 9 11
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9."! Aast coast site has the larger total cost (N1?). et up the spreadsheet to calculateH values in 81 million. Girst% per#orm 6/ on west coast site since donothing isan option. =t is justi#ied. *hen use incremental values to evaluate Aast versusHest. =t is also justi#ied since O(6/) 7 ".0!. elect east coast site.
9."5 Girst compare program 1 to donothing (>).
/ost6household6mo 7 850(6P%0.!-%50) 7 50(0.01933) 7 81.156/1 7 1."!61.15 7 1.0C Aliminate >
/ompare program " to program 1.
Ocost 7 !00(6P%0.!-%50) , 50(6P%0.!-%50) 7 (!00 , 50)(0.01933) 7 8C.!1
Obene#its 7 C , 1."! 7 85.?!
/hapter 9 1"
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=ncr 6/" 7 5.?!6C.!17 0.?9 Aliminate program "
*he utility should underta&e program 1.
9."? @sing the capital recovery costs% solar is the more costly alternative.
Ocost 7 (+%!00%000 , "%000%000)(6P%0.?!-%?") , (1!0%000 , 0)(6G%0.?!-%?")
7 "%!00%000(0.01C03) , 1!0%000(0.010!3) 7 8+3%+95
Obene#its 7 !0%000 , 10%000 7 8+0%000
=ncr 6/ 7 +0%0006+3%+95 7 0.9"
elect the conventional system.
9."C (a) Focation AH 7 / 7 3%000%000(0.1") !0%000
7 8+10%000
evenue 7 7 8!00%000 per year isbene#its 7 7 830%000 per year
Focation HH 7 / 7 ?%000%000 (0.1") 5!%000 "!%000
7 8CC0%000
evenue 7 7 8?00%000 per year isbene#its 7 7 8+0%000 per year
6/ ratio #or location A2( , )6/ 7 (!00%000 , 30%000)6+10%000
7 1.1!
Focation A is economically justi#ied. Focation H is now incrementallycompared to A.
Ocost o# H 7 CC0%000 , +10%000 7 8+?0%000
/hapter 9 13
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9."C (cont) Obene#its o# H 7 ?00%000 , !00%000 7 8"00%000
=ncr disbene#its o# H 7 +0%000 , 30%000 7 810%000
=ncr 6/ 7 ( , )6/ 7 ("00%000 , 10%000)6+?0%000 7 0.+0
ince incr( , )6/ L 1% H is not justi#ied. elect location A.
(b) Focation A 7 !00%000 , 30%000 , !0%000 7 8+"0%000/ 7 3%000%000 (0.1") 7 8350%000'odi#ied 6/ 7 +"0%0006350%000 7 1.1?
Focation A is justi#ied.
Focation HO 7 8"00%000O 7 810%000O/ 7 (? million , 3 million)(0.1")
7 8+C0%000O'K4 7 (5!%000 , "!%000) , !0%000 7 810%000 >ote that 'K4 is now an incremental cost advantage #or H.
'odi#ied O6/ 7 "00%000 , 10%000 10%000 7 0.+"+C0%000
H is not justi#ied select location A.
9."9 et up the spreadsheet to #ind H o# costs% per#orm the initial 6/ analyses usingcell re#erence #ormat. /hanges #rom part to part needed should be the estimatesand possibly a switching o# which options are incrementally justi#ied. ll 3analyses are done on a rolling spreadsheet shown below.(a) ob2 /ompare 1 vs >% then " vs 1. elect option 1.(b) Nudy2 /ompare 1 vs >% then " vs 1. elect option ".
(c) /hen2 /ompare " vs >% then 1 vs ". elect option " without doing the O6/analysis% since bene#its minus disbene#its #or 1 are less% but this option has alarger H o# costs than option ".
/hapter 9 1+
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9."9 (cont)
9.30 Gind the H o# costs #or each technique% order them% and determine the =ncr 6/values.
H o# costs 7 installed cost(6P%1!-%10) 4/
/hapter 9 1!
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9.30 (cont)*echnique H o# cost calculationDD
1 1!%000(6P%1!-%10) 10%000 7 1!%000(0.199"!) 10%000
7 81"%9C9
" 19%000(6P%1!-%10) 1"%000 7 19%000(0.199"!) 1"%000 7 81!%?C5
3 "!%000(6P%1!-%10) 9%000 7 "!%000(0.199"!) 9%000 7 813%9C1
+ 33%000(6P%1!-%10) 11%000
7 33%000(0.199"!) 11%000 7 81?%!?!
4rder o# incremental analysis is2 >% 1% 3% "% +.*echnique 1 vs > (current)
6/ 7 1!%00061"%9C9 7 1.1! M 1.0 Aliminate >% &eep technique 1.
*echnique 3 vs 1
∆/ 7 13%9C1 , 1"%9C9 7 899"
∆ 7 19%000 , 1!%000 7 8+%000 ∆6/ 7 +%000699"
7 +.03 M 1 Aliminate technique 1% &eep 3.
*echnique " vs 3
∆/ 7 1!%?C5 13%9C1 7 81%C0!
∆ 7 "0%000 , 19%000 7 81%000
∆6/ 7 1%00061%C0!
7 0.!! L 1.0 Aliminate technique "% &eep 3.
*echnique + vs 3∆/ 7 1?%!?! 13%9C1 7 83%!9+
∆ 7 ""%000 19%000 7 83%000
∆6/ 7 3%00063%!9+
7 0.C3 L 1.0 Aliminate technique +% &eep 3eplace the current method with technique 3.
/hapter 9 15
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9.31 etermine the H o# costs #or each technique and calculate overall 6/. electall #our since all have 6/ M 1.0.
1 A B C D E2 Technique 1 2 3 434
Inst!!e" c#st 1$,000 1%,000 2$,00033,000
$ A&C 10,000 12,000 %,000 11,000' A # c#sts 12,%8% 1$,8' 13,%81 1,$$
8 Beneits 1$,000 20,000 1%,000 22,000%
10 B/C 1.1$ 1.2 1.3' 1.2$11
12 e!ect- es es es es
9.3" /ombine the investment and installation costs% di##erence in usage #ees de#ine bene#its. @se the procedure in ection 9.3 to solve. ene#its are the incrementalamounts #or lowered costs o# annual usage #or each larger siBe pipe.
1% ". 4rder o# incremental analysis2 iBe 130 1!0 "00 "30
*otal #irst cost% 8 9%?C0 11%310 1+%!C0 1?%3!03. nnual bene#its% 8 "00 500 300
+. >ot used since the bene#its are de#ined by usage costs.
!?. etermine incremental and / and select at each pairwise comparison o# de#ender vs challenger.
#$% &s #'% mm
∆/ 7 (11%310 , 9%?C0)(6P%C-%1!)
7 1%!30(0.115C3)7 81?C.?!
∆ 7 5%000 , !%C00
7 8"00
∆6/ 7 "0061?C.?!
7 1.1" M 1.0 Aliminate 130 mm siBe.
(%% &s #$% mm
∆/ 7 (1+%!C0 , 11%310)(6P%C-%1!)
7 3"?0(0.115C3)7 83C".03
∆ 7 !C00 , !"00
/hapter 9 1?
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A # c#sts = E$ T 1$,10,E4
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7 8500
∆6/ 7 50063C".03
7 1.!? M 1.0 Aliminate 1!0 mm siBe.
('% &s (%% mm
∆/ 7 (1?%3!0 , 1+%!C0)(6P%C-%1!) 7 "??0(0.115C3)
7 83"3.5"
∆ 7 !"00 , +900
7 8300
∆6/ 7 0.93 L 1.0 Aliminate "30 mm siBe.
elect "00 mm siBe.
9.33 /ompare to > since it is not necessary to select one o# the sites.
A &s )*H o# /ost 7 !0(6P%10-%!) 3
7 !0(0."53C0) 3 7 15.19
H o# ene#its 7 "0 , 0.! 7 19.!
6/ 7 19.! 15.19 7 1."0 M 1.0 Aliminate >.
B &s A
∆/ 7 (90 , !0)(6P%10-%!) (+ , 3)7 +0(0."53C0) 17 811.!!"
∆ 7 ("9 , "0) , (1.! , 0.!) 7 C
∆6/ 7 C611.!!"
7 0.59 L 1.0 Aliminate .
C &s A
∆/ 7 ("00 , !0)(6P%10-%!) (5 , 3)
7 1!0(0."53C0) 37 +".!?
∆ 7 (51 , "0) , (".1 , 0.!) 7 39.+
∆6/ 7 39.+6+".!?
/hapter 9 1C
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7 0.93 L 1.0 Aliminate /elect site
9.3+ (a) /alculate the 6/ o# each proposal #or initial screening (row 5). Gour locations are retained , G% % A and ;. >o need to compare G vs > since
one site must be selected. ite is the one selected.
(b) Gor independent projects% use the 6/ values in row 5 o# the A$cel solutionabove and select the largest three o# the #our with 6/ M 1.0. *hose selected
#or are2 % G% and A.
9.3! (a) n incremental 6/ analysis is necessary between and Q% i# these aremutually e$clusive alternatives.
(b) =ndependent projects. ccept and Q% since 6/ M 1.0.
9.35 + &s )*
6/ 7 1.10 M 1.0 Aliminate >.
, &s +
6/ 7 0.+0 L 1.0 Aliminate R.
/hapter 9 19
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" &s +
6/ 7 1.+" M 1.0 Aliminate N.
! to "
6/ 7 0.0C L 1.0 Aliminate '.
elect alternative F.
>ote2 R and ' can be eliminated initially because they have 6/ L 1.0.
9.3? (a) Projects are listed by increasing PH o# cost values. Girst #ind bene#its #or eachalternative and then #ind incremental 6/ ratios2
ene#its #or P1.1 7 P 610
P 7 11
ene#its #or S".+ 7 S6+0S 7 95
ene#its #or 1.+ 7 6!0 7 ?0
ene#its #or 1.! 7 6C0 7 1"0
=ncremental 6/ #or S vs P6/ 7 95 , 11 +0 , 10 7 ".C3
=ncremental 6/ #or vs P6/ 7 ?0 , 11 !0 , 10 7 1.+C
=ncremental 6/ #or vs P6/ 7 1"0 , 11 C0 , 10 7 1.!5
/hapter 9 "0
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=ncremental 6/ #or vs S6/ 7 ?0 , 95 !0 , +0 7 ".50
isregard due to less #or more /.
=ncremental 6/ #or vs S6/ 7 1"0 , 95 C0 , +0 7 0.50
=ncremental 6/ #or vs 6/ 7 1"0 , ?0 C0 , !0 7 1.5?
(b) /ompare P to > eliminate >./ompare S to P eliminate P./ompare to S disregarded./ompare to S eliminate .elect S.
-E Re&ie. Solutions
9.3C nswer is (d)
9.39 nswer is (b)
9.+0 nswer is (a)
9.+1 nswer is (c)
9.+" Project 6/ values are given. =ncremental analysis is necessary to select one alternative. nswer is (d)
9.+3 nswer is (c)
9.++ nswer is (a)
/hapter 9 "1
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Etended Eercise solution
1. *he spreadsheet shows the incremental 6/ analysis. *he truc& should be purchased. *he annual worth values #or each alternative are determined using theequations2
H payperuse 7 1!0%000(6P%5-%!) 10(3000) 3(C000) 7 8C9%509 (cell 1!)
Hown 7 C!0%000(6P%5-%1!) !00%000(6P%5-%!0) 1!("000) !(?000)7 81C+%"+0 (cell G1!)
/hapter 9 ""
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". *he annual #ee paid #or ! years now would have to be negative (cell !) in thatrewster would have to pay 'ed#ord a Tretainer #eeU% so to spea&% to possibly use theladder truc&. *his is an economically unreasonable approach.
A$cel 4FEA is used to #ind the brea&even value o# the initial cost when 6/ 7 1.0(cell G"1).
/hapter 9 "3
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3. *he building cost o# over 8"." million could be supported by the rewster proposal(in cell G?)% again #ound by using 4FEA. *his is also not an economicallyreasonable alternative.
/hapter 9 "+
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+. *he estimated sum o# premium and property loss would need to be 8!"3%?1+ or less (cell G1?% 4FEA). *his is not much o# a reduction #rom the current estimate o#
8500%000.
/hapter 9 "!
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Case Study Solution
1. =nstallation cost 7 (3%!00)[C?.C6(0.05?)(")]
7 (3%!00)(5!!) 7 8"%"9"%!00
nnual power cost 7 (5!! poles)(")(0.+)(1")(35!)(0.0C) 7 81C3%510
*otal annual cost 7 "%"9"%!00(6P%5-%!) 1C3%510 7 8?"?%C!0
=# the accident reduction rate is assumed to be the same as that #or closer spacing o#lights%
6/ 7 1%111%!006?"?%C!0
7 1.!3
". >ight6day deaths% unlighted 7 !63 7 1.5
>ight6day deaths% lighted 7 ?6+ 7 1.C
3. =nstallation cost 7 "%!00(C?.C60.05?) 7 83%"?5%000
*otal annual cost 7 3%"?5%000(6P% 5-% !) 35?%"19 7 81%1++%9+1
6/ 7 1%111%!0061%1++%9+1 7 0.9?
+. atio o# night6day accidents% lighted 7 C39 7 0.+05 "059
=# the same ratio is applied to unlighted sections% number o# accidents preventedwhere property damage was involved would be calculated as #ollows2
0.+05 7 no. o# accidents
3?9
no. accidents 7 1!+
no. prevented 7 199 ,1!+ 7 +!
/hapter 9 "5
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!. Gor lights to be justi#ied% bene#its would have to be at least 81%+!5%030 (instead o#81%111%!00). *here#ore% the di##erence in the number o# accidents would have to be2
1%+!5%030 7 (di##erence)(+!00)
i##erence 7 3"+
>o. o# accidents would have to be 7 10C5 , 3"+ 7 ?5"
>ight6day ratio 7 ?5" 7 0.35C "059
/hapter 9 "?
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