Ch 9 Solutions

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





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 +





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 !





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





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 ?



B/C = (2,800,000)/(100,000+30,000,000*(0.08))



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





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





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





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





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"





=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





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+





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!





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





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?



A # c#sts = E$ T 1$,10,E4



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





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





" &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





=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





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 ""





". *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





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 "+





+. *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 "!





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





!. 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 "?

PROPRIETAR !ATERIA". : *he 'c;raw<ill /ompanies% =nc. ll rights reserved. >o part o# this 'anual may be

Ch 9 Solutions

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Transcript of Ch 9 Solutions