Sol INDR202 WP4

Solution  for  INDR202/Problem#4  

(a)  To  construct  the  original  cash  flow,  we  need  the  interest  rate:    

3

PB( )

$800(1 ) $460

$500

$800

$160

20%

i

i

i

i

=

−

+ +

=

−

−

=

−

=

n

n

A

Project  

Balance  

0   -­‐$1,000   -­‐$1,000  

1   $100   -­‐$1,100  

2   $520   -­‐$800  

3   $460   -­‐$500  

4   $600   $0  

(b)  Based  on  part  (a),  interest  rate  is  i=20%    

3

PB( )

$800(1 ) $460

$500

$800

$160

20%

i

i

i

i

=

−

+ +

=

−

−

=

−

=

(c)  Yes,  because  i  is  greater  than  MARR.    

5.26)        

(a)      To  construct  the  original  cash  flow,  we  need  the  interest  rate:    

i=10%  

n

_A

0   -­‐$1,000   -­‐$1,000  

1   $200   -­‐$900  

2   $490   -­‐$500  

3   $550   $0  

4   -­‐$100   -­‐$100  

5   $200   $90  

(b)  

2

PB( )

$900(1 ) $490

$500

10%

PW(10%) $90( / ,10%,5) $55.88

i

i

i

P F

=

−

+ +

=

−

=

=

=

    5.33)       (a)    

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃_- = $10,000,000  

𝑅𝑒𝑛𝑜𝑣𝑎𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃₇=$1,000,000   𝐴 𝐹 , 5%, 10

0.05 = $1,590,000  

𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒  𝑐𝑜𝑠𝑡: 𝑃_?=$100,000

0.05 = $2,000,000  

𝐶𝐸 5% = 𝑃_-+ 𝑃₇+ 𝑃_?= $13,590,000  

  (b)  

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃_- = $10,000,000  

𝑅𝑒𝑛𝑜𝑣𝑎𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃7=

$1,000,000   𝐴 𝐹 , 5%, 15

0.05 = $926,000  

𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒  𝑐𝑜𝑠𝑡: 𝑃_?=$100,000

0.05 = $2,000,000  

  (c)    

•   10-­‐year  cycle  with  10%  of  interest:    

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃_- = $10,000,000  

𝑅𝑒𝑛𝑜𝑣𝑎𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃₇=$1,000,000   𝐴 𝐹 , 10%, 10

0.10 = $627,000  

𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒  𝑐𝑜𝑠𝑡: 𝑃_?=$100,000

0.10 = $1,000,000  

𝐶𝐸 10% = 𝑃_-+ 𝑃₇+ 𝑃_?= $11,627,000  

•   15-­‐year  cycle  with  10%  of  interest:   𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃_- = $10,000,000  

𝑅𝑒𝑛𝑜𝑣𝑎𝑡𝑖𝑜𝑛  𝑐𝑜𝑠𝑡: 𝑃₇=$1,000,000   𝐴 𝐹 , 10%, 15

0.10 = $315,000  

𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒  𝑐𝑜𝑠𝑡: 𝑃_?=$100,000

0.10 = $1,000,000  

𝐶𝐸 10% = 𝑃_-+ 𝑃₇+ 𝑃_?= $11,315,000  

As  interest  rates  increases,  CE  value  decreases.    

    5.34)    

𝐶𝑜𝑠𝑡  𝑡𝑜  𝑑𝑒𝑠𝑖𝑔𝑛  𝑎𝑛𝑑  𝑏𝑢𝑖𝑙𝑑   = $650,000   𝑅𝑒𝑤𝑜𝑟𝑘  𝑐𝑜𝑠𝑡 = $100,000    𝑒𝑣𝑒𝑟𝑦  10  𝑦𝑒𝑎𝑟𝑠  

𝑁𝑒𝑤  𝑡𝑦𝑝𝑒  𝑜𝑓  𝑔𝑒𝑎𝑟 = $50,000  𝑎𝑡  𝑡ℎ𝑒  𝑒𝑛𝑑  𝑜𝑓  5QR  𝑦𝑒𝑎𝑟  

𝐴𝑛𝑛𝑢𝑎𝑙  𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔  𝑐𝑜𝑠𝑡𝑠 = $30,000  𝑓𝑜𝑟  𝑡ℎ𝑒  𝑓𝑖𝑟𝑠𝑡  15  𝑦𝑒𝑎𝑟𝑠, $35,000  𝑡ℎ𝑒𝑟𝑒𝑎𝑓𝑡𝑒𝑟    

𝐶𝐸 8% = $650,000 +$100,000(𝐴 𝐹, 8%, 10)

0.08 + $50,000(𝑃 𝐹, 8%, 5) + $30,000(𝑃 𝐴, 8%, 15)  

+$35,000

0.08 𝑃 𝐹 , 8%, 15 = $1,165,019      

5.43)     •   Model  A:  

cycle

total

PW(12%)

$11,000 $7,500( / ,12%,1)

$8,000( / ,12%, 2) $5,000( / ,12%,3)

$5,633.35

PW(12%)

$5,633.35[1 ( / ,12%,3)]

$9,643.11

P F

P F

P F

P F

=

−

+

+

+

=

=

+

=

•   Model  B:    

cycle

total

PW(12%)

$25,000 $14,500( / ,12%,1)

$18,000( / ,12%, 2)

$2, 296.65

PW(12%)

$2, 296.65[1 ( / ,12%, 2)

( / ,12%, 4)]

$5,587.06

P F

P F

P F

P F

=

−

+

+

=

=

+

+

=

Model  A  is  preferred.  

5.45)  

(a)  Assuming  a  common  service  period  of  15  years   •   Project  A:  

cycle

total

PW(12%)

$12,000 $2,000( / ,12%,5)

$2,000( / ,12%,5)

$18,075

PW(12%)

$18,075[1 ( / ,76.23%, 2)]

$34,151

P A

P F

P A

=

−

−

+

=

−

=

−

+

=

−

Note : (1.12)

−

1 76.23%

=

•   Project  B:  

cycle

total

PW(12%)

$10,000 $2,100( / ,12%,3)

$1,000( / ,12%,3)

$14,332

PW(12%)

$14,332[1 ( / , 40.49%, 4)]

$40,642

P A

P F

P A

=

−

−

+

=

−

=

−

+

=

−

Note : (1.12)

−

1 40.49%

=

(b)    

•   Project  A  with  2  replacement  cycles:    

PW(12%)

$18,074 $18,074( / ,12%,5)

$28,329.67

P F

=

−

−

=

−

•   Project  B  with  4  replacement  cycles  where  the  4th  _{replacement  cycle  ends  at  the  end  of}   first  operating  year:  

PW(12%)

$14,332[1 ( / ,12%,3) ( / ,12%,6)]

[$10,000 ($2,100 $6,000)( / ,12%,1)]

( / ,12%,9)

$34,144.73

P F

P F

P F

P F

=

−

+

+

−

+

−

×

=

−

Project  A  is  still  a  better  choice.  

  5.50)    

(a)  Since  only  Model  A  is  repeated  in  the  future,  the  following  sequence  of  replacement  cycles  is   possible:  

•   Option  1:  Purchase  Model  A  now  and  repeat  Model  A  forever.  

•   Option   2:   Purchase   Model   B   now   and   replace   it   at   the   end   of   year   2   by   Model   A.   Then   repeat  Model  A  forever.  

𝑃𝑊 15% _W= −$6,600 + $3,500 𝑃 𝐴 , 15%, 3 + $1,000 𝑃 𝐹 , 15%, 2 + $2,000 𝑃 𝐹 , 15%, 3

= $3,462.46  

𝐴𝐸 15% _W= $3,462.46 𝐴 𝑃 , 15%, 3 = $1516.49  

𝑃𝑊 15% _Z = −$16,500 + $11,000 𝑃 𝐹 , 15%, 1 + $12,000 𝑃 𝐹 , 15%, 2 = $2,138.94  

𝐴𝐸 15% Z= $2,138.94 𝐴 𝑃 , 15%, 2 = $1,315.7  

(a)  

•   Option  1:  

𝑃𝑊 15% _WWW…=𝐴 𝑖 =

$1,516.49

0.15 = $10,109.93  

•   Option  2:  

𝑃𝑊 15% _ZWW… = $2,138.94 +$1,516.49

0.15 𝑃 𝐹 , 15%, 2 = $8,901.21  

(b)    

−  $6,600 + $3,500 𝑃 𝐹 , 15%, 1 + $4,500 + 𝑆 𝑃 𝐹 , 15%, 2 = $2,138.94  

𝑆𝑜𝑙𝑣𝑖𝑛𝑔  𝑓𝑜𝑟  𝑆  𝑦𝑖𝑒𝑙𝑑𝑠  

𝑆 = $3,032.25  

6.10)    

$100 $100 _{$100 $100}

$60 $60 $60 $60

8

2 3 4

$100 $100

$60

$60

PW(14%)

$240.69

1.14 1.14

1.14

1.14

AE(14%) $240.69( / ,14%, 4) $82.60

A P

=

+

+

+

=

=

=

6.17)    

a)  Capital  recovery  cost:  

CR(10%) ($40,000 $15,000)( / ,10%, 2) $15,000(0.10)

$15,905

A P

=

−

+

=

b)   Annual  savings:    

PW(10%) $28,000( / ,10%,1) $40,000( / ,10%, 2)

$58,512.40

P F

P F

=

+

=

1

AE(10%)

$58,512.40( / ,10%, 2)

$33,714.84

A P

=

=

AE(10%) $33,714.84 $15,905 $17,809.84

=

−

=

b)   savings  generated  per  machine-­‐hour    

AE(10%)

[4,000( / ,10%,1) 6,000( / ,10%, 2)]( / ,10%, 2)

4,952.46

17,809.84 / 4,952.46

$3.596 / hour

C

P F

P F

A P

C

C

=

+

=

=

=

6.21)    

•   Capital  cost:    

CR(10%) ($100,000)( / ,10%,12)

$14,680

A P

=

=

•  Annual  O&M  costs:  

$10,000 $20,000( / ,10%, 4) $14,310

+

A F

=

AE(15%) $14,680 $14,310 $28,990

=

+

=

6.34)    

•   Capital  costs:  

1

CR(7%)

($25,000 $2,000)( / ,7%,12)

(0.07)($2,000)

$3,036

A P

=

−

+

=

•   Annual  battery  replacement  cost:  

2

AEC(7%)

$3,000[( / ,7%,3) ( / ,7%,6)

( / ,7%,9)]( / ,7%,12)

$763.14

P F

P F

P F

A P

=

+

+

=

•   Annual  recharging  cost:  

3

AEC(7%)

=

($0.015)(20,000) $300

=

•   Total  annual  equivalent  costs:  

AEC(7%)=$3,036+$763.14+$300=$4099.14  

•   Cost  per  mile:  

Cost/mile=$4099.14/20,000=$0.205  

6.37)    

O&M

ma

Salvage Value: $1, 200,000( / ,5%, 25) $4,063,680

CR(12%) ($6,000,000 $4,063,680)( / ,12%, 25) (0.12)$4,063,680

$734,522.4

AEC

$100 12 40 $400,000 $448,000

AEC(12%) CR(12%)

$1,182,522.4 per year

AEC(

F P

A P

AE

=

=

−

+

=

=

×

×

+

=

=

+

=

Monthly

0.9489%)

$1,182,522.4( / ,0.9489%,12)

$93,506 per month

A F

=

=

6.45)    

  Model  A   Model  B  

First  Cost   $95,000   $120,000  

Annual  Maintenance   $3,000   $9,000   Salvage  value   $12,000   $25,000   Useful  life  in  years   3   6    

•   Model    A:  

AEC(10%) ($95,000 $12,000)( / ,10%,3) (0.1)($12,000)

$3,000

$37,574.3 per year

A P

=

−

+

+

=

•   Model    B:  

AEC(10%) ($120,000 $25,000)( / ,10%,6) (0.1)($25,000)

$9,000

$33,312 per year

A P

=

−

+

+

=

 Therefore,  select  Model  B  (The  ROT  8).  

6.51)    

  Assumption:  jet  fuel  cost

=

$4.80

•  System  A  :  Equivalent  annual  fuel  cost:  A1  =  ($4.80/gal)(40,000gals/1,000  hours)(2,000  hours)

$384,000

1

.

AEC(10%)

[$384,000( / ,6%,10%,3)]( / ,10%,3)

$405,981

AEC(10%)

($100,000 $10,000)( / ,10%,3)

(0.10)($10,000) $405,981

$443,170

fuel

sys A

P A

A P

A P

=

=

=

−

+

+

=

•  System  B  :  Equivalent  annual  fuel  cost:  A1  =  ($4.80/gal)(32,000gals/1,000  hours)(2,000  hours)

$307, 200

=

1

.

AEC(10%)

[$307, 200( / ,6%,10%,3)]( / ,10%,3)

$324,785

AEC(10%)

($200,000 $20,000)( / ,10%,3)

(0.10)($20,000) $324,785

$399,163

fuel

sys B

P A

A P

A P

=

=

=

−

+

+

=

•   Equivalent  operating  cost  (  including  capital  cost  )  per  hour:  

System

A

=

$443,170 / 2,000 $221.6

=

System

B

=

$399,163 / 2,000 $199.6

=

System  B  is  a  better  choice.