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ENGINEERING ECONOMIC ANALYSIS CHAPTER 8 PR 19: The CROC Co may add a new milling

ID: 2779087 • Letter: E

Question

ENGINEERING ECONOMIC ANALYSIS CHAPTER 8 PR 19:

The CROC Co may add a new milling machine from among three alternatives; each has a life of 10 years.

A

B

C

FIRST COST

$220,000

$125,000

$75,000

Annual Benefit

79,000

43,000

28,000

Maintenance and Operating

38,000

13,000

8,000

Salvage Cost

16,000

6,900

3,000

a) Construct a truth table for interest rates from 0-100%.

b) Using 15% MARR, which alternative should be selected?

************PLEASE PROVIDE EXCEL FORMULAS IF TABLE IS USED FOR CALCULATIONS*******

I am having trouble doing this....

A

B

C

FIRST COST

$220,000

$125,000

$75,000

Annual Benefit

79,000

43,000

28,000

Maintenance and Operating

38,000

13,000

8,000

Salvage Cost

16,000

6,900

3,000

Explanation / Answer

CROC CO All amount in $ Alternative Details Year 0 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 A Initial Cost                (220,000) Annual Benefit      79,000      79,000      79,000      79,000      79,000      79,000      79,000      79,000      79,000      79,000 Maint & Operating Cost    (38,000)    (38,000)    (38,000)    (38,000)    (38,000)    (38,000)    (38,000)    (38,000)    (38,000)    (38,000) Salvage Value      16,000 Total Cash Flow                (220,000)      41,000      41,000      41,000      41,000      41,000      41,000      41,000      41,000      41,000      57,000 Interest rate k (between 0-100%) Discount factor with interest rate k                               1 1/k 1/k^2 1/k^3 1/k^4 1/k^5 1/k^6 1/k^7 1/k^8 1/k^9 1/k^10 Present value of Cash Flow                (220,000) =41000*1/k =41000*1/k^2 =41000*1/k^3 =41000*1/k^4 =41000*1/k^5 =41000*1/k^6 =41000*1/k^7 =41000*1/k^8 =41000*1/k^9 =57000*1/k^10 CROC CO All amount in $ Alternative Details Year 0 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 B Initial Cost                (125,000) Annual Benefit      43,000      43,000      43,000      43,000      43,000      43,000      43,000      43,000      43,000      43,000 Maint & Operating Cost    (13,000)    (13,000)    (13,000)    (13,000)    (13,000)    (13,000)    (13,000)    (13,000)    (13,000)    (13,000) Salvage Value         6,900 Total Cash Flow                (125,000)      30,000      30,000      30,000      30,000      30,000      30,000      30,000      30,000      30,000      36,900 Interest rate k (between 0-100%) Discount factor with interest rate k                               1 1/k 1/k^2 1/k^3 1/k^4 1/k^5 1/k^6 1/k^7 1/k^8 1/k^9 1/k^10 Present value of Cash Flow                (125,000) =30000*1/k =30000*1/k^2 =30000*1/k^3 =30000*1/k^4 =30000*1/k^5 =30000*1/k^6 =30000*1/k^7 =30000*1/k^8 =30000*1/k^9 =36900*1/k^10 CROC CO All amount in $ Alternative Details Year 0 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 C Initial Cost                  (75,000) Annual Benefit      28,000      28,000      28,000      28,000      28,000      28,000      28,000      28,000      28,000      28,000 Maint & Operating Cost      (8,000)      (8,000)      (8,000)      (8,000)      (8,000)      (8,000)      (8,000)      (8,000)      (8,000)      (8,000) Salvage Value         3,000 Total Cash Flow                  (75,000)      20,000      20,000      20,000      20,000      20,000      20,000      20,000      20,000      20,000      23,000 Interest rate k (between 0-100%) Discount factor with interest rate k                               1 1/k 1/k^2 1/k^3 1/k^4 1/k^5 1/k^6 1/k^7 1/k^8 1/k^9 1/k^10 Present value of Cash Flow                  (75,000) =20000*1/k =20000*1/k^2 =20000*1/k^3 =20000*1/k^4 =20000*1/k^5 =20000*1/k^6 =20000*1/k^7 =20000*1/k^8 =20000*1/k^9 =23000*1/k^10 Year 0 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 b Discount factor with interest rate 15%                               1        0.870         0.756         0.658         0.572         0.497         0.432         0.376         0.327         0.284         0.247 Alternative A Total Cash Flow                (220,000)      41,000      41,000      41,000      41,000      41,000      41,000      41,000      41,000      41,000      57,000 Present value of Cash Flow                (220,000) =41000*1/k =41000*1/k^2 =41000*1/k^3 =41000*1/k^4 =41000*1/k^5 =41000*1/k^6 =41000*1/k^7 =41000*1/k^8 =41000*1/k^9 =57000*1/k^10 Present value of Cash Flow                (220,000)      35,652      31,002      26,958      23,442      20,384      17,725      15,413      13,403      11,655      14,090 NPV of cash flows                  (10,276) Alternative B Total Cash Flow                (125,000)      30,000      30,000      30,000      30,000      30,000      30,000      30,000      30,000      30,000      36,900 Present value of Cash Flow                (125,000) =30000*1/k =30000*1/k^2 =30000*1/k^3 =30000*1/k^4 =30000*1/k^5 =30000*1/k^6 =30000*1/k^7 =30000*1/k^8 =30000*1/k^9 =36900*1/k^10 Present value of Cash Flow                (125,000)      26,087      22,684      19,725      17,153      14,915      12,970      11,278         9,807         8,528         9,121 NPV of cash flows                     27,269 Alternative c Total Cash Flow                  (75,000)      20,000      20,000      20,000      20,000      20,000      20,000      20,000      20,000      20,000      23,000 Present value of Cash Flow                  (75,000) =20000*1/k =20000*1/k^2 =20000*1/k^3 =20000*1/k^4 =20000*1/k^5 =20000*1/k^6 =20000*1/k^7 =20000*1/k^8 =20000*1/k^9 =23000*1/k^10 Present value of Cash Flow                  (75,000)      17,391      15,123      13,150      11,435         9,944         8,647         7,519         6,538         5,685         5,685 NPV of cash flows                     26,117 So considering MARR of 15% , alternative B should be selected