Calculate annual worth in constant dollar.

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Answer 1

Question

Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

(a):

Expert Solution

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.12(1+0.12)5(1+0.12)5−1)−70,000+40,000(0.12(1+0.12)5−1)=−150,000(0.27741)−70,000+40,000(0.15741)=−41,611.5−70,000+6,296.4=−105,315

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.12)−5,000=−123,000−5,000=−128,000

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

(b):

To determine

Calculate annual worth in current dollars.

(b):

Expert Solution

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

if=i+f+i×f=0.12+0.07+0.12×0.07=0.19+0.0084=0.1984

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.1984(1+0.1984)5(1+0.1984)5−1)−70,000+40,000(0.12(1+0.1984)5−1)=−150,000(0.3332)−70,000+40,000(0.1348)=−49,980−70,000+5,392=−114,588

The annual worth is -$114,588.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.1984)−5,000=−203,360−5,000=−208,360

The annual worth is -$208,360. Since the annual worth of the cost is less for Machine A, select Machine A.

Annual worth can be calculated using the spreadsheet as follows:

EBK ENGINEERING ECONOMY, Chapter 14, Problem 56P

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Answer 2

Question

Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

(a):

Expert Solution

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.12(1+0.12)5(1+0.12)5−1)−70,000+40,000(0.12(1+0.12)5−1)=−150,000(0.27741)−70,000+40,000(0.15741)=−41,611.5−70,000+6,296.4=−105,315

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.12)−5,000=−123,000−5,000=−128,000

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

(b):

To determine

Calculate annual worth in current dollars.

(b):

Expert Solution

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

if=i+f+i×f=0.12+0.07+0.12×0.07=0.19+0.0084=0.1984

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.1984(1+0.1984)5(1+0.1984)5−1)−70,000+40,000(0.12(1+0.1984)5−1)=−150,000(0.3332)−70,000+40,000(0.1348)=−49,980−70,000+5,392=−114,588

The annual worth is -$114,588.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.1984)−5,000=−203,360−5,000=−208,360

The annual worth is -$208,360. Since the annual worth of the cost is less for Machine A, select Machine A.

Annual worth can be calculated using the spreadsheet as follows:

EBK ENGINEERING ECONOMY, Chapter 14, Problem 56P

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Answer 3

Question

Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

(a):

Expert Solution

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.12(1+0.12)5(1+0.12)5−1)−70,000+40,000(0.12(1+0.12)5−1)=−150,000(0.27741)−70,000+40,000(0.15741)=−41,611.5−70,000+6,296.4=−105,315

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.12)−5,000=−123,000−5,000=−128,000

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

(b):

To determine

Calculate annual worth in current dollars.

(b):

Expert Solution

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

if=i+f+i×f=0.12+0.07+0.12×0.07=0.19+0.0084=0.1984

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.1984(1+0.1984)5(1+0.1984)5−1)−70,000+40,000(0.12(1+0.1984)5−1)=−150,000(0.3332)−70,000+40,000(0.1348)=−49,980−70,000+5,392=−114,588

The annual worth is -$114,588.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.1984)−5,000=−203,360−5,000=−208,360

The annual worth is -$208,360. Since the annual worth of the cost is less for Machine A, select Machine A.

Annual worth can be calculated using the spreadsheet as follows:

EBK ENGINEERING ECONOMY, Chapter 14, Problem 56P

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Answer 4

Question

Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

(a):

Expert Solution

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.12(1+0.12)5(1+0.12)5−1)−70,000+40,000(0.12(1+0.12)5−1)=−150,000(0.27741)−70,000+40,000(0.15741)=−41,611.5−70,000+6,296.4=−105,315

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.12)−5,000=−123,000−5,000=−128,000

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

(b):

To determine

Calculate annual worth in current dollars.

(b):

Expert Solution

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

if=i+f+i×f=0.12+0.07+0.12×0.07=0.19+0.0084=0.1984

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.1984(1+0.1984)5(1+0.1984)5−1)−70,000+40,000(0.12(1+0.1984)5−1)=−150,000(0.3332)−70,000+40,000(0.1348)=−49,980−70,000+5,392=−114,588

The annual worth is -$114,588.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.1984)−5,000=−203,360−5,000=−208,360

The annual worth is -$208,360. Since the annual worth of the cost is less for Machine A, select Machine A.

Annual worth can be calculated using the spreadsheet as follows:

EBK ENGINEERING ECONOMY, Chapter 14, Problem 56P

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

(a):

Expert Solution

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.12(1+0.12)5(1+0.12)5−1)−70,000+40,000(0.12(1+0.12)5−1)=−150,000(0.27741)−70,000+40,000(0.15741)=−41,611.5−70,000+6,296.4=−105,315

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.12)−5,000=−123,000−5,000=−128,000

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

(b):

To determine

Calculate annual worth in current dollars.

(b):

Expert Solution

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

if=i+f+i×f=0.12+0.07+0.12×0.07=0.19+0.0084=0.1984

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.1984(1+0.1984)5(1+0.1984)5−1)−70,000+40,000(0.12(1+0.1984)5−1)=−150,000(0.3332)−70,000+40,000(0.1348)=−49,980−70,000+5,392=−114,588

The annual worth is -$114,588.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.1984)−5,000=−203,360−5,000=−208,360

The annual worth is -$208,360. Since the annual worth of the cost is less for Machine A, select Machine A.

Annual worth can be calculated using the spreadsheet as follows:

EBK ENGINEERING ECONOMY, Chapter 14, Problem 56P

Answer 6

Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

(a):

Expert Solution

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.12(1+0.12)5(1+0.12)5−1)−70,000+40,000(0.12(1+0.12)5−1)=−150,000(0.27741)−70,000+40,000(0.15741)=−41,611.5−70,000+6,296.4=−105,315

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.12)−5,000=−123,000−5,000=−128,000

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

Answer 7

Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

Answer 8

(a):

Expert Solution

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.12(1+0.12)5(1+0.12)5−1)−70,000+40,000(0.12(1+0.12)5−1)=−150,000(0.27741)−70,000+40,000(0.15741)=−41,611.5−70,000+6,296.4=−105,315

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.12)−5,000=−123,000−5,000=−128,000

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

Answer 9

(a):

Expert Solution

Answer 10

(a):

Expert Solution

Answer 11

Expert Solution

Answer 12

(b):

To determine

Calculate annual worth in current dollars.

(b):

Expert Solution

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

if=i+f+i×f=0.12+0.07+0.12×0.07=0.19+0.0084=0.1984

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.1984(1+0.1984)5(1+0.1984)5−1)−70,000+40,000(0.12(1+0.1984)5−1)=−150,000(0.3332)−70,000+40,000(0.1348)=−49,980−70,000+5,392=−114,588

The annual worth is -$114,588.

The annual worth of Machine B (AWBA) can be calculated as follows:

AWB=−FC×(i)−MO=−1,025,000(0.1984)−5,000=−203,360−5,000=−208,360

The annual worth is -$208,360. Since the annual worth of the cost is less for Machine A, select Machine A.

Annual worth can be calculated using the spreadsheet as follows:

EBK ENGINEERING ECONOMY, Chapter 14, Problem 56P

Answer 13

(b):

To determine

Calculate annual worth in current dollars.

Answer 14

(b):

Expert Solution

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

if=i+f+i×f=0.12+0.07+0.12×0.07=0.19+0.0084=0.1984

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

AWA=−FC(i(1+i)n(1+i)n−1)−MO+SV(i(1+i)n−1)=−150,000(0.1984(1+0.1984)5(1+0.1984)5−1)−70,000+40,000(0.12(1+0.1984)5−1)=−150,000(0.3332)−70,000+40,000(0.1348)=−49,980−70,000+5,392=−114,588