The machine should be chosen.

Question

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Answer 1

Question

Chapter 6, Problem 51P

To determine

The machine should be chosen.

Expert Solution & Answer

Answer to Problem 51P

The machine B is the best choice.

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$15000	$25000
Annual maintenance cost	$1600	$400
Annual benefits	$8000	$13000
Salvage value	$3000	$6000
Life	6	10

Calculation:

Calculate the EUAC for Machine A.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (II)

Substitute 12% for interest and 6 years in Equation (II).

(AP, i, n)=[0.12( 1+0.12)6( 1+0.12)6−1]=0.2432

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (III).

Substitute 12% for interest and 6 years in Equation (III).

(AF, i, n)=[0.12( 1+0.12)6−1]=0.1232

Substitute −$15000 for P, $3000 for F, $1600 for operation and maintenance cost, $8000 for annual benefits, 0.2432 for (AP, i, n), and 0.1232 for (AF, i, n) in Equation (I).

EUAC=[−($15000×0.2432)−$1600+$8000+($3000×0.1232)]=$3121.6

Calculate the EUAC for Machine B.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (V)

Substitute 12% for interest and 10 years in Equation (V).

(AP, i, n)=[0.12( 1+0.12)10( 1+0.12)10−1]=0.1770

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (VI).

Substitute 12% for interest and 10 years in Equation (VI).

(AF, i, n)=[0.12( 1+0.12)10−1]=0.0570

Substitute −$25000 for P, $6000 for F, $400 for operation and maintenance cost, $13000 for annual benefits, 0.1770 for (AP, i, n), and 0.0570 for (AF, i, n) in Equation (IV).

EUAC=[−($25000×0.1770)−$400+$13000+($6000×0.0570)]=$8517.

Conclusion:

Therefore the machine B is the best choice.

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Answer 2

Question

Chapter 6, Problem 51P

To determine

The machine should be chosen.

Expert Solution & Answer

Answer to Problem 51P

The machine B is the best choice.

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$15000	$25000
Annual maintenance cost	$1600	$400
Annual benefits	$8000	$13000
Salvage value	$3000	$6000
Life	6	10

Calculation:

Calculate the EUAC for Machine A.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (II)

Substitute 12% for interest and 6 years in Equation (II).

(AP, i, n)=[0.12( 1+0.12)6( 1+0.12)6−1]=0.2432

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (III).

Substitute 12% for interest and 6 years in Equation (III).

(AF, i, n)=[0.12( 1+0.12)6−1]=0.1232

Substitute −$15000 for P, $3000 for F, $1600 for operation and maintenance cost, $8000 for annual benefits, 0.2432 for (AP, i, n), and 0.1232 for (AF, i, n) in Equation (I).

EUAC=[−($15000×0.2432)−$1600+$8000+($3000×0.1232)]=$3121.6

Calculate the EUAC for Machine B.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (V)

Substitute 12% for interest and 10 years in Equation (V).

(AP, i, n)=[0.12( 1+0.12)10( 1+0.12)10−1]=0.1770

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (VI).

Substitute 12% for interest and 10 years in Equation (VI).

(AF, i, n)=[0.12( 1+0.12)10−1]=0.0570

Substitute −$25000 for P, $6000 for F, $400 for operation and maintenance cost, $13000 for annual benefits, 0.1770 for (AP, i, n), and 0.0570 for (AF, i, n) in Equation (IV).

EUAC=[−($25000×0.1770)−$400+$13000+($6000×0.0570)]=$8517.

Conclusion:

Therefore the machine B is the best choice.

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Answer 3

Question

Chapter 6, Problem 51P

To determine

The machine should be chosen.

Expert Solution & Answer

Answer to Problem 51P

The machine B is the best choice.

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$15000	$25000
Annual maintenance cost	$1600	$400
Annual benefits	$8000	$13000
Salvage value	$3000	$6000
Life	6	10

Calculation:

Calculate the EUAC for Machine A.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (II)

Substitute 12% for interest and 6 years in Equation (II).

(AP, i, n)=[0.12( 1+0.12)6( 1+0.12)6−1]=0.2432

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (III).

Substitute 12% for interest and 6 years in Equation (III).

(AF, i, n)=[0.12( 1+0.12)6−1]=0.1232

Substitute −$15000 for P, $3000 for F, $1600 for operation and maintenance cost, $8000 for annual benefits, 0.2432 for (AP, i, n), and 0.1232 for (AF, i, n) in Equation (I).

EUAC=[−($15000×0.2432)−$1600+$8000+($3000×0.1232)]=$3121.6

Calculate the EUAC for Machine B.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (V)

Substitute 12% for interest and 10 years in Equation (V).

(AP, i, n)=[0.12( 1+0.12)10( 1+0.12)10−1]=0.1770

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (VI).

Substitute 12% for interest and 10 years in Equation (VI).

(AF, i, n)=[0.12( 1+0.12)10−1]=0.0570

Substitute −$25000 for P, $6000 for F, $400 for operation and maintenance cost, $13000 for annual benefits, 0.1770 for (AP, i, n), and 0.0570 for (AF, i, n) in Equation (IV).

EUAC=[−($25000×0.1770)−$400+$13000+($6000×0.0570)]=$8517.

Conclusion:

Therefore the machine B is the best choice.

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Answer 4

Question

Chapter 6, Problem 51P

To determine

The machine should be chosen.

Expert Solution & Answer

Answer to Problem 51P

The machine B is the best choice.

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$15000	$25000
Annual maintenance cost	$1600	$400
Annual benefits	$8000	$13000
Salvage value	$3000	$6000
Life	6	10

Calculation:

Calculate the EUAC for Machine A.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (II)

Substitute 12% for interest and 6 years in Equation (II).

(AP, i, n)=[0.12( 1+0.12)6( 1+0.12)6−1]=0.2432

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (III).

Substitute 12% for interest and 6 years in Equation (III).

(AF, i, n)=[0.12( 1+0.12)6−1]=0.1232

Substitute −$15000 for P, $3000 for F, $1600 for operation and maintenance cost, $8000 for annual benefits, 0.2432 for (AP, i, n), and 0.1232 for (AF, i, n) in Equation (I).

EUAC=[−($15000×0.2432)−$1600+$8000+($3000×0.1232)]=$3121.6

Calculate the EUAC for Machine B.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (V)

Substitute 12% for interest and 10 years in Equation (V).

(AP, i, n)=[0.12( 1+0.12)10( 1+0.12)10−1]=0.1770

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (VI).

Substitute 12% for interest and 10 years in Equation (VI).

(AF, i, n)=[0.12( 1+0.12)10−1]=0.0570

Substitute −$25000 for P, $6000 for F, $400 for operation and maintenance cost, $13000 for annual benefits, 0.1770 for (AP, i, n), and 0.0570 for (AF, i, n) in Equation (IV).

EUAC=[−($25000×0.1770)−$400+$13000+($6000×0.0570)]=$8517.

Conclusion:

Therefore the machine B is the best choice.

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Answer 5

Chapter 6, Problem 51P

To determine

The machine should be chosen.

Expert Solution & Answer

Answer to Problem 51P

The machine B is the best choice.

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$15000	$25000
Annual maintenance cost	$1600	$400
Annual benefits	$8000	$13000
Salvage value	$3000	$6000
Life	6	10

Calculation:

Calculate the EUAC for Machine A.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (II)

Substitute 12% for interest and 6 years in Equation (II).

(AP, i, n)=[0.12( 1+0.12)6( 1+0.12)6−1]=0.2432

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (III).

Substitute 12% for interest and 6 years in Equation (III).

(AF, i, n)=[0.12( 1+0.12)6−1]=0.1232

Substitute −$15000 for P, $3000 for F, $1600 for operation and maintenance cost, $8000 for annual benefits, 0.2432 for (AP, i, n), and 0.1232 for (AF, i, n) in Equation (I).

EUAC=[−($15000×0.2432)−$1600+$8000+($3000×0.1232)]=$3121.6

Calculate the EUAC for Machine B.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (V)

Substitute 12% for interest and 10 years in Equation (V).

(AP, i, n)=[0.12( 1+0.12)10( 1+0.12)10−1]=0.1770

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (VI).

Substitute 12% for interest and 10 years in Equation (VI).

(AF, i, n)=[0.12( 1+0.12)10−1]=0.0570

Substitute −$25000 for P, $6000 for F, $400 for operation and maintenance cost, $13000 for annual benefits, 0.1770 for (AP, i, n), and 0.0570 for (AF, i, n) in Equation (IV).

EUAC=[−($25000×0.1770)−$400+$13000+($6000×0.0570)]=$8517.

Conclusion:

Therefore the machine B is the best choice.

Answer 6

Chapter 6, Problem 51P

To determine

The machine should be chosen.

Expert Solution & Answer

Answer to Problem 51P

The machine B is the best choice.

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$15000	$25000
Annual maintenance cost	$1600	$400
Annual benefits	$8000	$13000
Salvage value	$3000	$6000
Life	6	10

Calculation:

Calculate the EUAC for Machine A.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (II)

Substitute 12% for interest and 6 years in Equation (II).

(AP, i, n)=[0.12( 1+0.12)6( 1+0.12)6−1]=0.2432

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (III).

Substitute 12% for interest and 6 years in Equation (III).

(AF, i, n)=[0.12( 1+0.12)6−1]=0.1232

Substitute −$15000 for P, $3000 for F, $1600 for operation and maintenance cost, $8000 for annual benefits, 0.2432 for (AP, i, n), and 0.1232 for (AF, i, n) in Equation (I).

EUAC=[−($15000×0.2432)−$1600+$8000+($3000×0.1232)]=$3121.6

Calculate the EUAC for Machine B.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (V)

Substitute 12% for interest and 10 years in Equation (V).

(AP, i, n)=[0.12( 1+0.12)10( 1+0.12)10−1]=0.1770

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (VI).

Substitute 12% for interest and 10 years in Equation (VI).

(AF, i, n)=[0.12( 1+0.12)10−1]=0.0570

Substitute −$25000 for P, $6000 for F, $400 for operation and maintenance cost, $13000 for annual benefits, 0.1770 for (AP, i, n), and 0.0570 for (AF, i, n) in Equation (IV).

EUAC=[−($25000×0.1770)−$400+$13000+($6000×0.0570)]=$8517.

Conclusion:

Therefore the machine B is the best choice.

Answer 7

Chapter 6, Problem 51P

To determine

The machine should be chosen.

Answer 8

Expert Solution & Answer

Answer to Problem 51P

The machine B is the best choice.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$15000	$25000
Annual maintenance cost	$1600	$400
Annual benefits	$8000	$13000
Salvage value	$3000	$6000
Life	6	10

Calculation:

Calculate the EUAC for Machine A.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (II)

Substitute 12% for interest and 6 years in Equation (II).

(AP, i, n)=[0.12( 1+0.12)6( 1+0.12)6−1]=0.2432

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (III).

Substitute 12% for interest and 6 years in Equation (III).

(AF, i, n)=[0.12( 1+0.12)6−1]=0.1232

Substitute −$15000 for P, $3000 for F, $1600 for operation and maintenance cost, $8000 for annual benefits, 0.2432 for (AP, i, n), and 0.1232 for (AF, i, n) in Equation (I).

EUAC=[−($15000×0.2432)−$1600+$8000+($3000×0.1232)]=$3121.6

Calculate the EUAC for Machine B.

EUAC=[P(AP, i, n)−Operating and maintenance cost+Annual benefits+F(AF, i, n)] ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor (AP, i, n).

(AP, i, n)=[i(1+i)n(1+i)n−1] ...... (V)

Substitute 12% for interest and 10 years in Equation (V).

(AP, i, n)=[0.12( 1+0.12)10( 1+0.12)10−1]=0.1770

Calculate the factor (AF, i, n).

(AF, i, n)=[i(1+i)n−1] ...... (VI).

Substitute 12% for interest and 10 years in Equation (VI).

(AF, i, n)=[0.12( 1+0.12)10−1]=0.0570

Substitute −$25000 for P, $6000 for F, $400 for operation and maintenance cost, $13000 for annual benefits, 0.1770 for (AP, i, n), and 0.0570 for (AF, i, n) in Equation (IV).