The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Question

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Answer 1

Question

Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Expert Solution & Answer

Answer to Problem 132P

The correct option is (c) $22 per unit.

Explanation of Solution

Given:

First cost is $350,000.

Cost of the operation is $45,000.

Salvage value is $15,000.

Project life is 5 years.

Interest rate is 10%.

Material cost per unit is $8.50.

Annual production is 10,000 units.

Calculation:

Calculate the total annual material cost.

Annual material cost=$8.50×10,000=$85,000

Write the expression to calculate the present worth of making the part.

PWM=[−$350,000−$45,000(P/A,i,n)−$85,000(P/A,i,n)+$15,000(P/F,i,n)] ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute 10% for i and 5 for n in Equation (I).

PWM=[−$350,000−$45,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)−$85,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)+$15,000( 1 ( 1+0.1) 5)]=[−$350,000−$45,000(3.7908)−$85,000(3.7908)+$15,000(0.621)]=[−$350,000−$170,586−$322,218+$9,315]

PWM=−$833489 ....... (II)

Write the expression to calculate the present worth of buying the parts.

PWB=−A(P/A,i,n)=−A( ( 1+i )n −1 i ( 1+i )n ) ...... (III)

Substitute 10% for i and 5 for n in Equation (III).

PWB=−A( ( 1+0.1 )5 −1 0.1 ( 1+0.1 )5 )PWB=−A(3.7908) ....... (IV)

Equate Equation (II) and Equation (IV) and calculate the maximum annual amount.

PWM=PWB−$833489=−A(3.7908)A=$8334893.7908A=$219,871.53

Calculate the annual amount per unit.

Aper unit=$219,871.5310,000=$21.9≃$22.

Conclusion:

Therefore, the correct option is (c) $22 per unit.

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Answer 2

Question

Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Expert Solution & Answer

Answer to Problem 132P

The correct option is (c) $22 per unit.

Explanation of Solution

Given:

First cost is $350,000.

Cost of the operation is $45,000.

Salvage value is $15,000.

Project life is 5 years.

Interest rate is 10%.

Material cost per unit is $8.50.

Annual production is 10,000 units.

Calculation:

Calculate the total annual material cost.

Annual material cost=$8.50×10,000=$85,000

Write the expression to calculate the present worth of making the part.

PWM=[−$350,000−$45,000(P/A,i,n)−$85,000(P/A,i,n)+$15,000(P/F,i,n)] ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute 10% for i and 5 for n in Equation (I).

PWM=[−$350,000−$45,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)−$85,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)+$15,000( 1 ( 1+0.1) 5)]=[−$350,000−$45,000(3.7908)−$85,000(3.7908)+$15,000(0.621)]=[−$350,000−$170,586−$322,218+$9,315]

PWM=−$833489 ....... (II)

Write the expression to calculate the present worth of buying the parts.

PWB=−A(P/A,i,n)=−A( ( 1+i )n −1 i ( 1+i )n ) ...... (III)

Substitute 10% for i and 5 for n in Equation (III).

PWB=−A( ( 1+0.1 )5 −1 0.1 ( 1+0.1 )5 )PWB=−A(3.7908) ....... (IV)

Equate Equation (II) and Equation (IV) and calculate the maximum annual amount.

PWM=PWB−$833489=−A(3.7908)A=$8334893.7908A=$219,871.53

Calculate the annual amount per unit.

Aper unit=$219,871.5310,000=$21.9≃$22.

Conclusion:

Therefore, the correct option is (c) $22 per unit.

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Answer 3

Question

Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Expert Solution & Answer

Answer to Problem 132P

The correct option is (c) $22 per unit.

Explanation of Solution

Given:

First cost is $350,000.

Cost of the operation is $45,000.

Salvage value is $15,000.

Project life is 5 years.

Interest rate is 10%.

Material cost per unit is $8.50.

Annual production is 10,000 units.

Calculation:

Calculate the total annual material cost.

Annual material cost=$8.50×10,000=$85,000

Write the expression to calculate the present worth of making the part.

PWM=[−$350,000−$45,000(P/A,i,n)−$85,000(P/A,i,n)+$15,000(P/F,i,n)] ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute 10% for i and 5 for n in Equation (I).

PWM=[−$350,000−$45,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)−$85,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)+$15,000( 1 ( 1+0.1) 5)]=[−$350,000−$45,000(3.7908)−$85,000(3.7908)+$15,000(0.621)]=[−$350,000−$170,586−$322,218+$9,315]

PWM=−$833489 ....... (II)

Write the expression to calculate the present worth of buying the parts.

PWB=−A(P/A,i,n)=−A( ( 1+i )n −1 i ( 1+i )n ) ...... (III)

Substitute 10% for i and 5 for n in Equation (III).

PWB=−A( ( 1+0.1 )5 −1 0.1 ( 1+0.1 )5 )PWB=−A(3.7908) ....... (IV)

Equate Equation (II) and Equation (IV) and calculate the maximum annual amount.

PWM=PWB−$833489=−A(3.7908)A=$8334893.7908A=$219,871.53

Calculate the annual amount per unit.

Aper unit=$219,871.5310,000=$21.9≃$22.

Conclusion:

Therefore, the correct option is (c) $22 per unit.

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Answer 4

Question

Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Expert Solution & Answer

Answer to Problem 132P

The correct option is (c) $22 per unit.

Explanation of Solution

Given:

First cost is $350,000.

Cost of the operation is $45,000.

Salvage value is $15,000.

Project life is 5 years.

Interest rate is 10%.

Material cost per unit is $8.50.

Annual production is 10,000 units.

Calculation:

Calculate the total annual material cost.

Annual material cost=$8.50×10,000=$85,000

Write the expression to calculate the present worth of making the part.

PWM=[−$350,000−$45,000(P/A,i,n)−$85,000(P/A,i,n)+$15,000(P/F,i,n)] ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute 10% for i and 5 for n in Equation (I).

PWM=[−$350,000−$45,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)−$85,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)+$15,000( 1 ( 1+0.1) 5)]=[−$350,000−$45,000(3.7908)−$85,000(3.7908)+$15,000(0.621)]=[−$350,000−$170,586−$322,218+$9,315]

PWM=−$833489 ....... (II)

Write the expression to calculate the present worth of buying the parts.

PWB=−A(P/A,i,n)=−A( ( 1+i )n −1 i ( 1+i )n ) ...... (III)

Substitute 10% for i and 5 for n in Equation (III).

PWB=−A( ( 1+0.1 )5 −1 0.1 ( 1+0.1 )5 )PWB=−A(3.7908) ....... (IV)

Equate Equation (II) and Equation (IV) and calculate the maximum annual amount.

PWM=PWB−$833489=−A(3.7908)A=$8334893.7908A=$219,871.53

Calculate the annual amount per unit.

Aper unit=$219,871.5310,000=$21.9≃$22.

Conclusion:

Therefore, the correct option is (c) $22 per unit.

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Answer 5

Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Expert Solution & Answer

Answer to Problem 132P

The correct option is (c) $22 per unit.

Explanation of Solution

Given:

First cost is $350,000.

Cost of the operation is $45,000.

Salvage value is $15,000.

Project life is 5 years.

Interest rate is 10%.

Material cost per unit is $8.50.

Annual production is 10,000 units.

Calculation:

Calculate the total annual material cost.

Annual material cost=$8.50×10,000=$85,000

Write the expression to calculate the present worth of making the part.

PWM=[−$350,000−$45,000(P/A,i,n)−$85,000(P/A,i,n)+$15,000(P/F,i,n)] ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute 10% for i and 5 for n in Equation (I).

PWM=[−$350,000−$45,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)−$85,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)+$15,000( 1 ( 1+0.1) 5)]=[−$350,000−$45,000(3.7908)−$85,000(3.7908)+$15,000(0.621)]=[−$350,000−$170,586−$322,218+$9,315]

PWM=−$833489 ....... (II)

Write the expression to calculate the present worth of buying the parts.

PWB=−A(P/A,i,n)=−A( ( 1+i )n −1 i ( 1+i )n ) ...... (III)

Substitute 10% for i and 5 for n in Equation (III).

PWB=−A( ( 1+0.1 )5 −1 0.1 ( 1+0.1 )5 )PWB=−A(3.7908) ....... (IV)

Equate Equation (II) and Equation (IV) and calculate the maximum annual amount.

PWM=PWB−$833489=−A(3.7908)A=$8334893.7908A=$219,871.53

Calculate the annual amount per unit.

Aper unit=$219,871.5310,000=$21.9≃$22.

Conclusion:

Therefore, the correct option is (c) $22 per unit.

Answer 6

Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Expert Solution & Answer

Answer to Problem 132P

The correct option is (c) $22 per unit.

Explanation of Solution

Given:

First cost is $350,000.

Cost of the operation is $45,000.

Salvage value is $15,000.

Project life is 5 years.

Interest rate is 10%.

Material cost per unit is $8.50.

Annual production is 10,000 units.

Calculation:

Calculate the total annual material cost.

Annual material cost=$8.50×10,000=$85,000

Write the expression to calculate the present worth of making the part.

PWM=[−$350,000−$45,000(P/A,i,n)−$85,000(P/A,i,n)+$15,000(P/F,i,n)] ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute 10% for i and 5 for n in Equation (I).

PWM=[−$350,000−$45,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)−$85,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)+$15,000( 1 ( 1+0.1) 5)]=[−$350,000−$45,000(3.7908)−$85,000(3.7908)+$15,000(0.621)]=[−$350,000−$170,586−$322,218+$9,315]

PWM=−$833489 ....... (II)

Write the expression to calculate the present worth of buying the parts.

PWB=−A(P/A,i,n)=−A( ( 1+i )n −1 i ( 1+i )n ) ...... (III)

Substitute 10% for i and 5 for n in Equation (III).

PWB=−A( ( 1+0.1 )5 −1 0.1 ( 1+0.1 )5 )PWB=−A(3.7908) ....... (IV)

Equate Equation (II) and Equation (IV) and calculate the maximum annual amount.

PWM=PWB−$833489=−A(3.7908)A=$8334893.7908A=$219,871.53

Calculate the annual amount per unit.

Aper unit=$219,871.5310,000=$21.9≃$22.

Conclusion:

Therefore, the correct option is (c) $22 per unit.

Answer 7

Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Answer 8

Expert Solution & Answer

Answer to Problem 132P

The correct option is (c) $22 per unit.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

First cost is $350,000.

Cost of the operation is $45,000.

Salvage value is $15,000.

Project life is 5 years.

Interest rate is 10%.

Material cost per unit is $8.50.

Annual production is 10,000 units.

Calculation:

Calculate the total annual material cost.

Annual material cost=$8.50×10,000=$85,000

Write the expression to calculate the present worth of making the part.

PWM=[−$350,000−$45,000(P/A,i,n)−$85,000(P/A,i,n)+$15,000(P/F,i,n)] ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute 10% for i and 5 for n in Equation (I).

PWM=[−$350,000−$45,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)−$85,000( ( 1+0.1) 5−1 0.1 ( 1+0.1) 5)+$15,000( 1 ( 1+0.1) 5)]=[−$350,000−$45,000(3.7908)−$85,000(3.7908)+$15,000(0.621)]=[−$350,000−$170,586−$322,218+$9,315]