Engineering Economic Analysis
Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
Question
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Chapter 2, Problem 14P
To determine

(a)

Expert Solution
Check Mark

Answer to Problem 14P

Fixed cost system I = 1.0.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0

Fixed cost refers to the cost of production that does not vary with the level of output. It is also known as the sunk cost.

Calculation:

Fixed cost can be found by looking at the intercept of the total cost function. Or by putting the value of x=0,

Fixed Cost of System I = 1.0

Conclusion:

Fixed Cost of System I = 1.0.

To determine

(b)

Fixed cost of System II.

Expert Solution
Check Mark

Answer to Problem 14P

Fixed cost system II = 5.0.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0.

Fixed cost refers to the cost of production that does not vary with the level of output. It is also known as the sunk cost.

Calculation:

Fixed cost can be found by look at the intercept of the total cost function. Or by putting the value of x = 0,

Fixed Cost of System II = 5.0.

Conclusion:

Fixed Cost of System II = 5.0.

To determine

(c)

Variable cost of System I.

Expert Solution
Check Mark

Answer to Problem 14P

Variable cost of System I = 0.90.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0.

Variable cost is the cost that varies with the level of output. It is given by the slope or the coefficient of the total cost function.

Calculation:

Variable cost is the cost that varies with the level of output. It is given by the slope or the coefficient of the total cost function.

Variable Cost of System I = 0.90x.

Conclusion:

Variable Cost of System I = 0.90x.

To determine

(d)

Variable cost of System II.

Expert Solution
Check Mark

Answer to Problem 14P

Variable cost of System II = 0.10.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0.

Variable cost is the cost that varies with the level of output. It is given by the slope or the coefficient of the total cost function.

Calculation:

Variable cost is the cost that varies with the level of output. It is given by the slope or the coefficient of the total cost function.

Variable Cost of System II = 0.10.

Conclusion:

Variable Cost of System II = 0.10.

To determine

(e)

Find the break-even point, where the costs are equal in both the systems.

Expert Solution
Check Mark

Answer to Problem 14P

Break-even = 5000 maps dispensed per year.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0.

Break-even point refers to the point where firm earns just zero economic profits. Here, it will be the point where total cost of system I is equal to the total cost of system II.

Calculation:

Break-even point is given by the intersection of the two cost curves. It can be calculated by equating the two cost functions.

At break-even,

Total cost of System I = Total cost of System II

0.90x+1.0 = 0.10x+5.00.80x=4X= 5.

Conclusion:

Therefore, for 5 thousand maps dispensed per year the total annual cost of system I and II are equal.

To determine

(f)

To determine the range of output at which system I will be considered.

Expert Solution
Check Mark

Answer to Problem 14P

Between (0-5000) maps dispensed per year, system I will be recommended.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0.

System I will be recommended for range of maps between 0-5000 per year. It is between these levels of output that the total cost line of system I is below the total cost line of system II.

Conclusion:

Thus between 0-5000 maps dispensed per year, system I will be less costly.

To determine

(g)

To determine the range of output at which system II will be considered.

Expert Solution
Check Mark

Answer to Problem 14P

Between (5000-10,000) maps dispensed per year, system II will be recommended.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0.

System II will be recommended for range of maps between (5000-10,000) per year. It is between this level of output that the total cost line of system I is below the total cost line of system II.

ConclusionThus between (5000-10,000) maps dispensed per year, system II will be less costly.

To determine

(h)

Find the average and marginal cost when 3000 maps are dispensed per year.

Expert Solution
Check Mark

Answer to Problem 14P

Average cost of System I = 0.0903

Average cost of System II = 0.1016

Marginal cost of system I = 0.90

Marginal cost of system II = 0.10.

Explanation of Solution

Given information:

Total Cost of System I = 0.90x+1.0

Total Cost of System II = 0.10x+5.0.

Average cost is also known as the per-unit cost. It can be calculated by dividing total cost by the number of units produced.

Marginal cost is the change in total cost when one more unit of an output is produced. Here.

Calculation:

At 3000 maps dispensed per year,

Total Cost of System I is termed as TC I

Total Cost of System II is termed as TC II

TC I = 0.90x+ 1.0= 0.90(3000) + 1.0= 271TCII = 0.10x + 5.0= 0.10(3000) + 5.0= 305

Average Cost of System I is termed as AC I

Average Cost of System II is termed as AC II

AC I = 2713000= 0.0903AC II =3053000= 0.1016

Marginal cost is the change in total cost when one more unit of an output is produced. Here, as output increases by 1 units total cost increases by 0.90 for system I and by 0.10 for system II. So, marginal cost for the two systems will be,

Marginal cost of System I is termed as MC I

Marginal cost of System II is termed as MC II

MC I = 0.90MC II = 0.10.

Conclusion:

Average cost of System I = 0.0903

Average cost of System II = 0.1016

Marginal cost of system I = 0.90

Marginal cost of system II = 0.10.

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Engineering Economic Analysis

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