The most economic life.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 13, Problem 1P

To determine

The most economic life.

Expert Solution & Answer

Answer to Problem 1P

The most economic life is 5 years.

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

$EUAB=−First cost(A/P,10%,n)+Salvage value(A/F,10%,n)EUAB=−First cost×( i ( 1+i ) n ( 1+i ) n −1)+Salvage value×(i ( 1+i ) n −1)$

EUAC can be calculated with the help of below expression:

$EUAC=Maintenance cost+Maintenance and operating gradient(A/G,10%,n)=Maintenance cost+Maintenance and operating gradient×(1i−n ( 1+i ) n −1)$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $EUAB−EUAC$ ,

When $n=4$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 4 ( 1+0.10 ) 4 −1 )+$225,000×( 0.10 ( 1+0.10 ) 4 −1 ))−( $235,000+$75,000×( 1 0.10 − 4 ( 1+0.10 ) 4 −1 ))]=[( −$1,050,000×0.31547)+( $225,000×0.21547)−( $235,000+( $75,000×( 10−8.6188 ) ))]=−$331,243.5+$48,480.75−$338,590=−$621,352.75$

When $n=5$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 5 ( 1+0.10 ) 5 −1 )+$225,000×( 0.10 ( 1+0.10 ) 5 −1 ))−( $235,000+$75,000×( 1 0.10 − 5 ( 1+0.10 ) 5 −1 ))]=[( −$1,050,000×0.2638)+( $225,000×0.1638)−( $235,000+( $75,000×( 10−8.1898 ) ))]=−$276,990+$36,855−$370,765=−$610,900$

When $n=6$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 6 ( 1+0.10 ) 6 −1 )+$225,000×( 0.10 ( 1+0.10 ) 6 −1 ))−( $235,000+$75,000×( 1 0.10 − 6 ( 1+0.10 ) 6 −1 ))]=[( −$1,050,000×0.2296)+( $225,000×0.1296)−( $235,000+( $75,000×( 10−7.7764 ) ))]=−$241,080+$29,160−$401,770=−$613,690$

Want to see more full solutions like this?

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

Students have asked these similar questions

Consider a firm facing conventional production technology. The short run Production Function has a small range of increasing marginal product (increasing marginal returns) and then is subject to the Law of Diminishing Marginal Product (diminishing marginal returns). A. Putting quantity on the horizontal axis and dollars on the vertical axis, depict three important curves: Fixed Cost (FC), Variable Cost (VC), and Total Cost (TC). (Note that we are not asking you to depict average cost functions!) B. Please clearly indicate on this graph the range of quantities where the firm is experiencing (1) increasing marginal product and (2) diminishing marginal product. C. In a few sentences, please justify why you've made this specific classification of increasing/diminishing marginal product in part (b).

please answer the following questions: What is money, and why does anyone want it? Explain the concept of the opportunity cost of holding money . Explain why an increase in U.S. interest rates relative to UK interest rates would affect the U.S.-UK exchange rate. Suppose that a person’s wealth is $50,000 and that her yearlyincome is $60,000. Also suppose that her money demand functionis given by Md = $Y10.35 - i2Derive the demand for bonds. Suppose the interest rate increases by 10 percentage points. What is the effect on her demand for bonds?b. What are the effects of an increase in income on her demand for money and her demand for bonds? Explain in words

Driving Quiz X My Course G city place w x D2L Login - Univ X D2L Login - Univ x D2L Login - U acmillanlearning.com/ihub/assessment/f188d950-dd73-11e0-9572-0800200c9a66/4db68a5e-69bb-4767-8d6c-a12d +1687 pts /1800 © Macmillan Learning Question 6 of 18 > The graph shows the average total cost (ATC) curve, the marginal cost (MC) curve, the average variable cost (AVC) curve, and the marginal revenue (MR) curve (which is also the market price) for a perfectly competitive firm that produces terrible towels. Answer the three questions, assuming that the firm is profit-maximizing and does not shut down in the short run. What is the firm's total revenue? S What is the firm's total cost? $ What is the firm's profit? (Enter a negative number for a loss.) $ Price $320 $300 $200 $150 205 260 336 365 Quantity MC ATC AVC MR=P

Answer 2

Question

Chapter 13, Problem 1P

To determine

The most economic life.

Expert Solution & Answer

Answer to Problem 1P

The most economic life is 5 years.

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

$EUAB=−First cost(A/P,10%,n)+Salvage value(A/F,10%,n)EUAB=−First cost×( i ( 1+i ) n ( 1+i ) n −1)+Salvage value×(i ( 1+i ) n −1)$

EUAC can be calculated with the help of below expression:

$EUAC=Maintenance cost+Maintenance and operating gradient(A/G,10%,n)=Maintenance cost+Maintenance and operating gradient×(1i−n ( 1+i ) n −1)$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $EUAB−EUAC$ ,

When $n=4$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 4 ( 1+0.10 ) 4 −1 )+$225,000×( 0.10 ( 1+0.10 ) 4 −1 ))−( $235,000+$75,000×( 1 0.10 − 4 ( 1+0.10 ) 4 −1 ))]=[( −$1,050,000×0.31547)+( $225,000×0.21547)−( $235,000+( $75,000×( 10−8.6188 ) ))]=−$331,243.5+$48,480.75−$338,590=−$621,352.75$

When $n=5$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 5 ( 1+0.10 ) 5 −1 )+$225,000×( 0.10 ( 1+0.10 ) 5 −1 ))−( $235,000+$75,000×( 1 0.10 − 5 ( 1+0.10 ) 5 −1 ))]=[( −$1,050,000×0.2638)+( $225,000×0.1638)−( $235,000+( $75,000×( 10−8.1898 ) ))]=−$276,990+$36,855−$370,765=−$610,900$

When $n=6$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 6 ( 1+0.10 ) 6 −1 )+$225,000×( 0.10 ( 1+0.10 ) 6 −1 ))−( $235,000+$75,000×( 1 0.10 − 6 ( 1+0.10 ) 6 −1 ))]=[( −$1,050,000×0.2296)+( $225,000×0.1296)−( $235,000+( $75,000×( 10−7.7764 ) ))]=−$241,080+$29,160−$401,770=−$613,690$

Want to see more full solutions like this?

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

Answer 3

Question

Chapter 13, Problem 1P

To determine

The most economic life.

Expert Solution & Answer

Answer to Problem 1P

The most economic life is 5 years.

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

$EUAB=−First cost(A/P,10%,n)+Salvage value(A/F,10%,n)EUAB=−First cost×( i ( 1+i ) n ( 1+i ) n −1)+Salvage value×(i ( 1+i ) n −1)$

EUAC can be calculated with the help of below expression:

$EUAC=Maintenance cost+Maintenance and operating gradient(A/G,10%,n)=Maintenance cost+Maintenance and operating gradient×(1i−n ( 1+i ) n −1)$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $EUAB−EUAC$ ,

When $n=4$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 4 ( 1+0.10 ) 4 −1 )+$225,000×( 0.10 ( 1+0.10 ) 4 −1 ))−( $235,000+$75,000×( 1 0.10 − 4 ( 1+0.10 ) 4 −1 ))]=[( −$1,050,000×0.31547)+( $225,000×0.21547)−( $235,000+( $75,000×( 10−8.6188 ) ))]=−$331,243.5+$48,480.75−$338,590=−$621,352.75$

When $n=5$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 5 ( 1+0.10 ) 5 −1 )+$225,000×( 0.10 ( 1+0.10 ) 5 −1 ))−( $235,000+$75,000×( 1 0.10 − 5 ( 1+0.10 ) 5 −1 ))]=[( −$1,050,000×0.2638)+( $225,000×0.1638)−( $235,000+( $75,000×( 10−8.1898 ) ))]=−$276,990+$36,855−$370,765=−$610,900$

When $n=6$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 6 ( 1+0.10 ) 6 −1 )+$225,000×( 0.10 ( 1+0.10 ) 6 −1 ))−( $235,000+$75,000×( 1 0.10 − 6 ( 1+0.10 ) 6 −1 ))]=[( −$1,050,000×0.2296)+( $225,000×0.1296)−( $235,000+( $75,000×( 10−7.7764 ) ))]=−$241,080+$29,160−$401,770=−$613,690$

Want to see more full solutions like this?

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

Answer 4

Question

Chapter 13, Problem 1P

To determine

The most economic life.

Expert Solution & Answer

Answer to Problem 1P

The most economic life is 5 years.

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

$EUAB=−First cost(A/P,10%,n)+Salvage value(A/F,10%,n)EUAB=−First cost×( i ( 1+i ) n ( 1+i ) n −1)+Salvage value×(i ( 1+i ) n −1)$

EUAC can be calculated with the help of below expression:

$EUAC=Maintenance cost+Maintenance and operating gradient(A/G,10%,n)=Maintenance cost+Maintenance and operating gradient×(1i−n ( 1+i ) n −1)$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $EUAB−EUAC$ ,

When $n=4$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 4 ( 1+0.10 ) 4 −1 )+$225,000×( 0.10 ( 1+0.10 ) 4 −1 ))−( $235,000+$75,000×( 1 0.10 − 4 ( 1+0.10 ) 4 −1 ))]=[( −$1,050,000×0.31547)+( $225,000×0.21547)−( $235,000+( $75,000×( 10−8.6188 ) ))]=−$331,243.5+$48,480.75−$338,590=−$621,352.75$

When $n=5$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 5 ( 1+0.10 ) 5 −1 )+$225,000×( 0.10 ( 1+0.10 ) 5 −1 ))−( $235,000+$75,000×( 1 0.10 − 5 ( 1+0.10 ) 5 −1 ))]=[( −$1,050,000×0.2638)+( $225,000×0.1638)−( $235,000+( $75,000×( 10−8.1898 ) ))]=−$276,990+$36,855−$370,765=−$610,900$

When $n=6$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 6 ( 1+0.10 ) 6 −1 )+$225,000×( 0.10 ( 1+0.10 ) 6 −1 ))−( $235,000+$75,000×( 1 0.10 − 6 ( 1+0.10 ) 6 −1 ))]=[( −$1,050,000×0.2296)+( $225,000×0.1296)−( $235,000+( $75,000×( 10−7.7764 ) ))]=−$241,080+$29,160−$401,770=−$613,690$

Want to see more full solutions like this?

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

Answer 5

Chapter 13, Problem 1P

To determine

The most economic life.

Expert Solution & Answer

Answer to Problem 1P

The most economic life is 5 years.

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

$EUAB=−First cost(A/P,10%,n)+Salvage value(A/F,10%,n)EUAB=−First cost×( i ( 1+i ) n ( 1+i ) n −1)+Salvage value×(i ( 1+i ) n −1)$

EUAC can be calculated with the help of below expression:

$EUAC=Maintenance cost+Maintenance and operating gradient(A/G,10%,n)=Maintenance cost+Maintenance and operating gradient×(1i−n ( 1+i ) n −1)$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $EUAB−EUAC$ ,

When $n=4$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 4 ( 1+0.10 ) 4 −1 )+$225,000×( 0.10 ( 1+0.10 ) 4 −1 ))−( $235,000+$75,000×( 1 0.10 − 4 ( 1+0.10 ) 4 −1 ))]=[( −$1,050,000×0.31547)+( $225,000×0.21547)−( $235,000+( $75,000×( 10−8.6188 ) ))]=−$331,243.5+$48,480.75−$338,590=−$621,352.75$

When $n=5$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 5 ( 1+0.10 ) 5 −1 )+$225,000×( 0.10 ( 1+0.10 ) 5 −1 ))−( $235,000+$75,000×( 1 0.10 − 5 ( 1+0.10 ) 5 −1 ))]=[( −$1,050,000×0.2638)+( $225,000×0.1638)−( $235,000+( $75,000×( 10−8.1898 ) ))]=−$276,990+$36,855−$370,765=−$610,900$

When $n=6$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 6 ( 1+0.10 ) 6 −1 )+$225,000×( 0.10 ( 1+0.10 ) 6 −1 ))−( $235,000+$75,000×( 1 0.10 − 6 ( 1+0.10 ) 6 −1 ))]=[( −$1,050,000×0.2296)+( $225,000×0.1296)−( $235,000+( $75,000×( 10−7.7764 ) ))]=−$241,080+$29,160−$401,770=−$613,690$

Answer 6

Chapter 13, Problem 1P

To determine

The most economic life.

Expert Solution & Answer

Answer to Problem 1P

The most economic life is 5 years.

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

$EUAB=−First cost(A/P,10%,n)+Salvage value(A/F,10%,n)EUAB=−First cost×( i ( 1+i ) n ( 1+i ) n −1)+Salvage value×(i ( 1+i ) n −1)$

EUAC can be calculated with the help of below expression:

$EUAC=Maintenance cost+Maintenance and operating gradient(A/G,10%,n)=Maintenance cost+Maintenance and operating gradient×(1i−n ( 1+i ) n −1)$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $EUAB−EUAC$ ,

When $n=4$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 4 ( 1+0.10 ) 4 −1 )+$225,000×( 0.10 ( 1+0.10 ) 4 −1 ))−( $235,000+$75,000×( 1 0.10 − 4 ( 1+0.10 ) 4 −1 ))]=[( −$1,050,000×0.31547)+( $225,000×0.21547)−( $235,000+( $75,000×( 10−8.6188 ) ))]=−$331,243.5+$48,480.75−$338,590=−$621,352.75$

When $n=5$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 5 ( 1+0.10 ) 5 −1 )+$225,000×( 0.10 ( 1+0.10 ) 5 −1 ))−( $235,000+$75,000×( 1 0.10 − 5 ( 1+0.10 ) 5 −1 ))]=[( −$1,050,000×0.2638)+( $225,000×0.1638)−( $235,000+( $75,000×( 10−8.1898 ) ))]=−$276,990+$36,855−$370,765=−$610,900$

When $n=6$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 6 ( 1+0.10 ) 6 −1 )+$225,000×( 0.10 ( 1+0.10 ) 6 −1 ))−( $235,000+$75,000×( 1 0.10 − 6 ( 1+0.10 ) 6 −1 ))]=[( −$1,050,000×0.2296)+( $225,000×0.1296)−( $235,000+( $75,000×( 10−7.7764 ) ))]=−$241,080+$29,160−$401,770=−$613,690$

Answer 7

Chapter 13, Problem 1P

To determine

The most economic life.

Answer 8

Expert Solution & Answer

Answer to Problem 1P

The most economic life is 5 years.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

$EUAB=−First cost(A/P,10%,n)+Salvage value(A/F,10%,n)EUAB=−First cost×( i ( 1+i ) n ( 1+i ) n −1)+Salvage value×(i ( 1+i ) n −1)$

EUAC can be calculated with the help of below expression:

$EUAC=Maintenance cost+Maintenance and operating gradient(A/G,10%,n)=Maintenance cost+Maintenance and operating gradient×(1i−n ( 1+i ) n −1)$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $EUAB−EUAC$ ,

When $n=4$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 4 ( 1+0.10 ) 4 −1 )+$225,000×( 0.10 ( 1+0.10 ) 4 −1 ))−( $235,000+$75,000×( 1 0.10 − 4 ( 1+0.10 ) 4 −1 ))]=[( −$1,050,000×0.31547)+( $225,000×0.21547)−( $235,000+( $75,000×( 10−8.6188 ) ))]=−$331,243.5+$48,480.75−$338,590=−$621,352.75$

When $n=5$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 5 ( 1+0.10 ) 5 −1 )+$225,000×( 0.10 ( 1+0.10 ) 5 −1 ))−( $235,000+$75,000×( 1 0.10 − 5 ( 1+0.10 ) 5 −1 ))]=[( −$1,050,000×0.2638)+( $225,000×0.1638)−( $235,000+( $75,000×( 10−8.1898 ) ))]=−$276,990+$36,855−$370,765=−$610,900$

When $n=6$

$EUAB−EUAC=[( −$1,050,000×( 0.10× ( 1+0.10 ) 6 ( 1+0.10 ) 6 −1 )+$225,000×( 0.10 ( 1+0.10 ) 6 −1 ))−( $235,000+$75,000×( 1 0.10 − 6 ( 1+0.10 ) 6 −1 ))]=[( −$1,050,000×0.2296)+( $225,000×0.1296)−( $235,000+( $75,000×( 10−7.7764 ) ))]=−$241,080+$29,160−$401,770=−$613,690$

Answer 12

Ch. 13 - Prob. 1QTC Ch. 13 - Prob. 2QTC Ch. 13 - Prob. 3QTC Ch. 13 - Prob. 4QTC Ch. 13 - Prob. 5QTC Ch. 13 - Prob. 1P Ch. 13 - Prob. 2P Ch. 13 - Prob. 3P Ch. 13 - Prob. 4P Ch. 13 - Prob. 5P

The most economic life.

Videos

The most economic life.

Answer to Problem 1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions