The most economic life.

Question

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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

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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

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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

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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

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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.

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The most economic life.

Answer to Problem 1P

Explanation of Solution

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Chapter 13 Solutions