ENGR.ECONOMIC ANALYSIS
ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
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Chapter 13, Problem 1P
To determine

The most economic life.

Expert Solution & Answer
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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=Firstcost(A/P,10%,n)+Salvagevalue(A/F,10%,n)EUAB=Firstcost×( i ( 1+i ) n ( 1+i ) n 1)+Salvagevalue×(i ( 1+i ) n 1)

EUAC can be calculated with the help of below expression:

  EUAC=Maintenancecost+Maintenanceandoperatinggradient(A/G,10%,n)=Maintenancecost+Maintenanceandoperatinggradient×(1in ( 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 EUABEUAC ,

When n=4

  EUABEUAC=[( $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×( 108.6188 ) ))]=$331,243.5+$48,480.75$338,590=$621,352.75

When n=5

  EUABEUAC=[( $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×( 108.1898 ) ))]=$276,990+$36,855$370,765=$610,900

When n=6

  EUABEUAC=[( $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×( 107.7764 ) ))]=$241,080+$29,160$401,770=$613,690

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Students have asked these similar questions
A contractor has a 4-year concrete mixer whose first cost was $6,000, having 3 more years to live before being scrapped and sold at $801. Itcould now be sold for $11,922. It has an annual cost for operation and maintenance of $9,352. Its replacement is being proposed with a newmachine whose first cost will be $8,000 having a life of 9 years and salvage value $1,600. It has an operating cost of $800 per year andmaintenance cost of $320 per year. Ifthe interest is 20% cpd-a, what is the Annual Equivalent Cost of the Old Machine? 14,792
Problem 8 A colleague has completed the following set of estimated costs and salvage values for a proposed machine with an initial cost of $15,000. However, he doesn't know how to find the most economic useful life. To demonstrate, you compute the equivalent uniform annual cost (EUAC) for year eight (EUAC) using a MARR of 15% Useful Estimated Estimated Life End-of- Salvage (years) Year MX Value 1 $0 $10,000 2 $0 $9000 3 $300 $8000 4 $300 $7000 5 $800 $6000 6 $1300 $5000 7 $1800 $4000 8 $2300 $3000 9 $2800 $2000 10 $3300 $1000
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