ENGINEERING ECONOMIC ENHANCED EBOOK
ENGINEERING ECONOMIC ENHANCED EBOOK
14th Edition
ISBN: 9780190931940
Author: NEWNAN
Publisher: OXF
Question
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Chapter 6, Problem 49P
To determine

The more economical choice.

Expert Solution & Answer
Check Mark

Answer to Problem 49P

The most economical choice is, Diesel.

Explanation of Solution

The cars average distance covered by car is 80000km a year.

The interest is 6%.

Calculation:

Calculate the annual cost of each fuel for each case.

Annualcost=(DistancecoveredMileage(km/litre))×(Fuelcostperlitre) ...... (I)

Calculate the annual cost for diesel.

Substitute 80000km for distance covered and 16km/liter for mileage and 88% for fuel cost per liter in Equation (I).

Annualcost=80000km16(km/litre)×0.88=$4400

Calculate the annual cost for Gasoline.

Substitute 80000km for distance covered and 11km/liter for mileage and 92% for fuel cost per liter in Equation (I).

Annualcost=80000km11(km/litre)×0.92=$6691

Alternative 1: Diesel

Given:

Vehicle cost is $24000

Annual repairs $900

Annual insurance premium is $1000

Salvage value is $4000

Interest rate 6%

Time period is 5years.

EUAC=(PS)(AP,i,n)+(otherannualcost)+(tax)EUAC=(PS)(AP,i,n)+($4000×0.06)+$4400+$900+$1000 ...... (II)

Here, vehicle cost is P, salvage value is S.

Calculate the factor (AP,i,n).

(AP,i,n)=[i(1+i)n(1+i)n1] ...... (III)

Substitute 6% for interest and 5years in Equation (III).

(AP,i,n)=[0.06( 1+0.06)5( 1+0.06)51]=0.2374

Substitute 0.2374 for (AP,i,n), $24000 for P, and $4000 for S in Equation (II).

EUAC=($240000$4000S)×0.2374+($4000×0.06)+$4400+$900+$1000=$11288

Alternative 2: Gasoline

Given:

Vehicle cost is $19000

Annual repairs $700

Annual insurance premium is $1000

Salvage value is $6000

Interest rate 6%

Time period is 4years

EUAC=(PS)(AP,i,n)+otherannualcost+taxEUAC=(PS)(AP,i,n)+$6000×0.06+$6691+$700+$1000 ...... (IV)

Here, vehicle cost is P, salvage value is S.

Calculate the factor (AP,i,n).

(AP,i,n)=[i(1+i)n(1+i)n1] ...... (V)

Substitute 6% for interest and 4years in Equation (V).

(AP,i,n)=[0.06( 1+0.06)4( 1+0.06)41]=0.2886

Substitute 0.2886 for (AP,i,n), $19000 for P, and $6000 for S in Equation (IV).

EUAC=($19000$6000)×0.2886+$6000×0.06+$6691+$700=$12503.

Conclusion:

EUAC of diesel is less than the Gasoline. Thus Diesel should be selected.

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