Engineering Economy
Engineering Economy
8th Edition
ISBN: 9780073523439
Author: Leland T Blank Professor Emeritus, Anthony Tarquin
Publisher: McGraw-Hill Education
Question
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Chapter 14, Problem 29P

(a):

To determine

Calculate the present worth without considering the inflation.

(a):

Expert Solution
Check Mark

Explanation of Solution

Table-1 shows the cash flow of different alternatives.

Table-1

AlternativeAB
First cost (C)-31,000-48,000
Annual cost (AM) per year-28,000-19,000
Salvage value (SV)5,0007,000
Time period (n)55

The interest rate (i) is 10%, and the inflation rate (f) is 3%.

The present worth (PW) for Alternative A can be calculated as follows:

PW=C+AM((1+i)n1i(1+i)n)+SV(1(1+i)n)=31,00028,000((1+0.1)510.1(1+0.1)5)+5,000(1(1+0.1)5)=31,00028,000(3.7908)+5,000(0.6209)=31,000106,143.4+3,104.5=134,038.9

The present worth is -$134,038.9. The present worth for Alternate A can be calculated using a spreadsheet as follows:

= - PV(10%,5,-28000,5000) – 31000

The above spreadsheet function gives the value of -$134,37.42.

The present worth (PW) for Alternative B can be calculated as follows:

PW=C+AM((1+i)n1i(1+i)n)+SV(1(1+i)n)=48,00019,000((1+0.1)510.1(1+0.1)5)+7,000(1(1+0.1)5)=48,00019,000(3.7908)+7,000(0.6209)=48,00072,025.2+4,346.3=115,678.9.

The present worth is -$115,678.9. Since the present worth of the cost is lower for Alternate B, select Alternate B. The present worth for alternate B can be calculated using a spreadsheet as follows:

= - PV(10%,5,-19000,7000) - 48000

The above spreadsheet function gives the value of -$115,678.5.

(b):

To determine

Calculate the present worth with considering the inflation.

(b):

Expert Solution
Check Mark

Explanation of Solution

The real interest rate (if) can be calculated as follows:

if=i+f+i×f=0.1+0.03+0.1×0.03=0.13+0.003=0.133

The real interest rate is 13.3%.

The present worth (PW) for Alternative A can be calculated as follows:

PW=C+AM((1+if)n1if(1+if)n)+SV(1(1+if)n)=31,00028,000((1+0.133)510.133(1+0.133)5)+5,000(1(1+0.133)5)=31,00028,000(3.4916)+5,000(0.5356)=31,00097,764.8+2,678=126,086.8

The present worth is -$126,086.8. The present worth for Alternate A can be calculated using a spreadsheet as follows:

= - PV(13.3%,5,-28000,5000) - 31000

The above spreadsheet function gives the value of -$126,87.79.

The present worth (PW) for Alternative B can be calculated as follows:

PW=C+AM((1+if)n1if(1+if)n)+SV(1(1+if)n)=48,00019,000((1+0.133)510.133(1+0.133)5)+7,000(1(1+0.133)5)=48,00019,000(3.4916)+7,000(0.5356)=48,00066,340.4+3,749.2=110,591.2

The present worth is -$115,678.9. Since the present worth of the cost is lower for Alternate B, select Alternate B. The present worth for Alternate B can be calculated using a spreadsheet as follows:

= - PV(13.3%,5,-19000,7000) - 48000

The above spreadsheet function gives the value of -$110,591.83.

(c):

To determine

Calculate the interest rate that makes both the alternates equal.

(c):

Expert Solution
Check Mark

Explanation of Solution

The interest rate that makes both alternates’ present worth equal can be calculated using the spreadsheet as follows:

Engineering Economy, Chapter 14, Problem 29P

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