EBK ENGINEERING ECONOMY
EBK ENGINEERING ECONOMY
8th Edition
ISBN: 8220103675437
Author: Blank
Publisher: YUZU
Question
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Chapter 5, Problem 25P

(a):

To determine

Compare the projects based on the present worth.

(a):

Expert Solution
Check Mark

Explanation of Solution

Plan A: The initial cost (C) is $1,000,000 per year.

Plan B: It is a two-year contract. The cost (C) is $600,000 per semiannual starts with immediate. No payment required in the second year.

Plan C: It is a three-year contract. The cost (C) is $1,500,000 now and the second payment (C1) is $500,000 after two years.

To compare the projects, all projects’ life time should be equal. Thus, all the projects’ cash flow converts to 6 years.

Interest rate (i) is 6% per year that is compounded semiannually. Interest calculated time period (Ci) is 2. The effective interest rate per year (Ei) can be calculated as follows:

Ei=(1+iCi)Ci1=(1+0.062)21=1.06091=0.0609

The effective interest rate is 6.09%.

Plan A cash flow is each year pays $1,000,000. Since the payment is made at the beginning of each year, the time period (n) is five. The present worth of plan A (PWA) can be calculated as follows:

PWA=CC((1+Ei)n1Ei(1+Ei)n)=1,000,0001,000,000((1+0.0609)510.0609(1+0.0609)5)=1,000,0001,000,000(1.34391610.0609(1.343916))=1,000,0001,000,000(0.3439160.081844)=1,000,0001,000,000(4.202092)=1,000,0004,202,092=5,202,092

The present worth of the plan A is -$5,202,092.

Plan B cash flow is each semiannual pays $600,000. Since the payment is made at the beginning of each semiannual for the first year, the time period (n1) is two, time period 2 (n2) is three, and time period 3 (n3) is seven. Since the interest is compounded semiannually, the semiannual interest (i1) is 3%(62). The present worth of plan B (PWB) can be calculated as follows:

PWB=C(1+((1+i1)n11i1(1+i1)n1)+((1+i1)n21i1(1+i1)n2)(1+i1)n2+((1+i1)n31i1(1+i1)n3)(1+i1)n3)

=600,000(1+((1+0.03)210.03(1+0.03)2)+((1+0.03)310.03(1+0.03)3)(1+0.03)3+((1+0.03)310.03(1+0.03)3)(1+0.03)7)=600,000(1+(1.060910.03(1.0609))+(1.09272710.03(1.092727))1.092727+(1.09272710.03(1.092727))1.229873)=600,000(1+(0.06090.031827)+(0.0927270.032782)1.092727+(0.0927270.032782)1.229873)=600,000(1+(1.91347)+(2.828595)1.092727+(2.828595)1.229873)=600,000(1+(1.91347)+2.588565+2.299991)=600,000(7.802026)=4,681,215.6

The present worth of plan B is -$4,681,215.6.

Plan C’s cash flow is $1,500,000 at the beginning of the year and pays $500,000 at the end of the second year. Thus, the time period (n1) is four, time period 2 (n2) is six, and time period 3 (n3) is 10. Since the interest is compounded semiannually, the semiannual interest (i1) is 3%(62). The present worth of plan C (PWC) can be calculated as follows:

PWC=CC1(1+i1)n1C(1+i1)n2C1(1+i1)n3=1,500,000500,000(1+0.03)41,500,000(1+0.03)6500,000(1+0.03)10=1,500,000500,0001.0927271,500,0001.194052500,0001.3463916=1,500,000457,570.82971,256,226.1497371,362.872=3,555159.85

The present worth of plan C is -$3,555,159.85. Since the present worth of plan C is greater than other two plans, select plan C.

(b):

To determine

Spreadsheet function for calculating the present worth.

(b):

Expert Solution
Check Mark

Explanation of Solution

The interest rate is filled in B2, cash flows are filled in B5 to B10. The spreadsheet function for plan A is given below:

=NPV($B2,B6:B10)+B5

The above function gives the value of -$5,202,070.

The interest rate is filled in D2, cash flows are filled in D5 to D15. The spreadsheet function for plan A is given below:

= NPV($D2,D6:D15)+D5

The above function gives the value of -$4,681,182.

The interest rate is filled in E2, and cash flows are filled in E5 to E15. The spreadsheet function for plan A is given below:

= NPV($E2,E6:E15)+E5

The above function gives the value of -$3,572,517.

Since the present worth of plan C is greater than other two plans, select plan C.

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