Chapter 9
Mutually
Exclusive
Alternatives
9-1
Using a 10% interest rate, determine which alternative, if any, should be selected, based on net
present worth.
Alternative A B
First Cost $5,300 $10,700
Uniform Annual Benefit 1,800 2,100
Useful life 4 years 8 years
Solution
Alternative A:
NPW = 1,800(P/A, 10%, 8) - 5,300 - 5,300(P/F, 10%, 4)
= $683.10
Alternative B:
NPW = 2,100(P/A, 10%, 8) - 10,700
= $503.50
Select alternative A
9-2
Three purchase plans are available for a new car.
Plan A: $5,000 cash immediately
Plan B: $1,500 down and 36 monthly payments of $116.25
Plan C: $1,000 down and 48 monthly payments of $120.50
If a customer expects to keep the car five years and her cost of money is 18% compounded
monthly, which payment plan should she choose?
Solution
127
,128 Chapter 9 Mutually Exclusive Alternatives
i = 18/12 = 1½%
PWA = $5,000
PWB = 1,500 + 116.25(P/A, 1½%, 36) = $4,715.59
PWC = 1,000 + 120.50(P/A, 1½%, 48) = $5,102.18
Therefore Plan B is the best plan.
9-3
Given the following three mutually exclusive alternatives
Alternative
A B C
Initial Cost $50 $30 $40
Annual Benefits 15 10 12
Useful Life (years) 5 5 5
What alternative is preferable, if any, assuming i = 10%?
Solution
PWA = -50 + 15(P/A, 10%, 5) = $6.87
PWB = -30 + 10(P/A, 10%, 5) = $7.91
PWC = -40 + 12(P/A, 10%, 5) = $5.49
Choose C
9-4
Consider two investments:
1. Invest $1,000 and receive $110 at the end of each month for the next 10 months.
2. Invest $1,200 and receive $130 at the end of each month for the next 10 months.
If this were your money, and you want to earn at least 12% interest on your money, which
investment would you make, if any? Solve the problem by annual cash flow analysis.
Solution
Alternative 1: EUAW = EUAB - EUAC = 110 - 1,000(A/P, 1%, 10) = $4.40
Alternative 2: EUAW = EUAB - EUAC = 130 - 1,200(A/P, 1%, 10) = $3.28
Maximum EUAW, therefore choose alternative A.
9-5
A farmer must purchase a tractor using a loan of $20,000. The bank has offered the following
choice of payment plans each determined by using an interest rate of 8%. If the farmer's minimum
attractive rate of return (MARR) is 15%, which plan should he choose?
Plan A: $5,010 per year for 5 years
, Chapter 9 Mutually Exclusive Alternatives 129
Plan B: $2,956 per year for 4 years plus $15,000 at end of 5 years
Plan C: Nothing for 2 years, then $9,048 per year for 3 years
Solution
PWCA = 5,010(P/A, 15%, 5) = $16,794
PWCB = 2,956(P/A, 15%, 4) + 15,000(P/F, 15%, 5) = $15,897
PWCC = 9,048(P/A, 15%, 3)(P/F, 15%, 2) = $15,618
Plan C is lowest cost plan
9-6
Projects A and B have first costs of $6,500 and $17,000, respectively. Project A has net annual
benefits of $2,000 during each year of its 5-year useful life, after which it can be replaced
identically.
Project B has net annual benefits of $3,000 during each year of its 10-year life. Use present worth
analysis, and an interest rate of 10% to determine which project to select.
Solution
PWA = -6,500[1 + (P/F, 10%, 5)] + 2,000(P/A, 10%, 10) = $1,754.15
PWB = -17,000 + 3,000(P/A, 10%, 10) = $ 1,435.00
Select A because of higher present worth
9-7
A manufacturing firm has a minimum attractive rate of return (MARR) of 12% on new
investments. What uniform annual benefit would Investment B have to generate to make it
preferable to Investment A?
Year Investment A Investment B
0 - $60,000 - $45,000
1-6 +15,000 ?
Solution
NPW of A = - 60,000 + 15,000(P/A, 12%, 6)
= $1,665
NPW of B ≥ 1,665 = - 45,000 + A(P/A, 12%, 6)
∴ A = 11,351
Annual Benefit > $11,351 per year
Mutually
Exclusive
Alternatives
9-1
Using a 10% interest rate, determine which alternative, if any, should be selected, based on net
present worth.
Alternative A B
First Cost $5,300 $10,700
Uniform Annual Benefit 1,800 2,100
Useful life 4 years 8 years
Solution
Alternative A:
NPW = 1,800(P/A, 10%, 8) - 5,300 - 5,300(P/F, 10%, 4)
= $683.10
Alternative B:
NPW = 2,100(P/A, 10%, 8) - 10,700
= $503.50
Select alternative A
9-2
Three purchase plans are available for a new car.
Plan A: $5,000 cash immediately
Plan B: $1,500 down and 36 monthly payments of $116.25
Plan C: $1,000 down and 48 monthly payments of $120.50
If a customer expects to keep the car five years and her cost of money is 18% compounded
monthly, which payment plan should she choose?
Solution
127
,128 Chapter 9 Mutually Exclusive Alternatives
i = 18/12 = 1½%
PWA = $5,000
PWB = 1,500 + 116.25(P/A, 1½%, 36) = $4,715.59
PWC = 1,000 + 120.50(P/A, 1½%, 48) = $5,102.18
Therefore Plan B is the best plan.
9-3
Given the following three mutually exclusive alternatives
Alternative
A B C
Initial Cost $50 $30 $40
Annual Benefits 15 10 12
Useful Life (years) 5 5 5
What alternative is preferable, if any, assuming i = 10%?
Solution
PWA = -50 + 15(P/A, 10%, 5) = $6.87
PWB = -30 + 10(P/A, 10%, 5) = $7.91
PWC = -40 + 12(P/A, 10%, 5) = $5.49
Choose C
9-4
Consider two investments:
1. Invest $1,000 and receive $110 at the end of each month for the next 10 months.
2. Invest $1,200 and receive $130 at the end of each month for the next 10 months.
If this were your money, and you want to earn at least 12% interest on your money, which
investment would you make, if any? Solve the problem by annual cash flow analysis.
Solution
Alternative 1: EUAW = EUAB - EUAC = 110 - 1,000(A/P, 1%, 10) = $4.40
Alternative 2: EUAW = EUAB - EUAC = 130 - 1,200(A/P, 1%, 10) = $3.28
Maximum EUAW, therefore choose alternative A.
9-5
A farmer must purchase a tractor using a loan of $20,000. The bank has offered the following
choice of payment plans each determined by using an interest rate of 8%. If the farmer's minimum
attractive rate of return (MARR) is 15%, which plan should he choose?
Plan A: $5,010 per year for 5 years
, Chapter 9 Mutually Exclusive Alternatives 129
Plan B: $2,956 per year for 4 years plus $15,000 at end of 5 years
Plan C: Nothing for 2 years, then $9,048 per year for 3 years
Solution
PWCA = 5,010(P/A, 15%, 5) = $16,794
PWCB = 2,956(P/A, 15%, 4) + 15,000(P/F, 15%, 5) = $15,897
PWCC = 9,048(P/A, 15%, 3)(P/F, 15%, 2) = $15,618
Plan C is lowest cost plan
9-6
Projects A and B have first costs of $6,500 and $17,000, respectively. Project A has net annual
benefits of $2,000 during each year of its 5-year useful life, after which it can be replaced
identically.
Project B has net annual benefits of $3,000 during each year of its 10-year life. Use present worth
analysis, and an interest rate of 10% to determine which project to select.
Solution
PWA = -6,500[1 + (P/F, 10%, 5)] + 2,000(P/A, 10%, 10) = $1,754.15
PWB = -17,000 + 3,000(P/A, 10%, 10) = $ 1,435.00
Select A because of higher present worth
9-7
A manufacturing firm has a minimum attractive rate of return (MARR) of 12% on new
investments. What uniform annual benefit would Investment B have to generate to make it
preferable to Investment A?
Year Investment A Investment B
0 - $60,000 - $45,000
1-6 +15,000 ?
Solution
NPW of A = - 60,000 + 15,000(P/A, 12%, 6)
= $1,665
NPW of B ≥ 1,665 = - 45,000 + A(P/A, 12%, 6)
∴ A = 11,351
Annual Benefit > $11,351 per year
