Question

For the alternatives shown below, determine the incremental rate of return for cash flows for Q – P.

	Alternative P	Alternative Q
First cost, $	-50,000	-85,000
Annual operating cost, $ per year	-8,600	-2,000
Annual revenue, $ per year	22,000	45,000
Salvage value, $	3,000	8,000
Life, years	2	3

Answer 1

We have the following information

	Alternative P	Alternative Q
First cost, $	-50,000	-85,000
Annual operating cost, $ per year	-8,600	-2,000
Annual revenue, $ per year	22,000	45,000
Salvage value, $	3,000	8,000
Life, years	2	3

Alternative P

Initial Cost (P) = $50,000

Annual Operating Cost (C) = $8,600

Annual Revenue (R) = $22,000

Salvage Value (S) = $3,000

Life (n) = 2

Assuming rate of return of 0.5% or 0.005

Present Worth (PW) = – P – C(P/A, 0.5%, 2) + R(P/A, 0.5%, 2) + S(P/F, 0.5%, 2)

PW = – 50,000 – 8,600(P/A, 0.5%, 2) + 22,000(P/A, 0.5%, 2) + 3,000(P/F, 0.5%, 2)

PW = – 50,000 – 8,600[((1 + 0.005)² – 1)/0.005(1 + 0.005)²] + 22,000[((1 + 0.005)² – 1)/0.005(1 + 0.005)²] + 3,000/(1 + 0.005)²

PW = – 50,000 – (8,600 × 1.9851) + (22,000 × 1.9851) + (3,000 × 0.9901)

PW = – 50,000 – 17,071.86 + 43,672.2 + 2,970.3

PW = – 20,429.36

The rate of return of Alternative P is Zero.

Alternative Q

Initial Cost (P) = $85,000

Annual Operating Cost (C) = $2,000

Annual Revenue (R) = $45,000

Salvage Value (S) = $8,000

Life (n) = 3

Assuming rate of return of 1% or 0.01

Present Worth (PW) = – P – C(P/A, 1%, 3) + R(P/A, 1%, 3) + S(P/F, 1%, 3)

PW = – 85,000 – 2,000(P/A, 1%, 3) + 45,000(P/A, 1%, 3) + 8,000(P/F, 1%, 3)

PW = – 85,000 – 2,000[((1 + 0.01)³ – 1)/0.01(1 + 0.01)³] + 45,000[((1 + 0.01)³ – 1)/0.01(1 + 0.01)³] + 8,000/(1 + 0.01)³

PW = – 85,000 – (2,000 × 1.9704) + (45,000 × 1.9704) + (8,000 × 0.9803)

PW = – 85,000 – 3,940.8 + 88,668 + 7,842.4

PW = 7,569.6

Assuming rate of return of 5% or 0.05

Present Worth (PW) = – P – C(P/A, 5%, 3) + R(P/A, 5%, 3) + S(P/F, 5%, 3)

PW = – 85,000 – 2,000(P/A, 5%, 3) + 45,000(P/A, 5%, 3) + 8,000(P/F, 5%, 3)

PW = – 85,000 – 2,000[((1 + 0.05)³ – 1)/0.05(1 + 0.05)³] + 45,000[((1 + 0.05)³ – 1)/0.05(1 + 0.05)³] + 8,000/(1 + 0.05)³

PW = – 85,000 – (2,000 × 1.8594) + (45,000 × 1.8594) + (8,000 × 0.9070)

PW = – 85,000 – 3,718.8 + 83,673 + 7,256

PW = 2,210.2

Assuming rate of return of 10% or 0.1

Present Worth (PW) = – P – C(P/A, 10%, 3) + R(P/A, 10%, 3) + S(P/F, 10%, 3)

PW = – 85,000 – 2,000(P/A, 10%, 3) + 45,000(P/A, 10%, 3) + 8,000(P/F, 10%, 3)

PW = – 85,000 – 2,000[((1 + 0.1)³ – 1)/0.1(1 + 0.1)³] + 45,000[((1 + 0.1)³ – 1)/0.1(1 + 0.1)³] + 8,000/(1 + 0.1)³

PW = – 85,000 – (2,000 × 1.7355) + (45,000 × 1.7355) + (8,000 × 0.8264)

PW = – 85,000 – 3,471 + 78,097.5 + 6,611.2

PW = – 3,762.3

Therefore the rate of return for Alternative Q is

i = 5% + ((2210.2 – 0)/2210.2 – (– 3762.3)) × 5%

i = 5% + (2210.2/5972.5) × 5%

i = 6.85%