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Problem 12-3A (Part Level Submission) (a) Problem 12-3A (Part Level Submission)

ID: 2532215 • Letter: P

Question

Problem 12-3A (Part Level Submission)

(a)

Problem 12-3A (Part Level Submission)

Brooks Clinic is considering investing in new heart-monitoring equipment. It has two options. Option A would have an initial lower cost but would require a significant expenditure for rebuilding after 4 years. Option B would require no rebuilding expenditure, but its maintenance costs would be higher. Since the Option B machine is of initial higher quality, it is expected to have a salvage value at the end of its useful life. The following estimates were made of the cash flows. The company’s cost of capital is 5%.
Option A Option B Initial cost $189,000 $285,000 Annual cash inflows $71,600 $82,500 Annual cash outflows $28,200 $26,700 Cost to rebuild (end of year 4) $50,100 $0 Salvage value $0 $8,100 Estimated useful life 7 years 7 years
Click here to view PV table.

Explanation / Answer

Solution:-

1. Net present value method:-

Option A

Option B

PV of annual cash inflow(PVAF @ 5% for 7 years *Cash inflow) (PVAF=5.78637)

[5.78637*71600]= 414304

[5.78637*82500] =477375.53

(-) PV of cash annual cash outflow

[5.78637*28200] =163175.63

[5.78637*26700] =154496

(-) PV of cost to rebuild at end of 4 year (PVIF at end of 4th yr = 0.82270)

[50100*0.82270]

=41217.27

---------------------

(+) PV of salvage value (PVIF at end of 7th year = 0.71068)

--------

(8100*0.71068) =5756.51

Net cash inflow (PV)

209911.1

328636.04

(-) Initial cost

189000

285000

Net cash inflow/outflow

20911.1

43636.04

2. Profitability Index:-

   Profitability Index = PV of Future Net Cash Flows / Initial Investment Required

         

Option A

Option B

(A) Initial cost

189000

285000

(B) Net cash inflow (PV)

209911.1

328636.04

PI = B/A

1.11

1.15

3. Internal rate of Return:-

    Option A:-

PV of Cash Inflow = Pv of Cash outflow

(71600*PVAF of 7 yrs) – (28200*PVAF of 7 yrs ) – (50100*PVIF of end of 4th yr) = 189000

Let we try @ 8%

(71600*5.20637) – (28200*5.20637) – (50100*0.73503) = 189000

189131 = 189000

Option B:-

PV of Cash Inflow = Pv of Cash outflow

(82500*PVAF of 7 yrs) – (26700*PVAF of 7 yrs ) + (8100*PVIF of end of 7th yr) = 285000

Let we try @ 9%

(82500*5.03295) – (26700*5.03295) + (8100*0.54703) = 285000

285269 = 285000

Net Present Value

Profitability Index

IRR

Option A

20911.1

1.11

8%

Option B

43636.04

1.15

9%

Option A

Option B

PV of annual cash inflow(PVAF @ 5% for 7 years *Cash inflow) (PVAF=5.78637)

[5.78637*71600]= 414304

[5.78637*82500] =477375.53

(-) PV of cash annual cash outflow

[5.78637*28200] =163175.63

[5.78637*26700] =154496

(-) PV of cost to rebuild at end of 4 year (PVIF at end of 4th yr = 0.82270)

[50100*0.82270]

=41217.27

---------------------

(+) PV of salvage value (PVIF at end of 7th year = 0.71068)

--------

(8100*0.71068) =5756.51

Net cash inflow (PV)

209911.1

328636.04

(-) Initial cost

189000

285000

Net cash inflow/outflow

20911.1

43636.04