Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
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Question
Chapter 5, Problem 101P
To determine
(a)
When the higher payments catch up in total dollars received with the lower payment that begins earlier.
Expert Solution
Answer to Problem 101P
The higher payments catch up in total dollars received with the lower payment that begins earlier in the 18 Year in case of higher payments collected.
Explanation of Solution
Given information:
Option A: Receive $2,000 per month from age 66 − 85 i.e. 19 Years
Option B: Receive $1,500 per month from age 62 − 85 i.e. 23 Years.
Based on the above information, the following tables express the total cumulative dollars received.
Annual Receipts (Option 1) = $2,000 × 12 = $24,000
Annual Receipts (Option 2) = $1,500 × 12 = $18,000
| Option 1: $2,000 from Age 66 - 85 | Option 2: $1,500 from Age 62 - 85 | ||||
| Age | Annual Receipt | Cumulative Receipt | Age | Annual Receipt | Cumulative Receipt |
| 66 | $24,000.00 | $24,000.00 | 62 | $18,000.00 | $18,000.00 |
| 67 | $24,000.00 | $48,000.00 | 63 | $18,000.00 | $36,000.00 |
| 68 | $24,000.00 | $72,000.00 | 64 | $18,000.00 | $54,000.00 |
| 69 | $24,000.00 | $96,000.00 | 65 | $18,000.00 | $72,000.00 |
| 70 | $24,000.00 | $120,000.00 | 66 | $18,000.00 | $90,000.00 |
| 71 | $24,000.00 | $144,000.00 | 67 | $18,000.00 | $108,000.00 |
| 72 | $24,000.00 | $168,000.00 | 68 | $18,000.00 | $126,000.00 |
| 73 | $24,000.00 | $192,000.00 | 69 | $18,000.00 | $144,000.00 |
| 74 | $24,000.00 | $216,000.00 | 70 | $18,000.00 | $162,000.00 |
| 75 | $24,000.00 | $240,000.00 | 71 | $18,000.00 | $180,000.00 |
| 76 | $24,000.00 | $264,000.00 | 72 | $18,000.00 | $198,000.00 |
| 77 | $24,000.00 | $288,000.00 | 73 | $18,000.00 | $216,000.00 |
| 78 | $24,000.00 | $312,000.00 | 74 | $18,000.00 | $234,000.00 |
| 79 | $24,000.00 | $336,000.00 | 75 | $18,000.00 | $252,000.00 |
| 80 | $24,000.00 | $360,000.00 | 76 | $18,000.00 | $270,000.00 |
| 81 | $24,000.00 | $384,000.00 | 77 | $18,000.00 | $288,000.00 |
| 82 | $24,000.00 | $408,000.00 | 78 | $18,000.00 | $306,000.00 |
| 83 | $24,000.00 | $432,000.00 | 79 | $18,000.00 | $324,000.00 |
| 84 | $24,000.00 | $456,000.00 | 80 | $18,000.00 | $342,000.00 |
| 85 | $24,000.00 | $480,000.00 | 81 | $18,000.00 | $360,000.00 |
| 82 | $18,000.00 | $378,000.00 | |||
| 83 | $18,000.00 | $396,000.00 | |||
| 84 | $18,000.00 | $414,000.00 | |||
| 85 | $18,000.00 | $432,000.00 | |||
From the above table, it can be seen that the cumulative collection in Year 23 is $432,000 in case of Option 2 and in case of Option 1 the cumulative collection in Year 18 is $432,000.
Conclusion:
Hence the year in which higher payments catch up in total dollars received with the lower payment that begins earlier is determined.
To determine
(b)
Which alternative has the higher Present value.
Expert Solution
Answer to Problem 101P
The alternative with the higher Present value is Option 1 i.e. Collect payments of $2,000 per month from age 66 − 85.
Explanation of Solution
Given information:
Option A: Receive $2,000 per month from age 66 − 85
Option B: Receive $1,500 per month from age 62 − 85.
Based on the above information, the following tables express the total cumulative dollars received.
Annual Receipts (Option 1) = $2,000 × 12 = $24,000
Annual Receipts (Option 2) = $1,500 × 12 = $18,000
| Option 1: $2,000 from Age 66 - 85 | Option 2: $1,500 from Age 62 - 85 | ||||||
| Age | Annual Receipt | P.V.Factor @ 6% | Present Value of Amount | Age | Annual Receipt | P.V.Factor @ 6% | Present Value of Amount |
| 1 | $24,000.00 | 0.9434 | $22,641.60 | 1 | $18,000.00 | 0.9434 | $16,981.20 |
| 2 | $24,000.00 | 0.8900 | $21,360.00 | 2 | $18,000.00 | 0.8900 | $16,020.00 |
| 3 | $24,000.00 | 0.8396 | $20,150.40 | 3 | $18,000.00 | 0.8396 | $15,112.80 |
| 4 | $24,000.00 | 0.7921 | $19,010.40 | 4 | $18,000.00 | 0.7921 | $14,257.80 |
| 5 | $24,000.00 | 0.7473 | $17,935.20 | 5 | $18,000.00 | 0.7473 | $13,451.40 |
| 6 | $24,000.00 | 0.7050 | $16,920.00 | 6 | $18,000.00 | 0.7050 | $12,690.00 |
| 7 | $24,000.00 | 0.6651 | $15,962.40 | 7 | $18,000.00 | 0.6651 | $11,971.80 |
| 8 | $24,000.00 | 0.6274 | $15,057.60 | 8 | $18,000.00 | 0.6274 | $11,293.20 |
| 9 | $24,000.00 | 0.5919 | $14,205.60 | 9 | $18,000.00 | 0.5919 | $10,654.20 |
| 10 | $24,000.00 | 0.5584 | $13,401.60 | 10 | $18,000.00 | 0.5584 | $10,051.20 |
| 11 | $24,000.00 | 0.5268 | $12,643.20 | 11 | $18,000.00 | 0.5268 | $9,482.40 |
| 12 | $24,000.00 | 0.4970 | $11,928.00 | 12 | $18,000.00 | 0.4970 | $8,946.00 |
| 13 | $24,000.00 | 0.4688 | $11,251.20 | 13 | $18,000.00 | 0.4688 | $8,438.40 |
| 14 | $24,000.00 | 0.4423 | $10,615.20 | 14 | $18,000.00 | 0.4423 | $7,961.40 |
| 15 | $24,000.00 | 0.4173 | $10,015.20 | 15 | $18,000.00 | 0.4173 | $7,511.40 |
| 16 | $24,000.00 | 0.3936 | $9,446.40 | 16 | $18,000.00 | 0.3936 | $7,084.80 |
| 17 | $24,000.00 | 0.3714 | $8,913.60 | 17 | $18,000.00 | 0.3714 | $6,685.20 |
| 18 | $24,000.00 | 0.3503 | $8,407.20 | 18 | $18,000.00 | 0.3503 | $6,305.40 |
| 19 | $24,000.00 | 0.3305 | $7,932.00 | 19 | $18,000.00 | 0.3305 | $5,949.00 |
| 20 | $24,000.00 | 0.3118 | $7,483.20 | 20 | $18,000.00 | 0.3118 | $5,612.40 |
| 21 | $18,000.00 | 0.2942 | $5,295.60 | ||||
| 22 | $18,000.00 | 0.2775 | $4,995.00 | ||||
| 23 | $18,000.00 | 0.2618 | $4,712.40 | ||||
| 24 | $18,000.00 | 0.2470 | $4,446.00 | ||||
| $275,280.00 | $225,909.00 | ||||||
Net present value is the difference of Sum of Present values of cash inflows and Sum of Present values of
In the given scenario, Present values are calculated by calculating the present values of cash inflows in the form of annual receipts.
Present value factor is calculated as 1/1.06 ^ N where N is the year.
From the above table, it can be seen that the net present value of Option 1 far exceeds the net present value of option 2.
Conclusion:
Hence the option with the higher net present value is determined.
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Chapter 5 Solutions
Engineering Economic Analysis
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