Chap002 Solutions

Topics: Time value of money, Net present value, Perpetuity Pages: 16 (1629 words) Published: December 13, 2014
CHAPTER 2
How to Calculate Present Values

Answers to Problem Sets

1. If the discount factor is .507, then .507 x 1.126 = $1.

Est time: 01-05

2. DF x 139 = 125. Therefore, DF =125/139 = .899.

Est time: 01-05

3. PV = 374/(1.09)9 = 172.20.

Est time: 01-05

4. PV = 432/1.15 + 137/(1.152) + 797/(1.153) = 376 + 104 + 524 = $1,003.

Est time: 01-05

5. FV = 100 x 1.158 = $305.90.

Est time: 01-05

6. NPV = −1,548 + 138/.09 = −14.67 (cost today plus the present value of the perpetuity).

Est time: 01-05

7. PV = 4/(.14 − .04) = $40.

Est time: 01-05

8. a. PV = 1/.10 = $10.

b. Since the perpetuity will be worth $10 in year 7, and since that is roughly double the present value, the approximate PV equals $5.

You must take the present value of years 1–7 and subtract from the total present value of the perpetuity:

PV = (1/.10)/(1.10)7 = 10/2= $5 (approximately).

c. A perpetuity paying $1 starting now would be worth $10, whereas a perpetuity starting in year 8 would be worth roughly $5. The difference between these cash flows is therefore approximately $5. PV = $10 – $5= $5 (approximately).

d. PV = C/(r − g) = 10,000/(.10-.05) = $200,000.

Est time: 06-10

9. a. PV = 600,000/(1.055) = `470,115.7 (assuming the cost of the car does not appreciate over those five years).

b. The six-year annuity factor [(1/0.08) – 1/(0.08 x (1+.08)6)] = 4.623. You need to set aside (600,000 × six-year annuity factor) = 600,000 × 4.623 = `2,773,727.8.

c. At the end of six years you would have 1.086 × (3,000,000 – 2,773,727.8) = `359,065.55.

Est time: 06-10

10.a. FV = 1,000e.12 x 5 = 1,000e.6 = `1,822.12.

b. PV = 5e−.12 x 8 = 5e-.96 = `1.914 million.

c. PV = C (1/r – 1/rert) = 2,000(1/.12 – 1/.12e .12 x15) = `13,912.

Est time: 01-05

11.
a. FV = 10,000,000 x (1.06)4 = 12,624,770.

b. FV = 10,000,000 x (1 + .06/12)(4 x 12) = 12,704,892.

c. FV = 10,000,000 x e(4 x .06) = 12,712,492.

Est time: 01-05

12.
a.
PV = `100/1.0110 = `90.53.

b.
PV = `100/1.1310 = `29.46.

c.
PV = `100/1.2515 = `3.52.

d.
PV = `100/1.12 + `100/1.122 + `100/1.123 = `240.18.

Est time: 01-05

13.a.r1 = 0.1050 = 10.50%.

b.

c. AF2 = DF1 + DF2 = 0.905 + 0.819 = 1.724.

d. PV of an annuity = C  [annuity factor at r% for t years]. Here:
$24.65 = $10  [AF3]
AF3 = 2.465

e. AF3 = DF1 + DF2 + DF3 = AF2 + DF3
2.465 = 1.724 + DF3
DF3 = 0.741

Est time: 06-10

14.The present value of the 10-year stream of cash inflows is:

Thus:
NPV = –`4,000,000 + `4,433,698.30 = +`433,698.30
At the end of five years, the factory’s value will be the present value of the five remaining `850,000 cash flows:

Est time: 01-05
15.
These figures are in thousands of rupees.
Est time: 01-05

16.a.Let St = salary in year t.

b. PV(salary) x 0.05 = `855,745.35
Future value = `855,745.35 x (1.08)30 = `8,611,071.81
c.

Est time: 06-10

17.
Period

Present Value
0

20,000.00
1

2

+10,000/1.122 =
+7,971.94
3

+15,000/1.123 =
+10,676.60

Total = NPV = − `135,136.06

Est time: 01-05

18.We can break this down into several different cash flows, such that the sum of these separate cash flows is the total cash flow. Then, the sum of the present values of the separate cash flows is the present value of the entire project. (All dollar figures are in millions.) Cost of the ship is $8 million

PV = $8 million
Revenue is $5 million per year, and operating expenses are $4 million. Thus, operating cash flow is $1 million per year for 15 years.

Major refits cost $2 million each and will occur at times t = 5 and t = 10. PV = ($2 million)/1.085 + ($2 million)/1.0810 = $2.288 million. Sale for scrap brings in revenue of $1.5 million at t = 15.

PV = $1.5 million/1.0815 = $0.473 million.
Adding these present values gives the present value of the...
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