St 226 Week 1 Homework Solution

1. Force of interest 0.05 100 exp Effective rate of return 0.05 100 (1.05) 2 = 102.47 2. The value is
6 6
1

1 × 0.05 2

= 102.53

exp
0.5

δ (s) ds

= exp
0.5

(0.01s + 0.04) ds t=6 =

exp

0.01

t2 + 0.04t 2

= 1.49. t=0.5 3. 150 in 3 years accumulates to
3

150 exp
0

0.05 + 0.02s2 ds

= 150 exp

0.02

t3 + 0.05t 3

t=3

= 208.65 t=0 200 in 5 years accumulates to
3 5

200 exp
0

0.05 + 0.02s2 ds +
3 t=3

0.05 + 0.01s2 ds t3 + 0.05t 3 t=5 =

200 exp

0.02

t3 + 0.05t 3

exp t=0 0.01

= 426.23 t=3 250 in 8 years accumulates to
3 8

250 exp
0

0.05 + 0.02s2 ds +
3 t=3

0.05 + 0.01s2 ds t3 0.01 + 0.05t 3 t=8 =

250 exp

t3 0.02 + 0.05t 3

exp t=0 = 2248.73 t=3 To find the effective rate earned we have to solve 600 = 208.65 (1 + i)
−3

+ 426.23 (1 + i)

−5

+ 2248.73 (1 + i)

−8

Trying various values such as i = 0.1 and i = 0.2, we have LHS¡RHS. With i = 0.25, RHS=620 and with i = 0.26, RHS=596, so it is between 0.25 and 0.26 and closer to 0.26. Further investigation shows it is 0.258, i.e. 25.8% 4. Same exercise

1

5. We start with the present value
6 s 2 s

ρ (s) exp −
0 5 0

δ (u) du ds =
1 s

12 exp −
0 6

δ (u) du ds + s (10 + s) exp −
2 0 2

δ (u) du ds +
5 s

15 exp −
0

δ (u) du ds =

12 exp −
1 5 2 0

0.08 du ds + s (10 + s) exp −
2 6 2 0

0.08 du −
2 5

du 10 + u s ds +

15 exp −
5 2 0

0.08 du −
2 5

du − 10 + u

0.06 du ds =
5

12 exp (−0.08s) ds +
1 6 2

(10 + s) exp (−0.08 × 2) exp − ln 15 12
5

10 + s 12

ds +

15 exp (−0.08 × 2) exp − ln
5

exp (−0.06 (s − 5)) ds = 12 ds + 10 + s

12

exp (−0.08) − exp (−0.16) + 0.08
6

(10 + s) exp (−0.16)
2

12 exp (−0.16) exp (−0.06 (s − 5)) ds =
5

1 − exp (−0.06) exp (−0.08) − exp (−0.16) +12 (5 − 2) exp (−0.16)+12 exp (−0.16) = 51.25 0.08 0.06 The accumulated value is 12
2 5

51.25 exp
0

0.08 du +
2