102 Session 4

102 Session 4 increasing annuities

© KSES Exam questions are copyright Faculty & Institute of Actuaries & are used with their permission Source: www.actuaries.org.uk 105

“Trick” for increasing annuities 1.21 1.1 1

1

Value =

2

3

+ 1.1v2

1v

+ 1.12 v3

Make it look like something you recognise Value = 1/1.1*{ 1.1v = 1/1.1 {

x

+ 1.12v2

+ 1.13 v3 }

+ x2

+ x3 }

= 1/1.1 a3¬ @ what rate of interest j, say? 106

“Trick” for increasing annuities 1.21 1.1 1

1

Value =

2

3

+ 1.1v2

1v

+ 1.12 v3

Make it look like something you recognise Value = 1/1.1*{ 1.1v = 1/1.1 {

x

+ 1.12v2

+ 1.13 v3 }

+ x2

+ x3 }

= 1/1.1 a3¬ @ what rate of interest j, say? x = 1.1 / (1 + i) = 1/(1 + j) so 1 + j = 1 + i / (1.1)

107

Exercise Write down the value of 1v + 1.1v2 + 1.12 v3 if the interest rate is

21% 10% 0%

108

Exercise Write down the value of 1v + 1.1v2 + 1.12 v3 if the interest rate is 1 + i / 1.1

Value

= a3¬

/ 1.1

21%

1.1

2.26077 = 2.487 / 1.1

10%

1

2.7273 = 3

0%

0.90909

3.31

/ 1.1

= 3.641 / 1.1

109

(Ia)3¬ formula in tables Goal: show that (Ia)3¬ = (ä3¬ – 3v3) / i

3 2

Let X = (Ia)3¬ ie X is amount needed to invest now, in order to pay 1,2,3 at end of next 3 years

1

1

Value (X)

=

1v +

2v2 +

(1+i) X

=1+

2v +

3v2

Subtract, iX

=(1 +

v+

v2

= ä3¬

)

2

3

3v3

–

3v3

–

3v3 110

Checking (Ia)n¬ Write down the value of (Ia)n and n(n+1)/2 * (1+i)^-(2/3 * n)

(Ia)10¬ @ 2% (Ia)20¬ @ 6% Rough check is on amount of cash paid treated as if paid about 2/3 of way through term 1 2 3 4 4 3 2 1

111

Checking (Ia)n¬ Write down the value of (Ia)n and n(n+1)/2 * (1+i)^-(n/2)

(Ia)10¬ @ 2%

(Ia)n 47.9

Rough 48.2

(Ia)20¬ @ 6%

99

97

Rough check is on amount of cash paid treated as if paid about 2/3 of way through term 1 2 3 4 4 3 2 1

112

(Da)3¬ formula in tables Goal: show that (Da)3¬ = (3 - a3¬) / i

3 2

Let Y = (Da)3¬ ie Y is amount needed to invest now, in order to pay 3,2,1 at end of next 3 years

1

1

Value (Y)

=

(1+i) Y

=

Subtract, iY

=

2

3

= 113

(Da)3¬ formula in tables Goal: show that (Da)3¬ = (3 - a3¬) / i

3 2

Let Y = (Da)3¬ ie Y is amount needed to invest now, in order to pay 3,2,1 at end of next 3 years

1

1

Value (Y)

=

3v +

2v2 +

(1+i) Y

=3+

2v +

1v2

Subtract, iY

=3-(

v

v2 +

=3-

a3¬

+

2

3

1v3

v3)

114

(Ia)3¬ formula in tables Goal: show that (Ia)3¬ = (a3¬ – 3v3) / δ Let Z = (Ia)3¬ ie Z is amount needed to invest now, in order to pay at rate t during next 3 years.

3 2 1 0 0

Value (Z)

1

2

3

= Integral ( t vt ) dt = Integral ( t e -δt ) dt

(Integrate by parts to get result)

115

Apr 2000 Q13

116

Apr 2000 Q13(i)

Loan amount = Value = ……. a20¬ – …..(Ia)20¬ a20¬ = ………… (Ia)20¬ = ………… So loan = £……….. 1

3

5

7

9

11

13

15

17

19

Check total cash paid is approx ½ * (6000 + …..)* 20 = about 100,000. On average paid half way through. => guess value = 100k / …..10 = 42k (OK)

117

Apr 2000 Q13(i)

Loan amount = Value = 6200 a20¬ – 200(Ia)20¬ a20¬ = 9.12855 (Ia)20¬ = 70.9055 So loan = £42,416 1

3

5

7

9

11

13

15

17

19

Check total cash paid is approx ½ * (6000 + 4200)* 20 = about 100,000. On average paid half way through. => guess value = 100k / 1.0910 = 42k (OK)

118

Apr 2000 Q13(ii)

a7¬ = …………….. (Ia)7¬ = ……………. Value of payments in first 7 years = ……….. a7¬ – …….(Ia)7¬ = …………. PV of balance outstanding = loan – ……… = …………. Rolled up balance outstanding = ……….. * 1.09^…. = ………… 1

3

5

7

9

11

13

15

17

19

End year 8, ……… becomes …….* 1.09 = …….. less ……. payment to be …….. End year 9, ……… becomes ……. * 1.09 = …….. less ……. payment to be …….. 119

Apr 2000 Q13(ii)

a7¬ = 5.03295 (Ia)7¬ = 18.4075 Value of payments in first 7 years = 6200 a7¬ – 200(Ia)7¬ = 27,523. PV of balance outstanding = loan – 27,523 = 14,893 Rolled up balance outstanding = 14,893 * 1.09^7 = 27,225 1

3

5

7

9

11

13

15

17

19

End year 8, 27225 becomes 27225 * 1.09 = 29675 less 4600 payment to be 25075. End year 9, 25075 becomes 25075 * 1.09 = 27332 less 4400 payment to be 22932. 120

Apr 2000 Q13(iii)

Amount outstanding = ………… = Value of payments = (X+……) a…¬ – …..(Ia)..¬ 1

3

5

7

9

11

13

15

17

19

a…¬ @ …% = …….. (Ia)…¬ @…% = ……

Payments approx 11 * (3.9k – 1k) = 32k On average paid half way through. => guess value = 32k / 1.075 = 23k (OK)

So X = £………. 121

Apr 2000 Q13(iii)

Amount outstanding = 22,932 = Value of payments = (X+200) a11¬ – 200(Ia)11¬ 1

3

5

7

9

11

13

15

17

19

a11¬ @ 7% = 7.4987 (Ia)11¬ @7% = 39.965

Payments approx 11 * (3,924 – 1k) = 32k On average paid half way through. => guess value = 32k / 1.075 = 23k (OK)

So X = £3,924 122

Apr 2000 Q13 (i-ii) (solution)

123

Apr 2000 Q13 (iii) (solution)

124

Apr 2003 Q11

125

Sep 2000 Q9

126

Sep 2001 Q9

127

Sep 2002 Q11

128

Next session: Net present value

END 129