Answer to Question #127387 in Financial Math for Wachira Ann Wangari

Question

The force interest is given by
(t)={0.04,0<t</=1
0.05t-0.01,1<t</5
0.24,t>5
What is the accumulated value at anytime (t>0)of investment of 1 at times 0,4 and 6?
What is present value at time, t =0 of a payment stream paid at a rate of p(t)=5t-1 received between t=1 and t=5

Answer 1

we have to break down the times for force of interest changes. if we let A(t) represent the accumulated value of the total investments made to date at time t then :

0<t≤1 :

A(t) = e^0.04t

1<t≤5 :

A(t) = e^0.04 x exp[$\int_1^t$ 0.05t - 0.01dt ]

A(t) = e^0.04 x exp [ 0.025t² - 0.01t ]${_1^t}$

A(t) = e^0.04 x exp[0.025t²-0.01t-0.025(1)+0.01]

A(t) = e^0.04 x e$^{0.025t^2 -0.01t-0.025+0.01}$

A(t) = e$^{0.025t^2-0.01t+0.025}$

5<t :

A(t) = e$^{0.025*5^2-0.01*5+0.025}$ x exp [ $\int_5^t$ 0.24 dt ]

A(t) = e^0.06 e^0.24t-1.2

A(t) = e^{0.24t - 0.6}

Now calculte the present value at time t 1 and then discount back to time 0 . the present value at time t 1 is :

$\implies$

$\int_1^5$ (5t-1) exp[- $\int_1^t$ 0.05t - 0.01dt ] dt

= $\int_1^5$ (5t-1) exp-[0.025t²+0.01t]$_1^t$ dt

= $\int_1^5$ (5t-1) exp[-0.025t²+0.01t+0.025-0.01] dt

= $\int_1^5$ (5t-1) exp[-0.025t²+0.01t+0.015] dt

= -100[exp(0.025t²+0.01t+0.015)]$_1^5$

= -100 [ e$^{-0.625+0.05+0.015}$ - e$^{-0.025+0.01+0.015}$ ]

= -100 [e^-0.56-1]

=42.879

we than need to discount this back to time 0 :

PV = 42.879 x exp [-$\int_0^1$ 0.04dt]

PV = 42.879 e^-0.04

PV = 41.20