Answer to Question #157001 in Financial Math for Wachira Ann Wangari

Question

a) Repayment on rental property by the university of Nairobi is done by kshs 200,000 at time 5, kshs 190,000 at time 6, kshs 180,000 at time 15. Assuming an annual effective rate of interest of 3.5% calculate:

I) The present value of the repayment at time 4

II) The present value of the repayment at time 0

III) The accumulated value of the repayment at time 15

b) The force of interest is given by

F(t)= 0.04<1<\=1

0.05t-0.01,1<t<\=5

0.24, t>5

I) what is the accumulated value at any time (t>0) of investment of 1 at times 0,4 and 6?

II) what is the present value at time, t=0 of a payment atreay paid at a rate of p(t)=5t-1 received between t=1 t=5

Answer 1

Answer:

a)

Step 1

A concept that implies the future worth of the money is lower than its current value due to several factors such as inflation, and many more is term as the TVM (time value of money).

Step 2

Computation of the present value of repayment at time 4 and 0, and the accumulated value of repayment at time 15:

It is computed in the following manner:

Step 3

I) The present value of repayment at time 4 is kshs 493,893.98.

II) The present value of repayment at time 0 is kshs 430,400.07.

III) The accumulated value of repayment at time 15 is kshs 721,070.25.

b)

Step 1

Let A(t) represents the accumulated values of the total investments made to date at time t.

We have to break down the times for force of interest changes.

0<t≤1

A(t)= $e^{ 0.04t}$

0<t≤1

when 1<t≤5

A(t)=$e^{0.04}$ ×$e ^{[∫^t_1 (0.05t−0.01)dt]}$

A(t)=$e^{0.04}$ × $e^{ [0.025t^2 −0.01t]^t_1 }$

A(t)=$e^{0.04}$ × $e^{ [0.025t^2 −0.01t−(0.025(1)−0.01)]}$

A(t)=$e^{0.04}$ ×$e^{ 0.025t^2 −0.01t−0.025+0.01}$

A(t)= $e^{ 0.025t^2 −0.01t−0.025 }$

Step 2

t>5

A(t)= $e ^{ 0.025t^2 −0.01t−0.025 }$× $e ^{ [∫^t_5 0.24t]}$

A(t)= $e^{0.06}$$e^{0.24t−1.2}$

A(t)= $e^{ 0.24t−0.6}$

Now, we will calculate the present value 

$∫^5_1 (5t−1)$− $∫^5_1 [0.05t−0.01]dt$

$∫^5_1 (5t−1)$ $e ^{−[0.025t^2 −0.01t] ^5_1} dt$

$∫^5_1 (5t−1)$$e ^−[0.025(5)^2 −0.01(5)−(0.025−0.01)] dt$

=42.879

PV=42.879 $e[−∫^1_0 0.04dt]$

PV=41.20

Therefore, the present value is 41.20.