Difficulty in chapter 5 of CT1

Exam-style question
This question is present at the end of chapter 5 (discounting and accumulating of CT1). I am not able to understand the solution given. Please help me understand this.

This is a typical exam-style question on this chapter. Have a go at it before turning over
and having a look at the solution.
Question
The force of interest at any time t (measured in years) is given by:
delta of t is 0.04 for 0<t<=1
delta of t is 0.05t-0.01 for 1<t<=5
delta of t is 0.24 for t >5

(i) What is the total accumulated value at any time t ( > 0 ) of investments of 1 at
times 0, 4 and 6?

The solutions says following
case1: 0<t<=1
here for finding accumulated value integrate taking limits 0 to t

case 2: 1<t<=4
A(0,t)= A(0,1) A(1,t)

case 3: 4<t<=5
A(0,t)= A(0,4) A(4,t) + A (4,t) [ for investment at t=4]

case 4: 5<t<=6
A(0,t)= A(0,5) A(5,t)

case 5: t>=6
A(0,t) = A(0,6) A(6,t)+ A(6,t)

I still have a doubt. for case 4<t<=5 why do we need to add A(4,t) when we have already considered A(4,t) by multiplying it to A(0,4) and why do we need to consider A(6,t) for t>6 when we have already considered A(6,t) by multiplying it to A(0,6)

Thanks in advance for all the help,
Bijal

 

Good day,

This is one of the easier questions your likley to face on the exam so its worth mastering it, as there arnt usually many tricks involved, and will generally earn alot of marks.

What we are after here is the Accumulated value of 1 dollar at any time "t". Ie, we want A(0,t) as a function of t. As we have a piecewise force of interest, AND payments at three different times, we need new functions whereever there is a payment, OR a change of interest rate.

So for t between zero to 1, we have A(0,t) = exp(0.04t), pretty straight foward.

Now between 1 to 5, we must remember we have the initial 1 which is still getting accumulated at the new force of interest, but we get another payment of 1 at time 4 aswell.

So the accumulation of the first payment at time 1 is now exp(0.04). So as a function of t between 1 to 4 this will be exp(0.04)*exp(Integ(1->t) 0.05t - 0.01dt)) but since we get another payment at time 4, we need another function. This will be A(0,1)A(1,4)A(4,t) for the first payment at time 0, plus A(4,t), for the new payment we received at time 4, this will be a function of t, between 4 to 5. We then find the accumulation of these two payments at time 5, and repeat the process from times 5 to 6, then at time 6 we have a new payment, so this requires the addition of another A(6,t) to the function, and from here there are no more changes of rate, or no new payments so the function remains the same for t>=6.

Hope this helps, let me know if you don't understand this and i will post a more detailed solution.

 

Thanks a lot Devon. Now this makes complete sense to me. Thanks a lot for your help.