CAA MODULE 1 - chapter 6 method check.

[Kchapman22]

Hello
In CAA module 1 textbook, chapter six, could someone provide the answer to the following question please?
Question 6.7 answer in back of book is 275.96 however when I integrate the payment stream from t=1 to t=10 and Set the accumulation factor to A(t,10) and got 260e^0.1 which gives a different answer than what is listed in solution .
Please could you provide working for this to show how 275.96 is reached?

 

[Darrell Chainey]

Hopefully to point you in the right direction:
The force of interest is different (and zero) in the first first years. So first work out the AV at time 5 of payments in the first 5 years.
You should get 67.5. Accumulate this to time 10 using A(5.10).
Then work out the accumulated value of payments between times 5 and 10 and add it on. You'll need to integrate by parts between 5 and 10.

 

[Kchapman22]

Many Thanks for your reply, i will do the working on this and post it soon, if the answer still doesn't match up.
Cheers

 

[Kchapman22]

Hello
The method i did was:
1. 67.5*e^0.1 = 74.59903697
2. integrated payment stream between times 5 and 10 using A(5,t) then calculating between limits. The answer i got here was 203.549365.
3. adding these together im getting 278.148... which is still different to the back of the book answer of 275.96

Please could you point out where i'm going wrong in the method?

Many thanks

 

[Darrell Chainey]

Hello
The method i did was:
1. 67.5*e^0.1 = 74.59903697
2. integrated payment stream between times 5 and 10 using A(5,t) then calculating between limits. The answer i got here was 203.549365.
3. adding these together im getting 278.148... which is still different to the back of the book answer of 275.96

Please could you point out where i'm going wrong in the method?

Many thanks

I think you need A(t, 10). And not A(5,t). You want to accumulate each payment at time t up to time 10.

 

[Kchapman22]

Hello,
Many thanks for your help - that method worked. In the end when I added 67.5e^0.1 to the integral of the payment stream between times 5 and 10 using accumulation factor of A(t,10) the answer is 275.96 .
I was however under the impression that due to the payment stream being from t=0 to t=10, we can use these as the limits and integrate just the A(t,10) (just using force of interest of times 5 to 10) for the method to work as said in the chapter. But clearly this was wrong. Any advice on this would be appreciated and thanks for continued help on this!.

Cheers

 

[Kchapman22]

Hi ,
to clarify the above point - the method thought in the book deals with continuous payment streams which can be integrated with limits t=a to t=b and then accumulate using A(t,n) or discounted 1/A(0,t)- this is why i used the following formula at first which was incorrect: {0,10)∫(5t+1)e^(0.2−0.02t)dt (produces a 278..answer) - also i tried using {0,10}∫(5t+1)e^(0.1)dt and the e^0.1 came from doing the accumulation from 0 to 10 for the entire stream but when splitting integral since it is a 0 for first five years force of interest and 0.02 thereafter for next five years, it was e^(0+0.1) effectively and gave an answer of 287.344... (260e^0.1). Both these methods were incorrect.

Please could some one explain why this method is wrong? or clarify if there is an alternative method to the above correct one?

Thanks

 

[Darrell Chainey]

You can only use the generic formula for the whole period if you are given only one function for the rate of payment and one for the force of interest. In this question the force of interest changes at time 5 and so you have to split it up into the periods before and after the change.

If you don't you will almost certainly be using incorrect accumulation (or discount) factors.

 

[Kchapman22]

Thanks for assistance on this. The method makes sense after your explanation.