Various exam paper problems

[Catrina]

Hi,

I'm struggling to understand the answers of a few past exam paper Qs and I'd apprediate any help i could get.

September 2000 Q2:

Which formula is being used to calculate i(2) = 5.0949% ? The examiner's report appears to be using the formula: (1-d) = (1 + i(p)/p)^-p. I didn't realise this formula exists and i can't derive it using the other formulas that are provided. Is this formula true or have i misunderstood the answer?

April 1999 Q3:

Isn't the answer A as i(0.5) = 0.5[(i+i)^1/0.5 - 1] so i=11.36?

April 2000 Q8:

Why does the integral of 0.06 start at time t?
And how does 0integral6 100*exp(0.36-0.06t+0.6+0.1152-0.3-0.0144)dt become 213.999* 0Integral6 exp(-0.06t) dt?

Thanks to anyone who can help me with one two or all of these Qs!

 

[didster]

Sept 2000 Q2
The way I do these is to think about what the situation and (magically!!) put numbers together. If you can learn this method, it would serve you well but of course it is either you pick it up at the onset or you don't.

The principal is that you equate the the two interest (discount) rates over an equivalent period. In the examiner solution it's a quarter year(90 days)
and discount rates are used.

You express what you are given in terms of this (period and interest/discount) and likewise what you want, i(2).

If the magic doesn't work for you stick to trying any and every formulae you know. The formulae you quoted is true and is derived as:

(1-d)=v=1/(1+i)
= 1/ {(1+i(p)/p)^p}
= {1+i(p)/p}^(-p)

Apr 1999 q3

The formula you quoted is correct (although I would have written it the other way since we have i(.5) and want i.

The mistake I think you're making is that i(0.5) is not 12% but 6%.
i(p) is the annual rate payable p-thly. The 12% is paid for a 2 year period so the equivalent rate, i(0.5) is 6% per annum payable 2 yearly.


Apr 2000 Q8
First question
Basically, you are adding all the accumulated values of the payments or put another way adding all the payments times the interest to accumulate that payment to the time 12.

As both interest and payments are continuous you use integration

Accumulation factor for payment at time t =
integral from t to 12 of (delta(s)) ds
This gives you the interest from t to time 12.
Makes sense??

Because delta(s) is "different" for s < 6 and s>6 then you need to split the integral over the two periods to give you something to integrate algebraically.
the second part of the integral starts at 6 (to 12) instead of t, because we know t is less than 6.

This is probably confusing so put another way
lets call the accumulation from t to 12 , A(t,12)

for t<6
A(t,12) = exp (integral from t to 12 of delta(s))
= exp (integral from t to 6 of 0.06 ds plus
integral from 6 to 12 of {0.05 + 0.0002s^2} ds)

for t>6 (which is not needed for this question but in my opinion you can't understand the shortcuts without knowing the long route)
A(t,12) = exp (integral from t to 12 of delta(s))
exp (integral from t to 12 of {0.05 + 0.0002s^2} ds)

then to get the accumulated value you use
integral over period payments made of rate of payment times A(t,12) (and since t<6 you don't need to worry bout the A(t,12) for t>6)


next question.
Once you add all the constant terms inside the exponential then you get .7608 +.06t

then removing the .7608 from inside the exp as follows

100exp(.7608+.06t)
=100 * exp(.7608) * exp(.06t)
=213.999 exp(.06t)

 

[Catrina]

It all makes sense now, thank you so much!!