Problems in the Queue

A week left until the financial mathematics exam. The best thing that I can say is that I can honestly not have worked any harder at studying. In the last three months, I have studied at least three hours a day, and frequently I have done eight to ten hours a day. Much of that time has been absolutely focused. The question is: have I studied as smartly as possible? I won’t know the answer to that until after the examination.

I have done about two hours of problems today. I see that there are 19 problems remaining in the queue. These are problems that I have seen before, and that I still wish to spend some time with. I spend enough time with each problem that I feel I am confident that I understand several ways to get to a solution, or until I get sick of looking at it. If I get sick of looking at it, it comes up on the queue again later today or tomorrow. If I am fluent with it, I don’t look at it for a few days. Eventually, it disappears from the queue.

Earlier today, I already did some easy numerical exercises. Later today, I will probably do an exam, which is a whole other kind of problem solving, because of time pressure (5 minutes per problem.)

What I am going to do here is type in problems as they come up on my screen, then write up solutions. If I don’t understand the problem, you will get to watch me struggle.

Person X enters into a long forward contract. If the spot price at expiration were S, the payoff would be -20. If the spot price at expiration were 1.2S, the payoff would be X.

Person Y enters into a short forward contract. If the spot price at expiration were 0.8S, the payoff would be 40. If the spot price at expiration were 1.1S, the payoff would be Y.

The forward price on each contract is the same.

What is X+Y?

We aren’t going to do a diagram for this one. I tried, on paper, and since we don’t know the values of 0.8S, 1.1S, or 1.2S, it is hard to know where to place them in relation to the forward price. So, let’s go with straight algebra, and see if that leads us to a reasonable solution.

Payoff on Long Forward = S – F

Payoff on Short Forward = F – S

From Givens:

S-F = -20 \\    1.2S=S=X \\    F-0.8S=40 \\    F-1.1S = Y

Add equations 1 and 3 to get S=100. Then solve for F and get F=120.

Using equations 2 and 4:

X+Y = 1.2(100)-120+120-1.1(100) = 10

That wasn’t too bad. These derivative problems can be a little intimidating at first.

A 15 year bond with semiannual coupons has a redemption value of $100. It is purchased at a discount to yield 10% compounded semiannually. If the amount for accumulation of discount in the 27th payment is $2.25, which of the following is closest to the total amount of discount in the original purchase price?

We have all sorts of information here, so it should be easy to get to an answer. First, since the coupons are semiannual, we can just think entirely in 6 month terms. Since the bond is sold at a discount, we know that the coupon rate is less than the interest rate. We know that we can solve this problem by finding the original purchase price, and subtracting it from the redemption price. First, we need the coupon, which we can find using the premium discount formula.

Let’s start with that. Assuming that F=C, the amount of discount in the kth payment is:

F(i-r)v^{n-k+1}

Here are our givens:

n=30 \\    F=C=100 \\    i=0.1 \\    F(i-r)v^{n-1+k} = 2.25 \\    100 (0.05-r)1.05^{-(31-27)} = 2.25 \\    r=0.02265 \\

Now solve for the original sale price. We might as well continue with the premium discount formula:

P = C+(Fr-Ci) a_{\overline{n}\lvert } \\    P=100-0.2735\frac{1-1.05^{-30}}{0.05} \\    P=57.96

Discount in original price = 42.04.

Let’s just do one more.

To accumulate 8000 at the end of 3n years, deposits are made at the end of the each of the first n years and 196 at the end of each of the next 2n years.  (1+i)^n = 2.

What is n?

These problems are much simplified by visualizing them in the right way. The easiest way to think of it is payments of 98 from 1 to 3n years, plus payments of 98 in years 2n+1 through 3n.

In math:

98s_{\overline{3n}\lvert }+98_{\overline{2n}\lvert }=8000 \\    \frac{(1+i)^{3n}}{i} + \frac{(1+i)^{2n}}{i}=81.633 \\    \frac{2^3-1}{i} + \frac{2^2-1}{i} = 81.633 \\    i= 12.25%

I am going to post one more, because it is a real bear. To solve it, you trust that the math will lead you to the answer.

Given a k year bond with semiannual coupons, and a yield rate of 10% convertible semi-annually, sold at a price p.

If the coupon rate had been r-0.04, the price would be P-200.

Calculate the present value of a 3k year annuity immediate paying 100 at the end of each 6 month period, at a rate of 10% semiannually.

Start at the end. We need to find:

100 \frac{1-1.05^{-6k}}{0.005}

Which means that what we really need is k (although v^k will do).

Dive in:

P=1000\frac r 2 \frac{1-1.05^{-2k}}{0.05}+1000(1.05)^{-2k} \\    P-200 = 1000\frac{r-0.04}{2}\times \frac{1-1.05^{-2k}}{0.05}+1000(1.05)^{-2k} \\    \text{We can see that the redemption values will not be significant}\\    1000 \frac r 2 a_{\overline{2k}\lvert }=1000 \frac{r-0.04}{2}a_{\overline{2k}\lvert }+200 \\    1000 \frac r 2 a_{\overline{2k}\lvert }-1000 \frac{r-0.04}{2}a_{\overline{2k}\lvert }=200 \\    500a_{\overline{2k}\lvert }(r-r+0.04)=200 \\    a_{\overline{2k}\lvert } = 10 \\    k = 7.1 \\    100 \frac{1-1.05^{-6\times7.1}}{0.05} =1749.75

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