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The course outline has the basic information about the course. It is now updated to reflect the changes caused by the strike.

Problem sessions (optional help sessions) are: Mondays and Thursdays, 3:30-4:20 in N501 Ross.

Your first assignment is due Wednesday, October 11, 2000. You can bring it to class on Wednesday or you can put it in the assignment box on the 5th floor North of Ross, near the elevators any time before midnight Wednesday.

The first class test is Friday, October 20, in the usual classroom, S137R. The test will cover Chapters 1 and 2. You may bring a single 8.5" by 11" 'cheat sheet' with formulas, etc. on it, written on both sides. You will also need a calculator. Day numbers of dates will be provided on the test. There will be multiple versions of the test to discourage copying. Detected cases of copying will result in sanctions as described in the York University Senate Policy on Academic Dishonesty.

Here are the worked solutions to the first assignment (using K = 25) and the numerical solutions for all values of K. (In the latter case, #2a, #9 and #10 are omitted since the answers are the same for all values of K.)

January 12. There is a remediation period from January 12 to February 12. We will cover chapters 3 and 4 (and perhaps a bit from chapter 5).

Assignment #2 will be due Friday February 2. Here are the worked solutions to assignment #2 (using K = 37) and the numerical solutions for all values of K.

Test #2 on chapters 3 and 4 will be held on Friday February 9. You may bring a single 8.5" by 11" 'cheat sheet' with formulas, etc. on it, written on both sides. You will also need a calculator. There will be multiple versions of the test to discourage copying. Detected cases of copying will result in sanctions as described in the York University Senate Policy on Academic Dishonesty. Here are the answers to test #2. The class average was 35/50.

Assignment #3 is due Friday, March 23. Here are the worked answers for K=25 and the numerical answers for all values of K. (Question #9 did not depend on K.)

Test #3 on Chapters 5 and 6 is Friday March 30. You may bring a single 8.5" by 11" 'cheat sheet' with formulas, etc. on it, written on both sides. You will also need a calculator. There will be multiple versions of the test to discourage copying. Detected cases of copying will result in sanctions as described in the York University Senate Policy on Academic Dishonesty. Here are the answers to test #3.

Our approach to Chapters 8 and 9 is explained in these Notes on Chapters 8 and 9. You will need life tables of lx for female aggregate, female nonsmoker, female smoker, male aggregate, male nonsmoker, male smoker. To implement these life tables into a spreadsheet model, see these instructions.

Assignment #4 is due Wednesday, May 2. Here are the answers to Assignment 4.

Test #4 is on Monday, May 7. It covers Chapters 7, 8, and 9. You may bring the usual one-page double-sided 'cheat sheet'. Here are the answers to Test #4.

The final exam is Thursday, May 17 from 7 to 10 pm in the Tait Mackenzie Main Gym. You should bring a calculator and you may bring one 8.5 inch by 11 inch 'cheat sheet' written on both sides. You must also have York ID and a photo ID as per standard York University exam rules.

The unofficial final grades are now posted beside my office, N515R. Also, the 4th assignment is available there.

NOTE: The material below was posted during the CUPE 3903 strike and represents a summary of what was done in class during the strike.

During the CUPE 3903 strike, the class is meeting but we are not advancing past section 3.3 of the book. However, we have been doing questions from the book (and similar examples not from the book). For example, 3.2 part B #3a and #3b (not the algebraic proof part, just the illustration with a time diagram which is really a more conceptual kind of proof), #4, #5. On Nov. 8 we did 3.3 Part B #1b and applied it to two ways of paying off a loan: equal payments or interest payments plus deposits into a fund to accumulate the principal. We also did 3.3 Part B #13. Then we looked at the EXCEL function PV. We found that part of #13 could be done in EXCEL with the formula =PV(0.05,12,240000).

Here are the test #1 solutions.

Friday, November 10. We did 3.2 part B, #9. The answer (see the back of the book and/or the Solutions Manual) should be compared with the compound interest version which is: ((1+i)n - 1)/i = ((1 + ni + (n(n-1)/2)i2 + (n(n-1)(n-2)/6)i3 + ...) - 1)/i = n + (n(n-1)/2)i + (n(n-1)(n-2)/6)i2 + ... + in-1. In other words, the first two terms of the compound interest formula are the simple interest formula.
We also did 3.2 Part B #13, 3.3 Part B #15 and #16.

Monday, November 13. We did 3.3 Part B #18. And we moved on with NEW MATERIAL in section 3.4. We did Examples 1 and 2 on page 83 and also Example 4 on page 85. In doing these examples, the important point is NOT whether an annuity is 'due' or 'deferred'! Those are just names. The important point is where you want to evaluate the annuity. If you want to evalute an annuity on the date of its last payment (just after the payment) then you use sn|i. (It doesn't matter whether it is an ordinary annuity or an annuity due or a deferred annuity!) If you want to evalute it one payment interval after the last payment then use (1+i)sn|i instead. (It doesn't matter whether it is an ordinary annuity or an annuity due or a deferred annuity!) And if you want it one period before the first payment, use an|i. But if you want it on the date of the first payment, use (1+i)an|i.
Also, an alternative formula for (1+i)sn|i is sn+1|i - 1 (see 3.4 Part B #3) and an alternative formula for (1+i)an|i is an-1|i + 1 (see 3.4 Part B #4).

Wednesday, November 15. Test #2 is postponed. (The new date is not yet decided.) In class, we did 3.4 Part B #10a and #10b and #12. In the back of the book, you will find solutions; but note that a double dotted a-letter or s-letter means there is a factor of (1+i) to reflect the time shift forward of one period. For example, n|i means (1+i)an|i and the latter formula is what you would use to calculate it.

Friday, November 17. We did Example 1 page 91 and Example 2 page 92. We also did 3.5 Part B #3.

Monday, November 20. I did not make it to class on time because of the picket lines and there was no class.

Wednesday, November 22. We did Example 1 and Example 2 (pages 95, 96) in 3.6. The book's method uses some formulas to find an approximate value for the interest rate and then refines this value using linear interpolation. The book gives the approximation formula but does not show how that formula was derived. You are not responsible for the derivation of this formula; but if you want to see it, look here.

Note that Excel and some financial calculators can find the interest rate for an annuity. For example, to solve Example 1 using Excel, you would enter into any cell the formula:

=4*RATE(16,-250,0,5000)

Excel will display 0.110635. This is the correct value of the nominal annual rate j4. Similarly, for Example 2, enter into any cell the formula

=12*RATE(6,-900,5000,0)

and Excel will display 0.269306. The RATE function has arguments as follows:
=RATE(#payments,paymentsize,presentvalue,futurevalue)

November 24: We did 3.6 B #7. Let i be the monthly interest rate. Then we want to find i by solving the equation
10000 = 600a12|i(1+i)-6 + 500a12|i(1+i)-18
It helps to rewrite the rhs as
f(i) = 100a12|i(1+i)-6(6 + 5(1+i)-12)
and then start guessing values of i. The special formula for an initial guess does not apply in this case. So try i = 1%. You will find that f(0.01) = 11066.395. (Answer is too big since we want 10000.) Larger values of i will produce a smaller answer because we are discounting future payments. So try i = 2%. We find f(0.02) = 9336.5724. (Answer too small.) Now we can apply linear interpolation:
(i-1)/(2-1) = (10000-11066.395)/(9336.5724-11066.395) = 0.616476
So i = 1.616476%. But that value was found by linear interpolation across a wide gap: i = 1% means about 12% per year and i = 2% means about 24% per year.
So we had better check it. Compute f(0.01616476) = 9958.1866. (Answer too small.) We can now do 'linear interpolation' with the last two values, namely i = 2% and i = 1.616476%. (This will really be 'linear extrapolation' because the target of 10000 does NOT lie between 9336.5724 and 9958.1866. But the formula we use is exactly the same and the result should be valid.)
(i-1.616476)/(2-1.616476) = (10000-9958.1866)/(9336.5724-9958.1866)
We find i = 1.590678%. Computing the annual effective rate (that's what the question wants) we fine (1+i)12-1 = 0.20849. the answer in the book is 20.83% so our answer is pretty close.

November 27: We did some work in 4.1 and 4.2. In general annuities, we will always convert the interest rate so the interest period matches the payment interval. This is procedure 2 on page 107. For example in 4.1 A #14 we will NOT cover 14 (a) but only 14 (b). To do 14 (b) you convert the interest rate to a rate per half-year: i = sqrt(1.11)-1 = 0.053565375 (and store this result in calculator memory for further use). Then calculate 500a20|i to get $6046.96. We also did Example 2 on page 113.

November 29. We did 4.2 A #9. We started 4.3. Note that if you put an amount P into an account earning interest at rate i per period then at the end of each period you can withdraw the interest iP and leave the principal P in the account. Thus an amount P now is equivalent to an infinite annuity (perpetuity) of iP at the end of each period. Let iP = R. Then P = R/i. So R/i is the present value of a perpetuity of payments of R at interest rate i.

December 1. We did Example 3 on page 119.
We did the following example: A person wants to provide funds for an annual scholarship of $2000. How much should she deposit into an account earning j12 = 6% if the first award is made on the same day as her deposit?
Solution: We convert to the equivalent annual interest rate i = (1.005)12-1 = 0.061677812. Then we take the present value of the perpetuity. But the first payment is now. So we add that as a separate term. PV = 2000 + 2000/0.061677812 = $34,426.57.
We also considered some material from 4.4. Consider a perpetuity of payments of 1. Its present value, one period before the first payment, is 1/i. Divide this result by i. Thus a perpetuity with payments of 1/i has present value 1/i2. But each 1/i term of this perpetuity can be represented by a perpetuity of payments of 1. Thus we get layers and layers of perpetuities each with payments of 1 and each starting one period after the layer below. This gives a perpetuity with payments at times 1, 2, 3, 4, 5, ... of 0, 1, 2, 3, 4, ... and its present value is 1/i2. This is essentially the solution to 4.4 Part B #14.

December 4. I was unable to reach class on time so the class was cancelled.

Richard L.W. Brown
e-mail: mathrlwb@yorku.ca
Department of Mathematics and Statistics
York University