Selected Answers to Problem Set 8

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a) If her balance is to decrease, then her balance after 1 month should be less than the original balance. That is, P₀ > P₁.

P0 > [1 + (i/12)]P0 - M
P0 + M > [1 + i/12]P0
M > [1 + i/12]P0 - P0
M > [1 + i/12 - 1]P0
M > [i/12]P0

So, her first monthly payment should be greater than the interest on the first payment, which is what should be expected (if the interest is greater, then her new balance will increase).

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b) Letting c = 1 + (i/12) we can formulate P₂:

P2 = cP1 - M
   = c[cP0 - M] - M
   = c2P0 - cM - M
   = c2P0 - (1 + c)M

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c) Taking what we've seen from part (b), we can rewrite P_k for any k:

Pk = ckP0 - M(1 + c + c2 + ... + ck-1)

The sum (1 + c + c² + ... + c^k-1) is a geometric sum, which equates to (1-c^k)/(1-c).

Pk = ckP0 - M[(1-ck)/(1-c)]

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d) To find the number of monthly payments that are needed, we must solve for k (or N), given that p_k = 0:

ckP0 - M[(1-ck)/(1-c)] = 0
ckp0 = M[(1-ck)/(1-c)]
ckp0(1-c) = M[(1-ck)]
ckp0(1-c) = M - Mck
ckp0(1-c) + Mck = M
ck[p0(1-c) + M] = M
ck = M / [p0(1-c) + M]
ln(ck) = ln(M / [p0(1-c) + M])
k = ln(M / [p0(1-c) + M]) / ln(c)

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e) On this one, you simply solve for M, given that p_k = 0:

ckP0 - M[(1-ck)/(1-c)] = 0
ckp0 = M[(1-ck)/(1-c)]
M = ckp0 / [(1-ck)/(1-c)]
  = ckp0(1-c) / (1-ck)

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f) My HTML and Perl scripts can be found here. Because either the number of months or the monthly payment had to be zero, there are two separate forms, one for each situation.

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