# PSTAT 171 Solutions of Homework 5

Problem 5.1

A loan is to be repaid by level annual installments of \$R at the end of each year for twenty years. You are given:

(i) The total principal repaid in the first three years is \$290.35.

(ii) The total principal repaid in the last three years is \$408.55.

Determine the total amount of interest paid during the life of the loan.

Proof. The principal repaid at the end of the k-th year is Pk = Rv20−k+1. Therefore, the total principal repaid in the first three years is

290.35 = P1 + P2 + P3 = Rv20+ Rv19+ Rv18= Rv18(v2+ v + 1)

Similarly, the total principal repaid in the last three years is

408.55 = P18 + P19 + P20 = Rv3+ Rv2+ Rv = Rv(v2+ v + 1)

Dividing the two equations we get

290.35
408.55 = v17

Solving for v yields v = 0.9801 ⇒ i = 2.0293%. Use this value of v to solve for R.

408.55
R = 0.9801[(0.9801)2+ 0.9801 + 1] = \$141.75

The amount of the loan is

L = 141.75a200.020293 = 141.751 (1.020293)20 0.020293

= \$2, 311.23

The total amount of interest paid is the total amount of payments made minus the amount of the loan, i.e.

I = \$20(141.75) \$2, 311.23 = \$523.72

Problem 5.2

A loan of L dollars is repaid with n payments starting one year after the loan, at an annual effective rate i. The amount of the first payment is P dollars. If Q > 0, each subsequent payment is Q dollars greater than the previous payment. Similarly, if Q < 0, then subsequent payments are Q dollars smaller than the previous payment. Find

(a) Payment amount at the end of the k-th period, Rk

(b) Amount of loan, L

(c) Outstanding loan balance at the end of the k-th period, Bk

(d) Portion of interest paid at the end of the k-th period, Ik

PSTAT 171 — Winter 2015

Solutions of Homework 5

Due Friday, Feb 20, 2015

(e) Portion of principal repaid at the end of the k-th period, Pk

(f) Amount of interest, I

Proof.

Rk = P + (k − 1)Q

L = (P + nQ)ani − Q(Da)ni

Ik = (P + nQ)(1 − vn−k+1) − Q[(n − k + 1) − an−k+1i]

Pk = (P + nQ)vn−k+1− Qan−k+1i

Bk = (P + nQ)an−ki − Q(Da)n−ki

Problem 5.3

I = nP + (n − 1)n 2

Q − (P + nQ)ani + Q(Da)ni

A 30-year loan of \$1000 is repaid with payments at the end of each year. Each of the first ten payments is \$80. Each of the next ten payments equals 80% of the amount of interest due. Each of the last ten payments equals the amount of interest due plus \$X. The lender charges interest at an annual effective interest rate of 10%. Calculate X.

Proof. At the end of ten years, the outstanding loan balance can be computed using the retrospective method as follows:
B10 = 1000(1.1)10 80s100.1 = 1318.75
For k = 11, 12, . . . , 20, the interest due at the end of the k-th year is

Ik = 0.1Bk−1

The payment made at the end of the k-th year is, by assumption, 80% of the amount of interest due, i.e.

Rk = 0.8Ik = 0.08Bk−1 The principal repaid at the end of the k-th year is

Pk = Rk − Ik = 0.08Bk−1 0.1Bk−1 = 0.02Bk−1

Therefore, the balance due at the end of the k-th year satisfies the recursive equation

Bk = Bk−1 − Pk = Bk−1 (0.02)Bk−1 = 1.02Bk−1

PSTAT 171 — Winter 2015

Solutions of Homework 5

Due Friday, Feb 20, 2015

In other words, for k = 11, 12, . . . , 20, the outstanding loan balance follows a geometric progression with common ratio 1.02. Thus, at the end of twenty years, the outstanding loan balance is

B20 = 1318.75(1.02)10= 1607.55

For each of the last ten payments, the annual amount of principal repaid is \$X. Since the loan is paid off immediately after making the last payment, the total principal repaid in the last ten payments is B20 = 1607.55 dollars. Therefore,

X =B20 10 = 1607.55

= 160.76

Problem 5.4

A loan of 100,000 has payments at the end of each month for 12 years. For the first 6 years the payments are Z each month, and for the final 6 years the payments are 2Z each month. Interest is at a nominal annual rate of 12% compounded monthly. Find the outstanding balance at the end of the first year.

Proof. The initial amortization relationship is

100, 000 = Za720.01 + 2Zv72a720.01

Solving for Z results in Z = 988.89

The interest due at the end of the first month is 1000, but the payment is only 988.89. The shortfall of 11.11 is added to the loan amount so that B1 = 100, 011.11 This can also be seen algebraically from
B1 = B0(1 + i) − Z1 = 100, 011.11
The outstanding balance at the end of the first year can be expressed retrospectively as

B12 = 100, 000(1.01)12 988.89s120.01 = 100, 140.90

Problem 5.5

A loan is being repaid by 2n level payments, starting one year after the loan. Just after the nth payment the borrower finds that she still owe3 4of the original amount. What proportion of the next payment is interest?

Proof.

Then

Bn = an = 3 4a2n

vn= 1

3

Principal in the (n + 1)-th payment is

v2n−(n+1)+1 = vn= 1

Interest in the (n + 1)-th payment is2 3.

PSTAT 171 — Winter 2015

Solutions of Homework 5

Due Friday, Feb 20, 2015

Problem 5.6

Copernicus borrows \$L and repays the principal by making ten annual payments at the end of the year into a sinking fund which earns an annual effective rate of 8%. The interest earned on the sinking fund in the third year is \$85.57. Determine L.

Proof. The interest earned on the sinking fund (in dollars) in the third year is

85.57 = 008L

s20.08 = 0.08 (1.08)2 1 (1.08)10 1L

.s100.08

L = 85.57[(1.08)10 1] 0.08[(1.08)2 1]

= 7, 449.61

Problem 5.7

Ron borrows \$20,000 for 20 years at an annual rate of interest of 10% convertible semiannually. He repays \$500 in interest at the end of each six months. The principal and the remaining accrued interest are to be paid at the end of 20 years by equal deposits at the end of six months to a sinking fund that accumulates interest at 8% per annum convertible semiannually. How much must Ron deposit in the sinking fund each six months?

Proof. Interest payments to lender must be \$20, 000 × 0.05 = \$1000 every six months, but he only pays \$500. Thus, the sinking fund must accumulate to

\$20, 000 + 500s400.05 = \$80, 400

at the end of 20 years. Alternatively, the balance on the loan at the end of 20 years can also be computed as
20000(1.05)40 500s400.05 = 80400
If X is the semiannual sinking fund deposit, we have

Xs400.04 = 80, 400 ⇒ X = \$846.09

Problem 5.8

Quetzalcoatl and Tonantzin each take out a 17-year loan of \$L. Quetzalcoatl repays his loan using the amortization method, at an annual effective rate of i. He makes an annual payment of \$500 at the end of each year. Tonantzin repays her loan using the sinking fund method. She pays interest annually, also at an annual effective interest rate of i. In addition, Tonantzin makes level annual deposits at the end of each year for 17 years into a sinking fund. The annual effective rate on the sinking fund is 4.62%, and she pays off the loan after 17 years. Tonantzin’s total payment each year is equal to 10% of the original loan amount. Calculate L.

PSTAT 171 — Winter 2015

Solutions of Homework 5

Due Friday, Feb 20, 2015

Proof. For Tonantzin, the direct loan interest is iL dollars, and the sinking fund deposit is

L=
0.0462L

= 0.04L

s170.0462

(1.0462)17 1

Therefore, Tonantzin’s total annual payment is equal to

0.10L = iL + 0.04L

which gives
i = 6%

Finally, the equation of value on Quetzalcoatl’s loan is

L = 500a170.06 = 5001 (1.06)17 0.06

= 5238.63

Problem 5.9

A company agrees to repay a loan over five years. Interest payments are made annually and a sinking fund is built up with five equal annual payments made at the end of the year. Interest on the sinking fund is compounded annually.

(i) The amount in the sinking fund immediately after the first payment is X.

(ii) The amount in the sinking fund immediately after the second payment is Y .

(iii) Y = 2.09X

(iv) The net amount on the loan immediately after the fourth payment is \$3,007.87.

Calculate the sinking fund annual payment.

Proof. Let j be the rate of interest earned on the sinking fund.

(i) This tells us that the sinking fund deposit is X.

(ii) Y is the accumulated value of the sinking fund balance at time 1 plus the deposit made at time 2, i.e. Y = X(1 + j) + X = X(2 + j)

(iii) X(2 + j) = 2.09X ⇒ j = 9%
(iv) The net amount on the loan (i.e. the outstanding balance) immediately after the fourth payment is the original loan amount, L, minus the accumulated value of the sinking fund deposits made up

to time 4, i.e.

L=s50.09 − s40.09=(1.09)4

= 3 00787

L −s50.09

s40.09 L

s50.09
Ls50.09

,.

Thus, the amount of the SF payment is

L

= 3, 007.87 = 2, 131

s50.09

(1.09)4 ,

PSTAT 171 — Winter 2015

Solutions of Homework 5

Due Friday, Feb 20, 2015

Problem 5.10

Show that the borrower’s periodic outlay for a standard sinking fund method repayment at rate j is larger than the level outlay under amortization method with the interest rate i, if i > j.

Proof. This follows from1

= 1
sni

+

If i > j then

anii
1

>1
sni

and thus

Li + 1snj+ 1> L1
= L� 1 ani1 snj

as desired.

snjsniani