# Assignment

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```					                                  010.212 Interest Theory (3)
Fall 2001 – Monday and Wednesday – 4:00PM to 5:15PM
Assignment 7
Due: November 5th, 2001
Each question is worth .2 marks

1) A 12-year loan has an initial balance of 120,000. The effective annual interest rate is 7%, and
payments are made at the end of each year. What is the outstanding balance on the loan
immediately after the 7th payment? What was the interest portion of the 5th payment?

120000 PV 12 N 7 %i 0 FV CPT PMT = 15,108.24
Balance after 7th = 7 BAL 61,946.76 or 15,108.24  4.1002 = same
Interest in 5th = 5 INT 6,315.11 or .07  5.9713  15,108.24
2) A 200,000 loan involves payments at the end of each year. The effective annual interest rate is 5%.
If the interest portion of the 4th payment is 9,046.60, what is the balance immediately after the 11th
payment?

(200,000  1.053  P  s3|.05 )  .05  9,046.60
P  (200,000  1.053  20  9,046.60) / 3.1525  16,048.53
B11  200,000  1.0511  16,048.53  s11|.05  114,069.74

3) You take out a 30-year 250,000 mortgage, at an effective annual interest rate of 7%. Immediately
after the 5th payment, you refinance your mortgage with a new 15-year mortgage at a 5.5%
effective annual interest rate. Both mortgages require annual year-end payments. What is the size
of each payment under the 15-year refinanced mortgage?

250000 PV 30 N 7 %i 0 FV CPT PMT = Payment = 20,146.60
To get balance outstanding after 5th payment 5 BAL 234,780.09
234780.09 PV 15 N 5.5 %i 0 FV CPT PMT = New = 23,390.11
4) A 25-year loan involves payments of 5,000 at the end of each year. If the amount of principal
included in the 14th payment is 2,904.04, what is the effective annual interest rate on the loan?

Pk = \$2,904.04 = \$5,000 ∙ v n+1-k ==>> (2,904.04/5000) = v12
i = (5,000 / 2,904.04) 1/12 – 1 = 4.63%

2001A07                                                                                   October 27, 2001
5) An n-year loan involves payments of 1,400 at the end of each year. The effective annual interest
rate is 9%. If the interest paid in the 9th installment is 634.15, calculate n.

Ik = Payment ∙ (1 – v n+1-k) ==>> \$634.15 = \$1,400 ∙ (1 - v n+1-9)
(1 - .4529643 ) = v n-8 ==>> (n-8) ∙ ln (1/1.09) = ln .5470357
n = 15
6) A 17-year loan has an initial balance of 150,000. The interest rate is 7.50%, and payments are
made at the end of each year. What is the total amount of interest paid over the life of the loan?

Payment = 150,000/ a17|.075 = 150,000/9.4339598 = 15,900.00
Total Interest = Total Payments – Original Loan
= 15,900.00 ∙ 17 - 150,000 = 120,300.00
7) A 5-year 12,000 car involves payments at the end of each month. If the annual interest rate is 9%,
convertible monthly, what is the amount of interest in the 53rd installment payment?

monthly i = 9%/12 = .75% and there are 60 total payments
Payment = 12,000 / a60|.0075 = 249.10
Ik = Payment ∙ (1 – v n+1-k) = 249.10 ∙ (1 - v 60+1-53) = 14.45
8) You invest 3,000 in a fund on January 1st,2001. On July 15th, 2001 your fund is worth 2,400, and
you add 1,500 to the fund. On October 1st, 2001, your fund is worth 4,000 and you remove 1,500
from the fund. On December 31st, 2001, your fund is worth 2,250. What is the annual time-
weighted rate of return on your investment from January 1, 2001 to December 31, 2001?

Annual Time Weighted Return (where t = total time studied) =
1
 B1   B2   B3 
                                       
Bn                           t

                                    B  C    1
  
  B  C    B  C                     
    
 B0  C0   1     1   2      2        n1    n 1  

 2400   4000   2250 
Time Weighted Return=  3000    3900    2500   1 = -26.15%
                        

2001A07                                                                                   October 27, 2001
9) You invest X in a fund on January 1st, 1999. On April 1st, 1999, your fund is worth 4,000, and you
add 2,500 to the fund. On July 1st, 1999 your fund is worth 7,000 and you withdraw 2,000 from
your fund. On December 31, 1999, your fund is worth 6,000. If the time-weighted rate of return on
your investment from January 1, 1999 to December 31, 1999 was 10%, find X.

 4000   7000   6000 
Time Weighted Return =  X    6500    5000   1 = 10%
                     
X = 5,169.23
10) You invest 4,000 in a fund on January 1, 2001. On May 1,2001, you deposit an additional 1,000
into the fund. On September 1, 2001, you withdraw 1,500 from the fund. On November 1, 2001,
you deposit 750 into the fund. On December 31, 2001, your fund is worth 4,750. What was the
annual dollar-weighted rate of return on your investment?

Find i such that
4,000∙(1+i) + 1,000∙(1+i).66667 –1,500∙(1+i).333 + 750∙(1+i).167 = 4,750
can use 5.13 or 5.14 as a starting estimate (5.13 also gives the final
i  \$500 / (\$4,000 + 1,000 ∙ 2/3 - \$1,500 ∙ 1/3 +\$750 ∙ 1/6) = .1165
Using iteration or spreadsheet IRR you find i = .1165 (full marks as
long as they show the cash flows used to do the IRR)

2001A07                                                                               October 27, 2001

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