• Monthly payment:
55.965$)360%,75.0,/(000,120$ == PAA
• Balance at the end of 5 years ( 60 months):
4.71
Given Data: interest rate = 0.75% per month, each individual has the identical
4.72
Given Data: loan amount = $130,000, point charged = 3%, N = 360 months,
interest rate = 0.75% per month, actual amount loaned = $126,100:
• Monthly repayment:
4.73
(b)
P = $50,000
Page | 20
End of month
Principal
Payment
Remaining
Balance
0
$0.00
$50,000.00
4.74
Given Data: r = 7% compounded daily, N = 25 years
• The effective annual interest rate is
4.75
(a) The dealer’s interest rate to calculate the loan repayment schedule.
(b)
• Required monthly payment under Option A:
1
$45,956.87
2
$39,134.21
3
$29,339.85
4
$16,368.34
5
Page | 21
(c) The dealer’s interest rate is only good to determine the required monthly
payments. The interest rate to be used in comparing different options should
Add–on Loans
4.76
• The total amount you paid in interest
4.77
(a)
(b)
$189.58( / ,1.789%,12) $2,031.10P PA= =
Page | 22
Loans with Variable Interest Rates
4.79
(a) Amount of dealer financing = $15,458(0.90) = $13,912
A=$13,912(A/P,11.5% /12,60) =$305.96
(b) Assuming that the remaining balance will be financed over 56 months,
4.80
Given: purchase price = $155,000, down payment = $25,000
2
(a) For the second mortgage, the monthly payment will be
Page | 23
(b) Monthly payment
(c) Total interest payment
(d) Equivalent interest rate:
Loans with Variable Payments
4.81
4.82
Given:
i=9% / 12 =0.75%
per month, deferred period = 6 months, N = 36
monthly payments, first payment due at the end of 7th month, the amount of
initial loan = $15,000
(a) First, find the loan adjustment required for the 6-month grace period.
(b) Since there are 10 payments outstanding, the loan balance after the 26th
payment is
Page | 24
(c) The effective interest rate on this new financing is
4.83
(a)
55.965$)360%,75.0,/(000,120$ == PAA
4.84
Let
i
A
be the monthly payment for
th
i
year and
j
B
the balance of the
loan at the end of
th
j
month. Then,
37.705$)360,12/%125.8,/(000,95$
1
== PAA
Page | 25
(a) The monthly payments over the life of the loan are
(c)
$95,000 =$705.37(P/A,i,12) +$840.17
(P/A,i,12)(P/F,i,12)
i
a
=(1+0.010058)
−1=12.76% per year
Investment in Bonds
4.85
Given: Par value = $1,000, coupon rate = 6%, paid as $30 semiannually, N = 60
semiannual periods. Note that the maturity date is December 31, 2044.
(a) Find YTM
Page | 26
(c)
• Sale price after 2 ½ years later = $922.38, the YTM for the new investors:
4.86
Given: Purchase price = $1,010, par value = $1,000, coupon rate = 9.5%, bond
4.87
Given: Par value = $1,000, coupon rate = 8%, $40 bond interest paid
4.88
• Option 1: Given purchase price = $513.60, N = 10 semiannual periods, par
value at maturity = $1,000
4.89
Given: Par value = $1,000, coupon rate = 15%, or $75 interest paid
semiannually, purchase price = $1,298.68, N = 24 semiannual periods
4.90
Given: Par value = $1,000,
%5.4=i
semiannually,
2,30 == BA NN
semiannual periods
4.91
Given: Par value = $1,000, coupon rate = 8.75%, or $87.5 interest paid annually,
N = 4 years
(a) Find YTM if the market price is $1,108:
Page | 28
4.92
Given: Par value = $1,000, coupon rate = 12%, or $60 interest paid every 6
months, N = 30 semiannual periods
4.93
Given: Par value = $1,000, coupon rate = 10%, paid as $50 every 6 months,
Short Case Studies
ST 4.1
(a)
• Bank A:
%27.201)0155.01( 12 =−+=
a
i
Page | 29
(c) Assume that Anita makes either the minimum 5% payment or $20,
whichever is larger, every month. It will take 59 months to pay off the
loan. The total interest payments are $480.37.
Month Beginning Interest Total Minimum Remaining
(n)Unpaid Balance Charged Outstanding Payment Balance
01,500.00$
11,500.00$ 20.63$ 1,520.63$ 76.03$ 1,444.59$
21,444.59$ 19.86$ 1,464.46$ 73.22$ 1,391.23$
31,391.23$ 19.13$ 1,410.36$ 70.52$ 1,339.85$
11 1,029.52$ 14.16$ 1,043.68$ 52.18$ 991.49$
12 991.49$ 13.63$ 1,005.13$ 50.26$ 954.87$
13 954.87$ 13.13$ 968.00$ 48.40$ 919.60$
14 919.60$ 12.64$ 932.25$ 46.61$ 885.63$
15 885.63$ 12.18$ 897.81$ 44.89$ 852.92$
Page | 30
26 585.40$ 8.05$ 593.45$ 29.67$ 563.78$
27 563.78$ 7.75$ 571.53$ 28.58$ 542.95$
28 542.95$ 7.47$ 550.42$ 27.52$ 522.90$
38 372.27$ 5.12$ 377.39$ 20.00$ 357.39$
39 357.39$ 4.91$ 362.30$ 20.00$ 342.30$
40 342.30$ 4.71$ 347.01$ 20.00$ 327.01$
41 327.01$ 4.50$ 331.50$ 20.00$ 311.50$
42 311.50$ 4.28$ 315.79$ 20.00$ 295.79$
43 295.79$ 4.07$ 299.85$ 20.00$ 279.85$
44 279.85$ 3.85$ 283.70$ 20.00$ 263.70$
45 263.70$ 3.63$ 267.33$ 20.00$ 247.33$
57 51.65$ 0.71$ 52.36$ 20.00$ 32.36$
58 32.36$ 0.44$ 32.81$ 20.00$ 12.81$
59 12.81$ 0.18$ 12.98$ 12.98$ –$
Page | 31
ST 4.2
(a) Annual percentage rate:
o Loan solicitation offer:
(b) Total finance changes:
o Loan solicitation option :
ST 4.3
(a) It is a quite deceiving claim that you would save $146,985.60 over 10
years, as the various loans have different maturity dates assumed. For
Interest
Rate
Outstanding
Balance
Monthly
Payment
Loan Life
Assumed
Credit Card
21%
$12,000
$448.62
37 months
Page | 32
Total Current
Debt
$220,000
$2,543.90
New
Homeowner’s
6%
$220,000
$1,319.02
360 months
(b) It is true that whenever a borrower can refinance the current debt at a
lower rate of interest, it would be advantageous to the borrower.
However, in this example, the loan maturity for each of debt is
$170,000
$1,247.40
Page | 33
ST 4.4
The primary risk for consumers purchasing with ARM loans is that
interest rates will rise and then their payments will increase to an amount
that they are no longer able to afford. A hybrid loan can be a ticking time
bomb for borrowers who plan on holding the loan for the long term.
ST4.5
(a)
$60,000( / ,13% /12,360) $663.70A AP= =
Solving for i by trial and error yields
Page | 34
Comments: With Excel, you may enter the loan payment series and use
the IRR (range, guess) function to find the effective interest rate.
(c) Compute the mortgage balance at the end of 5 years:
Conventional mortgage:
(d) Compute the total interest payment for each option:
Conventional mortgage:
(e) Compute the equivalent present worth cost for each option at
6% /12 0.5%i= =
per month:
Page | 35
$522.95( / ,0.5%,12)
P PA
=