IENG 323 Tutorial Chapter 04:
2- Western
Energy makes quarterly deposits into an account reserved for
purchasing new equipment 2 years from now. The interest paid on the deposits is
12% per year, compounded monthly. (a) Identify the interest period, compounding
period, and compounding frequency in the interest period. (b) Calculate the
effective annual interest rate, that is, the APY.
Solution:
(a) Interest period = year; CP = month; m = 12
(b) ia = (1 + r/m)^m -1 = (1 + 0.12/12)^12 – 1 = 0.12683
Equivalence When PP ≥ CP
3- specializes in online security software development. It wants to have $65
million available in 3 years to pay stock dividends. How much money must the
company set aside now in an account that earns interest at a rate of 12% per year,
compounded quarterly?
Solution:
P = 65,000,000(P/F,3%,12)
= 65,000,000(0.7014) = $45,591,00
4- In
an effort to have enough money to finance his passion for top-level off-road
racing, a mechanical engineering graduate decided to sell some of his “toys." At
time t = 0, he sold his chopper motorcycle for $18,000. Six months later, he sold
his Baja pre-runner SUV for $26,000. At the end of year 1, he got $42,000 for his
pro-light race truck. If he invested all of the money in a high-risk commodity
hedge fund that earned 24% per year, compounded semiannually, how much did he
have in the account at the end of 2 years?
Solution:
F = 18,000(F/P,12%,4) + 26,000(F/P,12%,3) + 42,000(F/P,12%,2)
= 18,000(1.5735) + 26,000(1.4049) + 42,000(1.2544)
= $117,535
Equivalence When PP < CP
5- The Fairfold family decided to buy a super ski and water sports boat. They took
out an $80,000, 5-year, 6% per year, compounded semiannually loan with monthly
payments from First Bank and Trust (FB&T). After making only two payments, a
banker friend offered to make them a better deal: a 5-year, 4.2% per year,
compounded semiannually loan with no transfer fee to his bank and a complete
repayment no-fee-required of the remaining principal to FB&T. The principal on
the new loan will be the remaining principal from the current loan. Answer the
following questions for the Fairfolds as they deliberate this new offer.
(a) What is the current monthly payment on the$80,000 loan?
(b) What is the current principal due on the current loan?
(c) How much interest have they already paid in the first two payments?
(d) What is the amount of the new monthly payment starting with month 3, if the
new loan offer is accepted?
Solution:
PP = month; CP = month; use i = 0.005 per month
(a) A = 80,000(A/P,0.5%,60) = 80,000(0.01933) = $1546.40 per month
(b) Current principal balance is $77,701.47 (see table below)
(c) Sum of interest paid: 400.00 + 394.27 = $794.27
(d) Now i = 0.042/12 = 0.0035 per month
A = 77,701.47(A/P,0.35%,60) = 77,701.47(0.01851) = $1438.25 per month
Month
(1)
0
1
2
3
i per month Interest
(2)
owed
(3)=(7
prior)(2)
Total owed
(4)=(7
prior)+(3)
Monthly
payment
(5)
Principal
reduction
(6)=(5)-(3)
0.005
0.005
0.0035
80,400.00
79,247.87
77,973.43
1546.40
1546.40
1438.25
(d)
1146.40
1152.13
1166.29
400.00
394.27
271.96
Principal
remaining after
payment
(7)=(7 prior)-(6)
80,000
78,853.60
77,701.47 (b)
76,535,18
6- Corrosion problems and manufacturing defects rendered a gasoline pipeline
between El Paso, TX, and Phoenix, AZ, subject to longitudinal weld seam failures.
The pressure was reduced to 80% of the design value. If the reduced pressure
resulted in delivery of less product valued at $100,000 per month, what will be the
value of the lost revenue after a 2-year period at an interest rate of 15% per year,
compounded continuously?
Solution:
r = 15% per year = 1.25% per month; need effective i per month
i = e0.0125 – 1 = 0.0126 (1.26% per month)
F = 100,000(F/A,1.26%,24)
= 100,000{[(1 + 0.0126)24 –1]/0.0126}
= 100,000(27.8213)
= $2,782,130