Download Report

SEEM2440 Engineering Economics
Tutorial 2
Guo Weiwei
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
1 / 12
Interest Rate
Simple interest rate
I =P ×N ×i
Compound interest rate
I = P((1 + i)N − 1)
Nominal interest rate: r
Effective interest rate: i
M compounding period in a year
i = (1 +
Guo Weiwei (CUHK)
r M
) −1
M
SEEM2440 Engineering Economics
2 / 12
Factor symbols, formulas and names
Symbols
(F/P,i,N)
Formula
(1 + i)N
Name
future value factor or compound factor
(P/F,i,N)
1
(1+i)N
present value factor or discounting factor
(F/A,i,N)
(1+i)N −1
i
annuity future value factor
−N
(P/A.i.N)
1−(1+i)
i
annuity present value factor
(A/F,i,N)
i
(1+i)N −1
F to A conversion factor
(A/P,i,N)
i
1−(1+i)−N
P to A conversion factor
(F/G,i,N)
(P/G,i,N)
N
1 (1+i) −1
i[
i
1 1−(1+i)
i[
i
Guo Weiwei (CUHK)
− N]
−N
−
N
]
(1+i)N
arithmetic gradient future value factor
arithmetic gradient present value factor
SEEM2440 Engineering Economics
3 / 12
Question 1
A good stock-based mutual fund should earn at least 10% per year over a long
period of time. Consider the case of Barney and Lynn, who were overheard
gloating about how well they had done with their mutual fund investment. ”We
turned a $25,000 investment of money in 1982 into $100,000 in 2007.”
a. What annual return (interest rate) did they really earn on their investment.
Should they have been bragging about how investment-savvy they were?
b. Instead, if $1,000 had been invest each year for 25 years to accumulate
$100,000, what return did Barney and Lynn earn?
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
4 / 12
Solution
Mutual Fund
Definition: A mutual fund is a type of professionally managed collective
investment scheme that pools money from many investors to purchase
securities.
Advantages:
Increased diversification
Daily liquidity
Professional investment management
Government oversight
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
5 / 12
Solution
(a)
N = 2007 − 1982 = 25
F = P(F /P, i, N) ⇒ 100, 000 = 25, 000(1 + i)25
i = 5.7%
So Barnet and Lynn did not really do that well at all.
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
6 / 12
Solution
(a)
N = 2007 − 1982 = 25
F = P(F /P, i, N) ⇒ 100, 000 = 25, 000(1 + i)25
i = 5.7%
So Barnet and Lynn did not really do that well at all.
(b)
100, 000 = 1, 000(F /A, j, 25)
j = 10.1%
In this situation they did pretty well on their mutual fund investment.
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
6 / 12
Question 2
Automobiles of the future will most likely be manufactured largely with carbon
fibers made from recycled plastics, wood pulp, and cellulose. Replacing half the
metals in current automobiles could reduce a vehicle’s weight by 60% and fuel
consumption by 30%. One impediment to using carbon fibers in cars is cost.
If the justification for the extra price of carbon-fiber cars is solely based on fuel
savings, how much extra price can be justified over a six-year life span if the
carbon-fiber car would average 39 miles per gallon of gasoline compared to a
conventional car averaging 30 miles per gallon?
Assume that gasoline costs $3.00 per gallon, the interest rate is 20% per year,
and 117,000 miles are driven uniformly over six years. Assume fuel costs occur at
the end of each year.
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
7 / 12
Solution
Extra price is the present value of the fuel cost saving in the life span of six
years.
Miles covered each year:
117, 000
= 19, 500
6
Annual fuel cost saving of carbon-fiber car over conventional car:
3.00 × (
19, 500 19, 500
−
) = 450
30
39
P = A(P/A, 20%, 6) = 1496.5
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
8 / 12
Question 3
John Smith took out a student loan to complete his four year engineering degree.
He borrowed $5000, four years ago when the interest rate was 5% per year. A
further $6000 was borrowed 3 years ago at 3% per year. Two years ago he
borrowed $6000 at 6% and last year $7000 was borrowed at 8% per year.
If Smith makes annual payments to repay his total debt payments up to 10 years
with 8% fixed annual interest rate.(Assume repayment starts from now) What is
the amount of each payment?
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
9 / 12
Solution
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
10 / 12
Solution
F−3 = 5000(F /P, 5%, 1) + 6000 = 11250
F−2 = 11250(F /P, 3%, 1) + 6000 = 17587.5
F−1 = 17587.5(F /P, 6%, 1) + 7000 = 25642.75
P = F0 = 25642.75(F /P, 8%, 1) = 27694.17
A = F0 (A/P, 8%, 10) = 4126.43
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
10 / 12
Question 4
An auto dealership is running a promotional deal whereby they will replace your
tires free of charge for the life of the vehicle when you purchase your car from
them.
You expect the original tires to last for 30,000 miles, and then they will need
replacement every 30,000 miles thereafter. Your driving mileage averages 15,000
miles per year. A set of new tires costs $400.
If you trade in the car at 150,000 miles with new tires then, what is the
lump-sum present value of this deal if your personal interest rate is 12% per year?
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
11 / 12
Solution
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
12 / 12
Solution
The interest rate for 2 years j satisfies 1 + j = (1 + 12%)2 ⇒ j = 25.44%
P = $400(P/A, j, 5) = $1066
Guo Weiwei (CUHK)
SEEM2440 Engineering Economics
12 / 12