This document contains 4 engineering economy problems involving calculations of present and future values, compound interest rates, and cash flows. Problem 1 involves calculating average costs, marginal costs, and break-even points for organizing a class trip. Problem 2 involves calculating unknown cash flows from diagrams with zero present value. Problem 3 involves calculating accumulated balances and withdrawals over time with changing interest rates. Problem 4 involves comparing effective interest rates for banks with different compounding periods and calculating accumulated deposits.

1. Sample_EXAM 1.
Engineering Economy: EIN 4354
Solutions

2. Problem 1 (25 points = 9(a) + 8(b) + 8(c) )
You are planning to organize a trip to Bush Gardens in Orlando for your classmates. You
have your own van and you can also order a bus. The van has places for 5 people plus for
you as a driver. If the number of participating people does not exceed 5 people, you will
use the van, if you have more than 5 people participating you decide to use both the van
(for 5 people) and the bus (for the rest of people). The variable cost (including food and
ticket price) per person is 60$. The fixed cost of using the van is $30, the fixed cost (gas
and money to bus driver) of ordering the bus is $50. You decide to charge $65 per
person.
a). What is the average cost of the trip if you have 20 people participating?
Average cost = ($30+$50+$60*20)/20 =$64
b). Suppose that currently you have 5 people signed up for the trip. What is the marginal
cost (the cost of having one more person) in this case?
Marginal cost = $50+$60 = $110.
c). Currently you have 10 people participating. How many additional participants do you
need to break even?
Total cost = $30 + $50 + $60*10 + $60*X = 680 + 60X
Total benefit = $65*10 + $65*X = 650 + 65X
Break even equation:
680 + 60X = 650 + 65X
30 = 5X
X=6
Answer: 6 more people to break even.

3. Problem 2 (26 points = 6(a)+5(b)+5(c)+5(d)+5(e) )
The following diagrams below show cash flow streams having zero present value.
Suppose that 5% is a one-period interest rate. Compute X:
a).
X = 10(F/P,5%,5)+20(F/P,5%,3)=10*1.28+20*1.16=$36
X
$10
$20

4. b). The following diagram shows a cash flow stream having zero present value.
Suppose that 5% is a one-period interest rate. Compute X:
X + 20(P/A,5%,3) = 2X(P/F,5%,4)
X =20*2.72/(-1+2*0.82)=$85
2X
$20 $20
X
$20

5. c). The following diagram shows a cash flow stream having zero present value.
Suppose that 5% is a one-period interest rate. Compute X:
X[(F/A, 5%, 3)+(P/F, 5%, 2)] = 30 + 30(P/F, 5%, 1)
X = (30+30*0.95)/(3.15+0.91) = $14.4
$30
X X
$30
X XX

6. d). The following diagram shows a cash flow stream having zero present value.
Suppose that 5% is a one-period interest rate. Compute X:
i(2) = (1+5%)^2-1 = 10.25%
X(F/A, 10.25%, 3)(F/P, 5%, 1) + 10(F/A,10.25%,3)(F/P,5%,2) = 60
X = (60-10*3.32*1.1)/(3.32*1.05) = 6.73
$10$10
XX
$60
X
$10

7. e). The following diagram shows a cash flow stream having zero present value.
Suppose that 5% is a one-period interest rate. Compute X:
10[ 4(F/A,5%,4) - (F/G,5%,4)] = X[ (P/A,5%,3) + (P/G,5%,3)]
X = (40*4.31-10*6.02)/(2.72+2.63) = 20.97
2X
$10
$20
X
$30
$40
3X

8. Problem 3 (25 points = 9(a) + 8(b) + 8(c) )
Mr. Smith found a new job paying $500 every month. He decided to deposit all these
money to his bank account at the end of each month. Initially, he has $1000 in his
account. At that time the monthly interest rate in the bank is 3%.
6 months later (right after 6th
monthly deposit) the interest rate changes to 5%.
Another 6 months later (after 12th
deposit) Mr. Smith looses the job and stops making
deposits to the bank.
Assume monthly compounding.
a) How much money will Mr. Smith have in his bank account after making the last
(12th
) deposit?
b) Suppose the following. When Mr. Smith lost his job, interest rates in the bank
dropped to 1% per month. Mr. Smith decided to take all money from his account
by making 3 equal end-of month withdrawals (starting from the end of 13th
month). What is the value of the monthly withdrawal?
c) Suppose that instead of getting a new job Mr. Smith remained jobless. Originally
he has $1000 in the bank account and makes no additional deposits. If the bank
has 3% simple monthly interest for the first 6 months and 5% simple monthly
interest for the next 6 months, what interest would Mr. Smith accumulated in his
account by the end of the year?
Solution:
(a) F(total) = 1000(F/P,3%,6)( F/P,5%,6)+500[ (F/A,3%,6)(F/P,5%,6) +
(F/A,5%,6) ] = 1000*1.19*1.34+500[6.47*1.34 + 6.80] = $9,329.5
(b) A(P/A,1%,3) = F(total); A = 9329.5/2.94 = $3,173.3
(c) Total accumulated interest is 3%*$1000*6 + 5%*$1000*6 = $480

9. Problem 4 (25 points = 6(a) + 6(b) + 6(c) + 6(d))
Bank A has 6% nominal interest rate per 6-month period.
Bank B has 12% nominal interest rate per year.
Assume quarterly compounding for bank A and
monthly compounding for bank B.
a) What is the effective annual interest rate for bank A?
%55.121
2
%6
1
4
=−⎟
⎠
⎞
⎜
⎝
⎛
+
b) What is the effective interest rate for three-month period (quarterly rate) for
bank B?
%03.31
12
%12
1
3
=−⎟
⎠
⎞
⎜
⎝
⎛
+
c) Suppose you put $100 in bank A and $100 in bank B for 10 years? What is the
difference in the accumulated interest at the end of the 10th
year.
Accumulated interest for bank A = ( )( ) 2.226$1%31100
4*10
=−+
Accumulated interest for bank B = ( )( ) 0.230$1%11100
12*10
=−+
Difference = $230.0 - $226.2 = $3.8
d) Someone makes deposits of $100 to bank B every 6 months. Today he made the
8th
deposit. What is the value of all 8 deposits today?
Effective interest for 6 months is %15.61
12
%12
1
6
=−⎟
⎠
⎞
⎜
⎝
⎛
+
Value of deposits = $100*(F/A,6.15%,8) = 100*9.95 = $995