GENTECH 3EE3
MID-TERM REVIEW QUESTIONS
Review Problem Questions
GEN TECH 3EE3
WEEK 6 LECTURE
Administrative Items
• Mid Term Test:
o Scheduled for Wednesday, February 16th, 7:00 – 9:00 PM (Online)
o A2L Midterm Study Groups – 10 Discussion Forum Groups (self-enrol)
o Chapters 1-4 including bonds and mortgage questions
o This Exam contains 10 multiple choice questions (A2L) and 20 FIB
problems (A2L)
o This is a closed-book, double sided hand-written 8.5 X 11 crib sheet is
permitted, Midterm 3EE3 Formula Sheet and tables are provided at the
end of the Exam paper, and only ‘McMaster Approved’ calculators are
permitted
2
Capitalized Formula Problem
3
What is the PW of a cost to install= $2,500,000, infinite life,
maintenance of $300,000 every 10 years, annual i=5%?
SOLUTION
PW = 2,500,000 + 300,000(A/F,i,n)/i
PW = 2,500,000 + 300,000(A/F,5%,10)/0.05
PW = 2,500,000 + 300,000(0.07950)/0.05
PW = $2,977,000
• CB Electronix needs to expand its plant. It is
considering two alternative plans. Both will last
indefinitely. The MARR is 15%. Use an AW comparison
to select the preferred alternative.
Plan 1: Expand by 20 000 square feet now
First cost = $2 000 000
Annual Maintenance costs = $10 000 per year
Periodic maintenance costs = $15 000 every 15 years,
starting in 15 years
4
Capitalized Formula Problem
• Plan 1: Expand by 20 000 square feet now
First cost = $2 000 000
Annual maintenance costs = $10 000/yr
Periodic maintenance costs = $15 000 every 15
years, starting in 15 years.
5
Capitalized Formula Problem
• Plan 2: Expand by 12 500 square feet now and 7
500 square feet in 10 years.
First cost $1 250 000 $1 000 000
Annual maintenance cost $5 000 years 1-10 $11 000 starting in
year 11
Periodic maintenance costs N/A $15 000 every 15
years starting 15
years after 2nd
addition
6
Capitalized Formula Problem
Solution
Plan 1
Using the capitalized cost formula for long lived assets:
AW = (2 000 000 i) + 10 000 + 15 000(A/F,15%,15)
AW = 2 000 000(0.15) + 10 000 + 15 000(0.02102)
= $310 315
P = A/i ➔ A = Pi
7
Capitalized Formula Problem
Plan 2
Step AW Equation Answer
1 First cost 1 250 000(0.15) + 1 000 000
(P/F,15%,10) (0.15)
$224 578.50
2 Maintenance cost:
years 1 to 10
5000(P/A,15%,10)(0.15) $3 764.01
3 Maintenance cost:
years 11 and on
[(11 000)/(0.15)](P/F,15%,10) (0.15) $2 720.00
4 Periodic maintenance
costs
[15000(A/F,15%,15)/(0.15)](P/F,15
%,10)(0.15)
$77.90
Total $231 140
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Solution
Capitalized Formula Problem
Solution
AW of plan 1: $310 315
AW of plan 2: $231 140
Select plan 2, which has the lowest AW
9
Capitalized Formula Problem
Non Uniform Annuity Problem
10
Write the general mathematical expression that calculates A when
PW = 0 for the cash flow diagram shown below.
Calculate A if P = $50,000, X = $5,000, i = 5% and g = 2%.
0 1 2 3
P
98 10 11 12 13 38 39 40
A A A A A A
X
X(1+g)
X(1+g)2
X(1+g)27
X(1+g)28
X(1+g)29
Non Uniform Annuity Problem, Cont’d
11
SOLUTION
P +A(P/A,i,10) – x(P/A,g,i,30)(P/F,i,10) = 0
A = [x(P/A,g,i,30)(P/F,i,10) – P][1/(P/A,i,10)]
(A/P,i,10) = 1/(P/A,i,10), substitute into above expression
A = [x(P/A,g,i,30)(P/F,i,10) – P](A/P,i,10) or
(A/P,i,10) = (F/P,i,10)(A/F,i,10), substitute into above expression
A = [x(P/A,g,i,30)(P/F,i,10) – P](F/P,i,10)(A/F,i,10)
A = [x(P/A,g,i,30)(P/F,i,10)(F/P,i,10) – P(F/P,i,10)](A/F,i,10)
A = [x(P/A,g,i,30) – P(F/P,i,10)](A/F,i,10)
io = (1+i)/(1+g) – 1 = (1+0.05)/(1+0.02) – 1 = 0.029412
A = [5,000(P/A,2.94%,30)/(1+0.02) – 50,000(F/P,5%,10)](A/F,5%,10)
A = $1,222
PW Comparison Method Problem
12
A company is upgrading one of their oldest operating lines at their
packaging facility. Two options are available for upgrading; option A
and B.
Option A upgrading costs $65,000 now and yields annual savings of
$25,000 in the first year, $24,000 the second year, $23,000 in the
third year, and so on.
Option B costs $15,000 and saves $6,500 in the first year, with the
savings decreasing by 15% each year thereafter.
If the upgraded packaging line will last for eight years, which
upgrading option is better using the present worth method and
8% MARR? Draw a cash flow diagram for both options.
PW Comparison Method Problem, Cont’d
13
SOLUTION
OPTION A:
PWA = -65,000 + [(25,000 – 1,000(A/G,8%,8)](P/A,8%,8)
PWA = -65,000 + [(25,000 – 1,000(3.0985)](5.7466)
PWA = $60,859.16
OPTION B:
io = (1+i)/(1+g) – 1 = (1+0.08)/(1-0.15) – 1 = 0.270588
PWB = -15,000 + 6,500(P/A,g,i,n)/(1+g) = -15,000 + 6,500(P/A,i
o,8)/0.85
PWB = -15,000 + 24,100.36 = $9,100.36
Since PWA > PWB, select OPTION A!
Uniform Annuity Cash Flow Problem
14
You would like to save $10,000 for an upgrade to your business
computer system in 2 years time. If you deposit $350 every month
(starting today), what nominal interest rate is required in order for
you to reach your financial goal if the investment is compounded
monthly?
SOLUTION
Given F = 10,000, A = 350
n = 25 months
A = F(A/F,i,n) or F=A(F/A,i,n)
A = F(A/F,i,n) = F{i/[(1+i)^n-1]}
350 = 10,000{i/[(1+i)^n-1]}
i = 1%; A = 354.07
i = 2%; A = 312.20
Linear Interpolation:
i = 1+(2 – 1)[(350 – 354.07)/(312.20 – 354.07)]=1.097206
Therefore, r = mi = 12(1.097206) = 13.166% compounded monthly
Mortgage Problem
15
You have just bought a house for $300,000 and put $50,000 down.
The rest of the cost has been obtained from a mortgage. The
mortgage has a nominal interest rate of 3%, compounded monthly
with a 10-year amortization period. The term of the mortgage is 5
years. What is the monthly mortgage payment? How much will you
owe in 5 years?
SOLUTION
P = 300,000 – 50,000 = 250,000
i = 0.25%
n = 120 months
A = P(A/P,i,n) = 250,000(A/P,0.25%,120) = 250,000(0.009656)
A = $2,414.02
F5 = P(F/P,i,n) – A(F/A,i,n)
F5 = 250,000(F/P,0.25%,60) – 2,414.02(F/A,0.25%,60)
F5 = 250,000(1.161617) – 2,414.02(64.647)
F5 = $134,345.83
Bond Problem 1
16
A man buys a corporate bond from a bond brokerage house for
$925. The bond has a face value $1,000 and pays 4% of the face
value each year. If the bond is paid off at the end of 10 years, what
rate of return will the man receive?
SOLUTION
PW of Cost = PW of Benefits
$925 = $40 (P/A, i%, 10) + $1,000 (P/F, i%, 10)
Try i = 5%
$925 = $40 (7.722) + $1,000 (0.6139) = $922.78 (i is too high)
Try i = 4.5%
$925 = $40 (7.913) + $1,000 (0.6439) = $960.42 (i is too low)
i* = 4.97%
Bond Problem 2
17
At what coupon interest rate will a $20,000 bond yield a nominal 12%
interest compounded quarterly if the purchaser pays $18,000 and the
bond becomes due in 20 years? Assume the bond interest is
payable quarterly.
SOLUTION
x = coupon rate per quarter
NPW = 0 = ($20,000* x) (P/A, 3%, 80) + $20,000 (P/F, 3%, 80) - $18,000
($20,000* x) (30.201) = $18,000 – $20,000 (0.0940)
x = ($18,000 – $20,000 (0.0940))/($20,000 × 30.201) = 0.0267 = 2.67%
Coupon rate = 2.67%/qtr × 4 qtr/yr = 10.68%
Supplementary Problem 1
18
A corporation sold 12% bonds with a face value of $200,000. These
bonds mature in in six years and interest is paid semi-annually. The
bonds were yield 10%. What is the issue price?
SOLUTION
PW=200,000(P/F,5%,12) + 200,000(0.12/2)(P/A,5%,12)
PW=200,000(0.556837) + 200,000(0.06)(8.863252)
PW=$217,726.50
Supplementary Problem 2
19
A company issued 5-year, $1 million bonds with a coupon rate of 6%
and yield of 5%. The bonds pay interest semi-annually. What is the
issue price of the bond?
SOLUTION
PW=1,000,000(P/F,2.5%,10) + 1,000,000(0.06/2)(P/A,2.5%,10)
PW=1,000,000(0.7811984) +1,000,000(0.03)(8.7520639)
PW=$1,043,760.32
Supplementary Problem 3
20
How are bonds priced?
SOLUTION
Bonds are priced at the present value of all future cash flows (i.e.
interest paid out at compounding period and final face value paid out
in the future) at the bond yield. Bond yield is defined as the amount
of return that an investor will realize on a bond.
Therefore, bonds are prices at the present value of all future
cash flows at yield.
Supplementary Problem 4
21
A $92,000 mortgage has an interest rate of 9% compounded semi-
annually, for a 5-year term. The mortgage is amortized over 20-years.
What is the monthly payment?
SOLUTION
ie=(1+r/m)^m - 1 = (1+i)^6 - 1 = 0.09/2 = 4.5%
solve for i (monthly interest rate); (1+i)^6 = 1.045; therefore, i =
(1.045)^(1/6) - 1 = 0.0073631
A=92,000(A/P,0.73631%,240)
A=$818.05
Supplementary Problem 4
22
On February 1, 2013, Sawa Inc. issued 5-year, $1 million bonds with a
coupon rate of 6% and yield of 5%. The bonds pay interest semi-
annually on August 1 and February 1 each year. What is the issue price
of the bond?
SOLUTION
P = 1000000(P/F, 2.5%, 10) + 30000(P/A, 2.5%, 10)
P = $1,043,760
NOTE: Because the coupon rate is paid as a semi-annually cash
flow, use 5%/2 = 2.5% semi-annually interest rate as the yield.
Questions Period
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