A drafting company employs 10 drafters at $800/week each. The CEO
considers
three alternatives:
1.Buy 8 low-end workstations at $2 000 each. Give two drafters 1 year’s
notice.
At the end of the year they each get $5 000 severance pay. Train the
remaining eight in AutoCAD. The first training course is available in a
year,
and costs $2 000 per participant. After completing the course, each
drafter
gets a $100/week raise.
2.Buy 5 high-end workstations at $5 000 each. All the drafters get a
year’s notice and $5 000 severance pay at the end of the year. Five new
graduates are hired at $1 200 per week. They are trained in ProEngineer; to keep current, they will each need a $5 000 retraining session
every six months.
3.Do nothing.
A drafting company employs 10 drafters at $800/week each. The CEO considers three alternatives:
1.Buy 8 low-end workstations at $2 000 each. Give two drafters 1 year’s notice. At the end of the year
they each get $5 000 severance pay. Train the remaining eight in AutoCAD. The first training course is
available in a year, and costs $2 000 per participant. After completing the course, each drafter gets a
$100/week raise.
2. Buy 5 high-end workstations at $5 000 each. All the drafters get a year’s notice and $5 000
severance pay at the end of the year. Five new graduates are hired at $1 200 per week. They are trained
in Pro-Engineer; to keep current, they will need a $5 000 retraining session every six months.
3. Do nothing.
Any of these options would allow the company to service its current customers. Any money saved can
be invested at 10%, which is also the cost of borrowing money. What should they do?
What time frame should we use?
How do we represent `Do nothing’?
Sketch the cash-flow diagrams.
What non-monetary factors would matter?
Option 1
$40,000
$16000
$16 000
$10 000
PW = -16,000 – (16,000+10,000)(P/F,0.1,1) + 40,000(P/A,0.1,N)(P/F,0.1,1)
$40,000
P=F(P/F,i,1)
$16 000
$10 000
PW = -16,000 - ….
$40,000
26,000(P/F,i,1
)
PW = -16,000 – 26,000(P/F,i,1)…
P’=40,000(P/A,i,4)
P’
PW = -16,000 – 26,000(P/F,i,1) + …
$40,000
P=F(P/F,i,1)
P’=40,000(P/A,i,4)
PW = -16,000 – 26,000(P/F,i,1) + …
P=40,000(P/A,i,4)(P/F,i,1)
PW = -16,000 – 26,000(P/F,i,1) + 40,000(P/A,i,4)(P/F,i,1)
$100 000
Option 2
$25 000
$50 000
PW ($000) = - 25 – …
$50,000
$100 000
P=F(P/F,i,1)
$50 000
PW ($000) = - 25 – …
$50,000
$100 000
$50,000
50,000(P/F,i,1
)
PW ($000) = - 25 – 50(P/F,i,1)
$50 000
PW ($000) = - 25 – 50(P/F,i,1) + …
P’=50,000(P/A,i,4)
$50 000
PW ($000) = - 25 – 50(P/F,i,1) + …
P=50,000(P/A,i,4)(P/F,i,1)
PW ($000) = - 25 – 50(P/F,i,1) + …
P=50,000(P/A,i,4)(P/F,i,1)
PW ($000) = - 25 – 50(P/F,i,1) + 50 (P/A,i,4)(P/F,i,1)
Some common variations on an annuity…
1. Your generous uncle gives you your age in dollars every
birthday.
2. The cost of maintaining your car increases by $100 a
month.
There is one formula that fits both these cases:
A = G(A/G,i,N)
Take careful note of the starting time
3G
2G
G
4G
3G
2G
G
A=G(A/G,i,N)
A
P=A(P/A,i,N)
A
3. You get a regular salary and a $100 raise every year
(or a $100 cut every year)
We use the same formula, and add in your base salary:
A = A/ + G(A/G,i,N)
Sample problem: I get to choose between a job in industry
with a starting salary of $50,000 and $5,000 raises every
year.
Or a job in academia where I get $70,000 right away, but
never get a raise.
If I can invest all my income at 5% interest, how long will
it be before the present value of the industrial job exceeds
the present value of the academic job?
4. You get a regular salary and a 10% raise every year
(or a 10% cut every year)
Too hard for Appendix A
Define a growth-adjusted interest rate io:
io = (1+i)/(1+g) - 1
Then (P/A, g, i, N) = (P/A, io, N)/(1+g)
And if i = g, P =NA/(1+g)
Sample problem: I get to choose between a job
in industry with a starting salary of $50,000 and
5% raises every year
Or a job in academia where I get $70,000 right
away, but never get a raise.
If I can invest all my income at 10% interest, how
long will it be before the present value of the
industrial job exceeds the present value of
the academic job?
5. A generous but eccentric uncle gives you $1 000 every leap year
Several alternatives to find present worth:
a)Use (P/F, i, N) on each payment (tedious if there are a lot)
b)Use j = (1+ i)4 – 1 to get a quadrennial interest rate, then use (P/A, j, N/4)
c)Use (A/F, i, 4) to turn the first payment to an annuity, then use (P/A,i,N)
Comparing projects based on Annual
Worth
If most of our cash flows are annual, it’s easier to convert one-time
expenses to their annual equivalent.
Also, you may want to know ``How much will I have to spend a
week?’’ rather than ``What is the present worth of my annual salary?’’
Comparing projects on this basis will always give the same result as
comparison based on present worth or future worth.
Example
A company wants to expand its capacity. It is considering two alternatives:
1.Construct a new building, at a cost of $2 000 000. Annual maintenance
costs will be $10 000. The building will need to be painted every 15 years
at a cost of $15 000.
2. Construct a smaller new building now, and another, smaller one in 10 years.
The first building costs $1 250 000 to build and $5 000 a year to maintain.
The addition will cost $1 000 000 to build, and once it’s built, the two
buildings together will cost $11 000 to maintain. Fifteen years after the
addition, and every fifteen years after that, the new buildings will be
painted at a cost of $15 000.
Assume an interest rate of 15%. Compare the annual cost of the two alternatives.
Summary of comparison methods we know so far:
Present worth
Future worth
Annual worth
Still to come:
Rate of return
Cost/benefit
Payback period
Comparing Alternatives
In comparing alternatives, any cost common to all the alternatives
can be ignored.
This will sometimes result in all the alternatives having negative
present worth. Don’t worry – just choose the least negative.
In many situations – for example, Question 2.5 – you will need to
be careful to set up the comparison fairly.
Study Period
Part of setting up the comparison fairly is to choose the same study
period for each alternative.
For example:
The Chopper lawnmower costs $100, requires $10 in fuel every year,
and will last 4 years. The LawnBoy lawnmower costs $120, uses $15
in fuel every year, and lasts 6 years. If I expect to be cutting lawns
for the rest of my life, which is the better buy?
PW(Chopper) = -100 - 10(P/A, i, 4)
PW(LawnBoy) = -120 - 15(P/A, i, 6)
This is wrong!
The Chopper lawnmower costs $100, requires $10 in fuel every year,
and will last 4 years. The LawnBoy lawnmower costs $120, uses $15
in fuel every year, and lasts 6 years. If I expect to be cutting lawns
for the rest of my life, which is the better buy?
Either determine a salvage value:
PW(Chopper) = -100 - 10(P/A, i, 4)
PW(LawnBoy) = -120 - 15(P/A, i, 4) + S(P/F, i, 4)
Or consider the lowest common multiple of lives:
PW(Chopper) = -100 - 10(P/A, i, 12) – 100(P/F,i,4) – 100(P/F,i,8)
PW(LawnBoy) = -120 - 15(P/A, i, 12) - 120(P/F, i, 6)
Sets of Alternatives
Alternatives may be independent, contingent, or exclusive
Independent: Buying socks and buying a tie
Contingent: Buying shoes and buying socks
Exclusive: Buying a Lamborghini and buying a Ferrari
In a complex situation, arrange the possibilities into mutually exclusive
subsets and pick the best subset.
Example: You can do A, B, C , D or E. If you do A, you can’t do B. To do
D, you must do either B or C. If you do E, you must do A. You can’t do
both B and C, and if you do D, you can’t do E.
What are the possible subsets?
Assignments
These appear weekly at
http://www.sfu.ca/~jones/ENSC201/index.html
They are due on Thursdays, the first one on Thursday September 12.
Majid will go through problems with slightly different numbers in the
Wednesday tutorial sessions, starting next Wednesday
Model answers will appear on the subsequent Friday.
Only three of these assignments will contribute to your final grade; you’ll be
warned of this in advance.