Lecture 4
Rates of Return
Summary of comparison methods we know so far:
Present worth
Future worth
Annual worth
Still to come:
Rate of return
Cost/benefit
Payback period
What is your interest rate?
The MARR
This is your Minimum Acceptable Rate of Return
If you’re in business at all, it’s because you think you
can make more money from your investment than the
bank can.
So, your MARR must be at least as great as the bank
rate.
The MARR
If all the initial investment in the business is yours,
your MARR is whatever you think it should be.
If you’ve borrowed some of the money from the bank,
your MARR must be enough to pay the interest on the
loan.
If your business is owned by its shareholders…
your rate of return must be high enough to keep them happy.
The Internal Rate of Return
One way of comparing projects is to calculate their
rates of return.
If the project’s rate of return is less than your MARR,
don’t do it.
If you have several projects, the one with the highest
rate of return might be the best…
The Internal Rate of Return
If your project requires a single present investment,
P, and yields a single future payout, F, in N years
time, then its rate of return is the solution to:
P = F(P/F, i, N)
The Internal Rate of Return
This can be re-stated as:
PW = P - F(P/F, i, N) = 0
More generally, the internal rate of return, or IRR,
is the interest rate which makes the total present
worth of the project cashflows equal to zero.
Example:
We build a bicycle factory at an initial cost
of $800 000. We hire workers at $100 000 a
year, payable at year’s end. In the first five
years of the project, our year-end sales are:
End of Year
Sales ($ 000)
1
80
2
200
3
500
4
1000
5
1200
What is the rate of return for the project?
A good way of solving these problems is to
build a spreadsheet.
We work out the value of:
PW = - 800 - 20(P/F,i,1) + 100(P/F,i,2)
+ 400(P/F,i,3) + 900(P/F,i,4) + 1100(P/F,i,5)
For different values of i and plot a graph.
-800
-800
-800
-800
-800
-800
-800
-800
-800
-800
-800
-20(P/F,i,1)
-20
-19
-18
-17
-17
-16
-15
-15
-14
-14
100(P/F,i,2)
100
91
83
76
69
64
59
55
51
48
400(P/F,i,3)
400
346
301
263
231
205
182
163
146
131
900(P/F,i,4)
900
740
615
515
434
369
315
271
234
204
1100(P/F,i,5)
1100
862
683
547
442
360
296
245
205
172
2000
1500
Present Worth
i
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
1000
500
0
0.00
-500
0.10
0.20
0.30
Rate of Return
0.40
0.50
PW
1680
1220
863
583
360
182
37
-81
-179
-260
This gives us a rate of return of about 32%, which
is pretty good.
So this is a good way of comparing investments,
apart from a couple of possible traps…
Trap 1
Suppose we have a million dollars, and we
can either build bicycles or unicycles.
The rate of return per unicycle is 50%, and the
rate of return per bicycle is 32%.
Solution: build unicycles.
Trap 1
Suppose we have a million dollars.
We can spend $600 000 on a gold mine, and get back
$900 000 in a year.
Or we can spend the whole million on a silver mine
and get back $1 400 000 in a year.
There are no other investment opportunities available
except putting money in the bank at 5% interest.
Trap 1
Gold Mine:
After a year we have $900 000 + $420 000 = $1 320 000
Rate of return on investment = 50%
Silver mine:
After a year we have $1 400 000.
Rate of return on investment = 40%
We can avoid this trap by considering the incremental
rate of return:
Consider three places we can put the $400,000 left over
from buying the gold mine:
Old sock under mattress: End up with $400,000, rate of return = 0%
Bank: End up with $420,000, rate of return = 20,000/400,000 = 5%
Cancel gold mine investment, use the $400,000 to upgrade to silver mine:
End up with additional $100,000*, rate of return = 100 000/400 000 = 25%
* $1,400,000 - $(900,000+400,000)
This is more than the MARR (5%), so we should do it.
General rule for evading Trap 1:
Arrange the possible projects from cheapest to most costly.
Choose the first one with a rate of return > MARR.
Then consider the incremental return from upgrading to the
next project. If this is greater than the MARR, choose that
next project.
Go on comparing the best candidate to the next most
costly until you run out of projects.
Trap 2
Example:
I have a small research company. A government agency
offers me a contract: I get $20 000 start-up funds now.
If all goes well, I put in $180 000 of my own money in three
year’s time to develop a prototype. If the prototype
works, I get $200 000 in five years time.
What is my rate of return if all goes well?
Solve:
PW ($000) = 20 – 180(P/F,i,3) + 200(P/F,i,5) = 0
Present Worth
50
40
30
20
10
0
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
-10
-20
Rate of Return
We calculate a graphical solution using a spreadsheet.
But it has two solutions, IRR = 16% and IRR = 80%.
Which is right?
The second solution, IRR =80%, assumes that I can
invest my start-up payment at 80% interest. But there
is no reason to think that this is possible.
To escape this trap, I have to assume that surplus
funds can only be invested at a realistic interest
rate, known as the auxiliary rate of return or explicit
rate of return. (Often taken as equal to the MARR.)
There are two ways of doing this, an exact way and
an approximate way. The rate of return for the project
that I calculate is called the external rate of return (exact
or approximate).
External Rate of Return (Exact)
Bring cash inflows forward at the auxiliary rate of return:
So after three years the initial $20 000 becomes
20 000 (F/P, 0.1, 3) = 26, 620
So the amount of my own cash I have to put in is
180 000 – 26 620 = 153 380
So my rate of return on this investment is ie, where
153 380(F/P, ie, 2) = 200 000
So ie = 14.2%
External Rate of Return (Approx)
Bring cash inflows to project’s end at the auxiliary rate of
return, bring cash outflows to project’s end at the (unknown)
approximate external rate of return, equate the future worth
to zero and solve for iea
20(F/P, 0.1, 5) – 180(F/P, iea, 2) + 200 = 0
(F/P, iea, 2) = 232/180
iea = 13.5%
The exact ERR is difficult to calculate if the cashflows
are complicated. Both the exact and the approximate
ERR’s are single-valued – we never get multiple
solutions.
Rate of return methods give the same ordering of
projects as PW, FW, and AW methods, but make it
easy to compare the profitability of projects of
different size.
Sunk Costs
Problem 1:
Quarmby Electronics makes mobile phones. 100 000 batteries will
be needed in the coming year. They can be produced in-house by
two workers. The salaries of these two workers will be $50 000
each, and the cost of raw materials is $1.00 per battery.
Alternatively, the batteries can be purchased from an external
supplier at a unit cost of $1.90 per battery.
Which option should they choose?
Sunk Costs
Problem 2:
Quarmby Electronics makes mobile phones. They have just
invested $500 000 in a battery-making machine. This machine
requires two workers to operate, and can produce 100 000 batteries
per year. The salaries of these two workers will be $50 000 each.
Raw materials for the batteries cost $1.00 per battery. The machine
has no salvage value.
Alternatively, the batteries can be purchased from an external
supplier at a unit cost of $1.90 per battery.
Which option should they choose?