Engineering Economics
The essential idea behind
engineering economics is that money generates money. You cannot compare $10.00 today to $10.00 a
year from now without adjusting for the investment potential. A simple example would be to take the $10.00
and put it in a savings account at 2% interests. After a year you have $10.20 instead of
$10.00.
You might be presented with
three options for being paid.
- A single payment right now.
- A single payment at some time in the future.
- A uniform annual payment over several years.
For example, consider a
lottery that is held in a fictional state (Rustiana). In this particular lottery, the winner is
given three different payment options.
The first is a lump sum payment immediately of $1,000,000. The second is a series of 21 annual payments
of $50,000. The third is a whopping
$2,000,000 to be paid ten years from now.
How does the winner choose which option to take? Of course, that will depend on the winners
personal needs, but beyond that, how does one compare the three options? The annual payment method provides $50,000
more than the immediate payment, and the deferred payment provides a cool
million more. For argument sake, let’s
say that the person who won the lottery knows that he can make 5% interest on
the money in some safe investment.
The notation is provided in
Table 1. We will call the immediate
payment the “Present Value” of the winnings.
The $50,000 annual payments will be called “Annual Value,” and the
$2,000,000 will be the “Final Value.”
The interest that we presume we can get on the money in some investment
is designated as “i”.
|
Symbol |
Meaning |
Amount (in example) |
|
P |
Present Value (What the
money is worth right now) |
$1,000,000 |
|
A |
Annual Value (What the
money is worth in annual payments) |
$50,000 |
|
F |
Final Value (What the
money will be worth at some future date) |
$2,000,000 |
|
i |
Interest (an estimate of
how fast the money can grow in some relatively safe investment). |
5% |
|
n |
Number of years (Duration
over which an investment is made). |
5 years |
These
parameters are related to each other through standard engineering formulas,
which are provided in Table 2. The
purpose of this document is to explain where the equations came from and how
they are used in engineering economics problems.
|
|
Find |
From |
Discrete Payments, Discrete Compounding |
Discrete Payments, Continuous Compounding |
Continuous Payments, Continuous Compounding |
|
|
Single Payment |
F |
P |
|
|
|
|
|
P |
F |
|
|
|
||
|
Equal-Payment Series |
F |
A |
|
|
|
|
|
A |
F |
|
|
|
||
|
P |
A |
|
|
|
||
|
A |
P |
|
|
|
||
|
Gradient Series |
A |
G |
|
|
|
|
|
P |
Fi |
|
|
|
||
Table 2 tells us how to
convert from one form of payment to another.
Consider the column under “Discrete Payments,
Discrete Compounding.” The equations are
designed to answer the following questions:
“If I know what the
present value is, what is the future value at a time n years from now given an interest of i? In our example, if we want to know the value
of the $1,000,000 in 10 years, it would be 
. Given this, the
delayed payment option doesn’t appear too bad!
“If I know what the
value of the funds will be in n
years, what are they worth now? In our
example, if we want to know what $2,000,000 is worth in today’s dollars, it
would be 
. This just confirms
what the previous calculation told us, i.e. that the value of the delayed
payment is grearter than that of the immediate
payment.
“If I know the annual
payments, what is the money worth n
years from now?” In our case, after 21
years the $50,000 per year will come out to be 
. This seems to be a
lot of money, but remember that this is in terms of the value of dollars 21
years from now. To compare that to the
immediate $1,000,000 we have to convert the $1,000,000 to 21 year old dollars via . Previously we used
10 for n, but now we need to use 21. The
answer is: 
. This is about a
million dollars more, which would argue for the immediate payment.
“If I know the future
value of the money, what is that equivalent to in annual payments?” So let’s say that we would like to convert
the $2,000,000 at ten years to annual payments over 10 years. We then obtain:
.
“If I am paid in
annual payments of A over n years (with the first payment starting
at the end of the first year), what does that correspond to in current
dollars?” In our case, 21 payments of $50,000
is equivalent to only 
, which is kind of a rip-off.
“If I could be paid in
a lump sum today, what must my annual payment be for me to recover the same
amount of money over n years?” A logical question to ask in our case is,
“How much money should I get if I am paid my winnings over 21 years?” The answer is: 
.
Exercise 1: The above analysis does not take into account
that the $1,000,000 will be taxed at a different rate from the $50,000 annual
payments. Assume that the tax on
$1,000,000 in one year is 50% and the tax on $50,000 per year is only 25%. This means that the present value of the
$1,000,000 is only $500,000, and that the annual payment of the $50,000 is only
$37,500. Calculate the equivalent annual
payment for the $500,000 and compare it to the $37,500. Also calculate the present value of the
$37,500 payment option.
The next two columns in the
table provide similar information, but for different ways of compounding the
interest and making the payments. We all
know that by compounding interest monthly we can earn greater return than if
the interest is compounded annually.
Notice that for the single payment rows the “Discrete Payment” and
“Continuous Payment” colums
are identical. This is because, by
definition, a single payment is discrete.
To see what happens to the Equal-payment series, consider the entry
under “Discrete Payments, Discrete Compounding” and in the “Compound amount”
row. This equation is:
. What if we were to
compound monthly rather than annually?
We would then have to replace “i” with “i/12”
(interest rate per month), and we would have to replace “n” with “12n” (number of
months). The amount of interest paid
would then be greater, as a result of compounding. On the other hand, if we are paid monthly
instead of annually, the value of our future worth is larger simply because we
get the money earlier. To understand
this completely, it is necessary to derive the equations in table 5.5.
Exercise 2: Your engineering firm needs a rapid
prototyping machine. The company gives
you two options. In Option 1 you
purchase the machine outright for $50,000, pay a maintenance contract of $1,000
per year, and expect to be able to resell the machine after 10 years at a
salvage value of $10,000. In Option 2,
you lease the machine at $7,000 per year and pay no maintenance, but receive no
salvage. Assume that you will be able to
take in $8,000 per year in income from this machine. Also assume that an additional option is not
to buy the machine at all, but to put the money in the bank at 5%
interest. Which option will be best for
the firm?
Origins of the Equations
The engineering economics
equations can be derived relatively simply.
Consider first the case of simple interest. At the end of one year, the principal amount
is worth its initial value, P, plus
an additional amount equal to 
, so the value is
dollars. At the end
of two years it is worth what its value at the end of the first year times 
, or 
each subsequent year
the value increases by a factor of , so by year n, the
value is simply 
dollars. If the intrest is
compounded monthly instead of annually, there are twelve times as many compoundings, so the exponent becomes 12n instead of n, but the interest at each compounding is 1/12th as
much (i.e. i/12). If we compound continuously, we speak of a
rate of interest, r, which is percent
per year. At any given time t, we have that the rate of change of
monetary value is that rate multiplied by the current value. In other words:
.
Note that 
is the value of 
at time 
, and that 
is the value of at time t. Since r
is a constant, this equation is easily integrated to 
. To obtain P in terms of F, it is sufficient to invert the relationships already
derived. For example, from we obtain
,
and from 
we obtain 
. This takes care of
the top section in the table from Ertas and Jones.
The next question is, “what
is the value of a “deal” where you are given a certain amount of money “A” each year. First, consider the present value (P) of the first year’s payment. From our relationship for simple interest, . If we consider A to be the future value after one year, we find that
. Similarly, the
present value of the second year’s payment is 
. Extending this to
the nth year’s payment, 
. Thus, the total
present worth of the entire “deal” is the sum of all P’s up to Pn. That is:
Okay, now let’s have a
little fun. Recall that
, which is a formula you may have seen in BIEN 225 and/or
BIEN 425 (in relation to z transforms).
And just where did that come from, you may ask? It goes like this. Let’s define 
. Let’s now take the
first term (the 
term) out of this
summation to get 
. Now make the simple substitution 
. This means that when 
, 
(in the lower limit),
and when 
, 
(in the upper limit),
and we can substitute 
for 
in the exponent of the summation. This gives us 
. Since
, and since x is
independent of k, we can write 
. We can now add in 
(i.e., add zero) to
obtain 
and notice that the
expression in square brackets is just equal to 
, which is the same as 
(because j and k are dummy variables), which is equal to 
, according to our original definition. It follows that 
. Thus, if we take to the left hand side,
factor out 
, and then divide by 
, we get 
, as promised. If we
substitute 1/x for x in this expression we get 
.
If we use this expression,
we get
. To obtain the
relationship for, given 
, we can either use this same method or use the expression
that converts to .
Exercise 3: This last expression for “P given A” is not
in the same form as the attached table.
Rewrite the expression as:
and simplify to obtain the form in the table.
Exercise 4: Use the
expression that converts to , along with the expression for given to obtain an
expression for given .
Exercise 5: Obtain the
expression for given directly, using the
summation method above.
The expression for given is for the case where
the first payment starts at the end of year 1.
While dividends will often be paid at the end of the year, other types
of transactions will start at the beginning of the first year. For example, if you are paying for
maintenance on a piece of equipment, the company will want to have their money
up front. They will not be receptive if
you wait until after your equipment breaks to tell them you would like to buy
the maintenance contract. A simple way
to determine P given A for this situation is to use P given A for n-1 years, instead of n years (to account for the payments at
the beginning of years 2-10) and add the payment for the first year. The result is
Recap:
Assume you want to know
whether Option 1 or Option 2 is better.
The approach is as follows:
- Decide what common time frame to use. For example, you can convert everything
to its value in current dollars (present worth), convert everything to its
value in future dollars (future worth), or convert everything to an annual
payment. For now we will assume
that we are using the present worth method, so we want to convert
everything to present worth.
- Anything that is already expressed in your time
frame does not need to be modified.
For example, if you are using the present worth method, there is no
need to convert the down payment or the initial investment to anything.
- For the present worth method, convert annual
payments to present dollars with
.
- For the present worth method, convert single
payments for any given year to present dollars with .
- Now that everything is in terms of the same time
frame, you can just add and subtract dollars.
Solution to Exercise 2 using the
Present Worth method:
Exercise 2: Your
engineering firm needs a rapid prototyping machine. The company gives you two options. In Option 1 you purchase the machine outright
for $50,000, pay a maintenance contract of $1,000 per year, and expect to be
able to resell the machine after 10 years at a salvage value of $10,000. In Option 2, you lease the machine at $7,000
per year and pay no maintenance, but receive no salvage. Assume that you will be able to take in
$8,000 per year in income from this machine.
Also assume that an additional option is not to buy the machine at all,
but to put the money in the bank at 5% interest. Which option will be best for the firm?
Option 1:
The
present worth of the initial 50,000 is just 50,000.
The
present worth of the maintenance contract (remembering that it must be paid at
the beginning of the year) is:
The
present worth of the income (8,000/year), remembering that income comes at the
end of the year, is:
The
present worth of the salvage ($10,000) is: 
.
Thus,
the profit is Income – Outlay = 61,774 – (50,000 + 8,107 + 6,139) = -$2,473.
Because
your profit is negative, you would be better off to put the money in the bank
at 5% interest, rather than going for Option 1.
Option 2:
The
present worth of the 7,000/year lease is:
We
already figured out that the present worth of the $8,000/year profit is
$61,774, thus the total profit by this option would be $61,774 - $56,754, or
$5,019. This may not sound like much,
but it’s better than having the money just sit in the bank at 5% interest.
Thus,
the correct choice is Option 2.