ECONOMICS
FUNDAMENTALS OF
ENGINEERING REVIEW
Lee L. Lowery, Jr., PhD, P.E.
Presented to Fort Hood
4/15/98
Moving the value of money
from the present to the future:
Say
we borrow $20,000 from the bank for a period of one year at an annual interest
rate of 6%. How much do we owe the bank after one year?
F =
$20,000 + 0.06($20,000)
= $20,000 (1 + 0.06)
= $20,000 (1 + 0.06)1
= $20,000 (1.06)
= $21,200
or,
in economics shorthand:
F =
P (F/P, i, n) = P(a factor from a
table) =
= P(Use F/P column, i% rate, n years or
periods)
= $20,000 (1.06)
= $21,200
Please
note that we will try to use 6% interest for all of the problems, since that
rate is listed in the reference manual.
Now
say we borrow the money for two years:
F =
$20,000 + 0.06($20,000) + 0.06($20,000) + 0.06(0.06 * $20,000)
= $20,000 (1 + 0.06)(1 + 0.06)
= $20,000 (1 + 0.06)2
= $20,000 (1.1236)
= $22,472
or,
in economics shorthand:
F =
P (F/P, i, n) = P(a factor from a
table) =
= P(Use F/P column, i% rate, n years or
periods)
= $20,000 (1.1236) = $22,472
Notice
how you know what factor you need to use from the tables by the fact that the
"P's" appear to cancel out of the factor, leaving you with F, which
is what you wanted. Thus, if you have money at M and you want to change it to
money at a different time (or type) Q, then you would write:
Q =
M(Q/M, i, n)
Notice
how the M's appear to cancel, leaving what you want - Q.
Similarly,
if I want to take a given Annuity or Annual amount and transfer it into a
Present amount, I would write:
What
I want = What I got times a factor from the tables:
P =
A(P/A, i, n)
To
transfer a given Future value into an Annuity or Annual amount I would write
A =
F(A/F, i, n)
To
GET the Gradient FROM an Annual amount I would write:
G =
A(G/A, i, n)
To
GET the Present value OF a (given) Future value, I write:
P =
F(P/F, i, n)
Note
also that sometimes the reference manual does not give a column, such as for
G/A. In that case:
G =
A(G/A, i, n)
= A / (A/G, i, n)
Also,
you may have to piece together what you want:
F =
G(F/G, i, n) (but F/G is not in the
reference manual)
= G(F/A, i, n)(A/G, i, n) (which are both in the manual.)
Now what if we borrow the money for 7 years:
F =
$20,000 (1+0.06)(1+0.06)(1+0.06)(1+0.06)(1+0.06)(1+0.06)(1+0.06)
= $20,000 (1 + 0.06)7
= $20,000 (1.5036)
= $30,072
or,
in economics shorthand:
F =
P (F/P, i, n) = P(a factor from a table) =
= P(Use F/P column, i% rate, n years or
periods)
= $20,000 (1.5036) = $30,072
Moving the value of money
from the future to the present:
Now
assume we would like to deposit a sum of money in the bank and have it earn
interest at 6% for one year, and that we desire the bank to return $20,000 to us at the end of that year. How
much must we invest?
Since
F = P(1+i)n then P = F(1+i)-n
So
the P you would need to deposit now to get back $20,000 at the end of one year
at 6% interest would be:
P =
$20,000(1+0.06)-1
= $18,868
or
in ecospeak:
P =
F(P/F, i, n) = Multiply the future value you want times a factor designed to
convert future money back to present money. Use the P/F column on the 6% page,
on the 1 year line.
P =
$20,000(0.9434) = $18,868
or,
for 7 years: P = $20,000(0.6651) = $13,302
Finding the Annual
contribution required to get some desired Future value:
How
much money would we have to invest annually during the next 7 years to be able
to withdraw $1678.76 at the end of that 7 year period? Now at this time some
discussion arises - namely, should we assume that in answering that question
the person intends to deposit the first installment now? Today? Immediately? Or
should we assume that having today decided to do this, it will take him a year
to get the first payment together? Either way is fine, but we don't want to
have a set of interest tables which say "7 year investment, first payment
made today" and a second set of tables which say "7 year investment,
first payment made one year from today." So we will pick one or the other,
and we have all agreed to pick the latter. Namely, that having decided today to
start investing money, we will have the first amount ready for investment at
the end of the first year. Thus the cash flow diagram will look:
Thus
in our case, A = F(A/F, i, n) where (A/F, i, n) can be found from the tables in
the Reference Manual as 0.1191,
Thus
A = F(A/F, i, n)
A = F(Use A/F column, 6% page, 7
year row)
A = $1678.76 (0.1191) = $199.94
Finding the Annual
withdrawal permitted given a Present contribution:
Now
what if we are given a gift of $1116.47 and wish to invest it in the bank at 6%
for 7 years, taking out an equal amount each year? How much would the bank let
us withdraw each year?
A =
P(A/P, i, n)
= $1116.47(Use A/P column, i% page, n year
row)
= $1116.47(0.1791)
= $ 199.96
This
is the same amount that we found on the previous problem, (within the decimals
of accuracy we are using), and we are in effect saying that $199.94 invested
for 7 years has the same value at the end of that 7 years, as it has at the
beginning of that 7 years, all in one lump sum.
Let's
check the accuracy of this statement:
Since
an annual investment of $199.94 for 7 years at 6% has an equivalent Present
value of $1116.47, and an equivalent Future value of $1678.76, is P equivalent
to F?
F =
P(F/P, i, n)
= $1116.47(Use F/P column, 6% page, 7 year
row)
= $1116.47(1.5036)
= $1678.72 which is the same as we found
earlier ($1678.76)
Finding F knowing A:
What
if we want to invest $199.96 for 7 years. How much will we be able to collect
at the end of 7 years?
F =
A(F/A, i%, n)
F =
$199.96(Use F/A column, 6% table, 7 year row)
= $199.96(8.3938)
= $1678.42 - same as before ($1678.76)
Finding P given A:
If
we wanted to determine how much money to put in a bank today at 6% so that we
could withdraw equal amounts of $199.96 for the next 7 years, starting at the
end of the first year:
P =
A(P/A, i, n)
= $199.96(Use P/A column on 6% table on 7
year row)
= $199.96(5.5824)
= $1116.15 (esentially the same as used
earlier)
Gradient cash flows:
Now
sometimes we have constantly increasing values of money as the years progress,
called a "gradient". The cash flow diagram must as before be started
from a commonly agreed on year, or no standard tables would be possible. The
agreement has been that gradient cash flows will look:
Could
we have agreed that the "G" started at year 1? Sure, we could have,
but we didn't. Could we have agreed that the "G" started at year
zero? Yep. But we didn't. So if you want to use our tables, you will have to
position your first non-zero value of "G" as having been deposited,
or withdrawn from the bank at the end of the second year. It will become
obvious why the "zero G" value was positioned at year 1 when we
combine Annual and Gradient cash flows.
Finding Present deposit to
yield a given Gradient:
Question:
How much money do we need to invest today, at 6%, to be able to withdraw $0 at
the end of year 1, $100 at the end of year 2, $200 at the end of year 3, ... ,
and $600 at the end of year 7?
P =
G(P/G, i%, n)
= $100(Use P/G column, 6% interest, 7 year
row)
= $100(15.4497)
= $1544.97
Note
also that:
P =
G(P/G, i, n)
=
G(P/A, i, n)(A/G, i, n)
= $100(5.5824)(2.7676)
= $1544.99 (Close enough)
Finding F given G:
Question:
How much money could we take out of the bank after 5 years if we were willing
to invest $100 today, $200 at the end of the first year, $300 at the end of the
second year, $400 at the end of the third year, $500 at the end of the fourth
year, and $600 at the end of the fifth year? Use 6% interest.
Note
that in this case the problem is not stated in the "standard" form
which we must use if we are to use the standard interest tables. The standard
tables are set up assuming that today you decide to make the investment. At the
end of the first year you invest 0G. At the end of the second year you invest
1G At the end of the third year you
invest 2g, etc. Thus in order to use the standard tables you will have to shift
time back as shown in the cash flow diagram shown above. Also note that the reference manual does not give F/G, so
you will have to combine F/A and A/G to get F/G:
F =
G(F/A, i, n)(A/G, i, n)
= $100(Use F/A column, 6% table, 7 yrs)(Use
A/G table, 6%, 7 yrs)
= $100(8.3938)(2.7676)
= $2323.07
Now,
let's check the equivalence of the previous two problems - i.e. is $1544.99
today the same as having $2323.07 at the end of 7 years, if the bank is paying
6% interest?
If
P = $1,544.99, what is F?
F =
P(F/P, i, n)
= $1544.99(Use F/P column, 6%, 7 year row)
= $1544.99(1.5036)
= $2323.05 OK
Find A given G:
Question:
If we were able to invest a Gradient = $100 at the end of each year for 7 years
at 6% interest (i.e. $0 at the end of year 1, $100 at the end of year 2, $200
at the end of year 3, ... , and $600 at the end of year 7, what Annual amount
or Annuity could we take out of the bank starting at the end of the first year
and continuing until the end of the seventh year?
A =
G(A/G, i, n)
= $100(Use A/G column, 6% interest, 7 year
row)
= $100(2.7676)
= $276.76
Find P given A and G:
Question:
How much money must you put in the bank TODAY, to be able to draw out $200 at
the end of the first year, $230 at the end of the second year, $260 at the end
of the next year, and so on, until taking out $380 at the end of the seventh
year?
P1
= A(P/A, i, n) = $200(P/A,6%,7 yrs) = $200(5.5824) = $1116.48
P2
= G{P/G, i, n) = $30(P/G,6%,7 yrs) = $30(15.4497) = $463.49
So
P = P1 + P2 = $1579.97
This
should demonstrate why the gradient STARTS ON THE 2nd YEAR!
Now
what if we were able to invest $600 today (note - this is not standard!), $550
at the end of the year 1, $500 at the end of year 2, and so on until finally
investing $250 at the end of year 7? How much would the money invested at 6% be
worth?
F1
= A(F/A,6%,8 years!!!!!) = $600(9.8975) = $5938.50
F2
= G(F/G,6%,8 years!!!!!) = G(F/A, 6%, 8 years)(A/G, 6%, 8 years)
= $50(9.8975)(3.1952) = $1581.22
Making
F = F1 + F2 = $4357.28
Compounding Non-Annually:
Now
what if you have a "nominal" interest rate of 6%, but if you will
bring your $6000 to my bank I will compound your interest every quarter (every
3 months) rather than only yearly. That means I am willing to pay you 6%/4 =
1.5% interest on your money every 3 months, which means you will make more
interest on your money because you will start getting "interest on
interest" sooner than at the end of each year. For example, at the end of
the first 3 months, I will put another $6000*0.015 = $90 in your account, which
will then earn interest for the next 9 months. Thus you now have $6090 in your
account. Then, after another 3 months, I will add to your account $6090*0.015 =
$91.35, for a total of $6181.35. At the end of the next 3 months I will add
$6181.35*0.015 = $92.72, for a total of $6274.07, and finally at the end of the
next 3 months I will add another $6274.07*0.015 = $94.11 for a total of
$6368.18. Compare this with the $6000(1+0.06) = $6360.00 you would have had had
I not compounded the interest quarterly. So you are $8.18 better off.
The
amount of interest earned if the bank is willing to compound more frequently
that yearly is given above. Thus we could have computed your $6000 account's
value using:
ieffective = (1+rate/m payment periods)m payment periods - 1
= (1 + 0.06/4 payment periods)4 - 1
= 0.06136
Thus
your account would be worth $6000(1+ ieffective) = $6000(1.06136) = $6368.18, which is what we
found before.
Also,
at the end of 7 years your account would be worth
$6000(1+ ieffective)7 = $6000(1+ 0.06136)7 = $6000(1.51719) =
$9103.12
Now
what if you could find a bank which would compound your interest bi-monthly
(every 2 months?)
Then
a $6000 deposit invested at a nominal rate of 6% compounded bi-monthly would be
worth (at the end of one year:
ieffective = (1+rate/m payment periods)m payment periods - 1
= (1 + 0.06/6 payment periods)6 - 1
= 0.06152
F =
Future Value = $6000(1+0.06152) = $6369.12
Well,
what if I am willing to compound your nominal interest rate of 6% bi-weekly
(every two weeks?)
F =
$6000(1+(0.06/26)26
- 1)) = $6370.58
And
finally, what if I am willing to compound your nominal interest rate
CONTINUOUSLY???
.
Then
the effective interest rate is given by ieffective = er - 1
For
us, ieffective
= er - 1 = 2.7180.06 - 1 = 0.0618365
and
your account value becomes $6000(1+0.0618365) = $6371.02
Mixed cash flows:
Question:
If you invest $600 today (!!!) and $600 more at the end of each year for a
total of 5 years, and then deposit $200 at the end of the next 4 years, how
much money will have accrued at the time of your last deposit at 6%?
F1
= A(F/A,6%,5 yr) = $600(5.6371) = $3382.26
F2
= F1(F/P,6%,4 yr) = $3382.26(1.2625) = $4270.10
F3
= A(F/A,6%,4 yr) = $200(4.3746) = $874.92
Thus
F = F2 + F3 = $5145.02
Continuous Compounding:
FIND:
The Future value of an investment given the Present value for 
Continuous
Compounding:
How
much would $20,000 invested today be worth at a nominal annual interest rate of
6% compounded continuously for 7 years? Note that in these equations the interest
rate is no longer labeled i - rather it is given the name r to show continuous
compounding.
F =
P(F/P, r%, n) = $20,000(ern ) = $20,000(e0.06*7 ) = $30,439.23
FIND:
The Present value needed to yield a Future value using Continuous Compounding:
How
much would have to be invested today to yield $30,439.23 under continuous
compounding for 7 years if the nominal annual interest rate is 6%?
P =
F(P/F, r%, n years) = $30,439.23(e-rn ) = $30,439.23(e-0.06*7 )
= $20,000
FIND:
The future value of a series of 7 uniform deposits of $200 each at the end of
years 1 through 7 (standard form), if the bank pays 6% annual nominal interest
rate compounded continuously.
F =
A(F/A, r%, n) = $200(ern - 1)/(er - 1)
= $200(e0.06*7 - 1)/(e0.06 - 1)
= $200(8.440988) = $1688.20
Book Value (All Methods of
Depreciation)
Book
value = Initial cost of the depreciable asset less the Sum of all depreciation
taken up to the present time.
Straight Line Depreciation
Dj = (C - Sn) / n
Where
Dj = depreciation taken in year j, C = Initial cost of the depreciable asset
used to make money, Sn = expected salvage value of the asset at the end of year
n, and n = expected life of the asset (years.)
So,
say you purchase a $30,000 machine to make income with this year. You think the
machine will be useful for 7 years before you will junk it. Also, at that time
you expect to be able to sell it for $2500.
Then
D1 = ($30,000-$2,500) / 7 = $3928.57
This
is the amount that you can depreciate from your taxes at the end of each year,
and also the amount of money you should take from your profits to buy a new
$30,000 machine when this one wears out. (Perhaps a little less because of the
interest it makes.) Thus, the resulting BOOK VALUE at the end of each year will
be:
|
End of Year |
Depreciation |
Book Value |
|
0 |
0 |
$ 30,000.00 |
|
1 |
$ 3,928.57 |
$ 26,071.43 |
|
2 |
$ 3,928.57 |
$ 22.142.86 |
|
3 |
$ 3,928.57 |
$ 18,214.29 |
|
4 |
$ 3,928.57 |
$ 14,285.72 |
|
5 |
$ 3,928.57 |
$ 10,357.15 |
|
6 |
$ 3,928.57 |
$ 6,428.58 |
|
7 |
$ 3,928.57 |
$ 2,500.01 (salvage) |
Accelerated Cost Recovery
System (ACRS)
Dj
= (a factor from the table) Original Total Cost of the asset
Unlike
the straight line method, you do not need to estimate salvage. You get to
depreciate the entire cost. The property is classed as being of a certain class
life, either 3, 5, 7, 10, 15, and 20 years. The reference manual only gives
tables for 3 through 10 years. The
property class will have to be stated in the problem, since the guidelines are
not listed in the manual.
Say
you purchase a $36,000 machine to make income with, and that the life has been
classed by the government for income tax purposes as being in the 5 year life
class. What is the depreciation you can take during each year, and what is the
book value at the end of each year?
|
Year |
Depreciation |
Book Value |
|
0 |
|
$ 36,000 |
|
1 |
0.20 * $36,000 = $ 7,200 |
$ 28,800 |
|
2 |
0.32 * $36,000 = $ 11,250 |
$ 17,280 |
|
3 |
0.192 * $36,000 = $ 6,912 |
$ 10,368 |
|
4 |
0.115 * $36,000 = $4,140 |
$ 6,228 |
|
5 |
0.115 * $36,000 = $ 4,140 |
$ 2,088 |
|
6 |
0.058 * $36,000 = $2,088 |
$ 0 |
|
|
Total Depreciation Above =
$ 36,000 |
|
The
method uses the $200% declining balance method, permitting you to depreciate
the asset much more quickly during its early life, but permitting you only half
of the first year's depreciation during the first year that you purchased the
asset, regardless of when during the year you purchased it. Thus if you
purchased the asset in December, you still get one-half year's depreciation the
first year. Also, if you purchase it in February, you unfortunately only get
1/2 of a year's depreciation that year. Further, at some time later in the life
of the asset, it would be to your advantage to switch from the 200% declining
balance method to straight line method, which is permitted by the IRS, and the
numbers listed in the ACRS table do this automatically.
Capitalized Costs
Project
I) How much money should we ask the taxpayers to give us if we need to repair a
dam for the next million years, and the yearly repair cost will be $120,000 a
year, and assuming that the money gleaned from the taxpayers will be invested
at 6% per year?
So,
P = A / i = $120,000/year / 0.06/year = $2,000,000
Why
it works:
If
you invest $2,000,000 at 6% compounded yearly for one year, at the end of that
year you will have your original $2,000,000 plus $2,000,000 * 0.06 = $120,000.
Thus you can take the $120,000 interest you made out of the bank to make the
repairs at the end of the year, and still have $2,000,000 remaining in the bank
to earn the next year's needed $120,000, forever and ever.
Project
II) Now what if we need $10,000 for repairs each month? At 6% nominal interest
rate, compounded monthly, how much do we need to extract from the taxpayers?
P =
A / i = $10,000/month / (0.06/year)(1 year/12 months)
= $10,000/month / 0.005/month =
$2,000,000 Total
Why
it works: If you invest $2,000,000 at a nominal interest rate of 6% per year
and can find a bank kind enough to compound your interest monthly, then you are
actually getting 0.06/12 = 0.005/month, which is a lot better than 6% a year,
and you will thus earn 0.005/month * $2,000,000 = $100,000/month to make your
repairs.
Note
that the Project I bank was only willing to pay us 6% compounded yearly, so we
had to wait until the end of the year to take out $120,000, whereas the Project
II bank was willing to compound monthly, permitting us to take out $10,000 a
month - a much better deal for us. As a matter of fact, the second bank is
giving us an effective interest rate of:
i
effective = (1 + r/m)m
- 1 = (1 + 0.06/12)12
- 1
= 0.0616778 = 6.16778% per year
Well
heck. If the second bank is a better deal, let's take Project I to him instead.
How much will we have to put in his bank to take out $120,000 at the end of
each year, forever?
P =
$120,000/year / 0.0616778 effective/year = $1,945,594.70
which
saves us $2,000,000 - $1,945,594,70 = $54,405.31 on our Project I bond issue.
But
wait! Bank 3 says they will give us 6% nominal interest rate compounded weekly:
i
effective = (1 + 0.06/52)52
- 1 = 6.17998 %, such that
P =
$120,000/year / 0.0617998 = $1,941,753.20
And
if we can find a bank which pays 6% nominal interest rate compounded
continuously:
i
effective = er - 1 = e0.06 - 1 = 0.0618365, such that
P =
$120,000/year / 0.0618365 = $1,940,000
Project
III) A&M has been told that they may win their bid to house the Bush
Library, if they can guarantee that they can fund the upkeep and repairs in
perpetuity (i.e. forever.) The library is expected to cost $2000 every two
months for light bulbs, plus $100,000 every 6 months for minor repairs, plus an
additional $5,000,000 every 5 years for a complete refurbishing of the
facility. How much must be deposited today to guarantee the funding of these expenses?
Assume a nominal interest rate of 6%, compounded monthly.
Now
the easiest way to work problems like this is to treat each annuity value as a
future value needed somewhere down the road and change it into an equivalent
annuity based on the compounding period of your loan, or rate. For example, how
much money would you have to invest each month (use the compounding period) to
have $100,000 at the end of 12 months? Easily found by taking the future value
needed ($100,000) and transforming it into an equivalent monthly annuity:
i
per payment period = 0.06/12 = 0.005%
A2
= F2(A/F,i per payment period, n)
= $100,000 (Use A/F column, 0.005 table,
12 payments)
= $100,000*0.0811 = $ 8110
This says that if you will invest $8110
dollars each month as an annuity (you must follow the rules - first deposit
goes in at the end of the first month,) for a total of 12 months, you will have
$100,000 at the end of the 12 month period. You can then withdraw this to make
your necessary repairs, and then continue depositing $8110 each month forever,
and you will always be able to withdraw $100,000 at the end of every year.
Now let's change the a3 into a monthly (the
bank's compounding period) annuity. Setting the value needed ($2000 every two
months) to the Future value needed, and realizing that we are paid 0.005% per
period, and that we have two periods to get the money:
A3
= a3(A/F,i per payment period, n)
= $2,000 (Use A/F column, 0.005 table, 2
payments)
= $2,000*0.4988 = $ 997.60
This says that if you will invest $997.60
dollars each month as an annuity (you must follow the rules - first deposit
goes in at the end of the first month,) for a total of 2 months, you will have
$2,000 at the end of the 2 month period. You can then withdraw this to make
your necessary repairs, and then continue depositing $997.60 each month
forever, and you will always be able to withdraw $2,000 at the end of every 2
month period.
Now let's change the A1 into a monthly (the
bank's compounding period) annuity. Setting the value needed ($5,000,000 every
60 months) to the Future value needed, and realizing that we are paid 0.005%
per period, and that we have 60 periods to get the money:
A1
= F3(A/F,i per payment period, n)
= $5,000,000 (Use A/F column, 0.005
table, 60 payments)
= $5,000,000*0.0143 = $ 71,500
This says that if you will invest $71,500
dollars each month as an annuity (you must follow the rules - first deposit
goes in at the end of the first month,) for a total of 60 months, you will have
$5,000,000 at the end of the 60 month period. You can then withdraw this to
make your necessary repairs, and then continue depositing $71,500 each month
forever, and you will always be able to withdraw $5,000,000 at the end of every
60 month period.
Thus,
in total, you have been asked to scrape up a total of $80,587.60 each and every
month forever, to fund the library. If instead I wish to make a single deposit
today, I can find this by:
P =
A / i = 80,587.60/month / 0.005/month = $ 16,117,520
Bond Value
Bond
value equals the present worth of the payments the purchaser or holder of the
bond receives during the life of the bond. Bonds are sold for some number of
years, n, during which the seller normally pays some monthly or yearly amount
to the purchaser. Then, at the end of n years, or months, the seller repays the
original bond cost back to the buyer. The Bond Yield equals the computed
interest rate of the bond value when compared with the bond cost.
The
owner of a bond is normally paid back for his purchase in two ways. First they
are paid a series of periodic payments monthly, or quarterly, or semi-annually,
or annually, until the bond is retired. These payments are usually r (the bond
rate = nominal interest rate) times the cost of the bond. Thus if you buy a 6%
bond for $1000 and are paid quarterly, you would receive $1000*(0.06/4) =
$15/quarter every quarter until the bond became due. When the bond is retired
you would then be paid back the cost of the bond. Thus the Bond Value of the
bond (to you) will be the Present value of the monthly (or quarterly, or
yearly) annuity payments, plus the Present value of the final repayment F, both
based on the desired rate of interest you require.
For
example, determine how much you would be willing to pay (or the "Bond
Value") of a 10 year bond sold for $1000 and paying a nominal interest
rate of 6%. Assume that the bond pays semi-annually. Also assume that you will
not buy the bond as an investment unless it actually yields 8%.
The
bond's periodic payments will be $1000 (0.06 / 2) = $30 every 6 months. Thus
the value of this stream of income to you, since you want 8% interest is:
P =
A (P/A, i desired, n payments)
= $30 (Use P/A column, 4% per payment
period, 20 payments)
= $30 (13.5903) = $ 407.71
Now
the final repayment of the value of the bond will be $1000, and the present
value of this, since you want 8% on your money, will be:
P =
F ( P/F, i desired, n payments)
= $ 1000 (Use P/F column, 4% per payment
period, 20 payments)
= $1000 (0.4564) = $456.40
Thus
to actually return 8% interest you would have to pay no more than $407.71 + $
456.40 = $ 864.11 for the bond, which would be its face value to you. Note that
the seller of the bond might suggest that the face value of his bond should be
based on his 6% nominal interest rate. In that case his computations for what
he would call the face value of the bond would substitute 3% per payment period
for the 4% values you used above.
The Bond Yield equals the computed interest rate of the bond value when
compared with the bond cost. Thus if you purchased the bond discussed above for
$864.11, the Bond Yield would indeed be 8%. If, however, someone wanted to cash
in that bond bad enough, and asked you if you would buy it for $656 cash on the
barrel head, you would probably say, well, yes, I guess I might, possibly. Then
to calculate the actual yield of your new bond, you have:
P =
Value of bond returned in 20 periods (P/F, i yield, 20) +
Periodic payments A (P/A, i yield,
20)
where
the Value of the bond to be returned = $1000, and
the Periodic payments were $30 each
payment period.
Thus P = $1000(P/F, i yield, 20) + $30(P/A, i
yield, 20)
Now
in the real world this might be rather messy, and you would have to resort to a
trial and error method to solve for i yield. However, on the FE exam it will be
a snap, since they will give you 5 possible answers, each of which you quickly
plug in and see if that's the answer. For example, say they give you the 5 choices
a) 4%, b) 6%, c) 8%, d) 12%, e) 16% nominal rate per year.
Well,
4% and 6% and 8% sure aren't correct, since you know 8% was the yield if you
paid the full $864.11. So, let's try 12%:
P =
$1000(P/F, i yield, 20) + $30(P/A, i yield, 20)
$
656 ?=? $1000(P/F, 0.06, 20) + $30(P/A, 0.06, 20)
$
656 ?=? $1000(0.3118) + $30(11.4699)
$
656 ?=? $311.80 + $344.10 = $ 655.90
CLOSE ENOUGH!
Rate of Return
The
Minimum Attractive Rate of Return is the highest rate of return that you can
obtain in alternative investments, and hence the minimum rate that you would
accept. The actual rate of return you receive on an investment would then
hopefully be greater than this MARR. The acutal rate of return could be
computer by setting the Present Worth Benefits = Present Worth Cost, or the
Uniform Annual Benefits = Uniform Annual Costs, or the Future Worth Benefits =
Future Worth Costs. All would be equivalent. By setting the benefits = costs,
you can compute an interest rate which would actually cause them to be equal,
thus telling you what interest rate you will be receiving, and whether that
rate exceeds your MARR or not. This rate is called the internal rate of return.
Example:
Find the internal rate of return for the following cash flow:
|
End of Year |
Receipt/Disbursement |
|
0 |
-$1000 |
|
1 |
$250.50 |
|
2 |
$250.50 |
|
3 |
$250.50 |
|
4 |
$250.50 |
|
5 |
$250.50 |
Answers:
(a) 2% (b) 4% (c) 6%
(d) 8% (e) 10%
Set
PWB = PWC
$250.50
(P/A, IRR%, 5) = $1000
(P/A,
IRR, 5) = $1000/$250.50 = 3.992
(Look
in P/A column, look in 6% table, look on 5 year row) = 3.992?
Nope,
too big = 4.2124. Try the 8% table:
(Look
in P/A column, look in 8% table, look on 5 year row) = 3.992?
Yep,
that's it, or close enough. Answer is d.
Note
I started in the middle of the possible answers, since that way I won't have to
try anymore than two others if (c) is not correct.
Example:
Determine the internal rate of return on a new piece of equipment that will
cost $25,000, and will increase productivity by $5050 each year for 5 years.
Further, the equipment will have a resale salvage value of $5000 at the end of
the 5 years. Answers: (a) 2% (b) 4%
(c) 6% (d) 8% (e) 10%
In
this case there are two terms which contain the IRR%, such that the IRR cannot
be directly found. However, since they give you 5 choices, just try them on
until you get one that fits. Start with the middle one:
Set
Present Benefit = Present Cost
A (
P/A, IRR%, n) + F ( P/F, IRR%, n) = $25,000
$5050
( Look in P/A, IRR%, 5) + $5000 ( Use P/F, IRR%, 5) = $25,000
Try
6%:
$5050
( Look in P/A, 6%, 5 yr) + $5000 ( Use P/F, 6%, 5 yr) = $25,000?
$5050*4.2124
= $5000*0.7473 = $25,009 Close enough.
Answer
is (c)
Break-Even Analysis
The
break-even point is found by varying one of the decision variables in a study,
until it causes two alternative investments to become equal. For example, you
might have two pumps available for purchase, each costing different amounts for
initial cost and yearly costs, wherein one is cheaper if you only need to pump
small amounts, and the other is cheaper if you have to pump large quantities.
Then you could find a break-even point by varying the amount to be pumped until
the costs of the two pumps were equal.
Example:
Assume that a pump is needed to remove water from excavations. Further, assume
that Pump A initially costs $1,800, costs $1.10 per hour to operate, requires
$360/year in maintenance and will have a salvage value of $600 at the end of
its 4 year life. Pump B has an initial
cost of $550, costs $2.35 per hour to operate, requires no maintenance, and has
no salvage value at the end of 4 years. Use an interest rate of 6%.
Then,
the annual equivalent fixed costs will be:
Initial
Cost:
Pump A: A = $1,800 (A/P, 6%, 4 years) =
$1,800*0.2886 = $519.48
Pump B: A =$ 550 (A/P, 6%, 4 years) = $
550*0.2886 = $158.73/year
Salvage
Value:
Pump A: A = $600 (A/F, 6%, 4 years) =
-$600*0.2286 = -$137.16/year
Pump B: $0/year
Maintenance
Cost:
Pump A: $360/year
Pump B: $0
Variable
Yearly Costs:
Pump A: $1.10/hour * X hours/year
Pump B: $2.35/hour * X hours/year
Then
setting the costs of the two alternatives equal to each other, we can solve for
the number of hours of pumping which divides when each pump should be selected:
Pump A annual equivalent cost = Pump B
annual equivalent cost
$519.48-$137.16+$360 + $1.10X =
$158.73-$0+$0 + $2.3X
X = 486.3 hours per year.
Thus
if the pumping requirements of the next job are less than 486.3 hours, Pump B
would be the better selection. However, if the pump will operate more than
486.3 hours per year, Pump A would be better.
A
second use of break-even analysis is to determine at what level a manufacturing
plant must operate in order to break even, considering its fixed and variable
costs, and profit.
Example:
A company is trying to determine at what level they must operate their plant
just to break even. Their fixed costs are $7500/month and their variable costs
are $3.00 per unit. Revenue is expected to be $10/unit.
As
shown above, the Fixed costs are plotted as $7500/month, and on top of this the
Variable costs of $ 3.00/unit are added. Then the income is plotted, started
from zero since if there is no production the only cost will be the fixed
monthly cost.
Solving
for X, the number of units required to break even:
Costs
= Benefits
$7500
+ $3.00X = $ 10.00X
X =
1071 units
Thus
we must make at least 1071 units, just to break even.
Payback Period
The
payback period is commonly defined as the time required to recover an initial
investment from the net cash flow from that investment. It can include the
effect of interest, or often simply ignores the interest. Thus, you might say
that an investment of $10,000, which pays back $5000/year at the end of the
next two years has a payback period of 2 years.
Example:
Find the payback period for the following investment, assuming an interest rate
of 6%:
|
End of Year |
Investment A |
|
0 |
- $10,000 |
|
1 |
$ 0 |
|
2 |
$ 2,000 |
|
3 |
$ 4,000 |
|
4 |
$ 4,000 |
|
5 |
$ 2,270 |
|
6 |
$ 4,000 |
Thus
at the end of the 0th year, we are $ 10,000 in the hole.
For
year 1, we add an additional present value of $0 * (P/F, 6%,1) = $0
For
the end of year 2, we would add $2000 * (P/F, 6%,2) = $1780
At
the end of year 3, we would add $4000 * (P/F, 6%,3) = $3358
At
the end of year 4, we add $4000 * (P/F, 6%,4) = $3168
At
the end of year 5, add $2270 * (P/F, 6%,5) = $1696
At
the end of year 6, add $4000 * (P/F, 6%,6) = $2820
Thus,
adding the additional present values year by year, we see that after the 5th
year, we would have our money back, including the effect of interest and the fact
that someone else was enjoying our money instead of us.
If
the quiz problem does not give an interest rate, then they are using the more
common definition of payback period. For the problem above, excluding interest,
the payback period would be the end of the 4th year.
Inflation
The
reference manual uses the equation d =
i + f + (i*f) to handle inflation, where
d = the stated interest rate paid by the
bank, or paid on the bond, or stated by the lender. This is the
"combined" interest rate - i.e. for the lender it combines the real
rate of return he desires, with the inflation rate. It is also known as the
"market" rate, since this is usually what the market charges for
money.
f = the inflation rate.
i = the "real" interest rate -
i.e. the "real" increase in purchasing value the lender is getting by
loaning his money to the borrower.
Example:
What is the actual present value (to you) of a bond which pays $6000/year at
the end of each year for 10 years, if the purchase price of $20,000 would be
paid back at the end of the 10th year. Assume that you could get 6%
interest on other investments, and that this is your MARR (minimum acceptable
rate of return.)
P =
A(P/A,IRR%,n) + F(P/F,IRR%,n)
= $6000(Use P/A column, 6%, 10 years) +
$20,000(P/F,6%,10 yr)
= $6000(7.3601) + $20,000(0.5584)
= $ 55,328.60
Thus
you should not be willing to pay more than this for the bond. This is the
"value" of the bond (to you.)
Now
what if you think that the inflation rate for the next 10 years will be 4%? How
will this influence the value of the bond? Well, first you will realize that
the dollars with which you will be repaid will not have the same purchasing
power, or value. Since you will still want a "true" rate of return of
6%, you will not be willing to pay as much for the bond. Its value will
decrease. What is its value (to you?)
(a)
$44,030 (b) $44,580 (c) $55,330 (d) $58,620 (d) $61,424
Using
d combined = i + f + (i*f) = 0.06 + 0.04 + (0.06*0.04) = 0.1024
Now
to get a quick and dirty answer, let's assume this came out d = 10%. This would
allow us to use the tables (there aren't any 10.24% tables.)
P =
A(P/A,i,n) + F(P/F,i,n)
= $6000(Use P/A column, 10%,10) +
$20,000(P/F,10%,10)
= $6000(6.1446) + $20,000(0.3855)
= $44,578
Now
this looks like answer (b), but that must be a trick, since we aren't really
using the correct interest rate. So my guess would be that answer (a) is the
correct answer. But, let's check it.
P =
A(P/A,i,n) + F(P/F,i,n)
= A((1+i)n-1)/(i(1+i)n + F(1+i)-n
=
$6000((1+0.1024)10-1)/(0.1024(1+0.1024)10+$20,000(1+0.1024)-10
= $6000(6.0817) + $20,000(0.37723)
= $44,034.83 Yep. Answer is (a).
Example:
A bank lends you $10,000 today, to be repaid in a lump sum at the end of 10
years at a combined, market, stated interest rate of 10% compounded annually.
Make sure you understand that this is the combination of what the bank hopes to
gain, plus the influence of inflation. If the rate of inflation is 8%, what is
the bank's true rate of return, and how much money will they really make in 10
years?
Remember
that the reference manual calls "d" the "combined" rate of
return, which means the interest rate found by adding both the "really
made me some money = i" interest rate to the inflation rate = f. The
"i" in the equation is the "real" interest rate. For
example, if you made me a loan at d = 6% interest, and the inflation rate was f
= 6%, you would actually not make any extra spendable money over the life of
the loan. So what you must do as a lender is ask for a higher rate on the loan
to give yourself some "real" rate of return "i." Be very
sure to understand the terminology used here. It is very likely different with
what you are accustomed.
Now
let's see what the bank's "real" rate of return will be:
Since
d = i + f + if, we can solve for i:
i = (d-f)/(1+f) = (0.10 - 0.08) / (1 +
0.08) = 0.0185 = 1.85%
So
the bank's real rate of return will be 1.85%. How much money they actually made
can be found by:
F =
P(F/P,ireal,n) = $10,000(Use F/P equation, 1.85%, 10 years)
= $10,000(1+i real)n = $12,011.86
So
the bank will really make an additional $2,011.86
Benefit-Cost Analysis
A
benefit-cost analysis attempts to determine if the Benefits > Costs, or if
Benefits/Costs > 1.0. This is normally done by computing either the Present
Worth of the Benefits and comparing them with the Present Worth of the Costs,
or by comparing their Annual Worths.
Example:
A project is being considered wherein a small airfield will be built. Land for
the project will cost $350,000, construction costs will be $800,000, and annual
maintenance will be $25,000/year. Annual benefits expected from this project =
$250,000 per year. Determine the Benefit-Cost ratio assuming a rate of 6% and a
life of 20 years.
Present
Worth of Benefits = A(P/A,i.n) = $250,000(P/A,6%,20)
= $250,000(11.4699) = $2,867,475
Present
Worth of Costs = $350,000 + $800,000 + A(P/A,i,n)
= $350,000 + $800,000 + $25,000(P/A,6%,20)
= $1,436,475
Then
B/C = $2,867,475/$1,436,475 > 1.0 and project should be considered.