Chapter 6
Economic Analysis
Why do we care about money?
What is our role as engineers as caretakers of the public
trust?
What is interest, inflation, social discount?
Time Value of Money
Money has a value associated with time. The simple notion
of interest rates suggests that $1.00 today is worth more than
$1.00 in the future because of our ability to invest it profitably.
To illustrate this we will define the following variables:
A - an annual payment
P - the present value of a payment
F - the future value of a payment
i - interest rate
n - number of years
P
F
A
0 1 2 3 4 5 6 7 8 9 10
The relationship between these values are:
o Present Value of a Future Payment
(Single Payment Present Worth Factor)
[P/F,i,n] : P=F/(1+i)n
You will receive $1000 seven years from today. Assuming an
interest rate of 12%, what is its present value?
P=F/(1+i)n=1000/(1.12)7=$452
o Future Value of a Present Payment
(Single Payment Compound Amount Factor)
[F/P,i,n] : F=P(1+i)n
You invest $1000 for seven years at 12% interest. What is its
value at maturity?
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F=P(1+i)n=1000(1.12)7=$2211
o Present Value of a Uniform Series of Payments
(Series Present Worth Factor)
[P/A,i,n]: P=A((1+i)n-1)/[i(1+i)n]
You wish to make annual payments of $1000 per year for seven
years. What amount must be invested if interest is 12%?
P=1000((1.12)7-1)/(.12(1.12)7)=$4563
o Future Value of a Uniform Series of Payments
(Series Compound Amount Factor)
[F/A,i,n] : F=A((1+i)n-1)/i
You invest $1000 per year for seven years at 12% interest. What
is its total value at the end of that time?
F=1000((1.12)7-1)/.12=$10089
In addition to these simple concepts is the notion of a gradient
where G is a payment made in year 1, 2G made in year 2, etc.
o Present Value of a Gradient Series
[P/G,i,n] : P=G[(1+i)n+1-(1+ni+i)]/[i2(1+i)n]
A mechanical device will cost $20,000 when purchased.
Maintainance will cost $1000 each year. The device will
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generate revenues of $5000 each year for 5 years after which the
salvage value is expected to be $7000. Draw and simplify the
cash flow diagram.
Discount Factors for Discrete Compounding
factor name
convert
symbol
formula
single payment
compound amount
P to F
(F/P,i%,n)
P(1+i)n
present worth
F to P
(P/F,i%,n)
F(1+i)-n
uniform series
F to A
(A/F,i%,n)
(F)i/[(1+i)n -1]
Sinking fund
capital recovery
P to A
(A/P,i%,n)
compound amount
A to F
(F/A,i%,n)
equal series
A to P
(P/A,i%,n) A((1+i)n-1)/(i(1+i)n)
present worth
uniform gradient
G to P
(P/G,i%,n) G[(1+i)n+1-(1+ni+i)]/[i2(1+i)n]
(P)i(1+i)n/[(1+i)n -1]
(A)((1+i)n -1)/i
Example 1
Investment A costs $10,000 today and pays back $11,500 two
years from now. Investment B costs $8000 today and pays back
$4500 each year for two years. If an interest rate of 5% is used,
which alternative is superior?
P(A) = -10,000 + 11,500(P/F,5%,2) = 431
P(B) = -8000 + 4500(P/A,5%,2) = 367
Alternative A is superior and should be chosen.
Example 2.
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How much should you put into a 10% savings account in order to
have $10,000 in 5 years? This problem could also be stated:
What is the equivalent present worth of $10,000 5 years from
now if money is worth 10%?
P = F/(1 +i)n = 10,000/(1 + 0.10)5 = 6209
The factor 0.6209 would usually be obtained from the tables.
Example 3
Maintenance costs for a machine are $250 each year. What is the
present worth of these maintenance costs over a 12 year period if
the interest rate is 8%?
Notice that : (P/A,8%,12) = (P/F,8%,1) + (P/F,8%,2) + ...
+(P/F,8%,12)
Then, P = A(P/A,i%,n) = -250(7.5361) = -1884
Example 4
Maintenance on an old machine is $100 this year but is expected
to increase by $25 each year thereafter. What is the present
worth of 5 years of maintenance? Use an interest rate of 10%.
In this problem, the cash flow must be broken down into parts.
Notice that the 5-year gradient factor is used even though there
are only 4 non-zero gradient cash flows.
P = A(P/A,10%,5) + G(P/G,10%,5)
= -100(3.7908) - 25(6.8618) = -551
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Evaluating Alternative Projects
The five most common means of evaluating alternatives
projects are:
1) Net Present Worth:
discount each project to same year
use identical discount rate
base each calculation on same time horizon
2) Capitalized Cost Method:
assumes present worth of infinitely lived project
use initial costs + annual costs/i
3) Annual Cost Method
similar to present cost method but translate values
to annual rather than present value
4) Benefit Cost Ratio:
ratio of present value of benefits and cost
problems with what are costs and what are negative
benefits
5) Internal Rate of Return:
calculate the interest rate at which the costs are equal
to the benefits
solve by trial and error
It is an important restriction that the use of present
worth calculations be on problems with similar analysis
periods. For instance, comparing a project with a life
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of 5 years with one of ten years is improper. In
situations like this you should use another technique,
such as annual cost, or reconstruct the problem so that
it has a similar period of analysis.
Example 5
Which of the following alternatives is superior over a 30 year
period if the interest rate is 7%?
A
type
brick
life
30 years
cost
$1800
maintenance $5/year
B
wood
10 years
$450
$20/year
EUAC(A) = 1800(A/P,7%,30) + 5 = 150
EUAC(B) = 450(A/P,7%,10) + 20 = 84
Alternative B is superior since its annual cost of operation is the
lowest. It is assumed that three wood facilities, each with a life
of 10 years and a cost of $450, will be built to span the 30 year
period.
Second Approach - Find Present value
PVc = 1800 + 5 (1.0730 - 1)
(.07)(1.0730)
PVc = 450 + 20(1.0710 - 1)
(0.7 )(1.07)10
= 1800 + 62
= 450 + 140.5 = 590.5
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= $1862
But what about year 11 and year 21?
PVc = 590.5 + 590.5 + 590.5
(1.07)10 (1.07)20
= 590.5 + 300.18 + 152.60
= $1043.28
Example 6
What is the return on invested capital if $1000 is invested now
with $500 being returned in year 4 and $1000 being returned in
year 8?
First, set up the problem as present worth calculation.
P = -1000 + 500(P/F,i%,4) + 1000(P/F,i%,8)
Arbitrarily select i = 5%. The present worth is then found to be
$88.15. Next take a higher value of i to reduce the present worth.
If i = 10%, the present worth is -$192. The ROR is found from
simple interpolation to be approximately 6.6%.
Rate and Period Changes
Nominal and Effective Interest
Nominal interest is the annual interest rate without considering
the affect of any compounding
Effective interest is the annual interest rate taking into account
the affect of any compounding
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Effective interest rate = (1+i)m - 1
where:
i is the interest rate for specified period
m is the number of compounding periods
or
i =(1+r/k)k -1
r = nominal rate
i = effective rate
k = number of compounding periods
If a bank advertizes a rate of 3% every three months, the
nominal interest rate per year is 12%, the effective interest rate is
what?
Effective Annual rate (1.03)4-1 = 12.55%
A bank charges 1.5% per month interest on the unpaid balance
for items that are charged. What is the nominal annual interest
rate? The effective annual interest rate?
Solution: Nominal annual rate = 1.5 x 12 = 18%
Effective annual rate = (1.015)12 - 1 = 19.56%
Capitalized Cost
o In special situations, the analysis period is infinite. The
present worth of the cost is called capitalized cost. ( Some
examples might be certain federal projects. ) In these cases the
fundamental relationship becomes:
9
A=Pi
Example 7
In their will a person wishes to establish a perpetual
trust to provide for the maintenance of a small local park. If
annual maintenance is $750 per year and the trust account can
earn 5%, how much money must be set aside?
P = A/i = 750/.05 = $15000
Example 8
A savings and loan offers 5.25% compounded daily. What is the
annual effective rate?
method 1: r = 0.0525, k = 365
i=( 1.0 +0.0525/365)365 - 1 = 0.0539
method 2: Assume daily compounding is the same as
continuous compounding.
i = (F/P) - 1
= e0.0525 - 1 = 0.0539
Depreciation
Depreciation involves the notion that in the process of producing
a product or goods the capital value of equipment or other
investments is consumed and must be replaced to maintain
production.
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Common Methods of Depreciation
Definitions:
Let
P - present value
Sn - salvage value of item in year n
j - year of concern
n - years of depreciation
1) Straight Line - an equal amount is subtracted for the value of
the equipment each year
Dj - Annual depreciation charge in year j
BVj - Book Value in year j
Dj = (P-Sn)/n
BVj = P-j(C-Sn)/n = C-jDj
2) Sum of Yearly Digits (SOYD)
Dj = (n-j+1)(C-Sn)/T
T=n(n+1)/2
3) Declining Balance - larger depreciation in early years, smaller
in later years book value is assumed to decrease at given
percentage
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R - set by law, value such as 2/n
Dj = C(1-R)j-1R
BVj = C(1-R)j
Example 9
An asset is purchased for $9000. Its estimated economic life is
10 years, after which it will be sold for $1000. Find the
depreciation in the first three years using SL, DDB, and SOYD.
SL: D = (9000-1000)
10
DDB:
= 800 each year
D1 = 2(9000) = 1800 in year 1
10
D2 = 2(9000-1800)
10
= 1440 in year 2
D3 = 2(9000-3240)
10
= 1152 in year 3
SOYD: T = 1/2(10)(11) = 55
D1 = (10/55)(9000 - 1000) = 1455 in year 1
D2 = (9/55)(8000) =
1309 in year 2
D3 = (8/55)(8000) =
1164 in year 3
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Example 10
For the asset described above, calculate the book value during the
first three years if SOYD depreciation is used.
The book value at the beginning of year 1 is $9000. Then,
BV1 = 9000 - 1455 = 7545
BV2 = 7545 - 1309 = 6236
BV3 = 6236 - 1164 = 5072
Example 11
For the asset described in example 2.9, calculate the after-tax
depreciation recovery (present worth of all depreciation over the
economic life) with SL and SOYD depreciation methods. Use
6% interest with 48% income taxes.
SL: D.R. = 0.48(800)(P/A,6%,10) = 2826
SOYD: The depreciation series can be thought of as a
constant 1,454 term with a negative 145 gradient.
D.R. = 0.48(1454)(P/A,6%,10)
- 0.48(145)(P/G,6%,10)
= 3076
Example 12
What is the after-tax present worth of the asset described in
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example 2.8 if SL, SOYD, and DDB depreciation methods are
used?
The after-tax present worth, neglecting depreciation, was
previously found to be -3766.
Using SL, the depreciation recovery is
D.R. = (0.53)(10,000-500)(P/A,9%,8)
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= 3483
Using SOYD, the depreciation recovery is calculated as follows:
T = 1/2(8)(9) = 36
Depreciation base = (10,000 - 500) = 9500
D1
= (8/36)9500) = 2111
G
= gradient = (1/36()9500)
= 264
D.R. = (0.53)[2111(P/A,9%,8) - 264(P/G,9%,8)]
= 3829
Using DDB, the depreciation recovery is calculated as follows:
d = 2/8 = 0.25
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z = 1.09/0.75 = 1.453
(P/EG) =((1.453)8 - 1 )/ (1.453)8(0.453) = 2.096
D.R.
= (0.53)[0.25(10,000/.75)2.096 +
(.758)(10,000)(P/F,9%,8)
= 3969
The after-tax present worths including depreciation recovery are:
SL:
Pt = -3766 + 3483 = -283
SOYD:
Pt = -3766 + 3829 = 63
DDB:
Pt = -3766 + 3969 = 203
Basic Tax Consideration
f%- Federal income tax on profits
s%- state income tax on profit
t%- composite tax rate
assume the feds recognize state taxes as expenses
t = s + f - sf
Example 13
15
A corporation which pays 53% of its revenue in income taxes
invests $10,000 in a project which will result in $3000 annual
revenue for 8 years. If the annual expenses are $700, salvage
after 8 years is $500, and 9% interest is used, what is the aftertax present worth? Disregard depreciation.
Pt = -10,000 + 3000(P/A,9%,8)(1 - 0.53) 700(P/A,9%,8)(1 - 0.53) + 500(P/F,9%,8)
= -3766
It is interesting that the alternative evaluated in example 2.8 is
undesirable if income taxes are considered but is desirable if
income taxes are omitted.
ADVANCED INCOME TAX CONSIDERATIONS
A. Investment Tax Credit
Investment tax credit is a credit against income tax
Based on initial investment
Refer to current tax law as this is changing
B. Gain on Sale of Depreciated Asset
If asset is sold for more than current book value, the
difference between selling price and book value is taxable
income.
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Taxed at capital gains rate
C. Capital Gains and Losses
Regular gain if item sold has been kept less than one year
Capital gain if item sold is kept longer than one year
Capital gains are taxed at taxpayer's usual rate, but 60% of
the gain is excluded
Reminder:
Income Taxes
Income taxes represent another of the various kinds of
disbursements encountered in an economic analysis. The starting
point in an after-tax computation is the before tax cash flow.
Generally, the before-tax cash flow contains three types of
entries.
1) Disbursements of money to purchase capital assets. These
expenditures create no direct tax consequences for they are the
exchange of one asset (cash) for another (capital equipment).
2) Period receipts and/or disbursements representing operating
income and/or expenses. These increase or decrease the year by
year tax liability of the firm.
3) Receipt of money from the sale of capital assests, usually in
the form of salvage value when equipment is removed. The tax
consequence depends on the relationship between the book cost
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(cost-depreciation taken) of the asset and its salvage value.
SituationTax
Salvage Value > Book Value
Salvage Value = Book Value
Salvage Value < Book Value
Consequence
Capital gain on difference
No Tax Consequence
Capital loss on difference
After the before-tax cash flow, the next step is to compute the
depreciation schedule for any capital assets. Taxable income is
the taxable component of the before-tax cash flow minus the
depreciation.
The income tax is the taxable income times the appropriate tax
rate. Finally, the after-tax cash flow is the before-tax cash flow
adjusted for income taxes.
Suggested arrangement:
Before Tax Taxable
Year Cash Flow Depreciation Income
Income
Taxes
After Tax
Cash Flow
Example
A corporation expects to receive $32,000. each year for 15 years
from the sale of a product. There will be an initial investment of
$150,000. Manufacturing and sales expenses will be $8067. per
year. Assume a straight line depreciation, a 15 year useful life
and no salvage value. Use a 46% income tax rate. Determine the
projected after-tax rate of return.
Solution:
Considering straight line depreciation: (P-S)/n
(150000 - 0)/15 = 10000/year
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Before Tax
Year Cash Flow
0 -150000
1 23933
2 23933
Taxable
Depreciation Income
Income
Taxes
10000
10000
13933
13933
-6409
-6409
After Tax
Cash Flow
-150000
17524
17524
15 23933
10000
13933
-6409
17524
Take the After Tax Cash Flow and compute the rate of return at
which PW of Cost equals PW of Benefits
150000 = 17524(P/A,i%,15)
(P/A,i%,15) = 150000/17524 = 8.559
i = 8%
Probabilistic Problems
Typically use expected cost approach
Example 2.14
Flood damage in any year is given according to the table below.
What is the present worth of flood damage for a 10-year period?
Use 6%.
Damage
0
$10,000
$20,000
$30,000
Probability
0.75
0.20
0.04
0.01
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The expected value of flood damage is
E(damage)
= (0)(0.75)+(10,000)(0.20)+ (20,000)(0.04)
+ (30,000)(0.01)
= 3100
present worth
= 3100(P/A,6%,10)
= 22,816
Example 2.15
A dam is being considered on a river which periodically
overflows and causes $600,000 damage. The damage is
essentially the same each time the river causes flooding. The
project horizon is 40 years. A 10% interest rate is being used.
Three different designs are available, each with different costs
and storage capacities.
design
maximum
alternative
cost
capacity
A
500,000
1.0unit
B
625,000
1.5units
C
900,000
2.0units
The U.S. Weather Service has provided a statistical analysis of
annual rainfall in the area draining into the river.
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units annual
rainfall
0
0.1-0.5
0.6-1.0
1.1-1.5
1.6-2.0
2.0 or more
probability
0.10
0.60
0.15
0.10
0.04
0.01
Which design alternative would you choose assuming the dam is
essentially empty at the start of each rainfall season?
The sum of the construction cost and the expected damage needs
to be minimized. If alternative A is chosen, it will have a
capacity of 1 unit. Its capacity will be exceeded (causing
$600,000 damage) when the annual rainfall exceeds 1 unit.
Therefore, the annual cost of A is:
EUAC(A) = 500,000(A/P,10%,40) + 600,000(0.10 + 0.04
+ 0.01)
= 141,150
Similarly,
EUAC(B) = 625,000(A/P,10%,40) + 600,000(0.04 + 0.01)
= 93,940
EUAC(C) = 900,000(A/P,10%,40) + 600,000(0.01)
= 98,070
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Alternative B should be chosen.
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