ISyE 3025
Engineering
Economy
Fundamentals of an Economic Decision
Fundamentals
of an
Economic
Decision
Jack R. Lohmann
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
School of Industrial and Systems Engineering
Georgia Institute of Technology
Fundamentals of an Economic Decision
Example
Overview
1
 Illustrate a fundamental
choice: Benefits vs. Costs
 Develop formal approach
and notation
 Introduce basic decision
criterion:
Prospective Benefit, Bj(i),
and Prospective Cost, Cj(i)
Fundamentals of an Economic Decision
2
Assume a 1-year magazine subscription costs
$20 and a 2-year subscription costs $35. As a
a special offer, these
prices are guaranteed for
the next four years if you
subscribe today.
Decisions based on differences
t
0
1
2
3
4
3
1-Yr 2-Yr Diff. (2Yr-1Yr)
-20 -35
-15
-20
+20
-20 -35
-15
-20
+20
Which would you prefer?
Fundamentals of an Economic Decision
Fundamental choice: Benefits, Costs
t
0
1
2
3
4
Benefit
Cost
Diff. (2-1) = Series - Series
-15
0
15
+20
20
0
-15
0
15
+20
20
0
FC
FB
Assume i = 0.03 per year
Fundamentals of an Economic Decision
4
A more formal approach
5
An economic decision
criterion includes both:
Measure of Worth
Decision Rule
FB = 20(1.03)2 + 20 = $41.22
FC = 15(1.03)3 + 15(1.03) = $31.84
Since FB > FC; choose 2-Yrs
ISyE 3025, 2015, Learning Cycle #2
1
Fundamentals of an Economic Decision
A special interest rate: MARR
Fundamentals of an Economic Decision
6
Interest Rate = 5%
An economical investment?
If MARR < 5%; Yes
If MARR > 5%; No
Fundamentals of an Economic Decision
Fundamentals of an Economic Decision
8
Some notation
Typically, the MARR for:
Ajt = net cash flow, where:
1) individuals represents the
min. attractive opportunities to invest in money
markets
j = index on opportunities
Ajt > 0 is a net receipt,
Ajt < 0 is a net expense, and
Ajt = 0 for t < 0 and t > N
2) corporations represents
the min. attractive opportunities to invest in the
company
Ajt = Bjt - Cjt, where:
Measure of Worth
N
Bj(i) =  Bjt
t=0
N
(1+i)T-t,
where
i = MARR
Cj(i) =  Cjt(1+i)T-t
t=0
Typically, T = 0 or N
Decision Rule
Accept (prefer) j if Bj(i) > Cj(i),
otherwise reject (not prefer) j
ISyE 3025, 2015, Learning Cycle #2
9
Bjt = Ajt if Ajt > 0, else Bjt = 0,
Cjt = - Ajt if Ajt < 0, else Cjt = 0
Fundamentals of an Economic Decision
Basic Criterion: Benefits and Costs
7
t
Ajt
0 - 100
1 105
i = minimum attractive rate
of return (MARR)
also known as -marginal growth rate,
discount rate,
cutoff rate,
hurdle rate,
yield,
among others
Values of the MARR
The role of the MARR
Fundamentals of an Economic Decision
10
An example
11
A manufacturer is considering
buying 10 robots to spray paint
its product on the assembly line.
Each robot costs $200,000 and
has an expected life of 9 years.
The cost to install all the robots
is $45,000. Each robot is expected to reduce labor costs by
$50,000 a year but will increase
energy costs by $15,000 a year.
If the MARR = 10% per year, are
the robots economical?
2
Fundamentals of an Economic Decision
. . . an example
t
Problem Data
0
10(-200K)- 45K
1
10(+50K-15K)
2
“
…
N=9
“
Fundamentals of an Economic Decision
12
. . . an example
13
t
ARt = BRt - CRt
0 - 2,045K
0 2,045K
1
350K 350K 0
2
350K 350K 0
…
N=9 350K 350K 0
ARt
- 2,045K
350K
350K
350K
BR(.1) = 350(F/A,.1,9) = $4752.8K
CR(.1) = 2045(F/P,.1,9) = $4822.0K
Reject; note only 1.4% difference!
Fundamentals of an Economic Decision
Summary
14
 A fundamental choice:
Benefits vs. Costs
ISyE 3025
Engineering
Economy
 Formal approach and
notation
 Criterion = MOW + DR
 MARR
Future Worth,
Present Worth,
Annual Worth
 Ajt, Bjt, Cjt
Jack R. Lohmann
 Basic decision criterion:
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
 Bj(i), Cj(i)
Future Worth, Present Worth, Annual Worth
Overview
Three classic economic
criteria based on worth:
School of Industrial and Systems Engineering
Georgia Institute of Technology
Future Worth, Present Worth, Annual Worth
1
Future Worth
Measure of Worth
FWj(i) = Bj(i) - Cj(i), where i = MARR
N
N
t=0
N
t=0
 Future Worth, FWj(i)
=  Bjt(1+i)T-t -  Cjt(1+i)T-t
 Present Worth, PWj(i)
=  [Bjt - Cjt](1+i)T-t
 Annual Worth, AWj(i)
ISyE 3025, 2015, Learning Cycle #2
2
t=0
N
=  Ajt(1+i)T-t ,
t=0
Typically T = N
Decision Rule
Accept j if FWj(i) > 0, else reject
3
Future Worth, Present Worth, Annual Worth
Robot example revisited
Future Worth, Present Worth, Annual Worth
3
FWR(.1) = - 2045(F/P,.1,9)
+ 350(F/A,.1,9) = - $69.20K
Future Worth, Present Worth, Annual Worth
Future Worth, Present Worth, Annual Worth
5
Measure of Worth
where i = MARR
=[  Ajt(1+i)N-t](1+i)-N
t=0
N
=  Ajt(1+i)-t
t=0
Decision Rule
Accept j if PWj(i) > 0,
else reject
Future Worth, Present Worth, Annual Worth
AR1,t
AR2,t
AR3,t
0 - 75K - 125K
- 100K
6
Route 1 Route 2 Route 3
Route length, miles
15
5
10
Construction cost/mile
$5K
$25K $10K
Annual maintenance/mile $200
$400
$250
Salvage value/mile
$3K
$5K
$3.5K
Annual power loss/mile
$500
$500
$500
Annual property taxes,
% of construction costs
2
2
2
Annual Revenue
$35K
$35K
$35K
N
t
An example
An engineer is evaluating 3 alternative routes for a
power line. For this decision, i = 0.08/year and N =
15 years. Relevant data:
PWj(i) = FWj(i)(1+i)-N,
. . . Example: Identify cash flows
4
t
ARt = BRt - CRt
0 - 2,045K
0
2,045K
1
350K 350K
0
2
350K 350K
0
…
N=9 350K 350K
0
FWR(.1) = 4752.8K - 4822.0K
= - $69.20K < 0; OR
A manufacturer is considering
buying 10 robots to spray paint
its product on the assembly line.
Each robot costs $200,000 and
has an expected life of 9 years.
The cost to install all the robots
is $45,000. Each robot is expected to reduce labor costs by
$50,000 a year but will increase
energy costs by $15,000 a year.
If the MARR = 10% per year, are
the robots economical?
Present Worth
. . . example revisited
Future Worth, Present Worth, Annual Worth
7
Identify incremental cash flows
t
AR1,t
AR2,t
AR3,t AR2-R1,t AR3-R1,t
0 - 75K - 125K - 100K
- 50K
- 25K
1-14
23K
28K
25.5K
1-14 23K
28K 25.5K
+5K +2.5K
15
68K
53K
60.5K
15 68K
53K 60.5K
- 15K - 7.5K
PWR1(.08) = - 75K + 23K(P/A,.08,14)
+ 68K(P/F,.08,15) = +136.1K
PWR2(.08) = +122.6K
PWR3(.08) = +129.3K
Choose R1?: PWR1 > PWR2, PWR3?
ISyE 3025, 2015, Learning Cycle #2
8
PWR1(.08) = +136.1K > 0
PWR2-R1(.08) = - 13.5K < 0
PWR3-R1(.08) = - 6.8K < 0
4
Future Worth, Present Worth, Annual Worth
An equivalent decision rule
Future Worth, Present Worth, Annual Worth
9
Annual Worth
Measure of Worth
For any two alternatives j = A1, A2:
AWj(i) = PWj(i)(A/P,i,N),
where i = MARR
PWA1-A2(i) = PWA1(i) - PWA2(i)
Thus, equivalent decision rule is:
= FWj(i)(A/F,i,N),
prefer A1 over A2 when
Decision Rule
Accept j if AWj(i) > 0,
else reject
Aj2 AjN
Aj1
PWA1(i) > PWA2(i)
and no capital rationing!
(Same is true for Future Worth)
Aj0
Future Worth, Present Worth, Annual Worth
Example, revisited (again)
t
AR1,t
AR2,t
28K
11
0
N
Summary
12
AR3,t AR2-R1,t AR3-R1,t
25.5K
+5K +2.5K
15 68K 53K 60.5K - 15K - 7.5K
AWR1(.08) = +15.9K > 0
AWR2(.08) = +14.3K > 0
AWR3(.08) = +15.1K > 0
AWR2-R1(.08) = - 1.6K < 0
AWR3-R1(.08) = - 0.8K < 0
AW also has equivalent decision rule
ISyE 3025
Engineering
Economy
A = AWj(i)
Future Worth, Present Worth, Annual Worth
0 - 75K - 125K - 100K - 50K - 25K
1-14 23K
10
Three classic economic
criteria based on worth:
 Future Worth, FWj(i)
 Present Worth, PWj(i)
 Annual Worth, AWj(i)
Internal Rate of Return
Overview
Internal Rate
of Return
1
 Definition
 Illustrative examples
 Potential problems unique
to this criterion
Jack R. Lohmann
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
School of Industrial and Systems Engineering
Georgia Institute of Technology
ISyE 3025, 2015, Learning Cycle #2
5
Internal Rate of Return
Internal Rate of Return
Internal Rate of Return
2
Example: An investment
Accept investment j
if ij* > MARR, else reject
Assume you can purchase a
photocopy machine for
$20,000. Its estimated life and
salvage value is 5 years and
$5,000, respectively. It will
save $7,000 a year in outside
vendor copying, but it will cost
$1,000 a year in maintenance,
repair, paper, ink, etc.
Accept borrowing j
if ij* < MARR, else reject
Assume the MARR is 9% per
year. What is the IRR?
Measure of Worth (most common definition)
ij* = {i | “PWj(i)” = 0}
or
ij* = {i | “FWj(i)” = 0}
Decision RuleS
Internal Rate of Return
Example: The net cash flows
Internal Rate of Return
4
t
ACt
0 - 20,000
1
6,000
2
6,000
3
6,000
4
6,000
5
11,000
PWC(.15) = 2598.81
PWC(.25) = - 2225.92
PWC(.199021) = +0.00089211
iC*  0.20 > 0.09
t
0
1
2
3
4
5
ALt
+15,000
- 3,561
- 3,561
- 3,561
- 3,561
- 3,561
PWL(i) = 15,000 - 3561(P/A,i,5)
iL*  0.06 < 0.09
Example: A borrowing
5
Assume you can finance the
copier with a loan from the
supplier. For a down payment of $5,000, you can
finance the remainder. The
loan is to be repaid in 5
years with annual payments
of $3,561.00.
What is the interest rate
(IRR) of the loan?
Internal Rate of Return
Example: The net cash flows
3
Internal Rate of Return
6
Investment? Borrowing?
t
0
1
2
3
4
1-Yr 2-Yr
- 20 - 35
- 20
- 20 - 35
- 20
A2Yr-1Yr
- 15
+20
- 15
+20
7
A1Yr-2Yr
+15
- 20
+15
- 20
PW2-1(i*) = -15+20(1+i*)-1-15(1+i*)-2+20(1+i*)-3 = 0
i2-1* = 0.335 > MARR = 0.03; Accept? Reject?
PW1-2(i*) = +15-20(1+i*)-1+15(1+i*)-2-20(1+i*)-3 = 0
i1-2* = 0.335 > MARR = 0.03; Accept? Reject?
ISyE 3025, 2015, Learning Cycle #2
6
Internal Rate of Return
Internal Rate of Return
What is an “investment?”
t
ACt 0
0 - 20,000
1
6,000
2
6,000
3
6,000
4
6,000
5 11,000
1
2
3
4
5 Time
0.0
- 9174
- 12655
- 9174(1.2)
- 15559
= -11000
- 17980
- 12655(1.2) = - 15174
- 20000
- 15559(1.2) = - 18655
Capital
- 17980(1.2) = - 21559
Invested - 20000(1.2) = - 23980
Capital remains invested at all times
Internal Rate of Return
t
0
1
2
3
4
Ajt
- 5,000
3,000
- 1,400
- 1,000 - 5000
8,400
ij* = 0.20
Capital
Borrowed
+15,000
t
ALt
0 +15,000
1 - 3,561
2 - 3,561
3 - 3,561
4 - 3,561
5 - 3,561
10
15000(1.06) = 15900
12339(1.06) = 13079
9518(1.06) = 10089
12339
6528(1.06)
= 6920
9518
3359(1.06)
= 3561
6528
3359
0.0
0
1
2
3
4
5 Time
Capital remains borrowed at all times
Internal Rate of Return
Accept or reject if MARR = 0.15?
0
What is a “borrowing?”
8
1
2
3
11
4
0.0
- 3000
- 3600
- 5000
- 6000 - 6000
- 7000
- 8400
j is an investment; ij* = 0.20 > 0.15, accept
Internal Rate of Return
Yet another example . . .
Accept or reject if MARR = 0.15? 12
t
0
1
2
3
4
Ajt
2,000
- 1,200
900
800
- 3,300
3000
2200
2000
2200
2000
1100
1000
ij* = 0.10
0.0
0
1
2
3
4
j is a borrowing; ij* = 0.10 < 0.15, accept
Internal Rate of Return
13
. . . the net cash flows
It is common in the oil industry to “acidize” a well to increase production. Acids are
pumped into the well to
cleanse the porous rock near
the bottom of the well to increase oil production.
t
A 0t
A 1t A 1-0,t
0
0 - 50K - 50K
1 125K 250K 125K
2 100K 100K
0
3 75K 50K - 25K
4 50K 25K - 25K
5 25K
0 - 25K
Assume j = 0 represents “do
nothing,” and j = 1 represents
acidizing. MARR = 20%.
i1-0* = 1.35 > 0.20; and
ISyE 3025, 2015, Learning Cycle #2
3300
14
i1-0* = 0.0 < 0.20
7
Internal Rate of Return
Internal Rate of Return
One final example . . . MARR = 0.15
15
t
AA1,t
AA2,t AA2-A1,t
0 --10,000
10,000 --20,000
20,000 - 10,000
1
5,500
- 5,500
2
5,500
- 5,500
3
5,500 40,000 +34,500
 Definition
 Two decision rules
 Illustrative examples
 Investment
 Borrowing
6,300 +3,742 > 0
 Mixed Investment/Borrowing
 No equivalent decision rule
Note: i *A1-A2  i *A1 - i *A2 ; IRR does
not have an equivalent decision rule!
Benefit/Cost Ratio, and Payback Period
ISyE 3025
Engineering
Economy
Benefit/Cost
Ratio, and
Payback
Period
Jack R. Lohmann
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
16
0.26 0.241 > 0.15
ij* = 0.30
PWj(0.15) = 2,558
Summary
School of Industrial and Systems Engineering
Georgia Institute of Technology
Measure of Worth
N
Bj(i) =  Bjt(1+i)T-t
t=0
BCRj(i) =
N
Cj(i) =  Cjt(1+i)T-t
t=0
where i = MARR & T = 0 or N
Decision Rule
Accept (prefer) j if BCRj(i) > 1.0,
otherwise reject (not prefer) j.
ISyE 3025, 2015, Learning Cycle #2
1
 Benefit/Cost Ratio
 Payback Period
(a secondary criterion)
Notes:
The presentation on Benefit/Cost Ratio is based on the
“strict” definition of benefits and costs. See text for the
“flexible” definition.
The definition of Payback Period is a conceptual one, and
some situations can require a more detailed definition.
Benefit/Cost Ratio, and Payback Period
Benefit/Cost Ratio, and Payback Period
Benefit/Cost Ratio
Overview
2
An example . . . MARR = 0.10
Note corrections
t
AA1,t
0 --2,000
2,000
1
900
2
900
3
900
4
900
5
900
BCRj(0.10) =
1.71
PWj(0.10) =
1,412
AA2,t
--3,000
3,000
1,200
1,200
1,200
1,200
1,200
1.52
AA2-A1,t
- 1,000
300
300
300
300
300
1.137 > 1
1,549
+137 > 0
3
Note: BCR has no equivalent decision rule!
8
Benefit/Cost Ratio, and Payback Period
Payback Period (Secondary Criterion)
Benefit/Cost Ratio, and Payback Period
4
Measure of “Worth” (actually, liquidity)
Nj*
N
= { N | [  Ajt
t=0
(1+i)N-t =
t
0
1
2
3
4
5
0] }
Decision Rule
Accept investment j if Nj* < Nmax,
else reject (for multiple alternatives, prefer the shortest
payback period).
t
AA2,t
0 - 10,000
1
4,600
2
4,600
3
4,600
4
4,600
5
4,600
- 10000
0
1
2
3
+2108
5
- 5275
- 1994(1.25) = - 2493
- 5275(1.25) = - 6594
- 7900(1.25) = - 9875
- 10000(1.25) = - 12500
3  NA2*  4
ISyE 3025
Engineering
Economy
3
- 4000
- 7200
4
5
0.0
- 4000(1.25) = - 5000
- 7200(1.25) = - 9000
- 9760(1.25) = - 12200
Summary
7
 Fundamental concepts
of a decision
 Benefit, Bj(i), & Cost, Cj(i)
 Five classic criteria:
 Future Worth, FWj(i)
 Present Worth, PWj(i)
 Annual Worth, AWj(i)
 Internal Rate of Return, ij*
 Benefit/Cost Ratio, BCRj(i)
 Popular secondary criterion:
 Payback Period, Nj*
Home Mortgage Refinancing - A Case Study
Overview
Home
Mortgage
Refinancing A Case Study
Gunter P. Sharp
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
2
Benefit/Cost Ratio, and Payback Period
- 1994
- 7900
1
5
NA1* = 3
6
4
0
AA1,t
- 9,760
5,000
5,000
5,000
5,000
5,000 - 9760
iA1* = 0.43
PWA1(.25) = 3,686
Benefit/Cost Ratio, and Payback Period
Payback Period
Payback Period
School of Industrial and Systems Engineering
Georgia Institute of Technology
ISyE 3025, 2015, Learning Cycle #2
1
 Homeowner’s situation
 Selecting a criterion
 Existing loan
 Refinancing option
 Defining a new alternative
 Analysis using break-even
internal rate of return
9
Home Mortgage Refinancing - A Case Study
Homeowner’s situation
Home Mortgage Refinancing - A Case Study
2
 GT faculty member recently bought a
home with a 30-year mortgage
 Refinancing opportunity available at
lower interest rate, 15-year period
 Question: Refinance ?
 Lower interest rate
– Transaction (closing) costs
– Higher monthly payment
because of shorter term
Home Mortgage Refinancing - A Case Study
Existing loan (Option 1)
4
 FW, PW, AW: what MARR?
 B/C ratio: classification of B?
 IRR: long periods, 180 and 360
Applied to 60-month period
 Another criterion:
Future cash amount
New alternative defined for
use with this criterion
Existing loan (Option 1)
5
P = $ 202,000
 N = (30)(12) = 360 months
 i = 9% / 12 = 0.75% per month
1 2 3 4 5
 Monthly payment for
principal and interest (P&I):
A = P(A/P, 0.75%, 360)
= 202,000(0.00804623)
= $ 1,625.34
360
i = 0.75% per month
A = $ 1,625.34/month
A includes principal and interest
Home Mortgage Refinancing - A Case Study
ISyE 3025, 2015, Learning Cycle #2
3
Home Mortgage Refinancing - A Case Study
 Initial balance, P = $ 202,000
New loan (Option 2)
The new loan would be for
the same amount as the
existing loan, but at a lower
interest rate of 0.625% per
month, and for a period of
15 years. Also, there would
be a $ 2,000 transaction fee
(closing cost).
Selecting a criterion
Home Mortgage Refinancing - A Case Study
6
New loan (Option 2)
7
10
Home Mortgage Refinancing - A Case Study
How to compare?
Home Mortgage Refinancing - A Case Study
8
Very few people live in a
house long enough to pay
off a 30-year mortgage.
P = $ 193,678
Home Mortgage Refinancing - A Case Study
10
1 2 3 4
1
61
120
180
i = 0.625% per month
A = $ 1,872.56/month
N = 120 months
 Slow decline is typical of
30-year mortgages
Home Mortgage Refinancing - A Case Study
ISyE 3025, 2015, Learning Cycle #2
11
P = $ 157,753
 After 15 years: $ 160,248
 After 15 years: zero !
Loan is completely paid off.
New loan (2), after 5 yrs.
P = $ 202,000
 After 10 years: $ 180,648
 After 10 years: $ 93,451
300
360
i = 0.75% per month
A = $ 1,625.34/month
N = 300 months
 After 5 years of payments
the remaining loan balance is
P = A(P/A, 0.75%, 300)
= $ 1,625.34(119.1616)
= $ 193,678
 After 5 years of payments
the remaining loan balance is
P = A(P/A, 0.625%, 120)
= $ 1,872.56(84.2447)
= $ 157,753
1
61
1 2 3 4
Home Mortgage Refinancing - A Case Study
Remaining loan balance (2)
9
P = $ 202,000
One way to compare the
two alternatives is to
calculate the remaining
loan balances after 5, 10,
and 15 years.
Remaining loan balance (1)
Existing loan (1), after 5 yrs.
Home Mortgage Refinancing - A Case Study
12
Apples and oranges (1 vs. 2) 13
The difficulty with the previous
comparisons is the difference in the cash
flows paid to the mortgage company.
Solution? Construct a hypothetical
alternative (Option 3) where the homeowner
keeps the existing mortgage contract but
replicates the cash flow of the new loan.
The excess payments reduce the principal
balance. Then, a comparison can be made
on remaining loan balances.
11
Home Mortgage Refinancing - A Case Study
Hypothetical alternative (3)
Home Mortgage Refinancing - A Case Study
14
Effect on loan balance (3)
15
 After 5 years the $ 2,000
would have the effect of
reducing the loan balance by
P = $ 202,000
F = P(F/P, 0.75%, 60)
= $ 2,000(1.56568) = $ 3,131
 The effect of the $ 247.22
per month extra payment is
N=?
1 2 3 4 5
$ 2,000 i remains 0.75% per month
A = $ 1,625.34 + 247.22
= $ 1,872.56/month
F = A(F/A, 0.75%, 60)
= $ 247.22(75.4241) = $ 18,646
Home Mortgage Refinancing - A Case Study
Effect on loan balance (3)
Home Mortgage Refinancing - A Case Study
16
 Effect of the $ 2,000 after
10 years: $ 4,903
15 years: $ 7,676
Compute loan balance (3)
17
 After 5 years: $ 193,678
$ 2,000  – 3,131
$ 247.22  – 18,646
Result: $ 171,901
 Effect of the $ 247.22 after
10 years: $ 47,841
15 years: $ 93,549
 Compare with $ 157,753
for new loan (Option 2)
 Next step: compute remaining
loan balances for hypothetical
alternative (Option 3)
 Conclusion: Homeowner is better off
with new loan, if he can make the
payments and stays at least 5 years.
Home Mortgage Refinancing - A Case Study
Home Mortgage Refinancing - A Case Study
Compute loan balance (3)
 After 10 yrs.
$ 180,648
– 4,903
– 47,841
$ 127,904
After 15 yrs.
$ 160,248
– 7,676
– 93,549
$ 59,023
 Compare with new loan (2)
$ 93,451
zero
 Conclusion: new loan is better
than hypothetical alternative.
ISyE 3025, 2015, Learning Cycle #2
18
Rate-of-return analysis
19
Define Option 4:
the homeowner deposits the
excess $ 2,000 and $ 247.22 per
month into an account that
grows in 5 years to $ 35,925,
which is the difference after 5
years between Options 1 and 2:
$ 193,678 – $ 157,753.
12
Home Mortgage Refinancing - A Case Study
Rate-of-return analysis
Home Mortgage Refinancing - A Case Study
20
F = $ 35,925
1 2 3 4 5
A = $ 247.22/mo.
$2,000
Rate-of-return = 2.084% per month
Home Mortgage Refinancing - A Case Study
22
 New criterion: future cash

Existing loan

New loan

Hypothetical alternative
21
Result: i* = 2.084% / month
Effective annual rate is
(1 + 0.0284)12 – 1 = 28%
This is a rather high rate
that might not be achieved
with a commercial account.
So the new loan is preferred.
60
Summary
Rate-of-return analysis
ISyE 3025
Engineering
Economy
Multiple
Alternatives
Part 1
 Rate-of-return analysis
for 5-year period on difference
between alternatives
Gunter P. Sharp
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
Multiple Alternatives - Part 1
Overview
 Define relationships possible
among investment alternatives
 Present typical examples
 Demonstrate selection process
for investment alternatives

PW for independent investments

Rate-of-return for mutually
exclusive investments
ISyE 3025, 2015, Learning Cycle #2
School of Industrial and Systems Engineering
Georgia Institute of Technology
Multiple Alternatives - Part 1
1
Definition
2
Independent alternatives:
Investment alternatives
where the selection of
one does not restrict the
selection of another.
13
Multiple Alternatives - Part 1
Example 1
Multiple Alternatives - Part 1
3
A Replace gas-fired boiler with
a more fuel-efficient unit
Example 1
4
Options are:
 Do Nothing
A Replace gas-fired boiler
B Replace plant electric
control system
(A + B) Replace both
B Replace plant electric control
system with “smart” system
that reduces peak load from
450 to 350 kW
Consider: , A, B, (A + B)
Multiple Alternatives - Part 1
Definition
Multiple Alternatives - Part 1
5
Investment alternatives
where the selection of
one precludes the
selection of another.
B Replace gas-fired boiler with
a more fuel-efficient unit from
Manufacturer Y
Multiple Alternatives - Part 1
Options are:
 Do Nothing
A Replace gas-fired boiler
with unit from Mfg. X
B Replace gas-fired boiler
with unit from Mfg. Y
6
A Replace gas-fired boiler with
a more fuel-efficient unit from
Manufacturer X
Mutually exclusive
alternatives:
Example 2
Example 2
Multiple Alternatives - Part 1
7
Definition
8
Contingent alternatives:
Investment alternatives
where the selection of
one depends on the
selection of another.
Consider: , A, B
ISyE 3025, 2015, Learning Cycle #2
14
Multiple Alternatives - Part 1
Example 3
Multiple Alternatives - Part 1
9
A Install stand-by 350 kW
diesel generator for
refrigeration system
A by itself doesn’t make much
sense, so A depends on B.
Multiple Alternatives - Part 1
Multiple Alternatives - Part 1
11
Two or more alternatives
are complementary when
the relative worth of their
combination is more than
the sum of the individual
relative worths.
Warehouse
Warehouse
Multiple Alternatives - Part 1
13
Example 4
14
(A + B) Company Z gains the right
to distribute its products in
zones A and B
Zone A
Warehouse
Zone B
12
Zone A
Multiple Alternatives - Part 1
B Company Z gains the right
to distribute its products in
zone B
Example 4
A Company Z gains the right
to distribute its products in
zone A
Complementary alternatives:
Example 4
10
Options are , B, (A + B)
 Do Nothing
B Install “smart” electric control
system that reduces peak load
from 450 to 350 kW (benefit from
rate reduction based on peak use)
(A + B) Install generator plus smart
control system (benefit from rate
reduction based on peak use
plus rate reduction based on
interruptible supply)
B Install “smart” electric control
system that reduces peak load
from 450 to 350 kW
Definition
Example 3
Zone B
We would expect some
savings in transport costs
when serving both zones.
ISyE 3025, 2015, Learning Cycle #2
15
Multiple Alternatives - Part 1
Example 4
Multiple Alternatives - Part 1
15
Consider these options:
Two or more alternatives
are competitive (but not
mutually exclusive) when
the relative worth of their
combination is less than
the sum of the individual
relative worths.
A Distribute products in zone A
B Distribute products in zone B
C Distribute products in both
zones A and B, with cost
reduction reflecting combined
transport operation
Multiple Alternatives - Part 1
Multiple Alternatives - Part 1
17
Example 6
A Replace gas-fired boiler for
heating system with a more
fuel-efficient unit
B Install system of sensors and
closers at doors, windows, and
other openings to reduce heat
loss during winter season
A + B technically feasible, but the
savings are not additive
Consider: , A, B, and C based on
A + B reflecting loss of savings
Independent alternatives A and B:
Alternative
A
B
Investment $ 300,000 $400,000
Project life
4 years
5 years
Annual net
cash flow $ 115,000 $125,000
MARR = 15%
No budget constraint
Use PW to make decisions
Which alternative(s) to select?
Multiple Alternatives - Part 1
Multiple Alternatives - Part 1
Example 6
Consider : , A, B, (A + B)
PW(15%) = 0 by definition
PWA(15%) = – 300,000
+ 115,000(P/A, 15%, 4)
= – 300,000 + 115,000(2.85498)
= $ 28,323 > 0, so A is acceptable
PWB(15%) = – 400,000
+ 125,000(P/A, 15%, 5)
= – 400,000 + 125,000(3.35216)
= $ 19,019 > 0, so B is acceptable
ISyE 3025, 2015, Learning Cycle #2
16
Competitive alternatives:
 Do Nothing
Example 5
Definition
19
Example 6
18
20
PWA+B(15%) = 28,323 + 19,019
= $ 47,342 > 0, so (A + B) is acceptable
Result: Select (A + B)
(Note: if mutually exclusive, select A)
Reminder: To compute PW of a
combination of two or more
independent investment alternatives,
simply add the PW values of the
individual investments.
16
Multiple Alternatives - Part 1
Example 7
Multiple Alternatives - Part 1
21
Mutually exclusive alternatives C and D:
Alternative
C
D
Investment $ 200,000 $260,000
Project life
6 years
6 years
Annual net
cash flow $ 85,000 $125,000
MARR = 15%
No budget constraint
Use Rate-of-return to make decisions
Which alternative to select?
Multiple Alternatives - Part 1
Summary
23
 Defined relationships possible
among investment alternatives
Independent
Mutually exclusive
Contingent
Complementary
Competitive
Example 7
22
Consider : , C, D in that sequence
PWC(i) = – 200,000 + 85,000(P/A, i, 6)
(P/A, i, 6) = 200,000/85,000 = 2.35294
so i* = 35.7% > 15%, so accept C for now
PWD-C(i) = – (260,000 – 200,000)
+ (125,000 – 85,000)(P/A, i, 6)
(P/A, i, 6) = 60,000/40,000 = 1.5
so i* = 63.1% > 15%, so accept D
Final choice is D
(If independent, select both since i*D = 42.3%)
ISyE 3025
Engineering
Economy





Multiple
Alternatives
Part 2
 Example of independent
alternatives, with PW criterion
Gunter P. Sharp
 Example of mutually exclusive
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
alternatives, with IRR criterion
School of Industrial and Systems Engineering
Georgia Institute of Technology
Copyright 1999. Georgia Tech Research Corporation. All rights reserved.
Multiple Alternatives - Part 2
Overview
 Present method for defining
combinations of independent
and mutually exclusive
investment alternatives
 Demonstrate selection process
for more involved situations
 Rate-of-return
 B/C ratio
ISyE 3025, 2015, Learning Cycle #2
Multiple Alternatives - Part 2
1
Procedure
2
Combination alternatives:
Construct logical combinations
of alternatives that are
technically feasible.
Eliminate those that exceed
the budget constraint, if any.
17
Multiple Alternatives - Part 2
Multiple Alternatives - Part 2
Example 1
3
A manufacturing company is
trying to expand its market into
some new areas: E, F, and G.
4
Example 1
Altern. Invest. Altern. Invest. Altern. Invest.
E, 
E, V
E, W
E, X
There are a number of potential
distributors for serving each of
the new areas:
area E: distributor V, W, and X
area F: distributor X, Y, and Z
area G: distributor V, W, and Z
Multiple Alternatives - Part 2
0
60,000
80,000
90,000
F, 
F, X
F, Y
F, Z
0
70,000
50,000
85,000
G, 
G, V
G, W
G, Z
0
45,000
65,000
75,000
Multiple Alternatives - Part 2
Example 1
5
Example 1
6
Constraint 1: There can be only
one distributor in an area.
Within an area, the choice of
a distributor is mutually exclusive.
Constraint 2: No distributor can
serve in two or more areas, to
prevent stretching sales staff.
Across areas, the choice of areas
for expansion is independent.
There are additional constraints
related to:
No distributor can have 2 areas.
Overall budget limit.
Constraint 3: There is a limit on
investment of $150,000 at the
inception of the expansion program.
There are 4x4x4 = 64 combinations.
Multiple Alternatives - Part 2
Multiple Alternatives - Part 2
7
Example 1 (a)
8
Example 1 (b)
Combin.
Inv.
OK?
Combin.
Inv.
OK?
Combin.
Inv.
OK?
Combin.
Inv.
E,F,G
0
Y
E,FY,G
50
Y
45
Y
E,FY,GV
95
Y
EV,F,G
60
Y
110
Y
--
N
EV,FY,GV
--
N
EW,F,G
80
Y
EV,FY,G
EW,FY,G
E,F,GV
EV,F,GV
130
Y
Y
EW,FY,GV
175
N
90
Y
140
Y
135
Y
EX,FY,GV
185
N
E,FX,G
70
Y
EX,FY,G
E,FZ,G
EW,F,GV
EX,F,GV
125
EX,F,G
85
Y
E,FX,GV
115
Y
E,FZ,GV
130
Y
EV,FX,G
130
Y
145
Y
EV,FX,GV
--
N
EV,FZ,GV
--
N
EW,FX,G
150
Y
EV,FZ,G
EW,FZ,G
165
N
EW,FX,GV
195
N
EW,FZ,GV
210
N
EX,FX,G
--
N
EX,FZ,G
175
N
EX,FX,GV
--
N
EX,FZ,GV
220
N
ISyE 3025, 2015, Learning Cycle #2
OK?
18
Multiple Alternatives - Part 2
Multiple Alternatives - Part 2
9
Example 1 (c)
10
Example 1 (d)
Combin.
Inv.
OK?
Combin.
Inv.
OK?
Combin.
Inv.
OK?
Combin.
Inv.
E,F,GW
65
Y
E,FY,GW
115
Y
75
Y
E,FY,GZ
125
Y
EV,F,GW
EW,F,GW
125
Y
EV,FY,GW
175
N
E,F,GZ
EV,F,GZ
OK?
135
Y
EV,FY,GZ
185
N
--
N
EW,FY,GW
--
N
N
EW,FY,GZ
205
N
155
N
EX,FY,GW
205
N
EW,F,GZ
EX,F,GZ
155
EX,F,GW
E,FX,GW
165
N
EX,FY,GZ
215
N
135
Y
E,FZ,GW
150
Y
E,FX,GZ
145
Y
E,FZ,GZ
--
N
EV,FX,GW
195
N
EV,FZ,GW
210
N
EV,FX,GZ
205
N
EV,FZ,GZ
--
N
EW,FX,GW
--
N
EW,FZ,GW
--
N
EW,FX,GZ
225
N
EW,FZ,GZ
--
N
EX,FX,GW
--
N
EX,FZ,GW
240
N
EX,FX,GZ
--
N
EX,FZ,GZ
--
N
Multiple Alternatives - Part 2
Example 1
Multiple Alternatives - Part 2
11
Summary of procedure:
Example 2
Form all logical combinations.
Consider 3 mutually exclusive
options:
Delete those that violate any
technical constraints.
A, B, and C, with cash flows
as shown in the table.
Delete those that violate any
budget constraint.
Apply Rate-of-return to
select the best option.
Apply PW to select the
best combination.
MARR = 15%
Multiple Alternatives - Part 2
Example 2
Project
Net
cash
flows
t
A
B
C
0 – 30,000 – 15,000 – 20,000
1
11,175
6,159
7,546
2
11,175
6,159
7,546
3
11,175
6,159
7,546
4
11,175
6,159
7,546
5
11,175
6,159
7,546
Sequence for analysis is B, C, A
ISyE 3025, 2015, Learning Cycle #2
12
Multiple Alternatives - Part 2
13
Example 2
14
PWB(i) = – 15,000 + 6,159(P/A, i, 5) = 0
(P/A, i, 5) = 15,000/6,159 = 2.43546
i* = 30% > 15%, so accept B for now
PWC-B(i) = – (20,000 – 15,000)
+ (7,546 – 6,159)(P/A, i, 5) = 0
(P/A, i, 5) = 5,000/1,387 = 3.60490
i* = 12.0% < 15%, so reject C
B remains the defender
19
Multiple Alternatives - Part 2
Multiple Alternatives - Part 2
Example 2
15
Example 3
PWA-B(i ) = – (30,000 – 15,000)
+ (11,175 – 6,159)(P/A, i, 5) = 0
Consider 2 mutually exclusive
options:
(P/A, i, 5) = 15,000/5,016 = 2.99043
A, and B, with cash flows
as shown in the table.
i* = 20.0% > 15%, so accept A
16
Apply Benefit/cost ratio to
select the best option.
Final selection is A
MARR = 15%
Sequence for analysis is A, B.
Multiple Alternatives - Part 2
Multiple Alternatives - Part 2
17
Example 3
Project
t
Outflows 0
1
2
3
Inflows 0
1
2
3
A
20,000
40,000
60,000
80,000
0
90,000
95,000
125,000
B
35,000
57,000
56,000
42,000
0
80,000
120,000
140,000
NOTE
Bj(i) and Cj(i) are
defined (slightly)
differently here
compared to their
initial definition (in
which either Bjt or
Cjt were equal to
zero). Both
definitions are used
in practice and
both are correct.
Multiple Alternatives - Part 2
Summary
 Method for defining combinations
of independent and mutually
exclusive investment alternatives:
enumeration, eliminate ineligible
combinations
19
Example 3
18
Consider BA(15%) = 232,284
Now CA(15%) = 152,753
so BCRA(15%) = 232,284/152,753
= 1.52 > 1, so accept A for now
Consider BB-A(15%)
= 252,355 – 232,284 = 20,071
Consider CB-A(15%)
= 154,525 – 152,753 = 1,772
so BCRB-A(15%) = 20,071/1,772
= 11.35 > 1, so accept B, final choice is B
ISyE 3025
Engineering
Economy
 Demonstrated selection process
for more involved situations
Rate-of-return, incremental method
 B/C ratio, incremental method

ISyE 3025, 2015, Learning Cycle #2
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved.
20