BUS 304 OPERATIONS RESEARCH
 Applications
of LP, Network, IP, NLP
 Applications of Decision A. and Project M.
 Inventory Modeling
 Queueing Analysis
 Markov Processes
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-1
Introduction to Optimization
and Linear Programming
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-2
Applications of
Mathematical Optimization
 Determining
Product Mix
 Manufacturing
 Routing and Logistics
 Financial Planning
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-3
Characteristics of
Optimization Problems
 Decisions
Variables
 Constraints
 Objectives
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-4
General Form of a
Linear Programming (LP) Problem
MAX (or MIN): c1X1 + c2X2 + … + cnXn
Subject to:
a11X1 + a12X2 + … + a1nXn <= b1
:
ak1X1 + ak2X2 + … + aknXn >=bk
:
am1X1 + am2X2 + … + amnXn = bm
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-5
An Example LP Problem
Blue Ridge Hot Tubs produces two types of hot
tubs: Aqua-Spas & Hydro-Luxes.
Pumps
Labor
Tubing
Unit Profit
Aqua-Spa
1
9 hours
12 feet
$350
Hydro-Lux
1
6 hours
16 feet
$300
There are 200 pumps, 1566 hours of labor,
and 2880 feet of tubing available.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-6
5 Steps In Formulating LP Models:
1. Understand the problem.
2. Identify the decision variables.
X1=number of Aqua-Spas to produce
X2=number of Hydro-Luxes to produce
3. State the objective function as a linear
combination of the decision variables.
MAX: 350X1 + 300X2
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-7
5 Steps In Formulating LP Models
(continued)
4. State the constraints as linear combinations
of the decision variables.
1X1 + 1X2 <= 200
} pumps
9X1 + 6X2 <= 1566
} labor
12X1 + 16X2 <= 2880 } tubing
5. Identify any upper or lower bounds on the
decision variables.
X1 >= 0
X2 >= 0
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-8
Summary of the LP Model for
Blue Ridge Hot Tubs
MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-9
Solving LP Problems:
An Intuitive Approach





Idea: Each Aqua-Spa (X1) generates the highest unit
profit ($350), so let’s make as many of them as possible!
How many would that be?
– Let X2 = 0
 1st constraint:
1X1 <= 200
 2nd constraint:
9X1 <=1566 or X1 <=174
 3rd constraint:
12X1 <= 2880 or X1 <= 240
If X2=0, the maximum value of X1 is 174 and the total
profit is $350*174 + $300*0 = $60,900
This solution is feasible, but is it optimal?
No!
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-10
Solving LP Problems:
A Graphical Approach
 The
constraints of an LP problem
defines its feasible region.
 The best point in the feasible region is
the optimal solution to the problem.
 For LP problems with 2 variables, it is
easy to plot the feasible region and find
the optimal solution.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-11
Plotting the First Constraint
X2
250
(0, 200)
200
boundary line of pump constraint
X1 + X2 = 200
150
100
50
(200, 0)
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-12
Plotting the Second Constraint
X2
(0, 261)
250
boundary line of labor constraint
9X1 + 6X2 = 1566
200
150
100
50
(174, 0)
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-13
Plotting the Third Constraint
X2
250
(0, 180)
200
150
boundary line of tubing constraint
12X1 + 16X2 = 2880
100
Feasible Region
50
(240, 0)
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-14
Plotting A Level Curve of the
Objective Function
X2
250
200
(0, 116.67)
objective function
150
350X1 + 300X2 = 35000
100
(100, 0)
50
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-15
A Second Level Curve of the
Objective Function
X2
250
(0, 175)
200
objective function
350X1 + 300X2 = 35000
objective function
350X1 + 300X2 = 52500
150
100
(150, 0)
50
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-16
Using A Level Curve to Locate
the Optimal Solution
X2
250
objective function
350X1 + 300X2 = 35000
200
150
optimal solution
100
objective function
350X1 + 300X2 = 52500
50
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-17
Calculating the Optimal Solution





The optimal solution occurs where the “pumps” and
“labor” constraints intersect.
This occurs where:
X1 + X2 = 200
(1)
and 9X1 + 6X2 = 1566
(2)
From (1) we have,
X2 = 200 -X1
(3)
Substituting (3) for X2 in (2) we have,
9X1 + 6 (200 -X1) = 1566
which reduces to X1 = 122
So the optimal solution is,
X1=122, X2=200-X1=78
Total Profit = $350*122 + $300*78 = $66,100
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-18
Enumerating The Corner Points
X2
250
obj. value = $54,000
(0, 180)
200
obj. value = $64,000
150
(80, 120)
obj. value = $66,100
(122, 78)
100
50
obj. value = $60,900
(174, 0)
obj. value = $0
(0, 0)
0
0
50
100
150
200
250
X1
Note: This technique will not work if the solution is unbounded.2-19
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
Example of Alternate Optimal
Solutions
X2
250
objective function level curve
450X1 + 300X2 = 78300
200
150
100
alternate optimal solutions
50
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-20
Example of a Redundant Constraint
X2
250
boundary line of tubing constraint
200
boundary line of pump constraint
150
boundary line of labor constraint
100
Feasible Region
50
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-21
Example of an Unbounded Solution
X2
1000
objective function
X1 + X2 = 600
800
-X1 + 2X2 = 400
objective function
X1 + X2 = 800
600
400
200
X1 + X2 = 400
0
0
200
400
600
800
1000
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-22
X2
Example of Infeasibility
250
200
X1 + X2 = 200
feasible region for
second constraint
150
100
feasible region
for first
constraint
50
X1 + X2 = 150
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-23
Modeling and Solving LP
Problems in a Spreadsheet
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-24
Introduction
 Solving
LP problems graphically is only
possible when there are two decision
variables
 Few real-world LP have only two
decision variables
 Fortunately, we can now use
spreadsheets to solve LP problems
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-25
Spreadsheet Solvers
 The
company that makes the Solver in
Excel, Lotus 1-2-3, and Quattro Pro is
Frontline Systems, Inc.
Check out their web site:
http://www.frontsys.com
 Other
packages for solving MP problems:
AMPL
CPLEX
LINDO
MPSX
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-26
The Steps in Implementing an LP
Model in a Spreadsheet
1. Organize the data for the model on the spreadsheet.
2. Reserve separate cells in the spreadsheet to
represent each decision variable in the model.
3. Create a formula in a cell in the spreadsheet that
corresponds to the objective function.
4. For each constraint, create a formula in a separate
cell in the spreadsheet that corresponds to the lefthand side (LHS) of the constraint.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-27
Let’s Implement a Model for the
Blue Ridge Hot Tubs Example...
MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1, X2 >= 0
} profit
} pumps
} labor
} tubing
} nonnegativity
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-28
Implementing the Model
See file Fig3-1.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-29
How Solver Views the Model
 Target
cell - the cell in the spreadsheet
that represents the objective function
 Changing cells - the cells in the
spreadsheet representing the decision
variables
 Constraint cells - the cells in the
spreadsheet representing the LHS
formulas on the constraints
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-30
Let’s go back to Excel and see
how Solver works...
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-31
Goals For Spreadsheet Design
 Communication - A spreadsheet's primary business
purpose is that of communicating information to managers.
 Reliability - The output a spreadsheet generates
should be correct and consistent.
 Auditability - A manager should be able to retrace the
steps followed to generate the different outputs from the
model in order to understand the model and verify results.
 Modifiability - A well-designed spreadsheet should be
easy to change or enhance in order to meet dynamic user
requirements.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-32
Spreadsheet Design Guidelines








Organize the data, then build the model around the data.
Do not embed numeric constants in formulas.
Things which are logically related should be physically
related.
Use formulas that can be copied.
Column/rows totals should be close to the columns/rows
being totaled.
The English-reading eye scans left to right, top to
bottom.
Use color, shading, borders and protection to distinguish
changeable parameters from other model elements.
Use text boxes and cell notes to document various
elements of the model.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-33
Sensitivity Analysis
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-34
Introduction
 When
solving an LP model we assume that
all relevant factors are known with certainty.
 Such certainty rarely exists.
 Sensitivity analysis helps answer questions
about how sensitive the optimal solution is to
changes in various coefficients in an LP
model.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-35
General Form of a
Linear Programming (LP) Problem
MAX (or MIN): c1X1 + c2X2 + … + cnXn
Subject to:
 How
a11X1 + a12X2 + … + a1nXn <= b1
:
ak1X1 + ak2X2 + … + aknXn <= bk
:
am1X1 + am2X2 + … + amnXn = bm
sensitive is a solution to changes in the ci, aij, and bi?
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-36
Approaches to Sensitivity Analysis


Change the data and re-solve the model!
– Sometimes this is the only practical approach.
Solver also produces sensitivity reports that can answer
questions about:
– amounts objective function coefficients can change
without changing the solution.
– the impact on the optimal objective function value of
changes in various constrained resources.
– the impact on the optimal objective function value of
forced changes in certain decision variables.
– the impact changes in constraint coefficients will have
on the optimal solution.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-37
Once Again, We’ll Use The
Blue Ridge Hot Tubs Example...
MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1, X2 >= 0
} profit
} pumps
} labor
} tubing
} nonnegativity
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-38
The Answer Report
See file Fig4-1.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-39
The Sensitivity Report
See file Fig4-1.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-40
X2
How Changes in Objective Coefficients
Change the Slope of the Level Curve
250
original level curve
200
new optimal solution
150
original optimal solution
100
new level curve
50
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-41
Changes in Objective Function
Coefficients
Values in the “Allowable Increase” and
“Allowable Decrease” columns for the
Changing Cells indicate the amounts by
which an objective function coefficient can
change without changing the optimal
solution, assuming all other coefficients
remain constant.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-42
Alternate Optimal Solutions
Values of zero (0) in the “Allowable
Increase” or “Allowable Decrease”
columns for the Changing Cells indicate
that an alternate optimal solution exists.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-43
Changes in Constraint RHS Values
 The
shadow price of a constraint indicates the
amount by which the objective function value
changes given a unit increase in the RHS value of
the constraint, assuming all other coefficients
remain constant.
 Shadow
prices hold only within RHS changes
falling within the values in “Allowable Increase” and
“Allowable Decrease” columns.
 Shadow
prices for nonbinding constraints are
always zero.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-44
Comments About Changes
in Constraint RHS Values
 Shadow
prices only indicate the changes that occur
in the objective function value as RHS values
change.
 Changing
a RHS value for a binding constraint also
changes the feasible region and the optimal
solution (see graph on following slide).
 To
find the optimal solution after changing a binding
RHS value, you must re-solve the problem.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-45
How Changing the RHS Value of a Constraint Can
Change the Feasible Region and Optimal Solution
X2
250
Suppose available labor hours
increase from 1,566 to 1,728
200
150
old optimal solution
old labor constraint
100
new optimal solution
50
new labor constraint
0
0
50
100
150
200
250
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
X1
2-46
Other Uses of Shadow Prices

Suppose a new Hot Tub (the Typhoon-Lagoon)
is being considered. It generates a marginal
profit of $320 and requires:
– 1 pump (shadow price = $200)
– 8 hours of labor (shadow price = $16.67)
– 13 feet of tubing (shadow price = $0)

Q: Would it be profitable to produce any?
A: $320 - $200*1 - $16.67*8 - $0*13 = -$13.33 = No!
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-47
The Meaning of Reduced Costs

The Reduced Cost for each product equals
its per-unit marginal profit minus the per-unit
value of the resources it consumes (priced at
their shadow prices).
Type of Problem
Optimal Value of
Decision Variable
Optimal Value of
Reduced Cost
Maximization
at simple lower bound
between lower & upper bounds
at simple upper bound
<=0
=0
>=0
Minimization
at simple lower bound
between lower & upper bounds
at simple upper bound
>=0
=0
<=0
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-48
Key Points

The shadow prices of resources equate the marginal
value of the resources consumed with the marginal benefit
of the goods being produced.

Resources in excess supply have a shadow price (or
marginal value) of zero.

The reduced cost of a product is the difference between
its marginal profit and the marginal value of the resources
it consumes.

Products whose marginal profits are less than the
marginal value of the goods required for their production
will not be produced in an optimal solution.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-49
Analyzing Changes in Constraint Coefficients
 Q:
Suppose a Typhoon-Lagoon required only
7 labor hours rather than 8. Is it now
profitable to produce any?
A: $320 - $200*1 - $16.67*7 - $0*13 = $3.31 = Yes!
 Q: What is the maximum amount of labor
Typhoon-Lagoons could require and still be
profitable?
A: We need $320 - $200*1 - $16.67*L3 - $0*13 >=0
The above is true if L3 <= $120/$16.67 = $7.20
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-50
Make vs. Buy Decisions:
The Electro-Poly Corporation
 Electro-Poly
is a leading maker of slip-rings.
 A $750,000 order has just been received.
Model 1
Model 2
Model 3
3,000
2,000
900
Hours of wiring/unit
2
1.5
3
Hours of harnessing/unit
1
2
1
Cost to Make
$50
$83
$130
Cost to Buy
$61
$97
$145
Number ordered
 The
company has 10,000 hours of wiring
capacity and 5,000 hours of harnessing
capacity.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-51
Defining the Decision Variables
M1 = Number of model 1 slip rings to make in-house
M2 = Number of model 2 slip rings to make in-house
M3 = Number of model 3 slip rings to make in-house
B1 = Number of model 1 slip rings to buy from competitor
B2 = Number of model 2 slip rings to buy from competitor
B3 = Number of model 3 slip rings to buy from competitor
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-52
Defining the Objective Function
Minimize the total cost of filling the order.
MIN: 50M1 + 83M2 + 130M3 + 61B1 + 97B2 + 145B3
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-53
Defining the Constraints
 Demand
Constraints
M1 + B1 = 3,000 } model 1
M2 + B2 = 2,000 } model 2
M3 + B3 =
 Resource
900 } model 3
Constraints
2M1 + 1.5M2 + 3M3 <= 10,000 } wiring
1M1 + 2.0M2 + 1M3 <= 5,000 } harnessing
 Nonnegativity
Conditions
M1, M2, M3, B1, B2, B3 >= 0
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-54
Implementing the Model
See file Fig3-17.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-55
An Investment Problem:
Retirement Planning Services, Inc.
 A client
wishes to invest $750,000 in the
following bonds.
Company
Return
Years to
Maturity
Acme Chemical
8.65%
11
1-Excellent
DynaStar
9.50%
10
3-Good
Eagle Vision
10.00%
6
4-Fair
Micro Modeling
8.75%
10
1-Excellent
OptiPro
9.25%
7
3-Good
Sabre Systems
9.00%
13
2-Very Good
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
Rating
2-56
Investment Restrictions
 No
more than 25% can be invested in
any single company.
 At least 50% should be invested in longterm bonds (maturing in 10+ years).
 No more than 35% can be invested in
DynaStar, Eagle Vision, and OptiPro.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-57
Defining the Decision Variables
X1 = amount of money to invest in Acme Chemical
X2 = amount of money to invest in DynaStar
X3 = amount of money to invest in Eagle Vision
X4 = amount of money to invest in MicroModeling
X5 = amount of money to invest in OptiPro
X6 = amount of money to invest in Sabre Systems
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-58
Defining the Objective Function
Maximize the total annual investment return.
MAX: .0865X1 + .095X2 + .10X3 + .0875X4 + .0925X5 + .09X6
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-59
Defining the Constraints
 Total
amount is invested
X1 + X2 + X3 + X4 + X5 + X6 = 750,000
 No
more than 25% in any one investment
Xi <= 187,500, for all i
 50%
long term investment restriction.
X1 + X2 + X4 + X6 >= 375,000
 35%
Restriction on DynaStar, Eagle Vision, and
OptiPro.
X2 + X3 + X5 <= 262,500
 Nonnegativity
conditions
Xi >= 0 for all i
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-60
Implementing the Model
See file Fig3-20.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-61
A Transportation Problem:
Tropicsun
Supply
Groves
Distances (in miles)
Capacity
21
Mt. Dora
275,000
Processing
Plants
1
Ocala
4
50
200,000
40
35
400,000
30
Eustis
Orlando
2
600,000
5
22
55
20
Clermont
300,000
3
Leesburg
25
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
6
225,000
2-62
Defining the Decision Variables
Xij = # of bushels shipped from node i to node j
Specifically, the nine decision variables are:
X14 = # of bushels shipped from Mt. Dora (node 1) to Ocala (node 4)
X15 = # of bushels shipped from Mt. Dora (node 1) to Orlando (node 5)
X16 = # of bushels shipped from Mt. Dora (node 1) to Leesburg (node 6)
X24 = # of bushels shipped from Eustis (node 2) to Ocala (node 4)
X25 = # of bushels shipped from Eustis (node 2) to Orlando (node 5)
X26 = # of bushels shipped from Eustis (node 2) to Leesburg (node 6)
X34 = # of bushels shipped from Clermont (node 3) to Ocala (node 4)
X35 = # of bushels shipped from Clermont (node 3) to Orlando (node 5)
X36 = # of bushels shipped from Clermont (node 3) to Leesburg (node 6)
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-63
Defining the Objective Function
Minimize the total number of bushel-miles.
MIN: 21X14 + 50X15 + 40X16 +
35X24 + 30X25 + 22X26 +
55X34 + 20X35 + 25X36
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-64
Defining the Constraints
 Capacity
constraints
X14 + X24 + X34 <= 200,000
X15 + X25 + X35 <= 600,000
X16 + X26 + X36 <= 225,000
 Supply
} Ocala
} Orlando
} Leesburg
constraints
X14 + X15 + X16 = 275,000
X24 + X25 + X26 = 400,000
X34 + X35 + X36 = 300,000
} Mt. Dora
} Eustis
} Clermont
 Nonnegativity conditions
Xij >= 0 for all i and j
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-65
Implementing the Model
See file Fig3-24.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-66
A Blending Problem:
The Agri-Pro Company
 Agri-Pro
has received an order for 8,000 pounds of
chicken feed to be mixed from the following feeds.
Percent of Nutrient in
Nutrient
Feed 1
Feed 2
Feed 3
Feed 4
Corn
30%
5%
20%
10%
Grain
10%
3%
15%
10%
Minerals
20%
20%
20%
30%
Cost per pound
$0.25
$0.30
$0.32
$0.15
 The
order must contain at least 20% corn, 15%
grain, and 15% minerals.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-67
Defining the Decision Variables
X1 = pounds of feed 1 to use in the mix
X2 = pounds of feed 2 to use in the mix
X3 = pounds of feed 3 to use in the mix
X4 = pounds of feed 4 to use in the mix
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-68
Defining the Objective Function
Minimize the total cost of filling the order.
MIN: 0.25X1 + 0.30X2 + 0.32X3 + 0.15X4
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-69
Defining the Constraints
 Produce
8,000 pounds of feed
X1 + X2 + X3 + X4 = 8,000
 Mix
consists of at least 20% corn
(0.3X1 + 0.5X2 + 0.2X3 + 0.1X4)/8000 >= 0.2
 Mix
consists of at least 15% grain
(0.1X1 + 0.3X2 + 0.15X3 + 0.1X4)/8000 >= 0.15
 Mix
consists of at least 15% minerals
(0.2X1 + 0.2X2 + 0.2X3 + 0.3X4)/8000 >= 0.15
 Nonnegativity
conditions
X1, X2, X3, X4 >= 0
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-70
A Comment About Scaling
Notice that the coefficient for X2 in the ‘corn’
constraint is 0.05/8000 = 0.00000625
 As Solver solves our problem, intermediate
calculations must be done that make coefficients
large or smaller.
 Storage problems may force the computer to use
approximations of the actual numbers.
 Such ‘scaling’ problems sometimes prevents
Solver from being able to solve the problem
accurately.
 Most problems can be formulated in a way to
minimize scaling errors...

Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-71
Re-Defining the Decision
Variables
X1 = thousands of pounds of feed 1 to use in the mix
X2 = thousands of pounds of feed 2 to use in the mix
X3 = thousands of pounds of feed 3 to use in the mix
X4 = thousands of pounds of feed 4 to use in the mix
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-72
Re-Defining the Objective
Function
Minimize the total cost of filling the order.
MIN: 250X1 + 300X2 + 320X3 + 150X4
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-73
Re-Defining the Constraints
 Produce
8,000 pounds of feed
X1 + X2 + X3 + X4 = 8
 Mix
consists of at least 20% corn
(0.3X1 + 0.5X2 + 0.2X3 + 0.1X4)/8 >= 0.2
 Mix
consists of at least 15% grain
(0.1X1 + 0.3X2 + 0.15X3 + 0.1X4)/8 >= 0.15
 Mix
consists of at least 15% minerals
(0.2X1 + 0.2X2 + 0.2X3 + 0.3X4)/8 >= 0.15
 Nonnegativity
conditions
X1, X2, X3, X4 >= 0
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-74
A Comment About Scaling
 Earlier
the largest coefficient in the
constraints was 8,000 and the smallest
is 0.05/8 = 0.00000625.
 Now the largest coefficient in the
constraints is 8 and the smallest is
0.05/8 = 0.00625.
 The problem is now more evenly scaled.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-75
The Assume Linear Model Option
The
Solver Options dialog box has an option
labeled “Assume Linear Model”.
When you select this option Solver performs some
tests to verify that your model is in fact linear.
These test are not 100% accurate & often fail as a
result of a poorly scaled model.
If Solver tells you a model isn’t linear when you
know it is, try solving it again. If that doesn’t work,
try re-scaling your model.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-76
Implementing the Model
See file Fig3-33.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-77
A Production Planning Problem:
The Upton Corporation

Upton is planning the production of their heavy-duty air
compressors for the next 6 months.
Month
1
2
3
4
5
6
Unit Production Cost
$240
$250
$265
$285
$280
$260
Units Demanded
1,000
4,500
6,000
5,500 3,500
4,000
Maximum Production 4,000
3,500
4,000
4,500 4,000
3,500
Minimum Production
1,750
2,000
2,250 2,000
1,750




2,000
Beginning inventory = 2,750 units
Safety stock = 1,500 units
Unit carrying cost = 1.5% of unit production cost
Maximum warehouse capacity = 6,000 units
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-78
Defining the Decision Variables
Pi = number of units to produce in month i, i=1 to 6
Bi = beginning inventory month i, i=1 to 6
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-79
Defining the Objective Function
Minimize the total cost production & inventory costs.
MIN: 240P1+ 250P2 + 265P3 + 285P4 + 280P5 + 260P6 +
3.6(B1+B2)/2 + 3.75(B2+B3)/2 + 3.98(B3+B4)/2 +
4.28(B4+B5)/2 + 4.20(B5+ B6)/2 + 3.9(B6+B7)/2
Note: The beginning inventory in any month is the same as the
ending inventory in the previous month.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-80
Defining the Constraints
 Production
levels
2,000 <= P1 <= 4,000 } month 1
1,750 <= P2 <= 3,500 } month 2
2,000 <= P3 <= 4,000 } month 3
2,250 <= P4 <= 4,500 } month 4
2,000 <= P5 <= 4,000 } month 5
1,750 <= P6 <= 3,500 } month 6
 Ending
Inventory (EI = BI + P - D)
1,500 <=
1,500 <=
1,500 <=
1,500 <=
1,500 <=
1,500 <=
B1 + P1 - 1,000 <= 6,000
B2 + P2 - 4,500 <= 6,000
B3 + P3 - 6,000 <= 6,000
B4 + P4 - 5,500 <= 6,000
B5 + P5 - 3,500 <= 6,000
B6 + P6 - 4,000 <= 6,000
} month 1
} month 2
} month 3
} month 4
} month 5
} month 6
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-81
Defining the Constraints
 Beginning
(cont’d)
Balances
B1 = 2750
B2 = B1 + P1 - 1,000
B3 = B2 + P2 - 4,500
B4 = B3 + P3 - 6,000
B5 = B4 + P4 - 5,500
B6 = B5 + P5 - 3,500
B7 = B6 + P6 - 4,000
Notice that the Bi can be computed directly from the Pi.
Therefore, only the Pi need to be identified as changing cells.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-82
Implementing the Model
See file Fig3-31.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-83
A Multi-Period Cash Flow Problem:
The Taco-Viva Sinking Fund - I



Taco-Viva needs to establish a sinking fund to pay $800,000
in building costs for a new restaurant in the next 6 months.
Payments of $250,000 are due at the end of months 2 and
4, and a final payment of $300,000 is due at the end of
month 6.
The following investments may be used.
Investment
A
B
C
D
Available in Month Months to Maturity Yield at Maturity
1, 2, 3, 4, 5, 6
1
1.8%
1, 3, 5
2
3.5%
1, 4
3
5.8%
1
6
11.0%
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-84
Summary of Possible Cash Flows
Cash Inflow/Outflow at the Beginning of Month
Investment
1
2
3
4
5
6
7
A1
-1
1.018
B1
-1 <_____> 1.035
C1
-1 <_____> <_____> 1.058
D1
-1 <_____> <_____> <_____> <_____> <_____> 1.11
A2
-1
1.018
A3
-1
1.018
B3
-1 <_____> 1.035
A4
-1
1.018
C4
-1 <_____> <_____> 1.058
A5
-1
1.018
B5
-1 <_____> 1.035
A6
-1
1.018
Req’d Payments $0
$0
$250
$0
$250
$0
$300
(in $1,000s)
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-85
Defining the Decision Variables
Ai = amount (in $1,000s) placed in investment A at the
beginning of month i=1, 2, 3, 4, 5, 6
Bi = amount (in $1,000s) placed in investment B at the
beginning of month i=1, 3, 5
Ci = amount (in $1,000s) placed in investment C at the
beginning of month i=1, 4
Di = amount (in $1,000s) placed in investment D at the
beginning of month i=1
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-86
Defining the Objective Function
Minimize the total cash invested in month 1.
MIN: A1
+ B1 + C1 + D1
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-87
Defining the Constraints
 Cash
Flow Constraints
1.018A1 – 1A2 = 0
1.035B1 + 1.018A2 – 1A3 – 1B3 = 250
1.058C1 + 1.018A3 – 1A4 – 1C4 = 0
1.035B3 + 1.018A4 – 1A5 – 1B5 = 250
1.018A5 –1A6 = 0
1.11D1 + 1.058C4 + 1.035B5 + 1.018A6 = 300
 Nonnegativity
} month 2
} month 3
} month 4
} month 5
} month 6
} month 7
Conditions
Ai, Bi, Ci, Di >= 0, for all i
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-88
Implementing the Model
See file Fig3-35.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-89
Network Modeling
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-90
Introduction
 A number
of business problems can be represented
graphically as networks.
 This chapter focuses on several types of network
flow problems:
–
–
–
–
–
Transshipment Problems
Shortest Path Problems
Maximal Flow Problems
Transportation/Assignment Problems
Generalized Network Flow Problems
 We
also consider a different type of network problem
called the Minimum Spanning Tree Problem
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-91
Characteristics of
Network Flow Problems
Network flow problems can be represented as a
collection of nodes connected by arcs.
 There are three types of nodes:
– Supply
– Demand
– Transshipment
 We’ll use negative numbers to represent supplies
and positive numbers to represent demand.

Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-92
A Transshipment Problem:
The Bavarian Motor Company
+100
Boston
2
$50
+60
$30
Newark
1
-200
Columbus
$40
3
$40
$35
+170
$30
Atlanta
5
Richmond
+80
4
$25
$45
$35
+70
Mobile
6
$50
$50
J'ville
7
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
-300
2-93
Defining the Decision Variables
For each arc in a network flow model
we define a decision variable as:
Xij = the amount being shipped (or flowing) from node i to node j
For example,
X12 = the number of cars shipped from node 1 (Newark) to node 2 (Boston)
X56 = the number of cars shipped from node 5 (Atlanta) to node 6 (Mobile)
Note: The number of arcs determine the number of variables
in a network flow problem!
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-94
Defining the Objective Function
Minimize total shipping costs.
MIN: 30X12 + 40X14 + 50X23 + 35X35
+40X53 + 30X54 + 35X56 + 25X65
+ 50X74 + 45X75 + 50X76
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-95
Constraints for Network Flow Problems:
The Balance-of-Flow Rules
For Minimum Cost Network
Flow Problems Where:
Total Supply > Total Demand
Apply This Balance-of-Flow
Rule At Each Node:
Inflow-Outflow >= Supply or Demand
Total Supply < Total Demand
Inflow-Outflow <=Supply or Demand
Total Supply = Total Demand
Inflow-Outflow = Supply or Demand
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-96
Defining the Constraints
 In
the BMC problem:
Total Supply = 500 cars
Total Demand = 480 cars
 So
for each node we need a constraint
of the form:
Inflow - Outflow >= Supply or Demand
 Constraint
for node 1:
–X12 – X14 >= -200
 This
(there is no inflow for node 1!)
is equivalent to:
+X12 + X14 <= 200
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-97
Defining the Constraints
 Flow
constraints
–X12 – X14 >= –200
+X12 – X23 >= +100
+X23 + X53 – X35 >= +60
+ X14 + X54 + X74 >= +80
+ X35 + X65 + X75 – X53 – X54 – X56 >= +170
+ X56 + X76 – X65 >= +70
–X74 – X75 – X76 >= –300

} node 1
} node 2
} node 3
} node 4
} node 5
} node 6
} node 7
Nonnegativity conditions
Xij >= 0 for all ij
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-98
Implementing the Model
See file Fig5-2.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-99
Optimal Solution to the BMC Problem
+100
Boston
2
$50
Newark
1
120
20
+60
$30
Columbus
80
3
-200
$40
$40
40
+170
Richmond
+80
4
Atlanta
5
$45
+70
Mobile
6
210
70
$50
J'ville
7
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
-300
2-100
Generalized Network Flow Problems
 In
some problems, a gain or loss occurs
in flows over arcs.
– Examples
 Oil
or gas shipped through a leaky pipeline
 Imperfections in raw materials entering a
production process
 Spoilage of food items during transit
 Theft during transit
 Interest or dividends on investments
 These
problems require some modeling
changes.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-101
Coal Bank Hollow Recycling
Recycling Process 1 Recycling Process 2
Material
Cost
Newspaper
$13
Mixed Paper
$11
White Office Paper $9
Cardboard
$13
Pulp Source
Recycling Process 1
Recycling Process 2
Demand
Yield
90%
80%
95%
75%
Cost
$12
$13
$10
$14
Yield
85%
85%
90%
85%
Supply
70 tons
50 tons
30 tons
40 tons
Newsprint
Packaging Paper
Print Stock
Cost Yield
$5 95%
$6 90%
60 tons
Cost Yield
$6 90%
$8 95%
40 tons
Cost Yield
$8 90%
$7 95%
50 tons
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-102
Network for Recycling Problem
-70
Newspaper
$13
1
$12
-50
Mixed
paper
2
-30
White
office
paper
3
$11
80%
5
75%
$9
85%
4
+60
7
90%
$8
90%
95%
Packing
paper
pulp
+40
8
$6
$10
$13
Cardboard
Newsprint
pulp
$6
85%
90%
Recycling
Process 2
6
-40
$5
Recycling
Process 1
95%
$13
95%
+0
90%
$8
90%
$7
85%
+0
95%
Print
stock
pulp
9
$14
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-103
+50
Defining the Objective Function
Minimize total cost.
MIN: 13X15 + 12X16 + 11X25 + 13X26
+ 9X35+ 10X36 + 13X45 + 14X46 + 5X57
+ 6X58 + 8X59 + 6X67 + 8X68 + 7X69
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-104
Defining the Constraints

Raw Materials
-X15 -X16 >= -70
-X25 -X26 >= -50
-X35 -X36 >= -30
-X45 -X46 >= -40

} node 1
} node 2
} node 3
} node 4
Recycling Processes
+0.9X15 +0.8X25 +0.95X35 +0.75X45 -X57 -X58 -X59 >= 0 } node 5
+0.85X16 +0.85X26 +0.9X36 +0.85X46 -X67 -X68 -X69 >= 0 } node 6

Paper Pulp
+0.95X57 +0.90X67 >= 60 } node 7
+0.90X57 +0.95X67 >= 40 } node 8
+0.90X57 +0.95X67 >= 50 } node 9
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-105
Implementing the Model
See file Fig5-17.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-106
Integer Linear Programming
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-107
An Employee Scheduling Problem:
Air-Express
Day of Week
Sunday
Workers Needed
18
Shift
Days Off
1
Sun & Mon
Wage
$680
Monday
27
2
Mon & Tue
$705
Tuesday
22
3
Tue & Wed
$705
Wednesday
26
4
Wed & Thr
$705
Thursday
25
5
Thr & Fri
$705
Friday
21
6
Fri & Sat
$680
Saturday
19
7
Sat & Sun
$655
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-108
Defining the Decision Variables
X1 = the number of workers assigned to shift 1
X2 = the number of workers assigned to shift 2
X3 = the number of workers assigned to shift 3
X4 = the number of workers assigned to shift 4
X5 = the number of workers assigned to shift 5
X6 = the number of workers assigned to shift 6
X7 = the number of workers assigned to shift 7
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-109
Defining the Objective Function
Minimize the total wage expense.
MIN: 680X1 +705X2 +705X3 +705X4 +705X5 +680X6 +655X7
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-110
Defining the Constraints
 Workers
required each day
0X1 + 1X2 + 1X3 + 1X4 + 1X5 + 1X6 + 0X7
0X1 + 0X2 + 1X3 + 1X4 + 1X5 + 1X6 + 1X7
1X1 + 0X2 + 0X3 + 1X4 + 1X5 + 1X6 + 1X7
1X1 + 1X2 + 0X3 + 0X4 + 1X5 + 1X6 + 1X7
1X1 + 1X2 + 1X3 + 0X4 + 0X5 + 1X6 + 1X7
1X1 + 1X2 + 1X3 + 1X4 + 0X5 + 0X6 + 1X7
1X1 + 1X2 + 1X3 + 1X4 + 1X5 + 0X6 + 0X7
 Nonnegativity
>= 18
>= 27
>= 22
>= 26
>= 25
>= 21
>= 19
} Sunday
} Monday
}Tuesday
} Weds.
} Thurs.
} Friday
} Saturday
& integrality conditions
Xi >= 0 and integer for all i
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-111
Implementing the Model
See file Fig6-14.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-112
Binary Variables
 Binary
variables are integer variables that
can assume only two values: 0 or 1.
 These
variables can be useful in a number
of practical modeling situations….
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-113
A Capital Budgeting Problem:
CRT Technologies
Expected NPV
Capital (in $000s) Required in
Project (in $000s)
Year 1 Year 2 Year 3 Year 4 Year 5
1
$141
$75
$25
$20
$15
$10
2
$187
$90
$35
$0
$0
$30
3
$121
$60
$15
$15
$15
$15
4
$83
$30
$20
$10
$5
$5
5
$265
$100
$25
$20
$20
$20
6
$127
$50
$20
$10
$30
$40
 The
company currently has $250,000 available to invest in new
projects. It has budgeted $75,000 for continued support for these
projects in year 2 and $50,000 per year for years 3, 4, and 5.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-114
Defining the Decision Variables
1, if project i is selected
X 
i  1,2,...,6
i  0, otherwise
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-115
Defining the Objective Function
Maximize the total NPV of selected projects.
MAX: 141X1 + 187X2 + 121X3 + 83X4 + 265X5 + 127X6
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-116
Defining the Constraints
 Capital
Constraints
75X1 + 90X2 + 60X3 + 30X4 + 100X5 + 50X6 <= 250 } year 1
25X1 + 35x2 + 15X3 + 20X4 + 25X5 + 20X6 <= 75 } year 2
20X1 + 0x2 + 15X3 + 10X4 + 20X5 + 10X6 <= 50
} year 3
15X1 + 0X2 + 15X3 + 5X4 + 20X5 + 30X6 <= 50
} year 4
10X1 + 30X2 + 15X3 + 5X4 + 20X5 + 40X6 <= 50 } year 5
 Binary
Constraints
All Xi must be binary
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-117
Implementing the Model
See file Fig6-17.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-118
Binary Variables & Logical Conditions

Binary variables are also useful in modeling a
number of logical conditions.
– Of projects 1, 3 & 6, no more than one may be
selected
 X1 + X3 + X6 <= 1
– Of projects 1, 3 & 6, exactly one must be selected
 X1 + X3 + X6 = 1
– Project 4 cannot be selected unless project 5 is
also selected
 X4 – X5 <= 0
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-119
The Fixed-Charge Problem

Many decisions result in a fixed or lump-sum cost
being incurred:
The cost to lease, rent, or purchase a piece of
equipment or a vehicle that will be required if a
particular action is taken.
The setup cost required to prepare a machine or
production line to produce a different type of
product.
The cost to construct a new production line or
facility that will be required if a particular decision
is made.
The cost of hiring additional personnel that will be
required if a particular decision is made.




Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-120
Example Fixed-Charge Problem :
Remington Manufacturing
Operation
Machining
Grinding
Assembly
Unit Profit
Setup Cost
Hours Required By:
Prod. 1 Prod. 2 Prod. 3
2
3
6
6
3
4
5
6
2
$48
$55
$50
$1000
$800
$900
Hours Available
600
300
400
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-121
Defining the Decision Variables
Xi = the amount of product i to be produced, i = 1, 2, 3
1, if Xi  0
Y 
i = 1, 2, 3
0, if X  0
i

i
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-122
Defining the Objective Function
Maximize total profit.
MAX: 48X1 + 55X2 + 50X3 – 1000Y1 – 800Y2 – 900Y3
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-123
Defining the Constraints

Resource Constraints
2X1 + 3X2 + 6X3 <= 600 } machining
6X1 + 3X2 + 4X3 <= 300 } grinding
5X1 + 6X2 + 2X3 <= 400 } assembly

Binary Constraints
All Yi must be binary

Nonnegativity conditions
Xi >= 0, i = 1, 2, ..., 6

Is there a missing link?
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-124
Defining the Constraints (cont’d)
 Linking
Constraints (with “Big M”)
X1 <= M1Y1
or
X1 - M1Y1 <= 0
X2 <= M2Y2
or
X2 - M2Y2 <= 0
X3 <= M3Y3
or
X3 - M3Y3 <= 0
If Xi > 0 these constraints force the
associated Yi to equal 1.
 If Xi = 0 these constraints allow Yi to equal 0
or 1, but the objective will cause Solver to
choose 0.
 Note that Mi imposes an upper bounds on Xi.
 It helps to find reasonable values for the Mi.

Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-125
Finding Reasonable Values for M1

Consider the resource constraints
2X1 + 3X2 + 6X3 <= 600 } machining
6X1 + 3X2 + 4X3 <= 300 } grinding
5X1 + 6X2 + 2X3 <= 400 } assembly
 What
is the maximum value X1 can assume?
Let X2 = X3 = 0
X1 = MIN(600/2, 300/6, 400/5)
= MIN(300, 50, 80)
= 50

Maximum values for X2 & X3 can be found
similarly.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-126
Summary of the Model
MAX: 48X1 + 55X2 + 50X3 - 1000Y1 - 800Y2 - 900Y3
Subject to:
2X1 + 3X2 + 6X3 <= 600
} machining
6X1 + 3X2 + 4X3 <= 300 } grinding
5X1 + 6X2 + 2X3 <= 400
} assembly
X1 - 50Y1 <= 0
X2 - 67Y2 <= 0
linking
X3 - 75Y3 <= 0
All Yi must be binary
Xi >= 0, i = 1, 2, 3
}
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-127
Potential Pitfall
 Do
not use IF( ) functions to model the
relationship between the Xi and Yi.
– Suppose cell A5 represents X1
– Suppose cell A6 represents Y1
– You’ll want to let A6 = IF(A5>0,1,0)
– This will not work with Solver!
 Treat
the Yi just like any other variable.
– Make them changing cells.
– Use the linking constraints to enforce the
proper relationship between the Xi and Yi.
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-128
Implementing the Model
See file Fig6-21.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-129
Nonlinear Programming
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-130
The Portfolio Optimization Problem

A financial planner wants to create the least risky
portfolio with at least a 12% expected return using
the following stocks.
Annual Return
Year
IBC
NMC
NBS
1 11.2%
8.0% 10.9%
2 10.8%
9.2% 22.0%
3 11.6%
6.6% 37.9%
4
-1.6% 18.5% -11.8%
5
-4.1%
7.4% 12.9%
6
8.6% 13.0% -7.5%
7
6.8% 22.0% 9.3%
8 11.9% 14.0% 48.7%
9 12.0% 20.5% -1.9%
10
8.3% 14.0% 19.1%
11
6.0% 19.0% -3.4%
12 10.2%
9.0% 43.0%
Avg 7.64% 13.43% 14.93%
IBC
NMC
NBS
Covariance Matrix
IBC
NMC
NBS
0.00258 -0.00025 0.00440
-0.00025 0.00276 -0.00542
0.00440 -0.00542 0.03677
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-131
Defining the Decision Variables
p1 = proportion of funds invested in IBC
p2 = proportion of funds invested in NMC
p3 = proportion of funds invested in NBS
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-132
Defining the Objective
Minimize the portfolio variance (risk).
n 1
n
MIN:

i =1
2 2
p
i i
2
n

i 1 j i 1
pp
ij i
j
 i2  the variance on investment i
 ij   ji = the covariance between investments i and j
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-133
Defining the Constraints
 Expected
return
0.0764 p1 + 0.1343 p2 + 0.1493 p3 >= 0.12
 Proportions
p1 + p2 + p3 = 1
p1, p2, p3 >= 0
p1, p2, p3 <= 1
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-134
Implementing the Model
See file Fig8-26.xls
Spreadsheet Modeling and Decision Analysis, 3e, by Cliff Ragsdale. © 2001 South-Western/Thomson Learning.
2-135