DECISION MODELING
WITH
MICROSOFT EXCEL
Chapter 3
Linear Optimization
Copyright 2001
Prentice Hall
Linear Optimization
A constrained optimization model takes the form of a
performance measure to be optimized over a range of
feasible values of the decision variables.
The feasible values of the decision variables are
determined by a set of inequality constraints.
Values of the decision variables must be chosen such
that the inequality constraints are all satisfied while
either maximizing or minimizing the desired
performance variable.
These models can contain tens, hundreds, or thousands
of decision variables and constraints.
Linear Optimization
Very efficient search techniques exist to optimize
constrained linear models.
These models are historically called linear programs
(LP). In this chapter we will:
1. Develop techniques for formulating LP models
2. Give some recommended rules for expressing LP
models in a spreadsheet that facilitates
application of Excel’s Solver
3. Use Solver to optimize spreadsheet LP models
Formulating LP Models
Every linear programming model has two important
features:
Objective Function
Constraints
A single performance
measure to be maximized
or minimized
(e.g., maximize profit,
minimize cost)
Constraints are limitations
or requirements on the set
of allowable decisions.
Constraints may be further
classified into physical,
economic, or policy
limitations or
requirements.
Formulating LP Models
The first step in model formulation is the development
of the constraints.
For example:
Investment decisions are restricted by the amount
of capital and government regulations.
Management decisions are restricted by plant
capacity and availability of resources.
Staffing and flight plans of an airline are restricted
by the maintenance needs of the planes and the
number of employees on hand.
The use of a certain type of crude oil in producing
gasoline is restricted by the characteristics of the
gasoline (e.g., octane rating, etc.)
Formulating LP Models
A constrained optimization model represents the
problem of allocating scarce resources in such a way as
to optimize an objective of interest.
To illustrate this, we will return to the Oak Products
model from Chapter 2. However, we will start with a
simplified model.
Oak Products, Inc.
Based on economic forecasts for the next week, it has
been determined that it will be possible to sell as many
Captains or Mates chairs as the firm can produce.
So, the question becomes: How many Captains and
Mates chairs should be produced given our constraints
in order to maximize next week’s profit contribution?
Formulating LP Models
Consider the following major factors:
1. The unit contribution margin (price minus unit
variable cost) is $56 for each Captain sold and
$40 for each Mate.
2. Long dowels, short dowels, legs and one of two
types of seats are needed to assemble the chairs.
3. We have a limited inventory of 1280 long dowels
and 1600 short dowels for next week’s
production.
Each Captain uses 8 long and 4 short dowels.
Each Mate uses 4 long and 12 short dowels.
Formulating LP Models
4. The inventory of legs is 760 units and each chair
produced of either type uses 4 legs.
5. The inventory of heavy and light seats is 140 and
120, respectively. Each Captain produced uses a
heavy seat and each Mate uses a light seat.
6. In order for management to honor an agreement
with the union, the total number of chairs
produced of both types cannot fall below 100.
Let’s summarize the information given so far in a table:
Given these considerations, now decide how many
Captains and Mates to produce next week.
This is called an optimal product mix problem or an
optimal production plan.
The first step in solving this is to identify the constraints
and objective function.
For notation purposes, let C = number of Captains
chairs and M = number of Mates chairs to be produced.
Start with the number of long dowels:
Both types of chairs require long dowels.
So, we can mathematically describe this as:
8(# Captains produced) + 4(# Mates produced) = 1280
or
8C + 4M = 1280
Once again, here is the equation for the long dowels:
8C + 4M = 1280
However, since only 1280 long dowels are available and
we can use at most, only that amount, we must make
this equation an inequality in order to satisfy this
restriction. In this case:
8C + 4M < 1280
We use the < (less than or equal to) inequality because,
we can use all 1280 long dowels, we can use less than
that amount, but we will never be able to use more than
1280.
This is called an inequality constraint.
Left hand side (LHS) is called a constraint function.
Right hand side (RHS) specifies the limitation.
Now let’s consider the rest of the constraints:
Short Dowels: We need 4 for each Captain and 12 for
each mate and we cannot exceed 1600 short dowels.
4C + 12M < 1600
Legs: Each chair requires 4 legs and we have 760 in
inventory.
4C + 4M < 760
Heavy Seat: These seats are only used for Captains
chairs and only 1 per chair is needed. 140 are available.
1C + 0M < 140
or
1C < 140
Light Seat: These seats are only used for Mates chairs
and only 1 per chair is needed. 120 are available.
1M < 120
We need to consider one other constraint not given in
the table.
Remember that there is a union agreement that we have
to honor: The total number of Captains and Mates
chairs cannot fall below 100.
This means that we can make exactly 100 chairs, more
than 100 chairs, but never less than 100 chairs.
Therefore, the inequality that we will use will be >
(greater than or equal to).
C + M > 100
Note that there are no coefficients in front of C or M
because we can make any combination of Captains or
Mates chairs – as long as the total number does not fall
below 100.
Nonnegativity Conditions
Since it does not make sense to produce a negative
number of Captain or Mate chairs, we must include two
additional conditions called Nonnegativity Conditions:
C > 0 and M > 0
Nonnegativity means that the resulting values can be 0
or positive, but not negative.
Summary of Constraints
Long Dowels
8C + 4M < 1280
Short Dowels
4C + 12M < 1600
Legs
4C + 4M < 760
Heavy Seat
C
Light Seat
< 140
M < 120
Union Requirement
C+
M > 100
Nonnegativity
C > 0, M > 0
Evaluating Various Decisions
The choice of values for the pair of variables (C,M) is
called a decision.
C and M are called decision variables because they are
quantities that you can control.
In this problem, a decision is a production mix.
The mix, or decision, is a combination of Captains and
Mates that will not violate one of the constraints.
Of the infinitely many nonnegative pairs of numbers
(C,M), including fractional values, some pairs, or
decisions, will violate at least one of the constraints,
and some will satisfy all the constraints.
Evaluating Various Decisions
Only nonnegative decisions that satisfy all the
constraints are allowable. These decisions are called
feasible decisions.
The Objective Function
Every linear programming model has a specific
objective as well as constraints. In this case, we would
like to maximize next week’s profit.
Profit contribution comes from the sale of two chairs:
Captains and Mates. In fact, we will receive $56 for
every Captain and $40 for every Mate sold. We can
mathematically state this as:
Total profit contribution = 56C + 40M
Evaluating Various Decisions
An Optimal Solution
An optimal solution is one that gives the largest total
profit contribution.
An optimal decision is a decision that will maximize the
total weekly profit contribution relative to the set of all
possible feasible decisions.
Since total profit contribution is a function of the
variables C,M, we refer to the mathematical expression
as the payoff or objective function.
The goal is to find feasible values of C and M that
optimize (i.e., maximize) the objective function:
Max 56C + 40M
Symbolic LP Model
Max 56C + 40M (objective function)
Subject to (s.t.)
8C + 4M < 1280 (Long Dowels Restriction)
4C + 12M < 1600 (Short Dowels Restriction)
C + M > 100 (Minimum Production)
4C + 4M < 760 (Legs Restriction)
C < 140 (Heavy Seats Restriction)
M < 120 (Light Seats Restriction)
C > 0 and M > 0 (Nonnegativity Conditions)
Symbolic
Linear Functions
LP Model
Max 56C + 40M (objective function)
Subject to (s.t.)
8C + 4M < 1280
4C + 12M < 1600
C+M
> 100
4C + 4M < 760
C
< 140
M
< 120
Notice that the
constraint functions
and the objective function
are linear functions of the two
decision variables.
The graph of a linear function
of the two variables is a
straight line.
C > 0 and M > 0
Variable X
Linear Functions
A linear function is one in which
each variable appears only once
there are no products or quotients of variables, no
exponents (other than 1), no logarithmic,
exponential or trigonometric terms, and no IF()
statements in Excel.
Examples of
Linear Equations
8C + 4M
4C + 12M
C+M
Examples of
Nonlinear Equations
14C + 12CM
9C2 + 8M
6 C +M
IF(), MAX(), MIN(), LN(),
ABS() Excel functions
Linear Programming
Remember,
1. A linear program always has an objective
function (to be either maximized or minimized)
and constraints).
2. All functions in the problem (objective and
constraints) are linear functions.
Integrality Considerations
In LP models where fractional solutions are not directly
meaningful, there are four possible recourses:
1. Add an “integrality condition” to the LP model,
which forces one or more decision variables to
take on only integer values. This results in an
Integer Optimization Model.
2. Solve the model as an ordinary LP and then round
any decision variable for which a fractional
answer cannot be implemented.
Integrality Considerations
3. Consider the one-week Oak Product model results
to be an average week’s production for an
ongoing multiple week situation. The whole units
are produced and the fraction is carried over to
the next week as a “work-in-process.”
4. Consider the one-week Oak Product model results
to be for planning purposes only and not
operational decisions to be implemented per se.
Take into account other information when making
the final decision.
The Art of LP Model Formulation
Create a verbal model to translate a managerial
situation into a symbolic model:
1. Express the objective and its performance
measure objective function in words.
2. Express each constraint in words.
3. Verbally identify the decision variables. Ask
yourself “What decisions must be made in order
to optimize the objective function?”
4. Express each decision variable in the constraint
using symbols.
5. Express each decision variable in the objective
function using symbols.
Finally, check your work for consistency of units of
measurement.
Sunk vs. Variable Cost
There are often two types of costs: sunk costs (fixed
costs) and variable costs.
Only the variable costs are relevant in optimization
models.
Unit contribution margin = per unit revenue – per unit variable cost
Sunk costs affect only the accounting report of income
or net profit in financial statements. They play no part
in future decisions.
Oak Product Spreadsheet Model
Here is a spreadsheet version of the simplified Oak
Product model containing the previously specified
objective function and constraints.
Note that
producing
110
Captains
and 90
Mates
violates
the Legs
constraint.
Labels are used to clarify the meaning of other entries.
The numbers represent the coefficients, parameters, and
decision values.
Formulas represent the objective function and constraints.
Slack is the difference between the constraint function and
the right-hand side computed so that it is non-negative.
More on Slack:
Ending Inventory is another
name for Slack.
The purpose of these
calculations is to give an
indication of how close any
constraint is to binding
(evaluated as an equality).
Zero Slack indicates a
binding constraint.
To calculate Slack:
For a < constraint, slack is the right-hand side minus
the left hand side.
For a > constraint, slack is the left-hand side minus
the right hand side.
Optional but useful, Slack tells us if any constraints are
violated, resulting in an infeasible production plan.
“What if?” Projections
You can use the spreadsheet model to perform “What
if?” analyses.
Type production values for the Captain and Mate chairs
and observe the resulting weekly Profit.
Also make sure that the slack is not negative.
Optimizing the Spreadsheet
Note that you can enter an infinite number of decision
values for C and M in a “What if?” fashion. However,
with Solver, you can transform any spreadsheet LP
model into an optimized model with just a few mouse
clicks. Here is the optimal model:
The LP Model and Spreadsheet Modeling
To produce an Excel LP model, follow these steps:
1. Write out the model on paper as a symbolic LP.
Examine your written formulation and look for
errors in the logic.
2. Use the symbolic LP model as a guide in creating
the Excel representation. Further debug the Excel
model by changing the decision variables and
checking for obvious errors.
3. Try to optimize the model with Solver. If there is a
problem, Solver will give an error message.
Overview of Solver
Solver is an add-in package to Excel.
Solver numerically optimizes constrained models
using a technique called a mathematical
programming algorithm (Simplex Method).
To use Solver, you must set up your spreadsheet
model in the appropriate form.
Solver can optimize both linear and nonlinear
models. We will focus only on linear models.
For LP optimization, every formula in your model
must be linear.
Solver
Flowchart
START EXCEL
BUILD OR RETRIEVE YOUR
OPTIMIZATION MODEL
SAVE YOUR WORKBOOK!!
CHOOSE “Solver…” IN THE “Tools” MENU
MODIFY MODEL
SPECIFY IN SOLVER DIALOG BOX:
1. CELL TO BE OPTIMIZED
2. CHANGING CELLS
3. CONSTRAINTS
IN OPTIONS DIALOG, CLICK “ASSUME LINEAR MODEL”
& CLICK THE “OK” BUTTON
CLICK ON “SOLVE” BUTTON TO BEGIN OPTIMIZATION
REVIEW SOLVER COMPLETION MESSAGE
Solver
Flowchart
NO
SOLVER FOUND
OPTIMUM SOLUTION?
YES
CLICK “KEEP SOLVER SOLUTION”
& CLICK “OK” BUTTON
YES
WANT TO CHANGE MODEL
AND RE-OPTIMIZE?
NO
SAVE FINAL MODEL AND
EXIT EXCEL
Overview of Solver
LP Modeling Terminology
Solver Terminology
Objective function
Set Cell
Decision variables
Changing Variable Cells
Constraints
Constraint functions (LHS)
RHS
LP Model
Constraints
Constraint Cell Reference
Constraint or Bound
Assume Linear Model or
Standard Simplex LP
NOTE: if you get a negative decision and it is not meaningful
to your model, be sure to specify the nonnegativity
constraints in your LP model before optimizing with Solver.
Optimizing with Solver
To optimize the Oak Products model with Solver, first
open the file in Excel and then choose Tools – Solver…
The Solver Parameters dialog box will appear.
By default, Max is selected (for maximization) and the cursor
is in the first edit field: Set Target Cell.
Look for the Premium button. If it is not there, then you have
not installed Premium Solver for Education (available from
the CD). Please install this version.
Clicking on Premium allows you to specify the type of
optimization that it will perform.
We will use the default Standard Simplex LP
optimization.
With your cursor in the Set Target Cell: edit field, specify
the cell to be optimized (i.e., your model’s performance
measure).
The easiest way to do this is to move the dialog (drag
the title bar) so that cell D4 is exposed and then click on
that cell.
You can also click on the
button in the edit field
of the dialog to collapse the dialog, click on the cell, and
then click on the
button to expand the dialog.
The Equal to: field allows you to specify the type of
optimization. You can either maximize or minimize the
performance variable or cause the Target Cell to be
equal to a value of your choosing (select Value of:).
Specify the Oak Product model’s decision variables
(cells B4:C4) in the By Changing Cells: edit field.
To specify the constraints, click on the Add button to
open the Add Constraint dialog.
For the LHS of the
constraint, specify
the cell ranges for
the Total LHS of
either one
constraint or a
group of similar
constraints (i.e.,
constraints with the
same inequality) in
the Cell Reference:
edit field of the Add
Constraint dialog.
For the RHS of the constraint, specify the cell ranges of
either one resource limitation or a group of similar
limitations in the Constraint: edit field.
Note that when
specifying many
constraints at the
same time, the
number of cells
referenced in the
LHS must equal
the number in the
RHS.
Click Add to add these constraints to Solver.
Finally, add the
Chair Production
constraint.
Note that the
inequality is > for
this constraint.
Here is the resulting Solver dialog after adding all of the
constraints:
Now, in the Solver Parameters dialog, click the Options
button to specify a linear model.
In the resulting Solver Options dialog, click on the
following options:
Assume Linear Model
(Specifies an LP
model, same as
Standard Simplex
LP)
Assume Non-Negative
(Apply nonnegativity
constraints)
Use Automatic Scaling
(to be discussed later)
Click OK to return to the Solver Parameters window and
then click Solve to start the optimization.
Remember that this is an iterative technique and may
take a few seconds or a few minutes depending on the
size of the model.
When completed, the Solver Results dialog will appear.
It is important to check to see if Solver found a solution and
if all constraints and optimality conditions were satisfied.
This information will be displayed in the first sentence in this
dialog. ALWAYS READ THIS SENTENCE!
Upon the successful completion of the Solver program,
you have the following options:
Keep the Solver Solution
Restore the Original Values (throw the solution away)
Receive up to three reports on the solution (each
formatted as a new worksheet added to the Excel
workbook.
NOTE: The Premium edition may also list an Infeasibility
Report and a Non-Linear Report if there is a problem.
Now, select the Answer report, accept the default Keep
Solver Solution, and click OK.
The resulting report will be placed in the Excel workbook
as a worksheet with the tab name: Answer Report 1
Note: this report has been edited to remove unneeded
blank lines.
Optimal decision values
Maximum Total Profit
Solver updates the spreadsheet to reflect the analysis results.
Note that these cells have now changed to reflect the
inventory consequences of the optimal production
decisions. Remember, a Slack (End. Inv.) = 0 indicates
the constraint is “binding at optimality.”
Recommendations for Solver
Three LP modeling habits you should develop for better
use of Solver:
1. Make sure the numbers in your LP model are
scaled such that the difference between the
smallest and largest numbers in the spreadsheet
is no more than 6 or 7 digits of precision (e.g., .05
and 10.00 is an acceptable range while .05 and
1,000,000 is not).
Recommendations for Solver
2. All RHS’s in the Constraints section of the Solver
Parameters dialog should contain cell references
(to the RHS cells in the spreadsheet model itself),
and
The RHS cells on the spreadsheet model itself
should contain constants and not formulas (or
more precisely, no formulas that involve decision
variables directly or indirectly).
Recommendations for Solver
3. Use Excel’s Range Naming commands to give
range names to the performance measure cell, its
decision cells, its constraint function (Total LHS)
cells, and its RHS cells.
These names will automatically be substituted for
the corresponding cell ranges in the Solver
Parameters dialog.
Solver is used on the more complex version of the Oak
Products model from Chapter 2. The optimal solution is
given below.
A Transportation Example
This is a class of LP problems called Network Problems
(notice later that all the coefficients in the constraints
are equal to one).
A company has 2 plants and 3 warehouse outlet stores.
Consider the following information:
Plant 1 can supply at most 100 units
Plant 2 can supply at most 200 units (of same
product)
Maximum sales quantities and price/unit sold are:
Warehouse Outlet 1: 150 units, $12/unit
Warehouse Outlet 2: 200 units, $14/unit
Warehouse Outlet 3: 350 units, $15/unit
A Transportation Example
The cost of manufacturing one unit at plant i and
shipping it to warehouse outlet j is as follows:
TO WAREHOUSE ($)
FROM PLANT
1
2
3
1
8
10
9
2
7
9
11
Given this information, determine how many units
should be shipped from each plant to each warehouse
so as to maximize profit.
A Transportation Example
The decision variables are:
Xij = number of units manufactured and sent from
plant i to warehouse outlet j
For each Xij (decision variable), the corresponding unit
profit is equal to :
Sales price/unit sold at warehouse j – cost of making
and shipping a unit from plant i to warehouse outlet j.
For example:
sales revenue = $12/unit, shipping cost = $8/unit,
so
Profit = $12 - $8 = $4/unit
Symbolic LP Model
Max 4X11 + 5X21 + 4X12 + 5X22 + 3X13 + 4X23
Subject to (s.t.)
X11 + X12 + X13 < 100 (supply capacity, plant 1)
X21 + X22 + X23 < 200 (supply capacity, plant 2)
X11 + X21 < 150 (sales limit, warehouse outlet 1)
X12 + X22 < 200 (sales limit, warehouse outlet 2)
X13 + X23 < 350 (sales limit, warehouse outlet 3)
Xij < 0 (for all i,j)
Here is the spreadsheet model associated with the
previous information.
Here are the spreadsheet formulas for this model:
Now, perform the Solver analysis to find the
optimal solution and maximum profit.
Solver Options
A Blending Example
Eastern Steel: A Blending Example.
In this example, we are concerned not with maximizing
profit, but with minimizing cost. Consider the following
information:
Iron ore from 4 different mines is blended to make a
metal alloy. In order to produce a blend with suitable
tensile qualities, the following minimum requirements
must be met on three basic elements (A, B, and C):
BASIC ELEMENT
A
B
C
MIN. REQUIREMENT PER TON OF BLEND
(pounds of each element)
5
100
30
A Blending Example
The amount of these elements in the ores from the three
different mines is given below:
MINE
(pounds per ton of each element)
BASIC ELEMENT
A
B
C
1
10
90
45
2
3
150
25
3
8
75
20
4
2
175
37
A Blending Example
We must now take into consideration the cost of ore
from each mine:
MINE
DOLLAR COST PER TON OF ORE
1
2
3
4
800
400
600
500
A Blending Example
Eastern Steel’s objective is to come up with a least-cost
feasible blend of the ores.
First, specify the decision variables:
T1 = fraction of a ton to be chosen from mine 1
T2 = fraction of a ton to be chosen from mine 2
T3 = fraction of a ton to be chosen from mine 3
T4 = fraction of a ton to be chosen from mine 4
A Blending Example
Now, specify the amounts of the basic elements in 1 ton
of blend:
Pounds of Element in 1 Ton of Blend
A = 10T1 + 3T2 + 8T3 + 2T4
B = 90T1 + 150T2 + 75T3 +
175T
4
C = 45T
+ 25T + 20T + 37T
1
2
3
4
Finally, create the constraints based on this information:
10T1 + 3T2 + 8T3 + 2T4 > 5
90T1 + 150T2 + 75T3 + 175T4 > 100
45T1 + 25T2 + 20T3 + 37T4 > 30
A Blending Example
There are two more constraints to consider:
Nonnegativity Constraint:
T1 , T2 , T3 , T4 > 0
Equality Constraint (also called a material balance
condition) restricts the values of the decision
variables in such a way that the left-hand side
exactly equals the right-hand side.
T1 + T2 + T 3 + T4 = 1
Note that the constraints in a linear programming
model can be equalities as well as inequalities.
A Blending Example
The objective function is to minimize cost of any blend
and is given by:
Cost of 1 ton of blend = 800T1 + 400T2 + 600T3 + 500T4
Now let’s put it all together and write the complete
symbolic model.
Symbolic LP Model
Min 800T1 + 400T2 + 600T3 + 500T4
Subject to (s.t.)
10T1 + 3T2 + 8T3 + 2T4 > 5
90T1 + 150T2 + 75T3 + 175T4 > 100
45T1 + 25T2 + 20T3 + 37T4 > 30
T1 + T2 + T3 + T4 = 1
T1 , T2 , T3 , T4 > 0
The spreadsheet model and resulting Solver analysis:
Note that this is a
minimization model.
Solver Options
The spreadsheet formulas used for this model: