Optimization Models
Module 9
OPTIMIZATION MODELS
EXTERNAL
INPUTS
OUTPUT
MODEL
DECISION
INPUTS
Optimization models answer the question, “What
decision values give the best outputs?”
SCARCITY
CONSTRAINTS
DECISION
INPUTS
Optimization models are necessary because
resources are always limited.
LINEAR PROGRAMMING
CONSTRAINTS
2x + 6y < 150
x + .8y < 40
x + 1.5y < 50
DECISION
INPUTS
x
y
OUTPUT
MODEL
(objective
function)
10x + 18y
Linear programming is a special type of
optimization model that sets up constraints as
linear equations and solves them simultaneously
while maximizing an objective function.
EXAMPLE 1: BOB’S MUFFLER SHOP
Bob manufactures 2 types of mufflers, family and sport. Each family
muffler nets him $10 and each sport muffler nets him $18. These
numbers are known as the products’ ‘contribution margin’.
His profit is determined by the function $10f + $18s, where f and s are
the total number of each type of muffler sold. This is his objective
function.
His decision variables are the number of family and sport mufflers to
manufacture each month.
In an unconstrained situation, he would want to sell as many sport
mufflers as because of their higher contribution. However…
EXAMPLE 1: BACKGROUND
In order to create a muffler, Bob must use brackets, labor hours, and
alloy. The resources needed for each type are:
Family
Sports
Brackets
2.00
6.00
Labor hours
1.00
0.80
1
1.5
Max Alloy
In any given month, he is limited in all three resources to these amounts:
Quantity
Brackets
150
Labor hours
40
Max Alloy
50
These are the problem constraints.
EXAMPLE 1: MODELING CONSTRAINTS AS LINEAR EQUATIONS
These constraints can be modeled as a system of linear equations:
2f +
6s ≤ 150
f +
.8s ≤ 40
50f + 33.33s ≤ 50
(brackets)
(labor)
(alloy)
Bob also has an outstanding contract that forces him to manufacture 5
sports style mufflers each month and knows, from historical
performance, that demand for family mufflers in a given month will not
exceed 35.
These constraints can be modeled as:
f ≤ 35
s ≥ 5
(maximum demand)
(contractual obligation)
The objective function (profit) is:
10f + 18s
EXAMPLE 1: GRAPHING A CONSTRAINT
This situation can be represented graphically. Consider the constraint
imposed by brackets: 2f + 6s ≤ 150.
If zero family mufflers are manufactured, the equation changes to
(0)f + 6s ≤ 150, which reduces to s ≤ 25. Thus a maximum of 25 sports
mufflers can be made if no family mufflers are made.
Similarly, 75 family mufflers can be made if there are no sports mufflers
manufactured.
Connecting these points gives a graphical representation of this
constraint.
EXAMPLE 1: RANGE OF FEASIBILITY
Points on the line are scenarios in which all available brackets are used.
The area under the line is called the Range of Feasibility because it
covers all product mixes that the constraint allows.
Any point above the line is infeasible because Bob does not have
enough brackets to create that combination of mufflers.
Infeasible
Point:
Range of
Feasibility
(30 sport,
40 family)
EXAMPLE 1: GRAPHING ALL CONSTRAINTS
The rest of the constraints can then be added in a similar fashion.
2f +
6s ≤
f + .8s ≤
f + 1.5s ≤
f
≤
s ≥
150
40
50
35
5
(brackets)
(labor)
(alloy)
(maximum demand)
(contractual obligation)
EXAMPLE 1: THE COMPOSITE RANGE OF FEASIBILITY
The Range of Feasibility for the model is the area enclosed by all lines.
2f +
6s ≤
f + .8s ≤
f + 1.5s ≤
f
≤
s ≥
Range of
Feasibility
150
40
50
35
5
(brackets)
(labor)
(alloy)
(maximum demand)
(contractual obligation)
EXAMPLE 1: POTENTIAL PROFIT LINES
The profit function will be a line with the slope of -5/9, the ratio of the
contribution margin of family mufflers to sports mufflers.
Each line corresponds to a particular level of total profit. The further out
the line is, the greater the amount of profit.
Total Profit: $100
$190
$280
$370
$460
$550
EXAMPLE 1: FEASIBLE PROFIT LINES
By combining the profit
lines with the range of
feasibility, you can
identify the highest
profit line that touches
a feasible point.
EXAMPLE 1: OPTIMAL SOLUTION VIA GRAPHICAL METHOD
The optimal point is a mix of 25 family mufflers and 16.67 sport
mufflers. This combination will net Bob $550.
Optimal
point
EXAMPLE 1: SPREADSHEET MODEL
The
situation
can be
modeled
in Excel
and an
optimal
solution
found
with
Solver.
EXAMPLE 1: SPREADSHEET MODEL
Contribution
Margins
Resources
needed
for each
type of
muffler
Available
resources
Max
demand
Contract
Decisions:
How many
of each
muffler to
make
Modeled constraints
Profit
EXAMPLE 1: SOLVER INPUTS
Select the
profit cell
and choose
solver from
the tools
menu.
EXAMPLE 1: SOLVER INPUTS
Objective
function
(profit)
Click
Decision
variables
Constraints
EXAMPLE 1: SOLVER RESULTS
Solver
finds a
solution.
Select the
answer and
sensitivity
reports for
further
analysis.
EXAMPLE 1: OPTIMAL SOLUTION
Solver
found the
same
optimal
solution
as the
graphical
method.
EXAMPLE 1: ANSWER REPORT
Brackets: 2f + 6s ≤ 150
Labor:
f + .8s ≤ 40
The total
resources
used in the
optimal
solution
‘Binding’
means
all available
resources are
used
Slack is
the amount
of leftover
resources
All available brackets are used in the optimal solution, making it a binding
constraint. There are 1.67 labor hours left over, or slack.
EXAMPLE 1: SENSITIVITY ANALYSIS
The shadow price is the amount
of profit that an additional unit of
available resources would yield.
A non-binding constraint will
have a shadow price of 0
because there is already an
excess of the resource.
An additional available bracket
(a total of 151) will yield $1 of
profit. An additional labor hour
will yield no profit because there
is already an excess.
The allowable increase is the
amount of resources that
would have to be available to
change the optimal solution.
If 50 additional brackets are
available, the optimal product
mix will change.
The allowable decrease is the
amount of fewer resources that
will change the optimal mix.
EXAMPLE 1: SENSITIVITY ANALYSIS
If an additional unit of resource
can be obtained for less than the
shadow price, then the
difference will be profit.
The allowable increase and
decrease tells you how
sensitive the decision inputs of
the optimal solution (final value
column) are to the right hand
sides of each constraint.
EXAMPLE 1: RANGE OF OPTIMALITY
Profit:
10 f + 18 s
The contribution margin
for each type of muffler
may be uncertain.
The range of optimality is the range of contribution margins (coefficients in
the objective function) in which a particular solution is the optimal solution.
EXAMPLE 1: RANGE OF OPTIMALITY
Profit:
8 ≤ contribution margin ≤ 14
Sports mufflers range:
10 f + 18 s
15 ≤ contribution margin ≤ 30
$18 + $12 = $30
$18 - $3 = $15
The range of optimality tells you how sensitive the decision inputs for the
optimal solution are to variation in the coefficients of the objective function.