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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. DECISION INPUTS. CONSTRAINTS.

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optimization models2

OPTIMIZATION MODELS

EXTERNAL

INPUTS

OUTPUT

MODEL

DECISION

INPUTS

Optimization models answer the question, “What decision values give the best outputs?”

scarcity

SCARCITY

DECISION

INPUTS

CONSTRAINTS

Optimization models are necessary because resources are always limited.

linear programming

LINEAR PROGRAMMING

MODEL

(objective

function)

10x +18y

DECISION

INPUTS

CONSTRAINTS

OUTPUT

2x + 6y < 150

x

x + .8y < 40

x + 1.5y < 50

y

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

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

EXAMPLE 1: BACKGROUND

In order to create a muffler, Bob must use brackets, labor hours, and alloy. The resources needed for each type are:

In any given month, he is limited in all three resources to these amounts:

These are the problem constraints.

example 1 modeling constraints as linear equations

EXAMPLE 1: MODELING CONSTRAINTS AS LINEAR EQUATIONS

These constraints can be modeled as a system of linear equations:

2f + 6s ≤ 150 (brackets)

f + .8s ≤ 40 (labor)

50f + 33.33s ≤ 50 (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 (maximum demand)

s ≥ 5 (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.

EXAMPLE 1: GRAPHING A CONSTRAINT

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.

EXAMPLE 1: RANGE OF FEASIBILITY

Infeasible

Point:

(30 sport,

40 family)

Range of

Feasibility

example 1 graphing all constraints

The rest of the constraints can then be added in a similar fashion.

EXAMPLE 1: GRAPHING ALL CONSTRAINTS

2f + 6s ≤ 150 (brackets)

f + .8s ≤ 40 (labor)

f + 1.5s ≤ 50 (alloy)

f ≤ 35 (maximum demand)

s ≥ 5 (contractual obligation)

example 1 the composite range of feasibility

The Range of Feasibility for the model is the area enclosed by all lines.

EXAMPLE 1: THE COMPOSITE RANGE OF FEASIBILITY

2f + 6s ≤ 150 (brackets)

f + .8s ≤ 40 (labor)

f + 1.5s ≤ 50 (alloy)

f ≤ 35 (maximum demand)

s ≥ 5 (contractual obligation)

Range of

Feasibility

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.

EXAMPLE 1: POTENTIAL PROFIT LINES

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: FEASIBLE PROFIT LINES

example 1 spreadsheet model

EXAMPLE 1: SPREADSHEET MODEL

The

situation

can be

modeled

in Excel

and an

optimal

solution

found

with

Solver.

example 1 spreadsheet model16

EXAMPLE 1: SPREADSHEET MODEL

Contribution

Margins

Available resources

Resources

needed

for each

type of

muffler

Profit

Max

demand

Contract

Decisions:

How many

of each

muffler to

make

Modeled constraints

example 1 solver inputs

EXAMPLE 1: SOLVER INPUTS

Select the profit cell and choose solver from the tools menu.

example 1 solver inputs18

Objective

function (profit)

EXAMPLE 1: SOLVER INPUTS

Click

Decision

variables

Constraints

example 1 solver results

EXAMPLE 1: SOLVER RESULTS

Solver finds a solution. Select the answer and sensitivity reports for further analysis.

example 1 optimal solution

EXAMPLE 1: OPTIMAL SOLUTION

Solver found the same optimal solution

as the graphical method.

example 1 answer report

EXAMPLE 1: ANSWER REPORT

‘Binding’ means

all available

resources are

used

Brackets: 2f + 6s ≤ 150

Labor: f + .8s ≤ 40

The total

resources

used in the

optimal

solution

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

EXAMPLE 1: SENSITIVITY ANALYSIS

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.

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.

example 1 sensitivity analysis23

EXAMPLE 1: SENSITIVITY ANALYSIS

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.

If an additional unit of resource can be obtained for less than the shadow price, then the difference will be profit.

example 1 range of optimality

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 optimality25

EXAMPLE 1: RANGE OF OPTIMALITY

Profit: 10 f + 18 s

8 ≤ contribution margin ≤ 14

15 ≤ contribution margin ≤ 30

Sports mufflers range:

$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.