Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik

Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik

Riset Operasiomal- dewiyani

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Methods to Solve LP Problems Graphical Method Simplex Method

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Resource Requirements

Product Labor (hr/unit)

Clay (lb/unit)

Profit ($/unit)

Bowl 1 4 40

Mug 2 3 50

Problem DefinitionA Maximization Model Example (1 of 3) Product mix problem - Beaver Creek Pottery

Company How many bowls and mugs should be produced to

maximize profits given labor and materials constraints?

Product resource requirements and unit profit:

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Problem DefinitionA Maximization Model Example (2 of 3)Resource 40 hrs of labor per dayAvailability: 120 lbs of clay

Decision x1 = number of bowls to produce per dayVariables: x2 = number of mugs to produce per day

Objective Maximize Z = $40x1 + $50x2

Function: Where Z = profit per day

Resource 1x1 + 2x2 40 hours of laborConstraints: 4x1 + 3x2 120 pounds of clay

Non-Negativity x1 0; x2 0 Constraints:

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Problem DefinitionA Maximization Model Example (3 of 3)Complete Linear Programming Model:

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x2 + 3x2 120 x1, x2 0

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A feasible solution does not violate any of the constraints: Example x1 = 5 bowls x2 = 10 mugs Z = $40x1 + $50x2 = $700

Labor constraint check: 1(5) + 2(10) = 25 < 40 hours, within constraint

Clay constraint check: 4(5) + 3(10) = 70 < 120 pounds, within constraint

Feasible Solutions

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An infeasible solution violates at least one of the constraints: Example x1 = 10 bowls x2 = 20 mugs Z = $1400

Labor constraint check: 1(10) + 2(20) = 50 > 40 hours, violates constraint

Infeasible Solutions

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Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty).

Graphical methods provide visualization of how a solution for a linear programming problem is obtained.

Graphical Solution of Linear Programming Models

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Coordinate AxesGraphical Solution of Maximization Model (1 of 13)

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.1Coordinates for Graphical Analysis

After defining the axes, begin drawing the constraints• What is the x or y intercept?• What is the slope?

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Labor ConstraintGraphical Solution of Maximization Model (2 of 13)

Figure 2.1Graph of Labor Constraint

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

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Labor Constraint AreaGraphical Solution of Maximization Model (3 of 13)

Figure 2.3Labor Constraint Area

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

After the labor constraint, let’s move onto the clay constraint• What is the x or y intercept?• What is the slope?

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Clay Constraint AreaGraphical Solution of Maximization Model (4 of 13)

Figure 2.4Clay Constraint Area

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

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Both ConstraintsGraphical Solution of Maximization Model (5 of 13)

Figure 2.5Graph of Both Model Constraints

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

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Don’t forget the non-negative constraintsGraphical Solution of Maximization Model (6 of 13)

Figure 2.6Feasible Solution Area

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0 X2 = 0

X1 = 0

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Feasible Solution AreaGraphical Solution of Maximization Model (7 of 13)

Figure 2.6Feasible Solution Area

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

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Objective Solution = $800Graphical Solution of Maximization Model (8 of 13)

Figure 2.7Objection Function Line for Z = $800

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Arbitrarily set the objective function as $800

X2 = 0

X1 = 0

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Alternative Objective Function Solution LinesGraphical Solution of Maximization Model (9 of 13)

Figure 2.8Alternative Objective Function Lines

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

X2 = 0

X1 = 0

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Optimal SolutionGraphical Solution of Maximization Model (10 of 13)

Figure 2.9Identification of Optimal Solution

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

What if we forgot to include the non-negative constraints?

Above and Beyond

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Optimal Solution CoordinatesGraphical Solution of Maximization Model (11 of 13)

Figure 2.10Optimal Solution Coordinates

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

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Corner Point SolutionsGraphical Solution of Maximization Model (12 of 13)

Figure 2.11Solution at All Corner Points

Maximize Z = $40x1 + $50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Why are corner points important?

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Optimal Solution for New Objective FunctionGraphical Solution of Maximization Model (13 of 13)

Figure 2.12Optimal Solution with Z = 70x1 + 20x2

Maximize Z = $70x1 + $20x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Resource Requirements

Product Labor (hr/unit)

Clay (lb/unit)

Profit ($/unit)

Bowl 1 4 70

Mug 2 3 20

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Problem DefinitionA Minimization Model Example (1 of 7)

Chemical Contribution

Brand Nitrogen (lb/bag)

Phosphate (lb/bag)

Super-gro 2 4

Crop-quick 4 3

Two brands of fertilizer available - Super-Gro, Crop-Quick.Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate.Super-Gro costs $6 per bag, Crop-Quick $3 per bag.Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ?

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Problem DefinitionA Minimization Model Example (2 of 7)Decision Variables: x1 = bags of Super-Gro

x2 = bags of Crop-Quick

The Objective Function:Minimize Z = $6x1 + 3x2

Where: $6x1 = cost of bags of Super-Gro $3x2 = cost of bags of Crop-Quick

Model Constraints:2x1 + 4x2 16 lb (nitrogen constraint)4x1 + 3x2 24 lb (phosphate constraint)x1, x2 0 (non-negativity constraint)

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Model Formulation and Constraint GraphA Minimization Model Example (3 of 7)

Minimize Z = $6x1 + $3x2

subject to: 2x1 + 4x2 16 4x1+ 3x2 24

x1, x2 0

Figure 2.14Graph of Both Model Constraints

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Feasible Solution AreaA Minimization Model Example (4 of 7)

Minimize Z = $6x1 + $3x2

subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.15Feasible Solution Area

After we determine the feasible solution area, what do we do to determine the optimal solution?

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Optimal Solution PointA Minimization Model Example (5 of 7)

Minimize Z = $6x1 + $3x2

subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.16Optimum Solution Point

Restaurant Case Study

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Angela and Zooey decided to open a French restaurant, a small town near the University they graduated from

Before they offer a full variety menu, they would like to do a “test run” to learn more about the market Offering two four-course meals each night, one with beef and one with fish as the main

course Experiment with different appetizers, soups, side dishes and desserts

Need to determine how many meals to prepare each night to maximize profit Impact ingredients purchasing, staffing and work schedule Cannot afford too much waste

Estimated about 60 meals per night. Available kitchen labor = 20 hours

Initial market analysis indicates that For health reasons, at least three fish dinners for every two beef dinners Regardless at least 10% of their customers will order beef dinners

Main Dish Time to Prepare ProfitFish 15 minutes $12Beef 30 minutes $16

Restaurant Case StudyUnderstand the Problem

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Need to determine how many meals to prepare each night to maximize profit

Estimated about 60 meals per night. Available kitchen labor = 20 hours For health reasons, at least three fish dinners

for every two beef dinners Regardless at least 10% of their customers will

order beef dinners

Main Dish Time to Prepare ProfitFish (x1) 15 minutes (1/4

hour)$12

Beef (x2) 30 minutes (1/2 hour)

$16

Objective

Constraints

Parameters

Restaurant Case Study Formulate the Problem

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Maximize Z = 12*x1 + 16*x2

Or 2*x1 – 3*x2 ≥ 0Or – 0.10*x1 + 0.90*x2 ≥ 0

subject to x1 + x2 ≤ 600.25*x1 + .50*x2 ≤ 20 x1 / x2 ≥ 3 / 2

x2 / (x1 + x2) >= .10 x1, x2 ≥ 0

Need to determine how many meals to prepare each night to maximize profit

Estimated about 60 meals per night. Available kitchen labor = 20 hours For health reasons, at least three fish dinners for every

two beef dinners Regardless at least 10% of their customers will order

beef dinnersMain Dish Time to Prepare ProfitFish (x1) 15 minutes (1/4

hour)$12

Beef (x2) 30 minutes (1/2 hour)

$16

Objective

Constraints

Parameters

Restaurant Case Study Graphical Solution

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Excel QM v4.lnk

# of meals Fish to Beef Ratio

Minimal Beef entrée %

Labor constraintObj. Fct.

Restaurant Case Study What-if Analysis (1 of 2)

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Maximize Z = 16*x1 + 16*x2

Or 2*x1 – 3*x2 ≥ 0Or – 0.20*x1 + 0.80*x2 ≥ 0

subject to x1 + x2 ≤ 600.25*x1 + .50*x2 ≤ 20 x1 / x2 ≥ 3 / 2

x2 / (x1 + x2) >= 0.20 x1, x2 ≥ 0

At least 20% of their customers will order beef dinners

Main Dish Time to Prepare ProfitFish (x1) 15 minutes (1/4

hour)$16

Beef (x2) 30 minutes (1/2 hour)

$16

Angela and Zooey wonders what would be the impact of increasing fish dish price to match profit with beef dish? Based on the initial market survey estimates, the demand for beef will increase from 10% of the sales to 20%.

Excel QM v4.lnk

What are the slopes?

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For some linear programming models, the general rules do not apply.

Special types of problems include those with: Multiple optimal solutions Infeasible solutions Unbounded solutions

Irregular Types of Linear Programming Problems

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Objective function is parallel to a constraint line.

Maximize Z=$40x1 + 30x2

subject to: 1x1 + 2x2 40 4x2 + 3x2 120 x1, x2 0Where:x1 = number of bowlsx2 = number of mugs

Figure 2.18Example with Multiple Optimal Solutions

Multiple Optimal SolutionsBeaver Creek Pottery Example

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An Infeasible Problem

Every possible solution violates at least one constraint:Maximize Z = 5x1 + 3x2

subject to: 4x1 + 2x2 8 x1 4 x2 6 x1, x2 0

Figure 2.19Graph of an Infeasible Problem

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Value of objective function increases indefinitely:Maximize Z = 4x1 + 2x2

subject to: x1 4 x2 2 x1, x2 0

An Unbounded Problem

Figure 2.20Graph of an Unbounded Problem