Pertemuan 2 – Menyelesaikan Formulasi Model Dengan Metode Grafik
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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
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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
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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
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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
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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.
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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