Pertemuan 2 – Menyelesaikan Formulasi Model
Dengan Metode Grafik
Riset Operasiomal- dewiyani
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Methods to Solve LP Problems

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Graphical Method
Simplex Method
Problem Definition
A 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:
Product
Resource Requirements
Labor
Clay
Profit
(hr/unit)
(lb/unit)
($/unit)
Bowl
1
4
40
Mug
2
3
50
3
4
Problem Definition
A Maximization Model Example (2 of 3)
Resource
Availability:
40 hrs of labor per day
120 lbs of clay
Decision
Variables:
x1 = number of bowls to produce per day
x2 = number of mugs to produce per day
Objective
Function:
Maximize Z = $40x1 + $50x2
Where Z = profit per day
Resource
Constraints:
1x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
Non-Negativity
Constraints:
x1  0; x2  0
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Problem Definition
A 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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Feasible Solutions
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
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Infeasible Solutions
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
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Graphical Solution of Linear Programming Models

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.
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Coordinate Axes
Graphical Solution of Maximization Model (1 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
After defining the axes, begin drawing
the constraints
• What is the x or y intercept?
• What is the slope?
Figure 2.1
Coordinates for Graphical Analysis
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Labor Constraint
Graphical Solution of Maximization Model (2 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.1
Graph of Labor Constraint
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Labor Constraint Area
Graphical Solution of Maximization Model (3 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
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?
Figure 2.3
Labor Constraint Area
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Clay Constraint Area
Graphical Solution of Maximization Model (4 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.4
Clay Constraint Area
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Both Constraints
Graphical Solution of Maximization Model (5 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.5
Graph of Both Model Constraints
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Don’t forget the non-negative constraints
Graphical Solution of Maximization Model (6 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
X2 = 0
X1 = 0
Figure 2.6
Feasible Solution Area
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Feasible Solution Area
Graphical Solution of Maximization Model (7 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.6
Feasible Solution Area
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Objective Solution = $800
Graphical Solution of Maximization Model (8 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
X2 = 0
Arbitrarily set the objective
function as $800
X1 = 0
Figure 2.7
Objection Function Line for Z = $800
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Alternative Objective Function Solution Lines
Graphical Solution of Maximization Model (9 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
X2 = 0
X1 = 0
Figure 2.8
Alternative Objective Function Lines
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Optimal Solution
Graphical Solution of Maximization Model (10 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Above and Beyond
What if we forgot to include the nonnegative constraints?
Figure 2.9
Identification of Optimal Solution
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Optimal Solution Coordinates
Graphical Solution of Maximization Model (11 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.10
Optimal Solution Coordinates
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Corner Point Solutions
Graphical Solution of Maximization Model (12 of 13)
Maximize Z =
subject to:
$40x1 + $50x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Why are corner points important?
Figure 2.11
Solution at All Corner Points
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Optimal Solution for New Objective Function
Graphical Solution of Maximization Model (13 of 13)
Product
Resource Requirements
Labor
Clay
Profit
(hr/unit)
(lb/unit)
($/unit)
Bowl
1
4
70
Mug
2
3
20
Maximize Z =
subject to:
$70x1 + $20x2
1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Figure 2.12
Optimal Solution with Z = 70x1 + 20x2
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Problem Definition
A Minimization Model Example (1 of 7)
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 ?
Chemical Contribution
Nitrogen
(lb/bag)
Phosphate
(lb/bag)
Super-gro
2
4
Crop-quick
4
3
Brand
23
24
Problem Definition
A 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 Graph
A Minimization Model Example (3 of 7)
Minimize Z =
subject to:
$6x1 + $3x2
2x1 + 4x2  16
4x1+ 3x2  24
x1, x2  0
Figure 2.14
Graph of Both Model Constraints
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Feasible Solution Area
A Minimization Model Example (4 of 7)
Minimize Z =
subject to:
$6x1 + $3x2
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
After we determine the feasible
solution area, what do we do to
determine the optimal solution?
Figure 2.15
Feasible Solution Area
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Optimal Solution Point
A Minimization Model Example (5 of 7)
Minimize Z =
subject to:
$6x1 + $3x2
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Figure 2.16
Optimum Solution Point
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Restaurant Case Study


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



Need to determine how many meals to prepare each night to maximize profit




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
Impact ingredients purchasing, staffing and work schedule
Cannot afford too much waste
Estimated about 60 meals per night. Available kitchen labor = 20 hours
Main Dish
Time to Prepare
Profit
Fish
15 minutes
$12
Beef
30 minutes
$16
Initial market analysis indicates that
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
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For health reasons, at least three fish dinners for every two beef dinners
Regardless at least 10% of their customers will order beef dinners
Restaurant Case Study
Understand the Problem

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
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Objective
Constraints
Main Dish
Time to Prepare
Profit
Fish (x1)
15 minutes (1/4 hour)
$12
Beef (x2)
30 minutes (1/2 hour)
$16
Parameters
Restaurant Case Study
Formulate the Problem

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
Profit
Fish (x1)
15 minutes (1/4 hour)
$12
Beef (x2)
30 minutes (1/2 hour)
$16
Maximize
Z = 12*x1 + 16*x2
subject to
x1 + x2 ≤ 60
0.25*x1 + .50*x2 ≤ 20
x1 / x2 ≥ 3 / 2
x2 / (x1 + x2) >= .10
x1, x2 ≥ 0
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Objective
Constraints
Parameters
Or
Or
2*x1 – 3*x2 ≥ 0
– 0.10*x1 + 0.90*x2 ≥ 0
Restaurant Case Study
Graphical Solution
# of meals
Fish to Beef Ratio
Minimal Beef entrée %
Obj. Fct.
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Labor constraint
Restaurant Case Study
What-if Analysis (1 of 2)
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%.

At least 20% of their customers will order beef dinners
Main Dish
Time to Prepare
Profit
Fish (x1)
15 minutes (1/4 hour)
$16
Beef (x2)
30 minutes (1/2 hour)
$16
Maximize
Z = 16*x1 + 16*x2
subject to
x1 + x2 ≤ 60
0.25*x1 + .50*x2 ≤ 20
x1 / x2 ≥ 3 / 2
x2 / (x1 + x2) >= 0.20
x1, x2 ≥ 0
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What are the slopes?
Or
2*x1 – 3*x2 ≥ 0
Or – 0.20*x1 + 0.80*x2 ≥ 0
Irregular Types of Linear Programming Problems

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
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Multiple Optimal Solutions
Beaver Creek Pottery Example
Objective function is parallel to a
constraint line.
Maximize Z=$40x1 + 30x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Where:
x1 = number of bowls
x2 = number of mugs
Figure 2.18
Example with Multiple Optimal Solutions
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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 2.19
Graph of an Infeasible Problem
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An Unbounded Problem
Value of objective function increases
indefinitely:
Maximize Z = 4x1 + 2x2
subject to: x1  4
x2  2
x1, x2  0
Figure 2.20
Graph of an Unbounded Problem
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