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Linear Programming
      Model Formulation Graphical Solution Method Linear Programming Model Simplex method Solution Solving Linear Programming Problems with Excel Dr A Lung  Student exercises Kingston University London

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Linear Programming (LP)
• A model consisting of linear relationships representing a firm’s objective and resource constraints • LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called constraints • Pioneered by George Dantzig in WWII, including Simplex Method

Learning objectives:  Identify Objective function and Constraints  Identify Feasible region  Apply Corner-point solution  Describe how to formulate linear models  Use Graphical method of LP 2

Types of LP

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Types of LP (cont.)

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Types of LP (cont.)

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LP Model Formulation
 Decision variables


mathematical symbols representing levels of activity of an operation a linear relationship reflecting the objective of an operation most frequent objective of business firms is to maximize profit most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost a linear relationship representing a restriction on decision making, e.g. an operator can work on m/c A or B, but what ever the mix he/she can only work 8 hours a day 6

 Objective function
  

 Constraint


LP Model Formulation (cont.)
Max/min
subject to: a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1 a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2 : am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm xj = decision variables bi = constraint levels cj = objective function coefficients aij = constraint coefficients

z = c1x1 + c2x2 + ... + cnxn

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LP Model: Example
RESOURCE REQUIREMENTS PRODUCT Bowl Mug Labor (hr/unit) 1 2 Clay (lb/unit) 4 3 Revenue ($/unit) 40 50

There are 40 hours of labor and 120 pounds of clay available each day Decision variables x1 = number of bowls to produce x2 = number of mugs to produce 8

LP Formulation: Example
Maximize Z = $40 x1 + 50 x2 Subject to x1 + 2x2 40 hr (labor constraint) 4x1 + 3x2 120 lb (clay constraint) x1 , x2 0 Solution is x1 = 24 bowls Revenue = $1,360 x2 = 8 mugs

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Graphical Solution Method
1. Plot model constraint on a set of coordinates in a plane 2. Identify the feasible solution space on the graph where all constraints are satisfied simultaneously 3. Plot objective function to find the point on boundary of this space that maximizes (or minimizes) value of objective function 10

Graphical Solution: Example
x2 50 –

40 –
30 –

4 x1 + 3 x2 120 lb

20 –
10 – 0– | 10 | 20

Area common to both constraints
x1 + 2 x2 40 hr
| 30 | 40 | 50 | 60 x1
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Computing Optimal Values
x2 40 – 30 – 20 – 10 – 8 0– | 10 | 24 | 20 30 | x1 40

From equations:

4 x1 + 3 x2 120 lb

x1 + 2x2 = 40 4x1 + 3x2 = 120 4x1 + 8x2 = 160 -4x1 - 3x2 = -120 5x2 = x2 = x1 + 2(8) = x1 = 40 8 40 24

x1 + 2 x2 40 hr

Z = $50(24) + $50(8) = $1,360
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Extreme Corner Points
x2
40 –

x1 = 0 bowls x2 =20 mugs Z = $1,000

30 –
20 – A 10 – 0– | 10 | 20

x1 = 224 bowls x2 =8 mugs x1 = 30 bowls Z = $1,360 x2 =0 mugs Z = $1,200

Graphical solution:

B
| C| 30 40

x1

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Objective Function
x2 40 –

4x1 + 3x2 120 lb

To evaluate when objectives change but constraints remain the same:

30 –

Z = 70x1 + 20x2

20 –A

Optimal point: x1 = 30 bowls x2 =0 mugs Z = $2,100

10 –

B | C 30

x1 + 2x2 40 hr
| 40 x1
14

0–

| 10

| 20

Minimization Problem
CHEMICAL CONTRIBUTION

Brand
Gro-plus Crop-fast

Nitrogen (lb/bag)
2 4

Phosphate (lb/bag)
4 3

X1 is Gro-plus, x2 is Crop-fast, Z is total cost Minimize Z = $6x1 + $3x2 subject to these given constraints from supplier 2x1 + 4x2  16 lb of nitrogen 4x1 + 3x2  24 lb of phosphate x 1, x 2  0 15

Graphical Solution
x2 14 –...
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