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# Operations Management - PowerPoint PPT Presentation

Operations Management. Module B – Linear Programming. PowerPoint presentation to accompany Heizer/Render Principles of Operations Management, 6e Operations Management, 8e . © 2006 Prentice Hall, Inc. Lecture Outline. More Examples Practice formulating Practice solving

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Operations Management

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#### Presentation Transcript

Operations Management

Module B – Linear Programming

PowerPoint presentation to accompany

Heizer/Render

Principles of Operations Management, 6e

Operations Management, 8e

### Lecture Outline

• More Examples

• Practice formulating

• Practice solving

• Practice interpreting the results

### Mixed Nuts

• Crazy Joe makes two blends of mixed nuts: party mix and regular mix.

• Crazy Joe has 10 lbs of cashews and 24 lbs of peanuts

### Ah, Nuts Formulation

• Let

• p = lbs of party mix to make

• r = lbs of regular mix to make

• z = total revenue

• Max z = 6p + 4r

• Subject to

• 0.6p + 0.9r< 24 (peanut constraint)

• 0.4p + 0.1r< 10 (cashew constraint)

• p, r> 0 (non-negativity constraints)

### Solving Minimization Problems

• Formulated and solved in much the same way as maximization problems

• In the graphical approach an iso-cost line is used

• The objective is to move the iso-cost line inwards until it reaches the lowest cost corner point

### Minimization Example

Let

X1 =number of tons of black-and-white chemical produced

X2 =number of tons of color picture chemical produced

Minimize total cost =2,500X1+3,000X2

Subject to:

X1≥ 30tons of black-and-white chemical

X2≥ 20tons of color chemical

X1 + X2≥ 60tons total

X1, X2≥ 0nonnegativity requirements

X2

60 –

50 –

40 –

30 –

20 –

10 –

X1 + X2= 60

X1= 30

X2= 20

|||||||

0102030405060

X1

Table B.9

Feasible region

b

a

### Minimization Example

Total cost at a=2,500X1+3,000X2

=2,500 (40)+3,000(20)

=\$160,000

Total cost at b=2,500X1+3,000X2

=2,500 (30)+3,000(30)

=\$165,000

Lowest total cost is at point a

Feed

ProductStock XStock YStock Z

A3 oz2 oz 4 oz

B2 oz3 oz 1 oz

C1 oz0 oz 2 oz

D6 oz8 oz 4 oz

### LP Applications

Diet Problem Example

### LP Applications

X1 = number of pounds of stock X purchased per cow each month

X2 = number of pounds of stock Y purchased per cow each month

X3 = number of pounds of stock Z purchased per cow each month

Minimize cost = .02X1 + .04X2 + .025X3

Ingredient A requirement:3X1 +2X2 +4X3≥ 64

Ingredient B requirement:2X1 +3X2 +1X3≥ 80

Ingredient C requirement:1X1 +0X2 +2X3≥ 16

Ingredient D requirement:6X1 +8X2 +4X3≥ 128

Stock Z limitation:X3≤ 80

X1,X2, X3≥ 0

Cheapest solution is to purchase 40 pounds of grain X

at a cost of \$0.80 per cow

### Multiple Optimal Solutions

• Often, real world problems can have more than one optimal solution

• When would this happen?

• What does the graph have to look like?

• Do want to have “ties”?

### No Solutions

• Can we ever have a problem without a feasible solution?

• When would this happen?

• What would the graph look like?

• Does this mean we did something wrong?

### The Simplex Method

• Real world problems are too complex to be solved using the graphical method

• The simplex method is an algorithm for solving more complex problems

• Developed by George Dantzig in the late 1940s

• Most computer-based LP packages use the simplex method