311_Session5_Chen

311_Session5_Chen - OperationsManagement...

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Operations Management Session 5:  Linear Programming
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Session 5 Operations Management 2 Previous Class Kristen Cookie Capacity Planning
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Session 5 Operations Management 3 Today’s Class Understand what a Linear Program is. Understand how to solve small LP’s graphically. Practice LP formulation
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Session 5 Operations Management 4 Production Example A toy company offers 2 different dolls: Beauty and  Beast.  The daily demand for each doll is 400  units. There are two types of machines. Everyday, M1  can make 10 units of Beauty and 2 units of Beast;   M2 can make 3 units of Beauty and 7 units of  Beast.
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Session 5 Operations Management 5 Production Example M1 costs $7 K and M2 costs $5 K. How many units of M1 and M2 should the  company purchase?
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Session 5 Operations Management 6 Production Example Demand                           Question      10 2 M1 3 7 M2 7 5 400 400 ? ?
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Session 5 Operations Management 7 Production Example What should be the objective of the toy company? Minimize the total purchasing cost Purchasing cost = 7*the number of M1  ( x 1 ) +                  5*the number of M2  ( x 2 ) This is the  objective function
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Session 5 Operations Management 8 Production Example Do we have any constraints? Number of M1 and M2 machines can’t be negative.  Is that all? We must have 400 Beauties and 400 Beasts 0 1 ξ 0 2
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Session 5 Operations Management 9 Production Example 400 * 3 * 10 2 1 + ξ Number of M1s Capacity for Beauties: 400 * 7 * 2 2 1 + The last 4 inequalities are called  constraints x 1  and  x 2  are called  decision variables Capacity for Beasts: Number of M2s
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Session 5 Operations Management 10 Production Example The linear program is: 0 1 ξ 0 2 400 * 3 * 10 2 1 + 400 * 7 * 2 2 1 + 2 1 5 7 Min x x + Objective function: Constraints:
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Example:  The feasible set    x 1 x 2 133.3 57.14 200 Feasible Set Optimal Solution (25,50) 10 x 1 +3 x 2 =400 2 x 1 +7 x 2 =400
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Session 5 Operations Management 12 Example:  The optimal  solution The optimal solution is: x 1  = 25 x 2  = 50 The minimum cost is: 25*7 + 50*5 = 425 Is it feasible? 25*10 + 50*3 = 400 25*2 + 50*7 = 400 Number of machines is positive
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Session 5 Operations Management 13 Linear Program Summary maximize or minimize a linear function of several variables subject to linear inequality constraints or linear equality constraints
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Session 5 Operations Management 14 What to produce? A computer manufacturer produces 2 types of  computers HS (High Speed) and LS (Low Speed).  The profit margin for HS computer is $100 per unit  and for LS computer is $75 per unit.  One assembling machine is used for both models.  The machine operates 600 minutes a day. A HS  unit needs 30 minutes, and A LS unit needs 10  minutes. 
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311_Session5_Chen - OperationsManagement...

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