{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

311_Session5_Chen

# 311_Session5_Chen - Session5:LinearProgramming...

This preview shows pages 1–15. Sign up to view the full content.

Operations Management Session 5:  Linear Programming

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Session 5 Operations Management 2 Previous Class Kristen Cookie Capacity Planning
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Session 5 Operations Management 6 Production Example Demand                           Question      10 2 M1 3 7 M2 7 5 400 400 ? ?
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 ξ
Session 5 Operations Management 9 Production Example 400 * 3 * 10 2 1 + ξ ξ Number of M1s Capacity for Beauties: 400 * 7 * 2 2 1 + ξ ξ The last 4 inequalities are called  constraints x 1  and  x 2  are called  decision variables Capacity for Beasts: Number of M2s

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Session 5 Operations Management 10 Production Example The linear program is: 0 1 ξ 0 2 ξ 400 * 3 * 10 2 1 + ξ ξ 400 * 7 * 2 2 1 + ξ ξ 2 1 5 7 Min x x + Objective function: Constraints:
Example:  The feasible set    x 1 x 2 40 133.3 57.14 200 Feasible Set Optimal Solution (25,50) 10 x 1 +3 x 2 =400 2 x 1 +7 x 2 =400

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Session 5 Operations Management 14 What to produce?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}