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Linear Programming Example 2 Alternate Optimal Solutions
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The Problem An electronics company can manufacture three styles of 30-inch televisions: –Standard (.2 hrs in production,.1 hr assembly,.05 hours inspection – profit $60) –Flat Screen (.2 hrs in production,.3 hr assembly,.075 hours insepction – profit $90) –Plasma (.4 hrs in production,.4 hr assembly,.15 hours inspection – profit $100) Weekly Time Allocation Production 600 man-hours Assembly 480 man-hours Inspection 160 man-hours
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Mathematical Model Decision Variables: X 1 = # Standard TV’s produced weekly X 2 = # Flat Screen TV’s produced weekly X 3 = # Plasma TV’s produced weekly Objective Max Total Weekly Profit MAX 60X 1 + 90X 2 + 100X 3 S.t. constraints Production:.2 X 1 +.2 X 2 +.4 X 3 ≤ 600 Assembly:.1 X 1 +.3 X 2 +.4 X 3 ≤ 480 Inspection:.05X 1 +.075X 2 +.15X 3 ≤ 160 Non-negativity: All X’s ≥ 0 Amount Used Amount available
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=SUMPRODUCT($C$4:$E$4,C6:E6) Drag down
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Optimal Weekly Production Schedule 1600 Standard 1066 2/3 Flat 0 Plasma Work in Progress Profit = 192000
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Indicates alternate optimal solutions
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Alternate Optimal Solutions An optimal solution that produces as many standard TV’s as possible. –Add a constraint that cell F6 must = 192000 –Then set objective function to MAX X 1 (cell C4) An optimal solution that produces as few standard TV’s as possible. –Add a constraint that cell F6 must = 192000 –Then set objective function to MIN X 1 (cell C4) An optimal solution that produces exactly twice the number of standard TV’s as flat screen TV’s. –Add a constraint X 1 = 2X 2.
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Maximum Standard TV’s MAX C4 Objective Function = 192000 Optimal Weekly Production Schedule 2600 Standard 400 Flat 0 Plasma
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Minimum Standard TV’s Objective Function = 192000 MIN C4 Optimal Weekly Production Schedule 1600 Standard 1066 2/3 Flat 0 Plasma
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Standard TV’s = 2* Flat Screen TV’s Optimal Weekly Production Schedule 1828.571 Standard 914.2857 Flat 0 Plasma Max Profit Standard = 2*(Flat Screen)
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Review Recognizing Alternate Optimal Solutions How to Generate Alternate Optimal Solutions that: –Maximizes a decision variable –Minimizes a decision variable –Establishes so relationship between the variables
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