Chapter 2 An Introduction to Linear Programming 18. a. Max 4A + 1B + 0S1 + 0S2 + 0S3 s.t. 10A + 2B + 1S1 = 30 3A + 2B + 1S2 = 12 2A + 2B + 1S3 = 10 A‚ B‚ S1‚ S2‚ S3 0 b. c. S1 = 0‚ S2 = 0‚ S3 = 4/7 23. a. Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced Max 2400E + 1800L s.t. 6E + 3L 2100 Engine time L 280 Lady-Sport maximum
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Case Problem: Textile Mill Scheduling Assuming‚ X1 = Yards of fabric 1 purchased X2 = Yards of fabric 1 on dobbie looms X3 = Yards of fabric 2 purchased X4 = Yards of fabric 2 on dobbie looms X5 = Yards of fabric 3 purchased X61 = Yards of fabric 3 on dobbie looms X62 = Yards of fabric 3 on regular looms X7 = Yards of fabric 4 purchased X81 = Yards of fabric 4 on dobbie looms X82 = Yards of fabric 4 on regular looms X9 = Yards of fabric
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Sensitivity Analysis Source: Introduction to Management Science 10 e‚ Anderson Sweeney Williams Example 1 Max s.t. 5x1 + 7x2 x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1‚ x2 > 0 x2 8 7 6 5 4 3 2 1 x1 + x2 < 8 Max 5x1 + 7x2 x1 < 6 Optimal: x1 = 5‚ x2 = 3‚ z = 46 2x1 + 3x2 < 19 x1 1 2 3 4 5 6 7 8 9 10 x2 8 7 6 5 4 3 2 1 5 5 Feasible Region 1 1 1 2 3 4 4 4 3 3 2 2 5 6 7 8 9 10 x1 Example 1 • Range of Optimality for c1 The slope of the objective
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An Example of Two Phase Simplex Method Consider the following LP problem. max z = 2x1 + 3x2 + x3 s.t. x1 + x2 + x3 · 40 2x1 + x2 ¡ x3 ¸ 10 ¡x2 + x3 ¸ 10 x1; x2; x3 ¸ 0 It can be transformed into the standard form by introducing 3 slack variables x4‚ x5 and x6. max z = 2x1 + 3x2 + x3 s.t. x1 + x2 + x3 + x4 = 40 2x1 + x2 ¡ x3 ¡ x5 = 10 ¡x2 + x3 ¡ x6 = 10 x1; x2; x3; x4; x5; x6 ¸ 0 There is no obvious initial basic feasible solution‚ and it is not even known whether there exists one
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Yu Liu 4.17.2012 Why? What? How? USCT‚ Dept. EEIS‚ Yu Liu 3 1. Image Matching in a common and important problem in computer vision. 2. Application in: ◦ ◦ ◦ ◦ ◦ Object or scene recognition 3D reconstruction Stereo correspondence Motion tracking Image Searching USCT‚ Dept. EEIS‚ Yu Liu 4 3. Traditional method: simple corner detectors is not stable when you have images of different scales and rotations. 4. We need a method can solve: ◦ ◦ ◦ ◦ ◦ ◦ Different
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This case presents some of the basic concepts of aggregate plan-ning by the transportation method. The case involves solving arather complex set of transportation problems. Four different con-figurations of operating plants have to be tested. The solutions‚ al-though requiring relatively few iterations to optimality‚ involvedegeneracy if solved manually. The costs are The lowest weekly total cost‚ operating plants 1 and 3 with 2closed‚ is $217‚430. This is $3‚300 per week ($171‚600 per year)or
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MB0048_MBA_Sem2_Fall/August 2012 Master of Business Administration - MBA Semester 2 MB0048 – Operations Research- 4 Credits Assignment Set- 1 (60 Marks) Note: Each question carries 10 Marks. Answer all the questions. Q.1 Maximise z = 3x1 + 4x2 Subject to constrains 5x1 + 4x2 200; 3x1 + 5x2 150; 5x1 + 4x2 100; 8x1 + 4x2 80‚ x1 0‚ x2 0 Q.2 State the ways in which customers in a queue are served. Q.3 Explain the use of simulation in networks? What are the advantages of using simulation? Q
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MSE 606A-OL: Engineering Operations Management Instructor: Ahmad R. Sarfaraz‚ Ph.D. Department: Manufacturing Systems Engineering and Management Faculty Office: EA 3305; (818) 677-6229 Email: sarfaraz@csun.edu Course Text: Bernard Taylor‚ Management Science‚ 8th Edition‚ Prentice Hall‚ 2004. Hamdy Taha‚ Operations Research‚ 9th Edition Both books will be in reserved room. Course Purpose: This course is designed to ensure that students gain: An ability to use
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Assignment Two Answers Problem: 3-2 a. Find the optimal product mix? The optimal product mix is as in screenshot above‚ fields of Number to produce: Basic XP VXP 514.285714 1200 28.57143 b. If you round the values in the changing cells to the nearest integer‚ is the resulting solution still feasible? No‚ the rounded number of Product to Produce does not generate feasible solution. In order to obtain feasible solution close to optimal‚ we need to set a new Integer constraint on Number to Produce
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Spreadsheet Modeling & Decision Analysis‚ 5ed Cliff T. Ragsdale Check figures for selected odd problems. Chapter 2 7. 9. 11. 13. 15. 17. 19. 21. 23. Optimal objective value = 10.55 Optimal objective value = 125 Optimal objective value = 154 Optimal objective value = 775 Optimal objective value = 32500 Optimal objective value = 0.75 Optimal objective value = 59300 Optimal objective value = 26000 Optimal objective value = 3.5 million 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35
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