Solution to Homework #3 4. a. Y 10 8 6 Optimal Solution X = 2.5‚ Y = 2.5 4 8 (2 2 .5) +1 2 (2 2 .5) =5 0 4 X 6 8 10 b. The value of the optimal solution to the revised problem is 8(2.5) + 12(2.5) = 50. Compared to the original problem‚ the value of the optimal solution has increased by 50 48 = 2. Thus‚ the dual value is 2. c. The right-hand side range for constraint 1 is 5 to 11. As long as the right-hand side stays within this range‚ the dual value of 2 is applicable. Since increasing
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CLICK TO DOWNLOAD MAT540 Week 8 Homework Chapter 4 14. Grafton Metalworks Company produces metal alloys from six different ores it mines. The company has an order from a customer to produce an alloy that contains four metals according to the following specifications: at least 21% of metal A‚ no more than 12% of metal B‚ no more than 7% of metal C and between 30% and 65% of metal D. The proportion of the four metals in each of the six ores and the level of impurities in each ore are provided
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IMSE2008 Operational Research Techniques Problem Set: Linear Programming Due: Monday‚ October 20‚ 2014 Please submit to the general office of the IMSE department (Ms Kate Lee HW8-17) on Oct 20 Late submission will be discounted by 20% on a daily basis Please e-mail me msong@hku.hk if you have any questions Problem 1. Work through the simplex method step by step to demonstrate that the following problem is unbounded. (5 marks) max 5x1 + x2 + 3x3 + 4x4 s.t. x1 – 2x2 + 4x3
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Vi Le MGM350C Professor Rohit RampalCase: Textile Mill Scheduling Let X3R = yards of fabric 3 on regular looms X4R = yards of fabric 4 on regular looms X5R = yards of fabric 5 on regular looms X1D = yards of fabric 1 on dobbie looms X2D = yards of fabric 2 on dobbie looms X3D = yards of fabric 3 on dobbie looms X4D = yards of fabric 4 on dobbie looms X5D = yards of fabric 5 on dobbie looms Y1 = yards of fabric 1 purchased Y2 = yards
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A. Formulate a linear programming model for Julia that will help you to advise her if she should lease the booth. Let‚ X1 =No. of pizza slices‚ X2 =No. of hot dogs‚ X3 = No. of barbeque sandwiches * Objective function co-efficient: The objective is to maximize total profit. Profit is calculated for each variable by subtracting cost from the selling price. For Pizza slice‚ Cost/slice=$4.5/6=$0.75 | X1 | X2 | X3 | SP | $1.50 | $1.60 | $2.25 | -Cost | 0.75 | $0.50 | $1.00 |
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Scottsville Textile Mill Case MGM 350 Production Schedule and loom assignment Decision Variable X1: Yards of fabric 1 on dobbie loom X2: Yards of fabric 2 on dobbie loom X3: Yards of fabric 3 on dobbie loom X4: Yards of fabric 4 on dobbie loom X5: Yards of fabric 5 on dobbie loom X6: Yards of fabric 3 on regular loom X7: Yards of fabric 4 on regular loom X8: Yards of fabric 5 on regular loom X9: Yards of fabric 1 purchased X10: Yards of fabric 2 purchased
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CHAPTER 1: INTRODUCTION TO OPERATIONS RESEARCH (OR) 1.1 INTRODUCTION Operation Research‚ an approach to decision making based on the scientific method‚ makes extensive use of quantitative approaches to decision making. In addition to operation research‚ two other widely known and accepted names are management sciences and decision science interchangeably. The scientific management revolution of the early 1900s‚ initiated by Frederic W. Taylor‚ provided the foundation for the use of quantitative
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Sample Paper for IEEE Sponsored Conferences & Symposia Derong Liu‚ Fellow‚ IEEE‚ and MengChu Zhou‚ Fellow‚ IEEE Abstract— The abstract goes here. What you need to do is to insert your abstract here. Please try to make it less than 150 words. We suggest that you read this document carefully before you begin preparing your manuscript. IEEE does not want conference papers to have keywords so you should remove your keyword list. Also‚ at this time‚ IEEE only has some general guidelines about
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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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