DAM assignment - LPP formulation, Graphical solution and Simplex Method

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Decision Analysis and Modeling Assignment - LPP formulation, Graphical solution and Simplex Method

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DAM assignment - LPP formulation, Graphical solution and Simplex Method

  1. 1. PT-MBA 1st year, 3rd Trimester, NMIMS Decision Analysis and Modeling Assignment LPP formulation, Graphical solution and Simplex Method Submitted to : Prof. Shailaja Rego Submitted by : Neha Kumar – A-029 Nikita Thakkar – B-029 3/23/2013
  2. 2. Overview: Johnson and Johnson Consumer Products Division is divided into 5 main product categories. Baby Care, Women’s Health Franchise (sanitary hygiene products), Skin Care (Neutrogena, Clean and Clear), Over-the-counter products (like Benadryl, Nicorette, etc), Wound care (Bandaid, Savlon) and Oral Care (Listerine). Women’s Health Franchise consists of sanitary protection products. Stayfree is a very important range of product under Women’s Health Franchise. It’s one of the highest revenue generating brands for consumer products division. Any decision pertaining to Stayfree is very crucial for the company given the contribution it generates for the company. Stayfree range consists of 7 variants depending on their features to cater to needs of women across SECs. The company has 2 products – product A and product B, both of which are manufactured on the same machine. One packet contains 8 pads each for both products. However, there are 2 major differences between the two products: 1. First difference is the cover which forms the top most surface of the napkin. Product A has a non-woven fabric cover which gives a soft feel. Product B has a dry net cover which gives a dry feel. 2. Product A is directly packed in polybags. Product B is first folded and then each napkin is packed in individually folded pouches before packing them in the polybag. Both these products are very important since they contribute significantly towards the turnover of the Stayfree portfolio. Both these products cater to two different consumer needs, namely, soft feel and dry feel. Therefore, demand for both is exclusive of each other. If Product A is discontinued then the consumers of Product A are not likely to shift to Product B and vice versa. The company therefore needs to continue manufacturing and marketing both products. Since there is only a cover difference between the two products, they are manufactured on the same machine with a change in only one raw material i.e. the top cover of the napkin. There is an additional process involved for Product B where it gets folded and packed in individual pouches. This process is done with the help of operating the machine along with an extension which increases the machine time in case of Product B. The volumes of both products are quite high. The gross profit margin is same for both at 40%. Details of the same are given in the following table. Product A Product B Selling Price (MRP per pack) 22 31 Less: Variable Cost 13.2 18.6 Contribution 8.8 12.4
  3. 3. Decision making: The company has one manufacturing unit which has 5 identical machines. Each machine can be operated for maximum 20 hours per day and gives the same amount of output. The machine hours, packaging time for both products are mentioned in the table below. The company is obliged to supply minimum 200,000 packs of Product A and 100,000 packs of product B to ensure that it does not lose its current market share. The company needs to maximize its profit by producing the optimal quantities of both products. Product A Product B Total Total Demand (packs per month) 350,000 250,000 600,000 Minimum requirement (packs per month) Machine time per pack (in minutes) – refer to working note 1 200,000 100,000 300,000 0.1 0.7 180,000 Packaging time per pack (in minutes) 0.5 (per month) 0.7 300,000 (per month) Linear Programming Problem Formulation: The company has to decide the optimal quantity to be produced of both products to maximize its profit. The company faces constraints with regards to the minimum quantity required of both products to maintain current market share, the machine hours and the packaging hours. Given this information, we need to formulate this problem into a linear programming problem: Let X1 be the number of packs of product A that need to be produced by the company. Let X2 be the number of packs of product B that need to be produced by the company. Objective function: Maximize Z (Profit Maximization) = 8.80 x1 + 12.40 x2 Subject to constraints: X1 ≥ 200,000 (Minimum requirement constraint for X1) X2 ≥ 100,000 (Minimum requirement constraint for X2) 0.10 X1 + 0.70 X2 ≤ 180000 (Machine hours constraint)
  4. 4. 0.50 X1 + 0.70 X2 ≤ 300000 (Packaging hours constraint) X1, X2 ≥ 0 (Non-negativity constraint) To solve this LPP formulation, we need to convert all these inequalities to equalities. We can enter objective (i.e. profit maximization in this case) and constraint values in QM for windows software and click on solve. We get the following output. Maximize Minimum Requirement X1 Minimum Requirement X2 Machine Hours Constraint Packaging Hours Constraint Constraint 5 Constraint 6 Solution-> X1 8.8 1 0 0.1 0.5 1 0 300000 X2 12.4 0 1 0.7 0.7 0 1 214285.7 RHS >= >= <= <= >= >= Dual 200000 0 100000 0 180000 0.1428564 300000 17.57143 0 0 0 0 $5,297,142.88 We also get the corner points and the area highlighted in pink is the feasible area. But as we can see in the above graph, the optimal solution is the point which is circled. We need to produce 300,000 packs of X1 (Product A) and 214,285 packs of X2 (Product B). The optimal solution will give us a profit of Rs. 5,297,143/-. The Z values for all 4 points of the feasible region are mentioned below.
  5. 5. Optimal solution - Range Table: Variable Value Reduced Cost Original Val Lower Bound Upper Bound X1 300000 0 8.8 1.77 8.86 X2 214285.7 0 12.4 12.32 61.6 Constraint Dual Value Slack/Surplus Original Val Lower Bound Upper Bound Minimum Requirement X1 0 100000 200000 -Infinity 300000 Minimum Requirement X2 0 114285.7 100000 -Infinity 214285.7 Machine Hours Constraint 0.1428564 0 180000 116000 220000 Packaging Hours Constraint 17.57143 0 300000 260000 620000 Constraint 5 0 300000 0 -Infinity 300000 Constraint 6 0 214285.7 0 -Infinity 214285.7
  6. 6. Iteration 8 0 0 0 Basic 8.8 12.4 surplus surplus 0 0 artfcl surplus Cj Variables X1 X2 0 artfcl 1 1 0 artfcl 2 2 0 slack 3 0 slack 4 0 artfcl 5 surplus 5 6 6 Quantity Iteration 8 0 surplus 5 0 0 0 0 0 0 -2.5 2.5 -1 1 0 0 300,000.00 0 surplus 6 0 0 0 0 0 0 1.7857 -0.3571 0 0 -1 1 214,285.72 0 surplus 2 0 0 0 0 -1 1 1.7857 -0.3571 0 0 0 0 114,285.72 0 surplus 1 0 0 -1 1 0 0 -2.5 2.5 0 0 0 0 100,000.00 8.8 X1 1 0 0 0 0 0 -2.5 2.5 0 0 0 0 300,000.00 12.4 X2 0 1 0 0 0 0 1.7857 -0.3571 0 0 0 0 214,285.72 zj 8.8 12.4 0 0 0 0 0.1429 17.5714 0 0 0 0 5,297,142.88 cj-zj 0 0 0 0 0 0 -0.1429 -17.5714 0 0 0 0 (Iterations 1 to 7 are included in the working note for reference) The optimal product mix is X1 = 300,000 and X2 = 214285.7 and total maximum profit contribution is Z = Rs. 5,297,143/The objective function coefficient of X1 is 8.8 and its allowable decrease is 1.77 and allowable increase is 8.86. The objective function coefficient of X2 is 12.4 and its allowable decrease is 12.32 and allowable increase is 61.6. If the value of the objective function coefficients goes below the allowable decrease or above the allowable increase then the optimal solution will change. The RHS range for machine hours is 116,000 to 220,000 and for packaging hours is 260,000 to 620,000. The shadow price value for machine hours is Rs. 0.1429 and packaging hours is Rs. 17.5714. Subject to the cost of increasing one unit of these resources, we should choose the resource which has a higher shadow price as it will give us a higher contribution per unit. Working Notes: 1. Calculation of Total available machine hours per month Number of Machines Maximum Number of Operating Hours Maximum Number of minutes Number of days per month Total Machine Hours per month (in minutes) 5 20 60 30 180000
  7. 7. 2. Iterations for simplex method 0 0 0 Basic 8.8 12.4 surplus surplus 0 0 artfcl surplus Cj Variables X1 X2 0 artfcl 1 1 0 artfcl 2 2 0 slack 3 0 slack 4 0 artfcl 5 surplus 5 6 6 Quantity Iteration 1 0 artfcl 1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000 0 artfcl 2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000 0 slack 3 0.1 0.7 0 0 0 0 1 0 0 0 0 0 180,000 0 slack 4 0.5 0.7 0 0 0 0 0 1 0 0 0 0 300,000 0 artfcl 5 1 0 0 0 0 0 0 0 1 -1 0 0 0 0 artfcl 6 0 1 0 0 0 0 0 0 0 0 1 -1 0 zj 6.8 10.4 0 1 0 1 0 0 0 1 0 1 300,000 cj-zj 2 2 0 -1 0 -1 0 0 0 -1 0 -1 Iteration 2 0 artfcl 1 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000 0 artfcl 2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000 0 slack 3 0 0.7 0 0 0 0 1 0 -0.1 0.1 0 0 180,000 0 slack 4 0 0.7 0 0 0 0 0 1 -0.5 0.5 0 0 300,000 8.8 X1 1 0 0 0 0 0 0 0 1 -1 0 0 0 0 artfcl 6 0 1 0 0 0 0 0 0 0 0 1 -1 0 zj 8.8 10.4 0 1 0 1 0 0 2 -1 0 1 300,000 cj-zj 0 2 0 -1 0 -1 0 0 -2 1 0 -1 Iteration 3 0 artfcl 1 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000 0 artfcl 2 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000 0 slack 3 0 0 0 0 0 0 1 0 -0.1 0.1 -0.7 0.7 180,000 0 slack 4 0 0 0 0 0 0 0 1 -0.5 0.5 -0.7 0.7 300,000 8.8 X1 1 0 0 0 0 0 0 0 1 -1 0 0 0 12.4 X2 0 1 0 0 0 0 0 0 0 0 1 -1 0 zj 8.8 12.4 0 1 0 1 0 0 2 -1 2 -1 300,000 cj-zj 0 0 0 -1 0 -1 0 0 -2 1 -2 1 Iteration 4 0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000 0 artfcl 2 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000 0 slack 3 0 0 -0.1 0.1 0 0 1 0 0 0 -0.7 0.7 160,000.00 0 slack 4 0 0 -0.5 0.5 0 0 0 1 0 0 -0.7 0.7 200,000 8.8 X1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000 12.4 X2 0 1 0 0 0 0 0 0 0 0 1 -1 0 zj 8.8 12.4 1 0 0 1 0 0 1 0 2 -1 100,000 cj-zj 0 0 -1 0 0 -1 0 0 -1 0 -2 1 Iteration 5 0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000 0 surplus 6 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000 0 slack 3 0 0 -0.1 0.1 -0.7 0.7 1 0 0 0 0 0 90,000.00 0 slack 4 0 0 -0.5 0.5 -0.7 0.7 0 1 0 0 0 0 130,000.00 8.8 X1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000 12.4 X2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000 zj 8.8 12.4 1 0 1 0 0 0 1 0 1 0 0 cj-zj 0 0 -1 0 -1 0 0 0 -1 0 -1 0 Iteration 6 0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000 0 surplus 6 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000 0 slack 3 0 0 -0.1 0.1 -0.7 0.7 1 0 0 0 0 0 90,000.00 0 slack 4 0 0 -0.5 0.5 -0.7 0.7 0 1 0 0 0 0 130,000.00 8.8 X1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000 12.4 X2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000 zj 8.8 12.4 8.8 -8.8 12.4 -12.4 0 0 0 0 0 0 3,000,000 cj-zj 0 0 -8.8 8.8 -12.4 12.4 0 0 0 0 0 0 Iteration 7 0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000 0 surplus 6 0 0 -0.1429 0.1429 0 0 1.4286 0 0 0 -1 1 228,571.43 0 surplus 2 0 0 -0.1429 0.1429 -1 1 1.4286 0 0 0 0 0 128,571.43

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