Day Date Weekday Daily Demand Weekend weekday Average daily demand weekdays 1 4/25/2016 Mon 297 0 sun 453.25 mon 2 4/26/2016 Tue 293 0 mon 301.38 tue CheckWed the formula 3 4/27/2016 327 0 tue 315.13 wed 4 4/28/2016 Thu 315 0 wed 342.75 thur 5 4/29/2016 Fri 348 0 thu 334 fri 6 4/30/2016 Sat 447 1 fri 380.38 mon 7 5/1/2016 Sun 431 1 sat 454.75 tue 8 5/2/2016 Mon 283 0 wed 9 5/3/2016 Tue 326 0 thur 10 5/4/2016 Wed 317 0 fri 11 5/5/2016 Thu 345 0 mon 12 5/6/2016 Fri 355 0 tue 13 5/7/2016 Sat 428 1 wed 14 5/8/2016 Sun 454 1 thur 15 5/9/2016 Mon 305 0 fri 16 5/10/2016 Tue 310 0 mon 17 5/11/2016 Wed 350 0 tue 18 5/12/2016 Thu 308 0 wed 19 5/13/2016 Fri 366 0 thur 20 5/14/2016 Sat 460 1 fri 21 5/15/2016 Sun 427 1 mon 22 5/16/2016 Mon 291 0 tue 23 5/17/2016 Tue 325 0 wed 24 5/18/2016 Wed 354 0 thur 25 5/19/2016 Thu 322 0 fri 26 5/20/2016 Fri 405 0 mon 27 5/21/2016 Sat 442 1 tue 28 5/22/2016 Sun 454 1 wed 29 5/23/2016 Mon 318 0 thur 30 5/24/2016 Tue 298 0 fri 31 5/25/2016 Wed 355 0 mon 32 5/26/2016 Thu 355 0 tue 33 5/27/2016 Fri 374 0 wed 34 5/28/2016 Sat 447 1 thur 35 5/29/2016 Sun 463 1 fri 36 5/30/2016 Mon 291 0 mon 37 5/31/2016 Tue 319 0 tue 38 6/1/2016 Wed 333 0 wed 39 6/2/2016 Thu 339 0 thur 40 6/3/2016 Fri 416 0 fri 41 6/4/2016 Sat 475 1 42 6/5/2016 Sun 459 1 43 6/6/2016 Mon 319 0 44 6/7/2016 Tue 326 0 45 6/8/2016 Wed 356 0 46 6/9/2016 Thu 340 0 600 500 47 48 49 50 51 52 53 54 55 56 6/10/2016 Fri 6/11/2016 Sat 6/12/2016 Sun 6/13/2016 Mon 6/14/2016 Tue 6/15/2016 Wed 6/16/2016 Thu 6/17/2016 Fri 6/18/2016 Sat 6/19/2016 Sun 395 465 453 307 324 350 348 384 474 485 0 1 1 0 0 0 0 0 1 1 600 500 400 300 200 100 0 daily demand 297 293 327 315 348 283 326 317 345 355 305 310 350 308 366 291 325 354 322 405 318 298 355 355 374 291 319 333 339 416 319 326 356 340 395 307 324 350 348 384 1 2 3 4 5 6 7 453 301 315 343 334 380 455 500 450 400 350 300 weekday Average daily demand 250 200 150 100 50 0 1 2 3 4 5 6 7 8 9 SUMMARY OUTPUT Regression Statistics Multiple R 0.27595 R Square 0.076148 Adjusted R 0.05904 Standard Er 59.3716 Observatio 56 ANOVA df Regression Residual Total Daily Demand 600 500 1 54 55 Coefficients Daily Demand Intercept 339.2896 X Variable 1.035578 600 500 400 300 200 100 0 Daily Demand RESIDUAL OUTPUT ObservationPredicted Y 1 340.3252 2 341.3608 3 342.3963 4 343.4319 5 344.4675 6 345.5031 7 346.5387 8 347.5742 9 348.6098 10 349.6454 11 350.681 12 351.7165 13 352.7521 14 353.7877 15 354.8233 16 355.8589 17 356.8944 18 357.93 19 358.9656 20 360.0012 21 361.0367 22 362.0723 23 363.1079 24 364.1435 25 365.179 26 366.2146 27 367.2502 28 368.2858 29 369.3214 30 370.3569 31 371.3925 32 372.4281 33 373.4637 34 374.4992 35 375.5348 36 376.5704 37 377.606 38 378.6416 39 379.6771 40 380.7127 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 381.7483 382.7839 383.8194 384.855 385.8906 386.9262 387.9618 388.9973 390.0329 391.0685 392.1041 393.1396 394.1752 395.2108 396.2464 397.282 SUMMARY OUTPUT Regression Statistics Multiple R 0.385132 R Square 0.148327 Adjusted R 0.125914 Standard Er 29.64367 Observatio 40 ANOVA df ay ge daily demand Regression Residual Total 1 38 39 Coefficients Intercept 313.3115 X Variable 1.044559 RESIDUAL OUTPUT SS MS F Significance F 15689.52 15689.52 4.450943 0.039532 190349.3 3524.987 206038.8 Standard Error t Stat P-value Lower 95%Upper 95%Lower 95.0% Upper 95.0% ObservatioPredicted Y 1 314.3561 2 315.4007 3 316.4452 4 317.4898 5 318.5343 6 319.5789 7 320.6235 8 321.668 9 322.7126 10 323.7571 11 324.8017 12 325.8462 13 326.8908 14 327.9354 15 328.9799 16 330.0245 17 331.069 18 332.1136 16.08265 21.09662 9.59E-028 307.0458 371.5334 307.0458 371.5334 0.490859 2.109726 0.039532 0.051465 2.01969 0.051465 2.01969 Residuals -43.32519 -48.36077 -15.39634 -28.43192 3.532502 101.4969 84.46135 -64.57423 -22.60981 -32.64539 -5.680964 3.283459 75.24788 100.2123 -49.82327 -45.85885 -6.894429 -49.93001 7.034416 99.99884 65.96326 -71.07232 -38.10789 -10.14347 -43.17905 38.78537 74.74979 85.71422 -51.32136 -72.35694 -16.39252 -17.42809 0.536329 72.50075 87.46517 -85.5704 -58.60598 -45.64156 -40.67714 35.28729 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 333.1582 334.2027 335.2473 336.2918 337.3364 338.381 339.4255 340.4701 341.5146 342.5592 343.6038 344.6483 345.6929 346.7374 347.782 348.8265 349.8711 350.9157 351.9602 353.0048 354.0493 355.0939 93.25171 76.21613 -64.81945 -58.85502 -29.8906 -46.92618 7.038243 76.00267 62.96709 -84.06849 -68.10407 -43.13964 -46.17522 -11.2108 77.75362 87.71805 SS MS F Significance F 5815.583 5815.583 6.618039 0.014126 33392.39 878.7472 39207.98 Standard Error t Stat P-value Lower 95%Upper 95%Lower 95.0% Upper 95.0% 9.552723 32.79814 1.75E-029 293.9731 332.65 293.9731 332.65 0.40604 2.572555 0.014126 0.222575 1.866543 0.222575 1.866543 Residuals -17.3561 -22.40066 10.55478 -2.489775 29.46567 -36.57889 5.376548 -4.668011 22.28743 31.24287 -19.80169 -15.84625 23.10919 -19.93537 37.02008 -39.02448 -6.069043 21.8864 -11.15816 70.79728 -17.24728 -38.29184 17.6636 16.61904 34.57448 -49.47008 -22.51463 -9.559193 -4.603752 71.35169 -26.69287 -20.73743 8.218011 -8.826548 45.12889 -43.91567 -27.96023 -3.004784 -6.049343 28.9061 SUMMARY OUTPUT Regression Statistics Multiple R 0.385132 R Square 0.148327 Adjusted R 0.125914 Standard Er 29.64367 Observatio 40 ANOVA df Regression Residual Total SS MS F Significance F 1 5815.583 5815.583 6.618039 0.014126 38 33392.39 878.7472 39 39207.98 Coefficients Standard Error t Stat P-value Lower 95%Upper 95%Lower 95.0% Upper 95.0% Intercept 313.3115 9.552723 32.79814 1.75E-029 293.9731 332.65 293.9731 332.65 X Variable 1.044559 0.40604 2.572555 0.014126 0.222575 1.866543 0.222575 1.866543 RESIDUAL OUTPUT ObservationPredicted Y Residuals 1 314.3561 -17.3561 2 315.4007 -22.40066 3 316.4452 10.55478 4 317.4898 -2.489775 5 318.5343 29.46567 6 319.5789 -36.57889 7 320.6235 5.376548 8 321.668 -4.668011 9 322.7126 22.28743 10 323.7571 31.24287 11 324.8017 -19.80169 12 325.8462 -15.84625 13 326.8908 23.10919 14 327.9354 -19.93537 15 328.9799 37.02008 16 330.0245 -39.02448 17 331.069 -6.069043 18 332.1136 21.8864 19 333.1582 -11.15816 20 334.2027 70.79728 21 335.2473 -17.24728 22 336.2918 -38.29184 23 337.3364 17.6636 24 338.381 16.61904 25 339.4255 34.57448 26 340.4701 -49.47008 27 341.5146 -22.51463 28 342.5592 -9.559193 29 343.6038 -4.603752 30 344.6483 71.35169 31 345.6929 -26.69287 32 346.7374 -20.73743 33 347.782 8.218011 34 348.8265 -8.826548 35 349.8711 45.12889 36 350.9157 -43.91567 37 351.9602 -27.96023 38 353.0048 -3.004784 39 354.0493 -6.049343 40 355.0939 28.9061 Upper 95.0% Daily DemaM2 M3 297 340.3252 314.3561 293 341.3608 315.4007 327 342.3963 316.4452 315 343.4319 317.4898 348 344.4675 318.5343 447 345.5031 319.5789 431 346.5387 320.6235 283 347.5742 321.668 326 348.6098 322.7126 317 349.6454 323.7571 345 350.681 324.8017 355 351.7165 325.8462 428 352.7521 326.8908 454 353.7877 327.9354 305 354.8233 328.9799 310 355.8589 330.0245 350 356.8944 331.069 308 357.93 332.1136 366 358.9656 333.1582 460 360.0012 334.2027 427 361.0367 335.2473 291 362.0723 336.2918 325 363.1079 337.3364 354 364.1435 338.381 322 365.179 339.4255 405 366.2146 340.4701 442 367.2502 341.5146 454 368.2858 342.5592 318 369.3214 343.6038 298 370.3569 344.6483 355 371.3925 345.6929 355 372.4281 346.7374 374 373.4637 347.782 447 374.4992 348.8265 463 375.5348 349.8711 291 376.5704 350.9157 319 377.606 351.9602 333 378.6416 353.0048 339 379.6771 354.0493 416 380.7127 355.0939 475 381.7483 459 382.7839 319 383.8194 326 384.855 356 385.8906 340 395 465 453 307 324 350 348 384 474 485 386.9262 387.9618 388.9973 390.0329 391.0685 392.1041 393.1396 394.1752 395.2108 396.2464 397.282 500 450 400 350 300 250 200 150 100 50 0 Daily Demand M2 M3 Part 1Regression and Forecasting Eli Orchid has designed a new pharmaceutical product, Orchid Relief, which improves the night sleep. Before initiating mass production of the product, Eli Orchid has been market-testing Orchid Relief in Orange County over the past 8 weeks. The daily demand values are recorded in the Excel file provided. Eli Orchid plans on using the sales data to predict sales for the upcoming week. An accurate forecast would be helpful in making arrangements for the company's production processes and designing promotions. Before a forecasting model is built and a forecast for the next week is generated, the COO of the company has asked the data analyst for an exploratory analysis of the demand. Specifically, the COO has asked the analyst1: 1. To provide a bar chart (with data labels rounded to two decimal points) showing the average demand for each week day (Sun., Mon., etc.) 500 450 400 350 300 250 Average daily demand 200 150 100 50 0 1 2 3 4 5 6 2. To fit a simple linear regression model to the data and to provide its equation (d = a*t + b), along with R2 d = 1.0356t + 339.29 R2= 0.0761 3. To fit a multiple regression model with a dummy variable representing the weekend, and to provide the regression equation (d = a*t + b*w + c), along with Adjusted R2. d = 0.7163t + 116.7679w + 315.0262 7 Adjusted R2= 0.8186 1 Round numbers to four decimal points (e.g. 0.1234), unless explicitly requested otherwise. 1 4. To provide a run-series plot of the actual demand with simple regression and multiple regression overlay. Daily Demand 600 500 400 Daily Demand 300 200 100 0 5. To write a short paragraph explaining the observations and providing general recommendations for the next seven days demand forecast. Note: this paragraph can be on page 2. The answers to previous questions must all fit on the first page. The intercept of the daily demand is 339.2896 while the x variable is given by 1.035578. Thus, using these values to forecast the demand for the next seven days will give a higher forecast. 6. To fit a new multiple regression model with dummy variables for weekdays (not the weekend), and to provide the regression equation (d = a*t + b1x1 + b2x2 + b3x3 + b4x4 + b5x5 + b6x6 + c), along with Adjusted R2. d = 313.3115*t - 17.3561x1-22.4007x2+10.5547x32.48977x4+29.4656x5-36.5789x6+1.044559 Adjusted R2= 0.125914 2 7. To use all three models: M1: d = 1.0356t + 339.29 M2: d = 0.7163t + 116.7679w + 315.02 62 M3: (the one considering weekdays) to predict the demand for seven days ahead (Mon, Tue, ..., Sun) and find the total weekly demand. 8. Take advantage of the fact that new demand data became available and use this new data to compare the forecasts using MAPE for days 57-63. 9. To provide a line chart with the actual demand (including the new data) and M2 and M3. 10. To choose the best model for forecasting daily demand at Orchid Relief for 7 days ahead and write a short paragraph explaining your choice. Note: this paragraph can be on page 2. The answers to previous questions must all fit on the first page. Mon. Tue. Wed. Thu. Fri. Sat. Sun. TOTAL: M1 651.399 665.639 694.24 685.18 379.036 810.2291 808.6757 4,694.40 M2 340.3252 341.3608 342.3963 343.4319 344.4675 345.5031 346.5387 2,404.024 M3 314.3561 315.4007 316.4452 317.4898 318.5343 319.5789 320.6235 2,222.43 New: M: 311 T: 341 W: 357 T: 363 F: 390 Sa: 490 Su: 492 MAPEM1: the values are quite higher MAPEM2: the values differs slightly MAPEM3: the values increases in a small range 500 450 400 350 300 250 200 150 100 50 0 Daily Demand M2 M3 The best model for forecasting daily demand at Orchid Relief for 7 days ahead is both M2 and M3. This is because the lines chart are straight as compared to the line chart of daily demand which is made up of curves. 3 4 Part 2Linear Programming Eli Orchid has designed a new pharmaceutical product, Orchid Relief, which improves the night sleep. Before initiating mass production of the product, Eli Orchid has been market-testing Orchid Relief in Orange County over the past 9 weeks. Now that the daily demand for Orange County can be predicted with reasonable accuracy using the M3 model, the COO of the company decided to use it to optimize the production of the new drug. The daily demand values and production process data are recorded in the Excel file provided. The new pharmaceutical product that the company wishes to introduce, Orchid Relief, uses two new ingredients. At this stage, Eli Orchid can procure limited amounts of each ingredient. The company has 4500 pounds of ingredient 1 and 3600 pounds of ingredient 2 available for this week. Eli Orchid can manufacture the new product using any of its three existing processes that have different capabilities. The production with each of the processes is done in batches (a batch typically represents one full run of a machine from when it starts a task until it finishes it). Each batch of production by each of the processes uses different amounts of ingredients 1 and 2, and results in different number of units of Orchid Relief produced (note the difference between a batch and units of Orchid Relief produced). The table below outlines the cost per batch, amounts of the two ingredients required, and the number of units of Orchid Relief yielded per batch. Process 1 $14,000 180 60 120 Cost of production per batch Ingredient 1 required per batch (pounds) Ingredient 2 required per batch (pounds) Orchid Relief yielded per batch (units) Process 2 $30,000 120 420 300 Process 3 $11,000 540 120 60 Eli Orchid needs to determine how many batches to produce with each process in the least costly way given the limited availability of the two ingredients. Also, the total production of Orchid Relief in units must be greater than or equal to the total forecasted demand (in units) for the following week. The COO of the company asked the analyst 1: 11. To use the new M3 model updated with week 9 data (d = 0.6568*Day -151.1703*Mon -136.2715*Tue -110.595*Wed -118.3629*Thu -74.7975*Fri + 1.7679*Sat + 434.5675) to predict the total demand (in units) for Week 10 (days 64-70). M3 Mon. Tue. Wed. Thu. Fri. Sat. Sun. 1 Round numbers to four decimal points (e.g. 0.1234), unless explicitly requested otherwise. 1 TOTAL: 12. To state if this is a maximization or a minimization optimization problem? 13. To provide the mathematical formulation of the objective function assuming that X1, X2, and X3 are the decision variables representing the number of batches of each process to be used. 14. To provide the mathematical formulation of the model constraints. 15. To use the \"Production\" tab of the Excel file and complete the setup by: - entering the forecasted total demand in the pink cell - entering formulas in the five grey cells based on the mathematical formulation 16. To set up Excel Solver (Assume Constraint Precision of 0.000001 and Integer Optimality (%) of 0) and provide the solution to the optimization problem. Supply of ingr. 1 Supply of ingr. 2 Units produced Non-negativity Integer X1, X2, X3 >= 0 X1, X2, X3 : Integer Excel Formulas: Cost of Production Supply of Ingr. 1 Unit Cost Number of batches Process 1 Process 2 Process 3 Cost of production (obj.) Unit cost ($###.##) 17. To label each constraint in the solution as binding or not-binding. Supply of ingr. 1 Supply of ingr. 2 Units produced 2 18. To consider a possible shortage of ingredients in the following week. What would the optimized production process look like if Eli Orchard could only procure 4320 pounds of Ingredient 1 and 1440 pounds of Ingredient 2? Process 1 Process 2 Process 3 19. To label each constraint in the new solution (for the shortage of ingredients) as binding or not-binding. Supply of ingr. 1 Supply of ingr. 2 Units produced 20. To make recommendations about the production processes and pricing of Orchid Relief. Number of batches Cost of production (obj.) Unit cost ($###.##) [write your paragraph here] Note: this paragraph must fit on page 3. The entire project report (with the original description) must fit on 3 pages. Part 3Simulation 3.1 The management of Brinkley Corporation is interested in using simulation to estimate the profit per unit for a new product. The selling price for the product will be $45 per unit. Probability distributions for the purchase cost, the labor cost, and the transportation cost are estimated as follows: Procurem Probabi Lab Probabi Transporta Probabi ent Cost lity or lity tion Cost lity ($) Cos ($) t ($) 10 0.25 20 0.10 3 0.75 3 Procurem Probabi Lab Probabi Transporta Probabi ent Cost lity or lity tion Cost lity ($) Cos ($) t ($) 11 12 0.45 0.30 22 24 25 0.25 0.35 0.30 5 0.25 1. 21. Compute profit per unit for base-case (most likely), worst-case, and best-case scenarios. Base Case using most likely costs Profit = Worst Case Profit = Best Case Profit = 2. 22. Construct a simulation model to estimate the mean profit per unit. The average profit from the simulation model should be approximately: 3. 23. Why is the simulation approach to risk analysis preferable to generating a variety of what-if scenarios? [write your paragraph here] 4. 24. Management believes that the project may not be sustainable if the profit per unit is less than $5. Use simulation to estimate the probability that the profit per unit will be less than $5. [write your paragraph here] 5. 3.2 Construct a spreadsheet simulation model to simulate 10,000 rolls of a die with the six sides numbered 1, 2, 3, 4, 5 and 6. 6. 25. Construct a histogram of the 10,000 observed dice rolls. [add chart here] 4 7. 26. For each roll of two dice, record the sum of the dice. Construct a histogram of the 10,000 observations of the sum of two dice. [add chart here] 8. 27. For each roll of three dice, record the sum of the dice. Construct a histogram of the 10,000 observations of the sum of three dice. [add chart here] 9. 28. Compare the histograms in parts (a), (b) and (c). What statistical phenomenon does this sequence of charts illustrate? (Hint: see Appendix 14.2) 10. 3.3 The management of Madeira Manufacturing Company is considering the introduction of a new product. The fixed cost to begin the production of the product is $30,000. The variable cost for the product is expected to be between $16 and $24, with a most likely value of $20 per unit. The product will sell for $50 per unit. Demand for the product is expected to range from 300 to 2,100 units, with 1,200 units the most likely. 11. 12. 13. 14. 29. Develop a what-if spreadsheet model computing profit for this product in the base-case, worst-case, and best-case scenarios. Profit for this product in the base-case: worst-case: best-case scenarios: 30. Discuss why simulation would be appropriate for this situation. Would simulation be a preferable approach to analyze this situation? Why or why not? [write your paragraph here] 15. 5 Day Date Weekday 1 4/25/2016 Mon 2 4/26/2016 Tue 3 4/27/2016 Wed 4 4/28/2016 Thu 5 4/29/2016 Fri 6 4/30/2016 Sat 7 5/1/2016 Sun 8 5/2/2016 Mon 9 5/3/2016 Tue 10 5/4/2016 Wed 11 5/5/2016 Thu 12 5/6/2016 Fri 13 5/7/2016 Sat 14 5/8/2016 Sun 15 5/9/2016 Mon 16 5/10/2016 Tue 17 5/11/2016 Wed 18 5/12/2016 Thu 19 5/13/2016 Fri 20 5/14/2016 Sat 21 5/15/2016 Sun 22 5/16/2016 Mon 23 5/17/2016 Tue 24 5/18/2016 Wed 25 5/19/2016 Thu 26 5/20/2016 Fri 27 5/21/2016 Sat 28 5/22/2016 Sun 29 5/23/2016 Mon 30 5/24/2016 Tue 31 5/25/2016 Wed 32 5/26/2016 Thu 33 5/27/2016 Fri 34 5/28/2016 Sat 35 5/29/2016 Sun 36 5/30/2016 Mon 37 5/31/2016 Tue 38 6/1/2016 Wed 39 6/2/2016 Thu 40 6/3/2016 Fri 41 6/4/2016 Sat 42 6/5/2016 Sun 43 6/6/2016 Mon 44 6/7/2016 Tue 45 6/8/2016 Wed 46 6/9/2016 Thu 47 6/10/2016 Fri Daily Demand Day 297 293 327 315 348 447 431 283 326 317 345 355 428 454 305 310 350 308 366 460 427 291 325 354 322 405 442 454 318 298 355 355 374 447 463 291 319 333 339 416 475 459 319 326 356 340 395 Mon 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 Tue 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 Wed 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 Thu 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 6/11/2016 Sat 6/12/2016 Sun 6/13/2016 Mon 6/14/2016 Tue 6/15/2016 Wed 6/16/2016 Thu 6/17/2016 Fri 6/18/2016 Sat 6/19/2016 Sun 6/20/2016 Mon 6/21/2016 Tue 6/22/2016 Wed 6/23/2016 Thu 6/24/2016 Fri 6/25/2016 Sat 6/26/2016 Sun 6/27/2016 Mon 6/28/2016 Tue 6/29/2016 Wed 6/30/2016 Thu 7/1/2016 Fri 7/2/2016 Sat 7/3/2016 Sun 465 453 307 324 350 348 384 474 485 311 341 357 363 390 490 492 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 Fri Sat 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 Decision variables Process 1 - X1 Process 2 - X2 Process 3 - X3 Cost of production per batch # batches produced Data for constraints Production requirements and output Ingredient 1 required per batch 2 Ingredient required per batch Orchid Relief yielded per batch Process 1 Process 2 Process 3 180 120 540 Supply of ingredient 1 60 420 120 Supply of ingredient 2 120 300 60 Units of Orchid Relief produced Unit Cost Data for the objective function Process 1 $14,000 Process 2 Process 3 $30,000 $11,000 Objective function Cost of production Constraints Left hand side of constraints Right hand side of constraints <= 4500 <= 3600 >= Calculations Pawel Kalczynski: Enter the total (sum of) forecasted demand for days 64-70