Mathematical Programming Assignment 1. James Beard makes Cheesecake and Black Forest cakes. During any month he can bake at most 700 cakes of either type. The cost for cakes and the demands for cakes, which must be met on time, are listed in the table below. It costs $1.50 to hold a Cheesecake and $1.20 to hold a Black Forest cake in inventory for a month. Determine how to minimize the total cost of meeting the next three months' demands. Month 1 Deman Cost/Cak Month 2 Deman Cost/Cak Month 3 Deman Cost/Cak Month 4 Deman Cost/Cak Cheesecak d 300 e $9 d 350 e $9.25 d 300 e $9.60 d 350 e $9.65 e Black 200 $7.30 450 $7.45 350 $7.55 400 $7.65 Forest 2. The Deckers Clothing Company produces men's pajamas and ladies' nightgowns. Pajamas are made in sets of shirts and pants while nightgowns are one-piece. Each pajama shirt requires 2 yards of cloth and each pair of pajama pants requires 3 yards of cloth, while each nightgown requires 4 yards of cloth. During the next three months the following demands for pajama shirt and pants pairs must be met on time: Month 1, 140,000 sets; Month 2, 155,000 sets; Month 3, 170,000. For nightgowns the monthly demands are 100,000, 140,000, and 120,000, respectively. In each month 1.3 million yards of cloth are available for purchase. Of course, Deckers can elect to purchase less than this amounts. In each month it costs $8 to make a set of pajamas on regular time and $11 to make a set on overtime. For nightgowns the regular and overtime costs are $6 and $9. Additionally, in each month the total number of nightgowns and pajama sets together can be no more than 150,000 produced on regular time, but an unlimited number can be produced on overtime. At the end of each month a holding cost of $1 per product set is assessed. Formulate and solve the mathematical programming model to determine how to meet demand for pajamas at the minimum possible cost. Assume there are 15,000 sets of pajamas and 9,000 nightgowns available in inventory at the beginning of Month 1. 3. Each year, Comfy Shoes faces demands (which must be met on time) for pairs of shoes as shown in the table below. Workers work three consecutive quarters and then receive one quarter off. For example, a worker might work during Quarters 3 and 4 of one year and Quarter 1 of the next year. During a quarter in which a worker works, he or she can produce 500 pairs of shoes. Each worker is paid $25,000 for each quarter worked. At the end of each quarter, a holding cost of $30 per pair of shoes is assessed. Additionally, no more than 16 workers can be employed at any given time. Determine how to minimize the cost per year (labor plus holding cost) of meeting the demand for shoes. To simplify matters, assume that at the end of the year, the ending inventory will be zero. (Hint: you may assume that a given worker will get the same quarter off during each year.) Quarter Demand Quarter 1 Quarter 2 Quarter 3 Quarter 4 7000 4500 8500 10000 4. Bibbins Manufacturing produces softballs and baseballs for youth recreation leagues. Each softball sells for $21, while each baseball sells for $24. The material and labor required to produce each item is listed here along with the availability of each resource. Formulate and solve the math programming model to determine the mix of products to make which will maximize profit while making sure each product comprises no less than 1/3 of the total unit production. In reading your solution state which of the resources below you have in excess over what is needed for the optimal mix of softballs and baseballs. Resource Leather Nylon Core General Labor Stitching Labor Amount Required For Softball Baseball 5 oz. 4 oz. 6 yds. 3 yds. 4 oz. 2 oz. 2.5 min. 2 min. 5 min. 4 min. Amount Available Cost 325 lbs. 5,400 yds. 250 lbs. 20 hours $6/lb. $15/100 yds. $9/lb. $10/hr. 50 hours $!8/hr. Registered Nurses Certified Nursin Monday Tuesday WednesdayThursday Friday Saturday Sunday Monday Tuesday 1500 1500 1500 1800 2100 2100 1800 150 150 Monday 1 1 1 0.5 Tuesday 1 1 1 0.5 Wednesday 1 1 1 Thursday 1 1 1 Friday 1 1 1 Saturday 1 1 1 Sunday 1 1 1 Monday 1 1 1 -1 Tuesday 1 1 1 -1 Wednesday 1 1 1 Thursday 1 1 1 Friday 1 1 1 Saturday 1 1 1 Sunday 1 1 1 Solution 21 0 5 18 0 0 18 38 18 Certified Nursing Assistants WednesdayThursday Friday Saturday Sunday 150 150 150 150 150 0.5 0.5 0.5 0.5 0.5 -1 -1 -1 -1 -1 24 22 14 16 16 126000 58 >= 48 >= 38 >= 34 >= 30 >= 26 >= 26 >= 1 >= 21 >= 2 >= 1 >= 9 >= 2 >= 2 >= 58 48 38 34 30 26 26 0 0 0 0 0 0 0 Wood Skilled Labor Unskilled Labor Ratio Ratio Limit Solution Dining Tables Chairs Coffee Tables End Tables 200 160 120 80 $ 5,482,080.00 18 9 12 4 376416 <= 1.5 2 1 1 60000 <= 4 2 3 2 114318 <= 1 -0.25 0 <= 1 -0.5 -1.5 <= 1 12000 <= 2526 10104 12000 24003 600000 60000 180000 0 0 12000 Brazilian ColombianPeruvian 3.55 4.26 4.97 Aroma -3 -18 7 Strength -1 4 2 Brazilian 1 Colombian 1 Peruvian 1 Total 1 1 1 Solution 1500 520 1980 17380.8 0 >= 4540 >= 1500 <= 520 <= 1980 <= 4000 >= 0 0 1500 1200 2000 4000 0.5 3.55 0.6 4.26 0.7 4.97 Restriction A Restriction B Restriction C Restriction D Restriction E Restriction F Restriction G Solution Certificates A Mortgages B Mortgages A Auto 0.02 0.04 0.05 1 1 1 -1 1 1 1 1 15,000,000 - 1 22,500,000 B Auto C Auto 0.03 1 0.05 1 0.12 1 -1 -1 -1 1 1 1 1 -1 -1 11,250,000 11,250,000 - D Auto A Personal B Personal C Personal D PersonalPayday 0.15 0.06 0.11 0.15 0.21 1 1 1 1 1 1 -1 1 1 1 -1 -1 - - 1 1 1 -1 30,000,000 30,000,000 0.38 1 1 1 -1 - 30,000,000 22,537,500 150000000 <= 7500000 >= 15000000 >= 30000000 <= 0 >= 60000000 <= 0 >= 0 >= 150,000,000 0 15,000,000.0 30,000,000.0 0 60,000,000.0 0 0 Let cake 1 is cheasecake and cake 2 is blackforest let Xt be the number of cakes i produced in month j, for (i = 1,2 and j = 1,2,3). let it be the inventory of cakes i at the end of month j, for (i = 1,2 and j = 1,2,3). the objective is to minimise the cost of meeting demands for the next 3 months. z = cost of producing + inventory cost z = (9X11 + 9.25X12 + 9.6X13 + 9.65X14) + (7.3X21 +7.45X22 + 7.55X23 + 7.65X24) + 1.5(i11 + i12 + i13 + i14) + 1.2 (i21 + i2 constraint 1: at the end of the month 1 for cake 1, the inventory will be: inventory at the end of the month = (number of cake 1 produced in month 1) - (demand during month 1) i11 = X11 - 300 X11 - i11 = 300 constraint 2: at the end of month 2 for cake 1: i12 = X12 - 350 + i11 i11 + X12 - i12 = 350 constraint 3: at the end of month 3 for cake 1: i12 + X13 - i13 = 300 constraint 4: at the end of moth 4 for cake 1: i13 + X14 - i14 = 350 constraint 5: at the end of month 1 for cake 2: X21 - i21 = 200 constraint 6: at the end of month 2 for cake 2: i21 + X22 - i22 = 450 constraint 7: at the end of month 3 for cake 2: i22 + X23 - i23 = 350 constraint 8: at the end of month 4 for cake 2: i23 + X24 - i24 = 400 constraint 9: at most 700 cake can be made during a month X11, X21, X12, X22, X13, X23, X14, X24 <= 700 Non-negative constraint: X11, X21, X12, X22, X13, X23, X14, X24 >= 0 + i13 + i14) + 1.2 (i21 + i22 + i23 + i24) nd during month 1) Microsoft Excel 14.0 Answer Report Worksheet: [aneeb9999 james beard Q1.xlsx]Sheet2 Report Created: 14/04/2017 02:57:46 Result: Solver found a solution. All Constraints and optimality conditions are satisfied. Solver Engine Engine: Simplex LP Solution Time: 0 Seconds. Iterations: 0 Subproblems: 0 Solver Options Max Time Unlimited, Iterations Unlimited, Precision 0.000001, Use Automatic Scaling Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume NonNegative Objective Cell (Min) Cell Name $D$23 z X11 Original Value 0 Final Value Variable Cells Cell Name $D$22 solution X11 $E$22 solution X12 $F$22 solution X13 $G$22 solution X14 $H$22 solution X21 $I$22 solution X22 $J$22 solution X23 $K$22 solution X24 $L$22 solution i11 $M$22 solution i12 $N$22 solution i13 $O$22 solution i14 $P$22 solution i21 $Q$22 solution i22 $R$22 solution i23 $S$22 solution i24 Original Value 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Final Value Constraints Cell Name $D$47 LHS $D$48 LHS $D$49 LHS $D$50 LHS $D$51 LHS $D$52 LHS $D$53 LHS $D$54 LHS Cell Value 0 Integer 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin 0 Contin Formula 0 $D$47<=$E$47 0 $D$48<=$E$48 0 $D$49<=$E$49 0 $D$50<=$E$50 0 $D$51<=$E$51 0 $D$52<=$E$52 0 $D$53<=$E$53 0 $D$54<=$E$54 Status Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Slack 300 350 300 350 200 450 350 400 $D$55 $D$56 $D$57 $D$58 $D$59 $D$60 $D$61 $D$62 $D$63 LHS LHS LHS LHS LHS LHS LHS LHS LHS 0 $D$55<=$E$55 0 $D$56<=$E$56 0 $D$57<=$E$57 0 $D$58<=$E$58 0 $D$59<=$E$59 0 $D$60<=$E$60 0 $D$61<=$E$61 0 $D$62<=$E$62 0 $D$63<=$E$63 Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding 700 700 700 700 700 700 700 700 700 e NonNegative Microsoft Excel 14.0 Sensitivity Report Worksheet: [aneeb9999 james beard Q1.xlsx]Sheet2 Report Created: 14/04/2017 02:57:56 Variable Cells Cell $D$22 $E$22 $F$22 $G$22 $H$22 $I$22 $J$22 $K$22 $L$22 $M$22 $N$22 $O$22 $P$22 $Q$22 $R$22 $S$22 Name solution X11 solution X12 solution X13 solution X14 solution X21 solution X22 solution X23 solution X24 solution i11 solution i12 solution i13 solution i14 solution i21 solution i22 solution i23 solution i24 Final Reduced Objective Allowable Allowable Value Cost Coefficient Increase Decrease 0 9 9 1.000E+030 9 0 9.25 9.25 1.000E+030 9.25 0 9.6 9.6 1.000E+030 9.6 0 9.65 9.65 1.000E+030 9.65 0 7.3 7.3 1.000E+030 7.3 0 7.45 7.45 1.000E+030 7.45 0 7.55 7.55 1.000E+030 7.55 0 7.65 7.65 1.000E+030 7.65 0 1.5 1.5 1.000E+030 1.5 0 1.5 1.5 1.000E+030 1.5 0 1.5 1.5 1.000E+030 1.5 0 1.5 1.5 1.000E+030 1.5 0 1.2 1.2 1.000E+030 1.2 0 1.2 1.2 1.000E+030 1.2 0 1.2 1.2 1.000E+030 1.2 0 1.2 1.2 1.000E+030 1.2 Constraints Cell $D$47 $D$48 $D$49 $D$50 $D$51 $D$52 $D$53 $D$54 $D$55 $D$56 $D$57 $D$58 $D$59 $D$60 $D$61 $D$62 $D$63 Name LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS LHS Final Shadow Constraint Allowable Allowable Value Price R.H. Side Increase Decrease 0 0 300 1.000E+030 300 0 0 350 1.000E+030 350 0 0 300 1.000E+030 300 0 0 350 1.000E+030 350 0 0 200 1.000E+030 200 0 0 450 1.000E+030 450 0 0 350 1.000E+030 350 0 0 400 1.000E+030 400 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 0 0 700 1.000E+030 700 MINIMISE z = (9X11 + 9.25X12 + 9.6X13 + 9.65X14) + (7.3X21 +7.45X22 + 7.55X23 + 7.65X24) + 1.5(i11 + i12 + i13 + i14) + 1 subject to 1 X11 - i11 = 300 2 i11 + X12 - i12 = 350 3 i12 + X13 - i13 = 300 4 i13 + X14 - i14 = 350 5 X21 - i21 = 200 6 i21 + X22 - i22 = 450 7 i22 + X23 - i23 = 350 8 i23 + X24 - i24 = 400 9 X11, X21, X12, X22, X13, X23, X14, X24 <= 700 10 X11, X21, X12, X22, X13, X23, X14, X24 >= 0 variables coefficients solution z X11 X12 9 0 0 CONSTRAINT X11 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 CONSTRAINTS 9.25 0 X12 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 LHS 1 2 X13 9.6 0 X13 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 RHS 0 0 X14 300 350 X21 9.65 0 X14 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 X22 7.3 0 X21 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 7.45 0 X22 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 300 350 200 450 350 400 700 700 700 700 700 700 700 700 700 ) + 1.5(i11 + i12 + i13 + i14) + 1.2 (i21 + i22 + i23 + i24) X23 X24 7.55 0 X23 i11 7.65 0 X24 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 i12 1.5 0 i11 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 i13 1.5 0 i12 -1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 i14 1.5 0 i13 0 -1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 i21 1.5 0 i14 0 0 -1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 i22 1.2 0 i21 0 0 0 -1 0 0 0 0 1 0 0 0 0 0 0 0 0 i23 1.2 0 i22 0 0 0 0 -1 1 0 0 1 0 0 0 0 0 0 0 0 i24 1.2 0 i23 0 0 0 0 0 -1 1 0 1 0 0 0 0 0 0 0 0 1.2 0 i24 0 0 0 0 0 0 -1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 = = = = = = = = <= <= <= <= <= <= <= <= <= 300 350 300 350 200 450 350 400 700 700 700 700 700 700 700 700 700 Microsoft Excel 14.0 Sensitivity Report Worksheet: [aneeb9999 DECKERS Q2.xlsx]Sheet1 Report Created: 14/04/2017 00:45:00 Variable Cells Cell $F$2 $G$2 $H$2 $F$3 $G$3 $H$3 $F$4 $G$4 $H$4 $F$5 $G$5 $H$5 Name number of panjama sets to produce (regular time) month 1 number of panjama sets to produce (regular time) month 2 number of panjama sets to produce (regular time) month 3 number of nightgowns to produce (regular time) month 1 number of nightgowns to produce (regular time) month 2 number of nightgowns to produce (regular time) month 3 number of panjama sets to produce (overtime) month 1 number of panjama sets to produce (overtime) month 2 number of panjama sets to produce (overtime) month 3 number of nightgowns to produce (overtime) month 1 number of nightgowns to produce (overtime) month 2 number of nightgowns to produce (overtime) month 3 Final Value 150000 134000 30000 0 16000 120000 14000 0 122000 91000 124000 0 Constraints Cell $F$15 $F$16 $F$17 $F$18 $F$19 $F$20 $F$21 $F$22 $F$23 $F$24 $F$25 $F$26 Name cloth availability month 1 month 1 cloth availability month 2 month 1 cloth availability month 3 month 1 Demand panjama sets (month 1) month 1 Demand panjama sets (month 2) month 1 Demand panjama sets (month 3) month 1 demand nightgowns (month 1) month 1 demand nightgowns (month 2) month 1 demand nightgowns (month 3) month 1 quantity that can be produced on regular time (month 1) month 1 quantity that can be produced on regular time (month 2) month 1 quantity that can be produced on regular time (month 3) month 1 Final Value 1229500 1300000 1300000 179000 173000 170000 100000 140000 120000 150000 150000 150000 Reduced Objective Allowable Allowable Cost Coefficient Increase Decrease 0 10 0 1.0E+030 0 9 0 0.111111 0 8 3 0 0 8 1.0E+030 0 0 7 0.111111 0 0 6 0 10.8 0 13 0.111111 0 0 12 1.0E+030 0 0 11 0 3 0 11 0 0.1 0 10 0 1 0 9 1.0E+030 0 Shadow Constraint Allowable Allowable Price R.H. Side Increase Decrease 0 1300000 1.0E+030 70500 -0.2 1300000 70000 70500 -0.4 1300000 70000 70500 0 140000 39000 1.0E+030 0 155000 18000 1.0E+030 13 170000 14100 14000 0.1 100000 15555.56 91000 0.1 140000 20000 17777.78 10.8 120000 15666.67 15555.56 -3 150000 14000 150000 -3 150000 124000 16000 -3 150000 122000 30000 Microsoft Excel 14.0 Answer Report Worksheet: [aneeb9999 DECKERS Q2.xlsx]Sheet1 Report Created: 14/04/2017 00:43:16 Result: Solver found a solution. All Constraints and optimality conditions are satisfied. Solver Engine Engine: Simplex LP Solution Time: 0.062 Seconds. Iterations: 17 Subproblems: 0 Solver Options Max Time Unlimited, Iterations Unlimited, Precision 0.000001, Use Automatic Scaling Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume NonNegative Objective Cell (Min) Cell $K$2 Name to minimise cost Variable Cells Cell Name number of panjama sets to produce (regular time) month 1 number of panjama sets to produce (regular time) month 2 number of panjama sets to produce (regular time) month 3 number of nightgowns to produce (regular time) month 1 number of nightgowns to produce (regular time) month 2 number of nightgowns to produce (regular time) month 3 number of panjama sets to produce (overtime) month 1 number of panjama sets to produce (overtime) month 2 number of panjama sets to produce (overtime) month 3 number of nightgowns to produce (overtime) month 1 number of nightgowns to produce (overtime) month 2 number of nightgowns to produce (overtime) month 3 Cell Name cloth availability month 1 month 1 cloth availability month 2 month 1 cloth availability month 3 month 1 Demand panjama sets (month 1) month 1 Demand panjama sets (month 2) month 1 Demand panjama sets (month 3) month 1 demand nightgowns (month 1) month 1 demand nightgowns (month 2) month 1 demand nightgowns (month 3) month 1 quantity that can be produced on regular time (month 1) month 1 quantity that can be produced on regular time (month 2) month 1 quantity that can be produced on regular time (month 3) month 1 $F$2 $G$2 $H$2 $F$3 $G$3 $H$3 $F$4 $G$4 $H$4 $F$5 $G$5 $H$5 Constraints $F$15 $F$16 $F$17 $F$18 $F$19 $F$20 $F$21 $F$22 $F$23 $F$24 $F$25 $F$26 Assume NonNegative Original Value -703000 Final Value 6840000 Original Value 0 0 0 0 0 0 0 0 0 0 0 0 Final Value Integer 150000 Contin 134000 Contin 30000 Contin 0 Contin 16000 Contin 120000 Contin 14000 Contin 0 Contin 122000 Contin 91000 Contin 124000 Contin 0 Contin Cell Value Formula 1229500 $F$15<=$H$15 1300000 $F$16<=$H$16 1300000 $F$17<=$H$17 179000 $F$18>=$H$18 173000 $F$19>=$H$19 170000 $F$20>=$H$20 100000 $F$21>=$H$21 140000 $F$22>=$H$22 120000 $F$23>=$H$23 150000 $F$24<=$H$24 150000 $F$25<=$H$25 150000 $F$26<=$H$26 Status Not Binding Binding Binding Not Binding Not Binding Binding Binding Binding Binding Binding Binding Binding Slack 70500 0 0 39000 18000 0 0 0 0 0 0 0 month 1 month 2 month 3 150000 134000 30000 0 16000 120000 14000 0 122000 91000 124000 0 140000 155000 170000 100000 140000 120000 1300000 1300000 1300000 15000 9000 1229500 <= 1300000 <= 1300000 <= 179000 >= 173000 >= 170000 >= 100000 >= 140000 >= 120000 >= 150000 <= 150000 <= 150000 <= 39000 0 18000 0 1300000 1300000 1300000 140000 155000 170000 100000 140000 120000 150000 150000 150000 objective to minimise 6840000 quarter demand labour at labor on ofproductionclosing invholding coslabor cost 1 7000 11 5 5500 200 6000 275000 2 4500 7 9 3500 500 15000 175000 3 8500 12 4 6000 -200 0 300000 4 10000 16 0 8000 0 0 400000 46 18 21000 1150000 total cost = 1171000