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(a)This is a variant of an assignment problem where the assignees are the students and the tasks are the classes.However,each student is to be assigned to two classes rather than just one and each class is to be assigned five students.Therefore,each student needs to be split into two assignees and each class needs to be split into five tasks.Furthermore,the objective here involves maximization rather than minimization of costs as assumed by the assignment problem.To convert to an equivalent problem involving minimization,we can multiply each bid by -1.To avoid negative costs,we then can add the largest bid (70)to each of these negative bids.(A number larger than 70 also could have been added to each of these negative bids.)Before splitting each student and each class into multiple assignees and multiple tasks,we now have the following cost table.Cost 11ea84c6_c853_fe2b_83dc_c7e38160bfbd_TB2462_00 After next splitting the students and classes into multiple assignees and tasks,we now have an appropriate cost table shown below for formulating this problem as an assignment problem. 11ea84c6_c854_253c_83dc_252fc5a4dd70_TB2462_00 (b)The formulation of the equivalent transportation problem begins in just the same way as in part (a)by converting the table of bids into the equivalent table of costs shown in the first table in part (a).However,it is no longer necessary to split the students and classes into multiple assignees and tasks because the supplies and demands in the parameter table provide the same information.Therefore,the appropriate parameter table is the one shown below. 11ea84c6_c854_4c4d_83dc_839fb1e553c7_TB2462_00 (c)This is a variant of an assignment problem.The assignees are the students and the tasks are the classes.Each student is assigned to take two classes.Each class has a limit of five students.Start by entering the data.The data for this problem are the bid points for each student for each class,where the corresponding data cells are assigned a range name of Points (C5:F14). 11ea84c6_c854_4c4e_83dc_bf7383ff2503_TB2462_00 The decisions to be made in this problem are whether or not to assign each student to each class.Therefore,a table of changing cells is created for each student and class combination in C18:F27,and given a range name of Assignment.The values in Assignment (C18:F27)will eventually be determined by the Solver.For now,arbitrary values of 0 and 1 are entered. 11ea84c6_c854_735f_83dc_dd8ceedd6b5b_TB2462_00 The goal is to maximize the total bid points of the assignments.Total Bid Points = SUMPRODUCT(Points,Assignment).This formula is entered into cell I29. 11ea84c6_c854_7360_83dc_e1a7f5afcfa8_TB2462_00 11ea84c6_c854_9a71_83dc_0f8932427dea_TB2462_00 The functional constraints in this problem are that each student must be assigned to two classes,each class is limited to five students,and each student can take each class at most once.The number of classes a student is assigned to is just the sum of the row of changing cells for each student.For example,for George it is =SUM(C18:F18).This formula is copied into G18:G27.The number of students assigned to a class is the sum of the column of changing cells for each class.For example,for Management Science it is =SUM(C18:C27).This formula is copied into C28:F28. 11ea84c6_c854_c182_83dc_195265a6809b_TB2462_00 11ea84c6_c854_c183_83dc_a751050a4a46_TB2462_00 11ea84c6_c854_e894_83dc_9be618397b85_TB2462_00 The Solver information and solved spreadsheet are shown below. 11ea84c6_c855_0fa5_83dc_a127c44a12bf_TB2462_00 11ea84c6_c855_0fa6_83dc_b7e13f2d5cd1_TB2462_00 11ea84c6_c855_36b7_83dc_8de1a0dd24a2_TB2462_00 11ea84c6_c855_36b8_83dc_d734d88654c2_TB2462_00 11ea84c6_c855_5dc9_83dc_b1d6c75f93c0_TB2462_00 11ea84c6_c855_84da_83dc_e9b06feab107_TB2462_00 11ea84c6_c855_84db_83dc_c1a260662289_TB2462_00 Thus,the 1's in Assignment (C18:F27)show the assignments that should be made,achieving a total of 705 points.(d)No.For example,Eric did not get into Management Science despite bidding 50 points,while Ann got in with only 45 points.Also,Eric got into classes worth only 45 total bid points to him while Liz got classes worth 100 bid points to her.(e)Perhaps maximizing the minimum total number of bid points achieved by each student.
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(a)This is a variant of an assignment problem where the assignees are the students and the tasks are the classes.However,each student is to be assigned to two classes rather than just one and each class is to be assigned five students.Therefore,each student needs to be split into two assignees and each class needs to be split into five tasks.Furthermore,the objective here involves maximization rather than minimization of costs as assumed by the assignment problem.To convert to an equivalent problem involving minimization,we can multiply each bid by -1.To avoid negative costs,we then can add the largest bid (70)to each of these negative bids.(A number larger than 70 also could have been added to each of these negative bids.)Before splitting each student and each class into multiple assignees and multiple tasks,we now have the following cost table.Cost 11ea84c6_c853_fe2b_83dc_c7e38160bfbd_TB2462_00 After next splitting the students and classes into multiple assignees and tasks,we now have an appropriate cost table shown below for formulating this problem as an assignment problem. 11ea84c6_c854_253c_83dc_252fc5a4dd70_TB2462_00 (b)The formulation of the equivalent transportation problem begins in just the same way as in part (a)by converting the table of bids into the equivalent table of costs shown in the first table in part (a).However,it is no longer necessary to split the students and classes into multiple assignees and tasks because the supplies and demands in the parameter table provide the same information.Therefore,the appropriate parameter table is the one shown below. 11ea84c6_c854_4c4d_83dc_839fb1e553c7_TB2462_00 (c)This is a variant of an assignment problem.The assignees are the students and the tasks are the classes.Each student is assigned to take two classes.Each class has a limit of five students.Start by entering the data.The data for this problem are the bid points for each student for each class,where the corresponding data cells are assigned a range name of Points (C5:F14). 11ea84c6_c854_4c4e_83dc_bf7383ff2503_TB2462_00 The decisions to be made in this problem are whether or not to assign each student to each class.Therefore,a table of changing cells is created for each student and class combination in C18:F27,and given a range name of Assignment.The values in Assignment (C18:F27)will eventually be determined by the Solver.For now,arbitrary values of 0 and 1 are entered. 11ea84c6_c854_735f_83dc_dd8ceedd6b5b_TB2462_00 The goal is to maximize the total bid points of the assignments.Total Bid Points = SUMPRODUCT(Points,Assignment).This formula is entered into cell I29. 11ea84c6_c854_7360_83dc_e1a7f5afcfa8_TB2462_00 11ea84c6_c854_9a71_83dc_0f8932427dea_TB2462_00 The functional constraints in this problem are that each student must be assigned to two classes,each class is limited to five students,and each student can take each class at most once.The number of classes a student is assigned to is just the sum of the row of changing cells for each student.For example,for George it is =SUM(C18:F18).This formula is copied into G18:G27.The number of students assigned to a class is the sum of the column of changing cells for each class.For example,for Management Science it is =SUM(C18:C27).This formula is copied into C28:F28. 11ea84c6_c854_c182_83dc_195265a6809b_TB2462_00 11ea84c6_c854_c183_83dc_a751050a4a46_TB2462_00 11ea84c6_c854_e894_83dc_9be618397b85_TB2462_00 The Solver information and solved spreadsheet are shown below. 11ea84c6_c855_0fa5_83dc_a127c44a12bf_TB2462_00 11ea84c6_c855_0fa6_83dc_b7e13f2d5cd1_TB2462_00 11ea84c6_c855_36b7_83dc_8de1a0dd24a2_TB2462_00 11ea84c6_c855_36b8_83dc_d734d88654c2_TB2462_00 11ea84c6_c855_5dc9_83dc_b1d6c75f93c0_TB2462_00 11ea84c6_c855_84da_83dc_e9b06feab107_TB2462_00 11ea84c6_c855_84db_83dc_c1a260662289_TB2462_00 Thus,the 1's in Assignment (C18:F27)show the assignments that should be made,achieving a total of 705 points.(d)No.For example,Eric did not get into Management Science despite bidding 50 points,while Ann got in with only 45 points.Also,Eric got into classes worth only 45 total bid points to him while Liz got classes worth 100 bid points to her.(e)Perhaps maximizing the minimum total number of bid points achieved by each student.
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The optimality test requires solving for the ui and vj such that cij = ui + vj for each (i,j)such that xij is basic.Since the number of unknowns (the ui and vj )is one larger than the number of equations,and since Source 1 has the largest number of basic variables,we can begin by setting ui = 0.We then solve for each of the remaining unknowns as outlined below.x12: 27 = u1 + v2.Set u1 = 0,so v2 = 27,x13: 28 = u1 + v3. v3 = 28,x15: 0 = u1 + v5. v5 = 0.x25: 0 = u2 + v5.Know v5 = 0,so u2 = 0.x24: 23 = u2 + v4.Know u2 = 0,so v4 = 23.x34: 21 = u3 + v4.Know v4 = 23,so u3 = -2.x31: 37 = u3 + v1.Know u3 = -2,so v1 = 39.We next calculate cij - ui - vj for each of the nonbasic variables.x11: c11 - u1 - v1 = 41 - 0 - 39 = 2 x14: c14 - u1 - v4 = 24 - 0 - 23 = 1 x21: c21 - u2 - v1 = 40 - 0 - 39 = 1 x22: c22 - u2 - v2 = 29 - 0 - 27 = 2 x23: c23 - u2 - v3 = M - 0 - 28 = M - 28 x32: c32 - u3 - v2 = 30 + 2 - 27 = 5 x33: c33 - u3 - v3 = 27 + 2 - 28 = 1 x35: c35 - u3 - v5 = 0 + 2 - 0 = 2 The optimality test states that a BF solution is optimal if and only if 11ea84c6_c84e_0aaa_83dc_f1e7ed172f19_TB2462_00 for every (i,j)such that xij is nonbasic.Since the above calculations verify that this test passes,the solution is optimal.
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The optimality test requires solving for the ui and vj such that cij = ui + vj for each (i,j)such that xij is basic.Since the number of unknowns (the ui and vj )is one larger than the number of equations,and since Source 1 has the largest number of basic variables,we can begin by setting ui = 0.We then solve for each of the remaining unknowns as outlined below.x12: 27 = u1 + v2.Set u1 = 0,so v2 = 27,x13: 28 = u1 + v3. v3 = 28,x15: 0 = u1 + v5. v5 = 0.x25: 0 = u2 + v5.Know v5 = 0,so u2 = 0.x24: 23 = u2 + v4.Know u2 = 0,so v4 = 23.x34: 21 = u3 + v4.Know v4 = 23,so u3 = -2.x31: 37 = u3 + v1.Know u3 = -2,so v1 = 39.We next calculate cij - ui - vj for each of the nonbasic variables.x11: c11 - u1 - v1 = 41 - 0 - 39 = 2 x14: c14 - u1 - v4 = 24 - 0 - 23 = 1 x21: c21 - u2 - v1 = 40 - 0 - 39 = 1 x22: c22 - u2 - v2 = 29 - 0 - 27 = 2 x23: c23 - u2 - v3 = M - 0 - 28 = M - 28 x32: c32 - u3 - v2 = 30 + 2 - 27 = 5 x33: c33 - u3 - v3 = 27 + 2 - 28 = 1 x35: c35 - u3 - v5 = 0 + 2 - 0 = 2 The optimality test states that a BF solution is optimal if and only if 11ea84c6_c84e_0aaa_83dc_f1e7ed172f19_TB2462_00 for every (i,j)such that xij is nonbasic.Since the above calculations verify that this test passes,the solution is optimal.
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We define the sources and destinations a...View Answer
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We define the sources and destinations a...
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(a)The above table already is close to being a parameter table for a transportation problem,but we still need to add the two unit costs shown for each combination of a plant and a warehouse,as well as identify the monthly production capacity as being the "supply" from each plant.Because the total supply (220)exceeds the total demand (210),we also need a dummy destination to receive the excess supply.The resulting parameter table for formulating this problem as a transportation problem is shown below. 11ea84c6_c84f_df6c_83dc_911987696080_TB2462_00 (b)This is a transportation problem as described in Sec.9.1 of the textbook.Start by entering the data.The data for this problem are the unit shipping cost,the unit production cost and monthly production capacity at each plant,and the monthly demand at each warehouse.The data are shown below,where range names of UnitProductionCost (H5:H6),MonthlyCapacity (H11:H12),and MonthlyDemand (C15:E15)are assigned to the corresponding data cells.Note that space has been reserved in the middle of the spreadsheet (C11:E12)for the changing cells. 11ea84c6_c850_067d_83dc_bbe395ede5bb_TB2462_00 The decisions to be made in this problem are how many units to ship from each plant to each warehouse.Therefore,we add a table of changing cells with range name UnitsShipped (C11:E12).The values in UnitsShipped (C11:E12)will eventually be determined by the Solver.For now,zeroes are entered. 11ea84c6_c850_067e_83dc_4d2d82ff5ef3_TB2462_00 The goal is to minimize the total cost,including production costs and shipping costs.The total production costs and the total shipping costs are calculated as follows.Total Production Cost = SUMPRODUCT(UnitProductionCost,UnitsShipped),Total Shipping Cost = SUMPRODUCT(ShippingCost,UnitsShipped).These formulas are entered into H16 and H17,with the overall total cost calculated by summing these two costs in H18,as shown in the spreadsheet below. 11ea84c6_c850_2d8f_83dc_a107f35cca60_TB2462_00 11ea84c6_c850_54a0_83dc_7de1651c3fc5_TB2462_00 The functional constraints in this problem are that each plant can't ship out more than its monthly capacity,and each warehouse needs to receive its monthly demand.The units shipped out of each plant are calculated by summing each row of changing cells,while the units shipped to each warehouse are caculated by summing each column of changing cells.These constraints are shown in the spreadsheet below. 11ea84c6_c850_54a1_83dc_313af93b601c_TB2462_00 11ea84c6_c850_7bb2_83dc_131cc8f7cce9_TB2462_00 11ea84c6_c850_a2c3_83dc_61c54433a1c8_TB2462_00 The Solver information and solved spreadsheet are shown below. 11ea84c6_c850_c9d4_83dc_b3c350bb11ac_TB2462_00 11ea84c6_c850_c9d5_83dc_b97eafd1688f_TB2462_00 11ea84c6_c850_f0e6_83dc_5ff147eeca40_TB2462_00 11ea84c6_c851_17f7_83dc_5551f0d8babd_TB2462_00 11ea84c6_c851_17f8_83dc_d97234ca9566_TB2462_00 11ea84c6_c851_3f09_83dc_d10964ae1a2e_TB2462_00 Thus,from Plant A they should ship 40 to Warehouse 1 and 60 units to Warehouse 2,from Plant B they should ship 40 units to Warehouse 1 and 70 units to Warehouse 3,giving an overall total cost of $132,790.
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(a)The above table already is close to being a parameter table for a transportation problem,but we still need to add the two unit costs shown for each combination of a plant and a warehouse,as well as identify the monthly production capacity as being the "supply" from each plant.Because the total supply (220)exceeds the total demand (210),we also need a dummy destination to receive the excess supply.The resulting parameter table for formulating this problem as a transportation problem is shown below. 11ea84c6_c84f_df6c_83dc_911987696080_TB2462_00 (b)This is a transportation problem as described in Sec.9.1 of the textbook.Start by entering the data.The data for this problem are the unit shipping cost,the unit production cost and monthly production capacity at each plant,and the monthly demand at each warehouse.The data are shown below,where range names of UnitProductionCost (H5:H6),MonthlyCapacity (H11:H12),and MonthlyDemand (C15:E15)are assigned to the corresponding data cells.Note that space has been reserved in the middle of the spreadsheet (C11:E12)for the changing cells. 11ea84c6_c850_067d_83dc_bbe395ede5bb_TB2462_00 The decisions to be made in this problem are how many units to ship from each plant to each warehouse.Therefore,we add a table of changing cells with range name UnitsShipped (C11:E12).The values in UnitsShipped (C11:E12)will eventually be determined by the Solver.For now,zeroes are entered. 11ea84c6_c850_067e_83dc_4d2d82ff5ef3_TB2462_00 The goal is to minimize the total cost,including production costs and shipping costs.The total production costs and the total shipping costs are calculated as follows.Total Production Cost = SUMPRODUCT(UnitProductionCost,UnitsShipped),Total Shipping Cost = SUMPRODUCT(ShippingCost,UnitsShipped).These formulas are entered into H16 and H17,with the overall total cost calculated by summing these two costs in H18,as shown in the spreadsheet below. 11ea84c6_c850_2d8f_83dc_a107f35cca60_TB2462_00 11ea84c6_c850_54a0_83dc_7de1651c3fc5_TB2462_00 The functional constraints in this problem are that each plant can't ship out more than its monthly capacity,and each warehouse needs to receive its monthly demand.The units shipped out of each plant are calculated by summing each row of changing cells,while the units shipped to each warehouse are caculated by summing each column of changing cells.These constraints are shown in the spreadsheet below. 11ea84c6_c850_54a1_83dc_313af93b601c_TB2462_00 11ea84c6_c850_7bb2_83dc_131cc8f7cce9_TB2462_00 11ea84c6_c850_a2c3_83dc_61c54433a1c8_TB2462_00 The Solver information and solved spreadsheet are shown below. 11ea84c6_c850_c9d4_83dc_b3c350bb11ac_TB2462_00 11ea84c6_c850_c9d5_83dc_b97eafd1688f_TB2462_00 11ea84c6_c850_f0e6_83dc_5ff147eeca40_TB2462_00 11ea84c6_c851_17f7_83dc_5551f0d8babd_TB2462_00 11ea84c6_c851_17f8_83dc_d97234ca9566_TB2462_00 11ea84c6_c851_3f09_83dc_d10964ae1a2e_TB2462_00 Thus,from Plant A they should ship 40 to Warehouse 1 and 60 units to Warehouse 2,from Plant B they should ship 40 units to Warehouse 1 and 70 units to Warehouse 3,giving an overall total cost of $132,790.
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