1. Resource Allocation Class 7: 3/9/11

2. 3.1 W HY N ETWORK P LANNING ? Find the right balance between inventory, transportation and manufacturing costs, Match supply and demand under uncertainty by positioning and managing inventory effectively, Utilize resources effectively by sourcing products from the most appropriate manufacturing facility

3. R ESOURCE A LLOCATION In operations and supply chain management there are several problems related to resource allocation: Production mix : how many units of each product or service should be produced given their profitability and constraints on available resources and their usage for each product Network location and sourcing : where to locate facilities, including manufacturing plants, distribution centers, and warehouses Given a network of facilities, how to best service my customer mix considering transportation and distribution costs (sourcing decision)

4. N ETWORK S CHEDULING E XAMPLE Single product Two plants p1 and p2 Plant p2 has an annual capacity of 60,000 units. The two plants have the same production costs. There are two warehouses w1 and w2 with identical warehouse handling costs. There are three markets areas c1,c2 and c3 with demands of 50,000, 100,000 and 50,000, respectively.

5. U NIT D ISTRIBUTION C OSTS Facility warehouse p1p1p2p2c1c1c2c2c3c3 w1w104345 w2w252212

6. H EURISTIC #1: C HOOSE THE C HEAPEST W AREHOUSE TO S OURCE D EMAND D = 50,000 D = 100,000 D = 50,000 Cap = 60,000 $5 x 140,000 $2 x 60,000 $2 x 50,000 $1 x 100,000 $2 x 50,000 Total Costs = $1,120,000

7. H EURISTIC #2: C HOOSE THE WAREHOUSE WHERE THE TOTAL DELIVERY COSTS TO AND FROM THE WAREHOUSE ARE THE LOWEST [C ONSIDER INBOUND AND OUTBOUND DISTRIBUTION COSTS ] D = 50,000 D = 100,000 D = 50,000 Cap = 60,000 $4 $5 $2 $3 $4 $5 $2 $1 $2 $0 P1 to WH1$3 P1 to WH2$7 P2 to WH1$7 P2 to WH 2$4 P1 to WH1$4 P1 to WH2$6 P2 to WH1$8 P2 to WH 2$3 P1 to WH1$5 P1 to WH2$7 P2 to WH1$9 P2 to WH 2$4 Market #1 is served by WH1, Markets 2 and 3 are served by WH2

8. D = 50,000 D = 100,000 D = 50,000 Cap = 60,000 Cap = 200,000 $5 x 90,000 $2 x 60,000 $3 x 50,000 $1 x 100,000 $2 x 50,000 $0 x 50,000 P1 to WH1$3 P1 to WH2$7 P2 to WH1$7 P2 to WH 2$4 P1 to WH1$4 P1 to WH2$6 P2 to WH1$8 P2 to WH 2$3 P1 to WH1$5 P1 to WH2$7 P2 to WH1$9 P2 to WH 2$4 Total Cost = $920,000 H EURISTIC #2: C HOOSE THE WAREHOUSE WHERE THE TOTAL DELIVERY COSTS TO AND FROM THE WAREHOUSE ARE THE LOWEST [C ONSIDER INBOUND AND OUTBOUND DISTRIBUTION COSTS ]

9. O PTIMIZATION A PPROACH The problem described earlier can be framed as a linear programming problem. A much better solution is found – total cost = $740,000! How does optimization work?

10. LINEAR PROGRAMMING LP deals with the problem of allocating limited resources among competing activities For example, consider a company that makes tables and chairs (competing activities) using a limited amount of large and small Legos (limited resources).

11. LINEAR PROGRAMMING The objective of LP is to select the best or optimal solution from the set of feasible solutions (those that satisfy all of the restrictions on the resources). Suppose profit for each table is $20 while profit for each chair is $16. We may choose to identify the number of tables and chairs to produce to maximize profit while not using more Legos than are available.

12. COMPONENTS OF AN LP Decision Variables: factors which are controlled by the decision maker. x1 = the number of tables produced per day x2 = the number of chairs produced per day Objective function: profit, cost, time, or service must be optimized. The objective may be to optimize profit.

13. COMPONENTS OF AN LP Constraints: restrictions which limit the availability and manner with which resources can be used to achieve the objective It takes 2 large and 2 small Legos to produce a table and 1 large and 2 smalls to produce a chair We may only have 6 large and 8 small Legos available each day

14. ASSUMPTIONS Linearity: Linear objective function and linear constraints. This implies proportionality and additivity. For example, it takes 2 large Legos to produce 1 table and 4 to produce 2 tables. It takes 3 large Legos to produce 1 table and 1 chair.

15. ASSUMPTIONS Divisibility: The decision variables can take on fractional values. The optimal solution may tell us to produce 2.5 tables each day. Certainty: The parameters of the model are known or can be accurately estimated. For example, we assume that the profitability information is accurate.

16. ASSUMPTIONS Non-negativity: All decision variables must take on positive or zero values.

17. LEGO PRODUCTS, INC. Lego Products, Inc. manufactures tables and chairs. Profit for each table is $20 while each chair generates $16 profit. Each table is made by assembling two large and two small legos. Each chair requires one large and two small legos. Currently, Lego Products has six large and eight small legos available each day.

18. TableChair LEGO EXAMPLE: I NTRODUCTION Pictures of a table and a chair are shown below.

19. How many tables and chairs should we produce to maximize daily profit? producing 3 tables generates a daily profit of $60, producing 4 chairs generates a daily profit of $64, however, producing 2 tables and 2 chairs generates the optimal daily profit of $72. LEGO EXAMPLE: O PTIMAL S OLUTION

20. U SING S OLVER Solver in Excel can be used to obtain the solution and will now be demonstrated The problem formulation is: MAX 20*X1+16*X2 Subject to: 2*X1+1*X2<=6 2*X1+2*X2<=8 How to enter this formulation into Solver?

21. Do we have any unused large or small legos for all of the solutions that you just found? There are no unused large or small legos for the optimal solution. There are 2 unused small legos if 3 tables are made. There are 2 unused large legos if 4 chairs are made. LEGO EXAMPLE: U NUSED LEGOS

22. The difference between the available resources and resources used is either slack or surplus. Slack is associated with each less than or equal to constraint, and represents the amount of unused resource. Surplus is associated with each greater than or equal to constraint, and represents the amount of excess resource above the stated level. LEGO EXAMPLE: S LACK AND S URPLUS

23. We have two slack values - one for large legos and one for small legos - and no surplus values. The slack or surplus section shows that both constraints have zero slack. Suppose we must produce at least one table (X1>=1). The original optimal solution is still the best. Since X1=2, we produce one surplus table. LEGO EXAMPLE: S LACK AND S URPLUS

24. R IGHT H AND S IDE C HANGES Now we are ready to illustrate the key concepts of sensitivity analysis. How much would you be willing to spend for one additional large Lego? One additional large Lego is worth $4. Original solution: X1=2, X2=2, profit = $72; New solution: X1=3, X2=1, profit = $76. You would be willing to spend up to $4 (76-72) for one additional large Lego.

25. How much would you be willing to spend for two additional large Legos? Two additional large Legos are worth $8. Original solution: X1=2, X2=2, profit = $72; New solution: X1=4, X2=0, profit = $80.. You would be willing to spend up to $8 (80- 72) for two additional large Legos. RIGHT HAND SIDE CHANGES

26. How much would you be willing to spend for three more large legos? The third large lego is not worth anything since the optimal solution remains unchanged. What happens if your supplier can only provide five large legos each day? The optimal solution is: X1=1, X2=3, profit=68, so we lose $4. RIGHT HAND SIDE CHANGES

27. What happens if your supplier can only provide four large legos each day? The optimal solution is: X1=0, X2=4, profit=64, so we lose another $4. What happens if your supplier can only provide three large legos each day? The optimal solution is: X1=0, X2=3, profit=48, so we lose an additional $16, and not $4. RIGHT HAND SIDE CHANGES

28. SHADOW PRICES This last set of exercises enables us to determine the shadow price for a resource constraint (large legos). The shadow price, for a particular constraint, is the amount the objective function value will increase (decrease) if the right hand side value of that constraint is increased (decreased) by one unit. We found that the shadow price for large legos is $4.

29. SHADOW PRICES What is the shadow price of the small legos? With two additional small legos the new solution: X1=1, X2=4, profit = $84. You would be willing to spend up to $12 (84-72) for two additional small legos, so the shadow price is $6 (12/2).

30. SHADOW PRICES In general, the shadow prices are meaningful if one right hand side (RHS) value of a constraint is changed, and all other parameters of the model remain unchanged.

31. What happens if the profit of tables increases to $35? The optimal solution is X1=3, X2=0, profit = $105. Note that no chairs are being produced. REDUCED COSTS

32. When a decision variable has an optimal value of zero, the allowable increase for the objective function coefficient is also called the reduced cost.. The reduced cost of a decision variable is the amount the corresponding objective function coefficient would have to change before the optimal value would change from zero to some positive value.

33. REDUCED COSTS The reduced cost for tables is zero in the original formulation. Why is this the case? We are already producing tables. What is the reduced cost for X2 and what does it mean? The reduce cost is -1.5 meaning that if I force production of 1 chair profit will drop by 1.5

34. SENSITIVITY ANALYSIS PROBLEM A manufacturing firm has discontinued production of a certain unprofitable product line thus creating considerable excess production capacity. Management is considering devoting this excess capacity to one or more of three products; call them products 1, 2, and 3. The available capacity on the machines that might limit output is summarized below:

35. AVAILABLE TIME MACHINE TYPE (machine hours / week) Milling machine500 Lathe350 Grinder150 SENSITIVITY ANALYSIS PROBLEM

36. The number of machine hours required for each unit of the respective products is: MACHINE TYPE P1P2P3 Milling machine 935 Lathe 540 Grinder 302 SENSITIVITY ANALYSIS PROBLEM

37. The sales department indicates that the sales potential for products 1 and 2 exceeds the maximum production rate and that the sales potential for product 3 is 20 units per week.. The unit profit would be $3000, $1200, and $900, respectively, for products 1, 2, and 3. SENSITIVITY ANALYSIS PROBLEM

38. Solver is used to determine the optimal solution a.What are the optimal weekly production levels for each of the three products? Product 1 = 45.23 Product 2 = 30.95 Product 3 = 0 SENSITIVITY ANALYSIS PROBLEM

39. b.What profit will be obtained if the optimal solution is implemented? $172,857.10 per week c.How much unused capacity exists on the milling machine, the lathe, and the grinder? Milling = 0; Lathe = 0; Grinder = 14.28 (See SLACK entries) SENSITIVITY ANALYSIS PROBLEM

40. d.How much would the objective function change if the amount of available time on the grinder increased from 150 hours per week to 250 hours? Will the objective function increase or decrease? Currently the grinder has 14.28 hours of slack time so its shadow price is 0. Increasing the available hours from 150 to 250 will not change the total profit. SENSITIVITY ANALYSIS PROBLEM

41. e.The profit for product 3 is $900 per unit and the current production level is zero. How much would the profit per unit have to change before it would be profitable to produce product 3's? The profit per unit would have to increase by its reduced cost of $528.57. SENSITIVITY ANALYSIS PROBLEM

42. f. The milling machine capacity can be increased at a cost of $160 per hour. Is it economic to increase capacity by 10 hours? The shadow price per hour (285.7143) is greater than the cost (160), so it is worth increasing milling capacity on the margin. However, since the shadow price might change with increasing capacity we need to rerun to see the full effect of increasing capacity by 10 hours. Profit does increase by 2857 (10*285.714) which is greater than the cost increase of 1600 (10*160). Therefore the milling capacity should be increased b y 10 hours. SENSITIVITY ANALYSIS PROBLEM

43. TRANSPORTATION PROBLEM Mathematical programming has been successfully applied to important supply chain problems. These problems address the movement of products across links of the supply chain (supplier, manufacturers, and customers).. We now focus on supply chain applications in transportation and distribution planning.

44. TRANSPORTATION PROBLEM A manufacturer ships TV sets from three warehouses to four retail stores each week. Warehouse capacities (in hundreds) and demand (in hundreds) at the retail stores are as follows: Capacity Demand Warehouse 1200Store 1100 Warehouse 2 150Store 2 200 Warehouse 3 300Store 3 125 650Store 4 225 650

45. The shipping cost per hundred TV sets for each route is given below: To FromStore 1 Store 2 Store 3 Store 4 warehouse 1$10 5 12 3 warehouse 24 9 15 6 warehouse 315 8 6 11 TRANSPORTATION PROBLEM

47. What are the decision variables? XIJ=number of TV sets (in cases) shipped from warehouse I to store J I is the index for warehouses (1,2,3) J is the index for stores (1,2,3,4) TRANSPORTATION PROBLEM

48. What is the objective? Minimize the total cost of transportation which is obtained by multiplying the shipping cost by the amount of TV sets shipped over a given route and then summing over all routes OBJECTIVE FUNCTION ; MIN = 10*X11+5*X12+12*X13+ 3*X14+ 4* X21+9*X22+15*X23+ 6*X24+ 15*X31+8*X32+ 6*X33+11*X34 TRANSPORTATION PROBLEM

49. How are the supply constraints expressed? For each warehouse the amount of TV sets shipped to all stores must equal the capacity at the warehouse X11+X12+X13+X14=200; SUPPLY CONSTRAINT FOR WAREHOUSE 1 X21+X22+X23+X24=150; SUPPLY CONSTRAINT FOR WAREHOUSE 2 X31+X32+X33+X34=300; SUPPLY CONSTRAINT FOR WAREHOUSE 3 TRANSPORTATION PROBLEM

50. How are the demand constraints expressed? For each store the amount of TV sets shipped from all warehouses must equal the demand of the store X11+X21+X31=100; DEMAND CONSTRAINT FOR STORE 1 X12+X22+X32=200; DEMAND CONSTRAINT FOR STORE 2 X13+X23+X33=125; DEMAND CONSTRAINT FOR STORE 3 X14+X24+X34=225; DEMAND CONSTRAINT FOR STORE 4 TRANSPORTATION PROBLEM

51. Since partial shipment cannot be made, the decision variables must be integer – valued However, if all supplies and demands are integer-valued, the values of our decision variables will be integer valued TRANSPORTATION PROBLEM

52. After solution in Solver: The total shipment cost is $3500, and the optimal shipments are: warehouse 1 ships 25 cases to store 2 and 175 to store 4; warehouse 2 ships 100 to store 1 and 50 to store 4, and warehouse 3 ships 175 to store 2 and 125 to store 3. The reduced cost of X11 is 9, so the cost of shipping from warehouse 1 to store 1 would have to be reduced by $9 before this route would be used TRANSPORTATION PROBLEM

53. UNBALANCED PROBLEMS Suppose warehouse 2 actually has 175 TV sets. How should the original problem be modified? Since total supply across all warehouses is now greater than total demand, all supply constraints are now “<=“ Referring to the original problem, suppose store 3 needs 150 TV sets. How should the original problem be modified? The demand constraints are now “<=“

54. RESTRICTED ROUTE Referring to the original problem, suppose there is a strike by the shipping company such that the route from warehouse 3 to store 2 cannot be used. How can the original problem be modified to account for this change? Add the constraint: X32=0

55. WAREHOUSE LOCATION Suppose that the warehouses are currently not open, but are potential locations. The fixed cost to construct warehouses and their capacity values are given as: WAREHOUSESFIXED COST CAPACITY Warehouse 1125,000300 Warehouse 2185,000525 Warehouse 3100,000325

56. How do we model the fact that the warehouses may or may not be open? Define a set of binary decision variables YI, I =1,2,3, where warehouse I is open if YI = 1 and warehouse I is closed if YI = 0 WAREHOUSE LOCATION

57. How must the objective function change? Additional terms are added to the objective function which multiply the fixed costs of operating the warehouse by YI and summing over all warehouses I: 125000*Y1+185000*Y2+100000*Y3 Why can’t we use the current capacity constraints? Product cannot be shipped from a warehouse if it is not open. Since the capacity is available only if the warehouse is open, we multiply warehouse 1’s capacity by Y1. WAREHOUSE LOCATIO N

58. Also, we must make the YI variables binary integer Total fixed and shipping costs are $289,100; warehouses 2 and 3 are open; warehouse 2 ships 100 to store 1, 225 to store 4; and warehouse 3 ships 200 to store 2 and 125 to store 4 WAREHOUSE LOCATION

59. F INALLY, BACK TO THE MOTIVATING PROBLEM...

60. N ETWORK S CHEDULING E XAMPLE Single product Two plants p1 and p2 Plant p2 has an annual capacity of 60,000 units. The two plants have the same production costs. There are two warehouses w1 and w2 with identical warehouse handling costs. There are three markets areas c1,c2 and c3 with demands of 50,000, 100,000 and 50,000, respectively.

61. U NIT D ISTRIBUTION C OSTS Facility warehouse p1p1p2p2c1c1c2c2c3c3 w1w104345 w2w252212

62. D = 50,000 D = 100,000 D = 50,000 Cap = 60,000 T HE N ETWORK

63. T HE O PTIMIZATION M ODEL This problem can be framed as the following linear programming problem: Let x(p1,w1), x(p1,w2), x(p2,w1) and x(p2,w2) be the flows from the plants to the warehouses. x(w1,c1), x(w1,c2), x(w1,c3) be the flows from the warehouse w1 to customer zones c1, c2 and c3. x(w2,c1), x(w2,c2), x(w2,c3) be the flows from warehouse w2 to customer zones c1, c2 and c3

64. O PTIMAL S OLUTION Facility warehouse p1p1p2p2c1c1c2c2c3c3 w1w1140,000050,00040,00050,000 w2w2060,0000 0 Total cost for the optimal strategy is $740,000