1. 1 Lecture 2 MGMT 650 Linear Programming Applications Chapter 4

2. 2 Possible Outcomes of a LP (Section 2.6)  A LP is either  Infeasible – there exists no solution which satisfies all constraints and optimizes the objective function  or, Unbounded – increase/decrease objective function as much as you like without violating any constraint  or, Has an Optimal Solution  Optimal values of decision variables  Optimal objective function value

3. 3 Infeasible LP – An Example  minimize 4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16 x33+5x34 Subject to  x11+x12+x13+x14=100  x21+x22+x23+x24=200  x31+x32+x33+x34=150  x11+x21+x31=80  x12+x22+x32=90  x13+x23+x33=120  x14+x24+x34=170  xij>=0, i=1,2,3; j=1,2,3,4 Total demand exceeds total supply

4. 4 Unbounded LP – An Example maximize 2x 1 + x 2 subject to - x 1 + x 2  1 x 1 - 2x 2  2 x 1, x 2  0 x 2 can be increased indefinitely without violating any constraint => Objective function value can be increased indefinitely

5. 5 Multiple Optima – An Example maximize x 1 + 0.5 x 2 subject to 2x 1 + x 2  4 x 1 + 2x 2  3 x 1, x 2  0 x 1 = 2, x 2 =0, objective function = 2 x 1 = 5/3, x 2 =2/3, objective function = 2

6. 6 Marketing Application: Media Selection  Advertising budget for first month = $30000  At least 10 TV commercials must be used  At least 50000 customers must be reached  Spend no more than $18000 on TV adverts  Determine optimal media selection plan Advertising Media# of potential customers reached Cost ($) per advertisement Max times available per month Exposure Quality Units Day TV100015001565 Evening TV200030001090 Daily newspaper15004002540 Sunday newspaper25001000460 Radio3001003020

7. 7 Media Selection Formulation  Step 1: Define decision variables  DTV = # of day time TV adverts  ETV = # of evening TV adverts  DN = # of daily newspaper adverts  SN = # of Sunday newspaper adverts  R = # of radio adverts  Step 2: Write the objective in terms of the decision variables  Maximize 65DTV+90ETV+40DN+60SN+20R  Step 3: Write the constraints in terms of the decision variables DTV<=15 ETV<=10 DN<=25 SN<=4 R 30 1500DTV+3000ETV+400DN+1000SN+100R<=30000 DTV+ETV>=10 1500DTV+3000ETV<=18000 1000DTV+2000ETV+1500DN+2500SN+300R>=50000 Budget Customers reached TV Constraints Availability of Media DTV, ETV, DN, SN, R >= 0 VariableValue DTV10 ETV0 DN25 SN2 R30 Exposure = 2370 units

8. 8 Production-Inventory Model  Nike produces footballs and must decide how many footballs to produce each month over the next 6 months  Starting inventory = 5000  Production capacity each month = 30000 footballs  Storage capacity = 10000 footballs  Inventory holding cost of a month = 5% of production cost of that month  Determine production schedule that minimizes production and holding cost  Assume for simplicity  Production occurs continuously  Demand occurs at month end Month 1Month 2Month 3Month 4Month 5Month 6 Demand100001500030000350002500010000 Unit cost ($)12.5012.5512.7012.8012.8512.95

9. 9 Production-Inventory Model Formulation  Step 1: define decision variables  P j = production quantity in month j  I j = end-of-month inventory in month j  Step 2: formulate objective function is terms of decision variables  Sum of production cost + inventory holding cost  Step 3: formulate objective function is terms of decision variables  I j-1 + P j = D j + I j  P j < = 30000  I j <= 10000  P j, I j >=0

10. 10 Production-Inventory Formulation in LINDO  min 12.50p1+12.55p2+12.70p3+12.80p4+12.85p5+12.95p6+0.625i1+0.6275i2+0.635i3+0.64i4+0.6425i5+0.6475i6 st  p1-i1=5000Production-inventory constraint for month 1  p2+i1-i2=15000Production-inventory constraint for month 2  p3+i2-i3=30000Production-inventory constraint for month 3  p4+i3-i4=35000Production-inventory constraint for month 4  p5+i4-i5=25000Production-inventory constraint for month 5  p6+i5-i6=10000Production-inventory constraint for month 6  p1<=30000  p2<=30000  p3<=30000Production capacity constraints  p4<=30000  p5<=30000  p6<=30000  i1<=10000  i2<=10000  i3<=10000Inventory storage constraints  i4<=10000  i5<=10000  i6<=10000 Month 1 Month 2 Month 3 Month 4 Month 5 Month 6 Demand 100001500030000350002500010000 Production 50002000030000 2500010000 Inventory 05000 000 Cost = 1,535,562.00

11. 11 Blending Problem – Self-study  Ferdinand Feed Company receives four raw grains from which it blends its dry pet food.  The pet food advertises that each 8-ounce packet meets the minimum daily requirements for vitamin C, protein and iron.  The cost of each raw grain as well as the vitamin C, protein, and iron units per pound of each grain are as follows:  Ferdinand is interested in producing the 8-ounce mixture at minimum cost while meeting the minimum daily requirements of 6 units of vitamin C, 5 units of protein, and 5 units of iron.

12. 12 Blending Problem Formulation  Define the decision variables x j = the pounds of grain j (j = 1,2,3,4) used in the 8-ounce mixture  Define the objective function in terms of decision variables Minimize the total cost for an 8-ounce mixture: MIN.75x 1 +.90x 2 +.80x 3 +.70x 4

13. 13 Blending Problem - Constraints  Define the constraints Total weight of the mix is 8-ounces (.5 pounds): (1) x 1 + x 2 + x 3 + x 4 =.5 Total amount of Vitamin C in the mix is at least 6 units: (2) 9x 1 + 16x 2 + 8x 3 + 10x 4 >= 6 Total amount of protein in the mix is at least 5 units: (3) 12x 1 + 10x 2 + 10x 3 + 8x 4 >= 5 Total amount of iron in the mix is at least 5 units: (4) 14x 2 + 15x 3 + 7x 4 >= 5 Nonnegativity of variables: x j > 0 for all j

14. 14 OBJECTIVE FUNCTION VALUE = 0.406 VARIABLE VALUE REDUCED COSTS X1 0.099 0.000 X2 0.213 0.000 X3 0.088 0.000 X4 0.099 0.000 Thus, the optimal blend is about .10 lb. of grain 1, .21 lb. of grain 2, .09 lb. of grain 3, and .10 lb. of grain 4.  The mixture costs 40.6 cents. Blending Problem – Optimal Solution

15. 15 Transportation Problem – Chapter 7  Objective:  determination of a transportation plan of a single commodity  from a number of sources  to a number of destinations,  such that total cost of transportation is minimized  Sources may be plants, destinations may be warehouses  Question:  how many units to transport  from source i  to destination j  such that supply and demand constraints are met, and  total transportation cost is minimized

16. 16 A Transportation Table Warehouse 47 7 1 100 12 3 8 8 200 8 1016 5 150 450 8090120160 1234 1 2 3 Factory Factory 1 can supply 100 units per period Demand Warehouse B’s demand is 90 units per period Total demand per period Total supply capacity per period

17. 17 LP Formulation of Transportation Problem  minimize 4x 11 +7x 12 +7x 13 +x 14 +12x 21 +3x 22 +8x 23 +8x 24 +8x 31 +10x 32 + 16x 33 +5x 34 Subject to  x 11 +x 12 +x 13 +x 14 =100  x 21 +x 22 +x 23 +x 24 =200  x 31 +x 32 +x 33 +x 34 =150  x 11 +x 21 +x 31 =80  x 12 +x 22 +x 32 =90  x 13 +x 23 +x 33 =120  x 14 +x 24 +x 34 =160  x ij >=0, i=1,2,3; j=1,2,3,4 Supply constraint for factories Demand constraint of warehouses Minimize total cost of transportation

18. 18 Solution in Management Scientist Total transportation cost = 4(80) + 7(0) + 7(10)+ 1(10) + 12(0) + 3(90) + 8(110) + 8(0) + 8(0) +10(0) + 16(0) +5 (150) = $2300

19. 19 Assignment Problem – Chapter 7  Special case of transportation problem  When # of rows = # of columns in the transportation tableau  All supply and demands =1  Objective: Assign n jobs/workers to n machines such that the total cost of assignment is minimized  Plenty of practical applications  Job shops  Hospitals  Airlines, etc.

20. 20 Cost Table for Assignment Problem 1234 1$1$4$6$3 2$9$7$10$9 3$4$5$11$7 4$8$7$8$5 Pilot (i) Aircraft (j) All assignment costs in thousands of $

21. 21 Formulation of Assignment Problem  minimize x 11 +4x 12 +6x 13 +3x 14 + 9x 21 +7x 22 +10x 23 +9x 24 + 4x 31 +5x 32 +11x 33 +7x 34 + 8x 41 +7x 42 +8x 43 +5x 44 subject to  x 11 +x 12 +x 13 +x 14 =1  x 21 +x 22 +x 23 +x 24 =1  x 31 +x 32 +x 33 +x 34 =1  x 41 +x 42 +x 43 +x 44 =1  x 11 +x 21 +x 31 +x 41 =1  x 12 +x 22 +x 32 +x 42 =1  x 13 +x 23 +x 33 +x 43 =1  x 14 +x 24 +x 34 +x 44 =1  x ij = 1, if pilot i is assigned to aircraft j, i=1,2,3,4; j=1,2,3,4 0 otherwise PilotAssigned to aircraft # Cost (`000 $) 111 2310 325 445 Optimal Solution: x 11 =1; x 23 =1; x 32 =1; x 44 =1; rest=0 Cost of assignment = 1+10+5+5=$21 (`000)

22. 22  Transshipment problems are transportation problems in which a shipment may move through intermediate nodes (transshipment nodes) before reaching a particular destination node. Transshipment Problem – Chapter 7 2 2 33 44 55 66 7 7 1 1 c 13 c 14 c 23 c 24 c 25 c 15 s1s1s1s1 c 36 c 37 c 46 c 47 c 56 c 57 d1d1d1d1 d2d2d2d2 Intermediate Nodes Sources Destinations s2s2s2s2 Demand Supply Network Representation

23. 23 Example: Goodyear Tires  The Detroit (1) and Akron (2) facilities of Goodyear supply three customers at Memphis, Pittsburgh, and Newark.  Distribution is done through warehouses located at Charlotte (3) and Atlanta (4).  Current weekly demands by the customers are 50, 60 and 40 units for Memphis (5), Pittsburgh (6), and Newark (7) respectively.  Both facilities at Detroit and Akron can supply at most 75 units per week.

24. 24 Transportation Costs  Network Representation ARNOLD WASH BURN ZROX HEWES 75 75 50 60 40 5 8 7 4 1 5 8 3 4 4 Detroit Akron Pittsburgh Memphis Charlotte Atlanta Newark

25. 25 Goodyear Tires Formulation  Define Decision Variables  x ij = amount shipped from manufacturer i to warehouse j  x jk = amount shipped from warehouse j to customer k  where  i = 1 (Detroit), i = 2 (Akron),  j = 3 (Charlotte), j = 4 (Atlanta),  k = 5 (Memphis), k = 6 (Pittsburgh), k = 7 (Newark)  Define Objective Function Minimize Overall Shipping Costs: Min 5x 13 + 8x 14 + 7x 23 + 4x 24 + 1x 35 + 5x 36 + 8x 37 + 3x 45 + 4x 46 + 4x 47

26. 26 Goodyear Tires Formulation  Define Constraints Amount Out of Detroit: x 13 + x 14 < 75 Amount Out of Akron: x 23 + x 24 < 75 Amount Through Charlotte: x 13 + x 23 - x 35 - x 36 - x 37 = 0 Amount Through Atlanta: x 14 + x 24 - x 45 - x 46 - x 47 = 0 Amount Into Memphis: x 35 + x 45 = 50 Amount Into Pittsburgh: x 36 + x 46 = 60 Amount Into Newark: x 37 + x 47 = 40 Non-negativity of variables: x ij, x jk > 0, for all i, j and k.

27. 27 Goodyear Tires Solutions Objective Function Value = 1150.000 Objective Function Value = 1150.000 Variable Value Reduced Costs Variable Value Reduced Costs X13 75.000 0.000 X13 75.000 0.000 X14 0.000 2.000 X14 0.000 2.000 X23 0.000 4.000 X23 0.000 4.000 X24 75.000 0.000 X24 75.000 0.000 X35 50.000 0.000 X35 50.000 0.000 X36 25.000 0.000 X36 25.000 0.000 X37 0.000 3.000 X37 0.000 3.000 X45 0.000 3.000 X45 0.000 3.000 X46 35.000 0.000 X46 35.000 0.000 X47 40.000 0.000 X47 40.000 0.000

28. 28 Goodyear Tires Solutions ARNOLD WASH BURN ZROX HEWES 75 75 50 60 40 5 8 7 4 1 5 8 3 4 4 Detroit Akron Pittsburgh Memphis Charlotte Atlanta Newark 75 75 50 25 35 40