1 Lecture 2 MGMT 650 Linear Programming Applications Chapter 4

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 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 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 Multiple Optima – An Example maximize x 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 Marketing Application: Media Selection  Advertising budget for first month = $30000  At least 10 TV commercials must be used  At least 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 TV Evening TV Daily newspaper Sunday newspaper Radio

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 DTV+3000ETV+400DN+1000SN+100R<=30000 DTV+ETV>= DTV+3000ETV<= DTV+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 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 = footballs  Storage capacity = 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 Demand Unit cost ($)

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 < =  I j <=  P j, I j >=0

10 Production-Inventory Formulation in LINDO  min 12.50p p p p p p i i i3+0.64i i i6 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 Production Inventory Cost = 1,535,562.00

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 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 x x x 4

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 x 2 + 8x x 4 >= 6 Total amount of protein in the mix is at least 5 units: (3) 12x x x 3 + 8x 4 >= 5 Total amount of iron in the mix is at least 5 units: (4) 14x x 3 + 7x 4 >= 5 Nonnegativity of variables: x j > 0 for all j

14 OBJECTIVE FUNCTION VALUE = VARIABLE VALUE REDUCED COSTS X X X X 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 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 A Transportation Table Warehouse 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 LP Formulation of Transportation Problem  minimize 4x 11 +7x 12 +7x 13 +x x 21 +3x 22 +8x 23 +8x 24 +8x x x 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 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 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 Cost Table for Assignment Problem $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 Formulation of Assignment Problem  minimize x 11 +4x 12 +6x 13 +3x x 21 +7x x 23 +9x x 31 +5x x 33 +7x x 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 $) Optimal Solution: x 11 =1; x 23 =1; x 32 =1; x 44 =1; rest=0 Cost of assignment = =$21 (`000)

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 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 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 Transportation Costs  Network Representation ARNOLD WASH BURN ZROX HEWES Detroit Akron Pittsburgh Memphis Charlotte Atlanta Newark

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 x x x x x x x x x 47

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 Goodyear Tires Solutions Objective Function Value = Objective Function Value = Variable Value Reduced Costs Variable Value Reduced Costs X X X X X X X X X X X X X X X X X X X X

28 Goodyear Tires Solutions ARNOLD WASH BURN ZROX HEWES Detroit Akron Pittsburgh Memphis Charlotte Atlanta Newark