1 Lecture 5 Linear Programming (6S) and Transportation Problem (8S)

2  George Dantzig –  Concerned with optimal allocation of limited resources such as  Materials  Budgets  Labor  Machine time  among competitive activities  under a set of constraints Linear Programming George Dantzig –

3 Product Mix Example (from session 1) Type 1Type 2 Profit per unit $60$50 Assembly time per unit 4 hrs10 hrs Inspection time per unit 2 hrs1 hr Storage space per unit 3 cubic ft ResourceAmount available Assembly time100 hours Inspection time22 hours Storage space39 cubic feet

4 Maximize 60X X 2 Subject to 4X X 2 <= 100 2X 1 + 1X 2 <= 22 3X 1 + 3X 2 <= 39 X 1, X 2 >= 0 Linear Programming Example Variables Objective function Constraints What is a Linear Program? A LP is an optimization model that has continuous variables a single linear objective function, and (almost always) several constraints (linear equalities or inequalities) Non-negativity Constraints

5  Decision variables  unknowns, which is what model seeks to determine  for example, amounts of either inputs or outputs  Objective Function  goal, determines value of best (optimum) solution among all feasible (satisfy constraints) values of the variables  either maximization or minimization  Constraints  restrictions, which limit variables of the model  limitations that restrict the available alternatives  Parameters: numerical values (for example, RHS of constraints)  Feasible solution: is one particular set of values of the decision variables that satisfies the constraints  Feasible solution space: the set of all feasible solutions  Optimal solution: is a feasible solution that maximizes or minimizes the objective function  There could be multiple optimal solutions Linear Programming Model

6 Another Example of LP: Diet Problem  Energy requirement : 2000 kcal  Protein requirement : 55 g  Calcium requirement : 800 mg FoodEnergy (kcal)Protein(g)Calcium(mg)Price per serving($) Oatmeal Chicken Eggs Milk Pie Pork

7 Example of LP : Diet Problem  oatmeal: at most 4 servings/day  chicken: at most 3 servings/day  eggs: at most 2 servings/day  milk: at most 8 servings/day  pie:at most 2 servings/day  pork: at most 2 servings/day Design an optimal diet plan which minimizes the cost per day

8 Step 1: define decision variables  x 1 = # of oatmeal servings  x 2 = # of chicken servings  x 3 = # of eggs servings  x 4 = # of milk servings  x 5 = # of pie servings  x 6 = # of pork servings Step 2: formulate objective function In this case, minimize total cost minimize z = 3x x x 3 + 9x x x 6

9 Step 3: Constraints  Meet energy requirement 110x x x x x x 6  2000  Meet protein requirement 4x x x 3 + 8x 4 + 4x x 6  55  Meet calcium requirement 2x x x x x x 6  800  Restriction on number of servings 0  x 1  4, 0  x 2  3, 0  x 3  2, 0  x 4  8, 0  x 5  2, 0  x 6  2

10 So, how does a LP look like? minimize 3x x x 3 + 9x x x 6 subject to 110x x x x x x 6  x x x 3 + 8x 4 + 4x x 6  55 2x x x x x x 6   x 1  4 0  x 2  3 0  x 3  2 0  x 4  8 0  x 5  2 0  x 6  2

11 Optimal Solution – Diet Problem Using LINDO 6.1  Cost of diet = $96.50 per day Food# of servings Oatmeal4 Chicken0 Eggs0 Milk6.5 Pie0 Pork2

12 Optimal Solution – Diet Problem Using Management Scientist  Cost of diet = $96.50 per day Food# of servings Oatmeal4 Chicken0 Eggs0 Milk6.5 Pie0 Pork2

13 Guidelines for Model Formulation  Understand the problem thoroughly.  Describe the objective.  Describe each constraint.  Define the decision variables.  Write the objective in terms of the decision variables.  Write the constraints in terms of the decision variables  Do not forget non-negativity constraints

14 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

15 LP Formulation of Transportation Problem  minimize 4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24 +8x31+10x32+16x33+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=160  xij>=0, i=1,2,3; j=1,2,3,4 Supply constraint for factories Demand constraint of warehouses Minimize total cost of transportation

16 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

17 Solution using LINDO Notice multiple optimal solutions!

18 Product Mix Problem Floataway Tours has $420,000 that can be used to purchase new rental boats for hire during the summer. The boats can be purchased from two different manufacturers. Floataway Tours would like to purchase at least 50 boats. They would also like to purchase the same number from Sleekboat as from Racer to maintain goodwill. At the same time, Floataway Tours wishes to have a total seating capacity of at least 200. Formulate this problem as a linear program

19 Maximum Expected Daily Boat Builder Cost Seating Profit Speedhawk Sleekboat $ $ 70 Silverbird Sleekboat $ $ 80 Catman Racer $ $ 50 Classy Racer $ $110 Product Mix Problem

20  Define the decision variables x 1 = number of Speedhawks ordered x 2 = number of Silverbirds ordered x 3 = number of Catmans ordered x 4 = number of Classys ordered  Define the objective function Maximize total expected daily profit: Max: (Expected daily profit per unit) x (Number of units) Max: 70x x x x 4 Product Mix Problem

21  Define the constraints (1) Spend no more than $420,000: 6000x x x x 4 < 420,000 (2) Purchase at least 50 boats: x 1 + x 2 + x 3 + x 4 > 50 (3) Number of boats from Sleekboat equals number of boats from Racer: x 1 + x 2 = x 3 + x 4 or x 1 + x 2 - x 3 - x 4 = 0 (4) Capacity at least 200: 3x 1 + 5x 2 + 2x 3 + 6x 4 > 200 Nonnegativity of variables: x j > 0, for j = 1,2,3,4 Product Mix Problem

22 Max 70x x x x 4 s.t. 6000x x x x 4 < 420,000 x 1 + x 2 + x 3 + x 4 > 50 x 1 + x 2 - x 3 - x 4 = 0 3x 1 + 5x 2 + 2x 3 + 6x 4 > 200 x 1, x 2, x 3, x 4 > 0 Product Mix Problem - Complete Formulation  Daily profit = $5040 Boat# purchased Speedhawk28 Silverbird0 Catman0 Classy28

23 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

24 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 Exposure = 2370 units VariableValue DTV10 ETV0 DN25 SN2 R30

25 Applications of LP  Product mix planning  Distribution networks  Truck routing  Staff scheduling  Financial portfolios  Capacity planning  Media selection: marketing

26 Possible Outcomes of a LP  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

27 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

28 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

29 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

30 Operations Scheduling Chapter 16

31  Establishing the timing of the use of equipment, facilities and human activities in an organization  Effective scheduling can yield  Cost savings  Increases in productivity Scheduling

32 High-Volume Systems  Flow system: High-volume system with Standardized equipment and activities  Flow-shop scheduling: Scheduling for high- volume flow system Work Center #1Work Center #2 Output

33 High-Volume Success Factors  Process and product design  Preventive maintenance  Rapid repair when breakdown occurs  Optimal product mixes  Minimization of quality problems  Reliability and timing of supplies

34 Scheduling Low-Volume Systems  Loading - assignment of jobs to process centers  Sequencing - determining the order in which jobs will be processed  Job-shop scheduling  Scheduling for low-volume systems with many variations in requirements

35 Gantt Load Chart  Gantt chart - used as a visual aid for loading and scheduling Figure 16.2

36 More Gantt Charts

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

38 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 $

39 Management Scientist Solution PilotAssigned to aircraft # Cost (`000 $)

40 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)

41 Sequencing  Sequencing: Determine the order in which jobs at a work center will be processed.  Workstation: An area where one person works, usually with special equipment, on a specialized job.  Priority rules: Simple heuristics used to select the order in which jobs will be processed.  FCFS - first come, first served  SPT - shortest processing time  Minimizes mean flow time  EDD - earliest due date In-class example

42 Performance Measures  Job flow time  Length of time a job is at a particular workstation  Includes actual processing time, waiting time, transportation time etc.  Lateness = flow time – due date  Tardiness = max {lateness, 0}  Makespan  Total time needed to complete a group of jobs  Length of time between start of first job and completion of last job

43 Scheduling Difficulties  Variability in  Setup times  Processing times  Interruptions  Changes in the set of jobs  No method for identifying optimal schedule  Scheduling is not an exact science  Ongoing task for a manager

44 Minimizing Scheduling Difficulties  Set realistic due dates  Focus on bottleneck operations  Consider lot splitting of large jobs