Linear Programming (6S) and Transportation Problem (8S) Lecture 3 Linear Programming (6S) and Transportation Problem (8S)

Linear Programming George Dantzig – 1914 -2005 Concerned with optimal allocation of limited resources such as Materials Budgets Labor Machine time among competitive activities under a set of constraints George Dantzig – 1914 -2005

Linear Programming Example Variables Objective function Maximize 60X1 + 50X2 Subject to 4X1 + 10X2 <= 100 2X1 + 1X2 <= 22 3X1 + 3X2 <= 39 X1, X2 >= 0 Constraints Non-negativity 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)

Linear Programming Model 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

Another Example of LP: Diet Problem Energy requirement : 2000 kcal Protein requirement : 55 g Calcium requirement : 800 mg Food Energy(kcal) Protein(g) Calcium(mg) Price per serving($) Oatmeal 110 4 2 3 Chicken 205 32 12 24 Eggs 160 13 54 Milk 8 285 9 Pie 420 22 Pork 260 14 80

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

Step 1: define decision variables x1 = # of oatmeal servings x2 = # of chicken servings x3 = # of eggs servings x4 = # of milk servings x5 = # of pie servings x6 = # of pork servings Step 2: formulate objective function In this case, minimize total cost minimize z = 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6

Meet energy requirement Meet protein requirement Step 3: Constraints Meet energy requirement 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 Meet protein requirement 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55 Meet calcium requirement 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6  800 Restriction on number of servings 0x14, 0x23, 0x32, 0x48, 0x52, 0x62

So, how does a LP look like? minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6 subject to 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6  800 0x14 0x23 0x32 0x48 0x52 0x62

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

Transportation Problem 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

A Transportation Table Table 8S.1 Warehouse 1 2 3 4 Factory 4 7 7 1 Factory 1 can supply 100 units per period 100 1 12 3 8 8 2 200 8 10 16 5 150 3 Total supply capacity per period 450 Demand 80 90 120 160 450 Warehouse B’s demand is 90 units per period Total demand per period

LP Formulation of Transportation Problem Minimize total cost of transportation 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

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

Assignment Problem 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.

Cost Table for Assignment Problem Aircraft (j) 1 2 3 4 6 9 7 10 5 11 8 Pilot (i)

LP Formulation of Assignment Problem minimize x11+4x12+6x13+3x14 + 9x21+7x22+10x23+9x24 + 4x31+5x32+11x33+7x34 + 8x41+7x42+8x43+5x44 subject to x11+x12+x13+x14=1 x21+x22+x23+x24=1 x31+x32+x33+x34=1 x41+x42+x43+x44=1 x11+x21+x31+x41=1 x12+x22+x32+x42=1 x13+x23+x33+x43=1 x14+x24+x34+x44=1 xij = 1, if pilot i is assigned to aircraft j, i=1,2,3,4; j=1,2,3,4 0 otherwise

Product Mix Problem Formulate this problem as a linear program 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

Product Mix Problem Maximum Expected Daily Boat Builder Cost Seating Profit Speedhawk Sleekboat $6000 3 $ 70 Silverbird Sleekboat $7000 5 $ 80 Catman Racer $5000 2 $ 50 Classy Racer $9000 6 $110

Product Mix Problem Define the decision variables x1 = number of Speedhawks ordered x2 = number of Silverbirds ordered x3 = number of Catmans ordered x4 = number of Classys ordered Define the objective function Maximize total expected daily profit: Max: (Expected daily profit per unit) x (Number of units) Max: 70x1 + 80x2 + 50x3 + 110x4

Product Mix Problem Define the constraints (1) Spend no more than $420,000: 6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000 (2) Purchase at least 50 boats: x1 + x2 + x3 + x4 > 50 (3) Number of boats from Sleekboat equals number of boats from Racer: x1 + x2 = x3 + x4 or x1 + x2 - x3 - x4 = 0 (4) Capacity at least 200: 3x1 + 5x2 + 2x3 + 6x4 > 200 Nonnegativity of variables: xj > 0, for j = 1,2,3,4

Product Mix Problem - Complete Formulation Max 70x1 + 80x2 + 50x3 + 110x4 s.t. 6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000 x1 + x2 + x3 + x4 > 50 x1 + x2 - x3 - x4 = 0 3x1 + 5x2 + 2x3 + 6x4 > 200 x1, x2, x3, x4 > 0

Marketing Application: Media Selection Advertising Media # of potential customers reached Cost ($) per advertisement Max times available per month Exposure Quality Units Day TV 1000 1500 15 65 Evening TV 2000 3000 10 90 Daily newspaper 400 25 40 Sunday newspaper 2500 4 60 Radio 300 100 30 20 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

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 >= 18000 1000DTV 2000ETV 1500DN 2500SN 300R 50000 Availability of Media Budget TV Constraints Customers reached DTV, ETV, DN, SN, R >= 0

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

Graphical Solution of LPs Consider a Maximization Problem Max 5x1 + 7x2 s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1, x2 > 0

Graphical Solution Example Constraint #1 Graphed x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 < 6 (6, 0) x1

Graphical Solution Example Constraint #2 Graphed x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 (0, 6 1/3) 2x1 + 3x2 < 19 (9 1/2, 0) x1

Graphical Solution Example Constraint #3 Graphed x2 (0, 8) 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 + x2 < 8 (8, 0) x1

Graphical Solution Example Combined-Constraint Graph x2 x1 + x2 < 8 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 < 6 2x1 + 3x2 < 19 x1

Graphical Solution Example Feasible Solution Region x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Feasible Region x1

Graphical Solution Example Objective Function Line x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 (0, 5) Objective Function 5x1 + 7x2 = 35 (7, 0) x1

Graphical Solution Example Optimal Solution x2 Objective Function 5x1 + 7x2 = 46 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Optimal Solution (x1 = 5, x2 = 3) x1

Graphical Linear Programming Set up objective function and constraints in mathematical format Plot the constraints Identify the feasible solution space Plot the objective function Determine the optimum solution

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

Infeasible LP – An Example 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=170 xij>=0, i=1,2,3; j=1,2,3,4 Total demand exceeds total supply

Unbounded LP – An Example maximize 2x1 + x2 subject to -x1 + x2  1 x1 - 2x2  2 x1 , x2  0 x2 can be increased indefinitely without violating any constraint => Objective function value can be increased indefinitely

Multiple Optima – An Example maximize x1 + 0.5 x2 subject to 2x1 + x2  4 x1 + 2x2  3 x1 , x2  0 x1= 2, x2=0, objective function = 2 x1= 5/3, x2=2/3, objective function = 2