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Lecture2 & 3

Linear Programmingand

Transportation Problem

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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

Linear ProgrammingLinear Programming

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Maximize 60X1 + 50X2

Subject to

4X1 + 10X2 <= 100

2X1 + 1X2 <= 22

3X1 + 3X2 <= 39

X1, X2 >= 0

Linear Programming ExampleLinear Programming ExampleVariables

Objective function

Constraints

What is a Linear Program?What is a Linear Program?

• A LP is an optimization model that hasA LP is an optimization model that has

• continuous variablescontinuous variables

• a single linear objective function, anda single linear objective function, and

• (almost always) several constraints (linear equalities or inequalities)(almost always) several constraints (linear equalities or inequalities)

Non-negativity Constraints

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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 ModelLinear Programming Model

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Another Example of LP: Diet Another Example of LP: Diet ProblemProblem

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 13

Milk 160 8 285 9

Pie 420 4 22 24

Pork 260 14 80 13

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Example of LP : Diet ProblemExample 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

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Step 1: define decision variablesStep 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

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Step 3: ConstraintsStep 3: Constraints

Meet energy requirement110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 Meet protein requirement4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55 Meet calcium requirement2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800 Restriction on number of servings0x14, 0x23, 0x32, 0x48, 0x52, 0x62

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So, how does a LP look like?So, how does a LP look like?

minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6

subject to110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000

4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55

2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800

0x14

0x23

0x32

0x48

0x52

0x62

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Guidelines for Model FormulationGuidelines 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

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Transportation ProblemTransportation 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

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A Transportation TableA Transportation Table

Warehouse

4 7 7 1100

12 3 8 8200

8 10 16 5150

450

45080 90 120 160

1 2 3 4

1

2

3

Factory Factory 1can supply 100units per period

Demand

Table 8S.1

Warehouse B’s demand is 90 units per period Total demand

per period

Total supplycapacity perperiod

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LP Formulation of Transportation ProblemLP Formulation of Transportation Problem minimize

4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32 +16x33+5x34Subject 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

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Assignment ProblemAssignment 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.

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Cost Table for Assignment ProblemCost Table for Assignment Problem

1 2 3 4

1 1 4 6 3

2 9 7 10 9

3 4 5 11 7

4 8 7 8 5

Worker (i)

Machine (j)

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LP Formulation of Assignment ProblemLP Formulation of Assignment Problem minimize x11+4x12+6x13+3x14 + 9x21+7x22+10x23+9x24 +

4x31+5x32+11x33+7x34 + 8x41+7x42+8x43+5x44subject 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 worker i is assigned to machine j, i=1,2,3,4; j=1,2,3,4 0 otherwise

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Product Mix ProblemProduct 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

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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 ProblemProduct Mix Problem

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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 ProblemProduct Mix Problem

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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 ProblemProduct Mix Problem

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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

Product Mix Problem - Complete FormulationProduct Mix Problem - Complete Formulation

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Applications of LPApplications of LP

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

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Graphical Solution of LPsGraphical Solution of LPs

Consider a Maximization ProblemConsider a Maximization Problem

Max 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1, x2 > 0

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Graphical Solution ExampleGraphical Solution Example

Constraint #1 GraphedConstraint #1 Graphed

xx22

xx11

xx11 << 6 6

(6, 0)(6, 0)

88

77

66

55

44

33

22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

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Graphical Solution ExampleGraphical Solution Example

Constraint #2 GraphedConstraint #2 Graphed

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

(0, 6 (0, 6 1/31/3))

(9 (9 1/21/2, 0), 0)

88

77

66

55

44

33

22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

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Graphical Solution ExampleGraphical Solution Example

Constraint #3 GraphedConstraint #3 Graphed

xx22

xx11

xx11 + + xx22 << 8 8

(0, 8)(0, 8)

(8, 0)(8, 0)

88

77

66

55

44

33

22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

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Graphical Solution ExampleGraphical Solution Example

Combined-Constraint GraphCombined-Constraint Graph

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

xx11 + + xx22 << 8 8

xx11 << 6 6

88

77

66

55

44

33

22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

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88

77

66

55

44

33

22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

Graphical Solution ExampleGraphical Solution Example

Feasible Solution RegionFeasible Solution Region

xx11

FeasibleFeasibleRegionRegion

xx22

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88

77

66

55

44

33

22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

Graphical Solution ExampleGraphical Solution Example

Objective Function LineObjective Function Line

xx11

xx22

(7, 0)(7, 0)

(0, 5)(0, 5)Objective FunctionObjective Function55xx11 + + 7x7x2 2 = 35= 35Objective FunctionObjective Function55xx11 + + 7x7x2 2 = 35= 35

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88

77

66

55

44

33

22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

Graphical Solution ExampleGraphical Solution Example

Optimal SolutionOptimal Solution

xx11

xx22

Objective FunctionObjective Function55xx11 + + 7x7x2 2 = 46= 46Objective FunctionObjective Function55xx11 + + 7x7x2 2 = 46= 46

Optimal SolutionOptimal Solution((xx11 = 5, = 5, xx22 = 3) = 3)Optimal SolutionOptimal Solution((xx11 = 5, = 5, xx22 = 3) = 3)

6s-31 Linear Programming

1. Set up objective function and constraints in mathematical format

2. Plot the constraints

3. Identify the feasible solution space

4. Plot the objective function

5. Determine the optimum solution

Graphical Linear ProgrammingGraphical Linear Programming

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Possible Outcomes of a LPPossible 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

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Infeasible LP – An ExampleInfeasible 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

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Unbounded LP – An ExampleUnbounded 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

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Multiple Optima – An ExampleMultiple 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

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Practice ExamplePractice Example