Linear Programming OPIM 310-Lecture 2 Instructor: Jose Cruz

Linear ProgrammingLinear Programming

OPIM 310-Lecture 2

Instructor: Jose Cruz

Model consisting of linear relationships Model consisting of linear relationships representing a firm’s objectives & resource representing a firm’s objectives & resource constraintsconstraints

Decision variables are mathematical Decision variables are mathematical symbols representing levels of activity of symbols representing levels of activity of an operationan operation

Objective function is a linear relationship Objective function is a linear relationship reflecting objective of an operationreflecting objective of an operation

Constraint is a linear Constraint is a linear relationship representing relationship representing a restriction on decision makinga restriction on decision making

Linear ProgrammingLinear Programming

General Structure of a Linear General Structure of a Linear Programming (LP) ModelProgramming (LP) Model

Max/minMax/min z = c z = c11xx11 + c + c22xx22 + ... + c + ... + cnnxxnn

subject to:subject to: aa1111xx11 + a + a1212xx22 + ... + a + ... + a11nnxxnn bb11 (or(or , =), =)

aa2121xx11 + a + a2222xx22 + ... + a + ... + a22nnxxnn bb22

::

aamm11xx11 + a + amm22xx22 + ... + a + ... + amnmnxxnn bbmm

xxjj = decision variables = decision variables

bbi i = constraint levels= constraint levels

ccj j = objective function coefficients= objective function coefficients

aaij ij = constraint coefficients= constraint coefficients

Linear Programming Linear Programming Model FormulationModel Formulation

LaborLabor ClayClay RevenueRevenuePRODUCTPRODUCT (hr/unit)(hr/unit) (lb/unit)(lb/unit) ($/unit)($/unit)

BowlBowl 11 44 4040

MugMug 22 33 5050

There are 40 hours of labor and 120 pounds of clay There are 40 hours of labor and 120 pounds of clay available each dayavailable each day

Decision variablesDecision variables

xx11 = number of bowls to produce = number of bowls to produce

xx22 = number of mugs to produce = number of mugs to produce

RESOURCE REQUIREMENTSRESOURCE REQUIREMENTS

Example S9.1Example S9.1

Objective Function Objective Function and Constraintsand Constraints

Maximize Maximize ZZ = $40 = $40 xx11 + 50 + 50 xx22

Subject toSubject to

xx11 ++ 22xx22 40 hr40 hr (labor constraint)(labor constraint)

44xx11 ++ 33xx22 120 lb120 lb (clay constraint)(clay constraint)

xx1 1 , , xx22 00

Solution is Solution is xx11 = 24 bowls = 24 bowls

xx2 2 = 8 mugs= 8 mugs

Revenue = $1,360Revenue = $1,360Example S9.1Example S9.1

Graphical Solution Graphical Solution MethodMethod

1.1. Plot model constraint on a set of Plot model constraint on a set of coordinates in a planecoordinates in a plane

2.2. Identify the feasible solution space on the Identify the feasible solution space on the graph where all constraints are satisfied graph where all constraints are satisfied simultaneouslysimultaneously

3.3. Plot objective function to find the point on Plot objective function to find the point on boundary of this space that maximizes (or boundary of this space that maximizes (or minimizes) value of objective functionminimizes) value of objective function

Graph of Pottery ProblemGraph of Pottery Problem

4 4 xx11 + 3 + 3 xx2 2 120 lb120 lb

xx11 + 2 + 2 xx2 2 40 hr40 hr

Area common toArea common toboth constraintsboth constraints

50 50 –

40 40 –

30 30 –

20 20 –

10 10 –

0 0 – |1010

|6060

|5050

|2020

|3030

|4040 xx11

xx22

Example 1Example 1

Plot Objective FunctionPlot Objective Function40 40 –

30 30 –

20 20 –

10 10 –

0 0 –

$800 = 40$800 = 40xx11 + 50 + 50xx22

Optimal pointOptimal point

BB

|1010

|2020

|3030

|4040 xx11

xx22

Example S9.2Example S9.2

Computing Optimal Computing Optimal ValuesValues

ZZ = $50(24) + $50(8) = $50(24) + $50(8)

ZZ = $1,360 = $1,360

xx11 ++ 22xx22 == 4040

44xx11 ++ 33xx22 == 120120

44xx11 ++ 88xx22 == 160160

-4-4xx11 -- 33xx22 == -120-120

55xx22 == 4040

xx22 == 88

xx11 ++ 2(8)2(8) == 4040

xx11 == 2424

AA

..88BB

CC

xx11 + 2 + 2xx2 2 = 40= 40

44xx11 + 3 + 3xx2 2 = 120= 120

|2020

|3030

|4040

|1010 xx11

xx22

40 40 –

30 30 –

20 20 –

10 10 –

0 0 –

Example 1Example 1

Extreme Corner PointsExtreme Corner Points

xx11 = 24 bowls = 24 bowls

xx2 2 ==8 mugs8 mugs

ZZ = $1,360 = $1,360 xx11 = 30 bowls = 30 bowls


ZZ = $1,200 = $1,200



ZZ = $1,000 = $1,000

AA

BBCC|

2020|

3030|

4040|

1010 xx11

xx22

40 40 –

30 30 –

20 20 –

10 10 –

0 0 –

Example 1Example 1

Objective Function Objective Function Determines Optimal SolutionDetermines Optimal Solution

44xx11 + 3 + 3xx2 2 120 lb120 lb

xx11 + 2 + 2xx2 2 40 hr40 hr

40 40 –

30 30 –

20 20 –

10 10 –

0 0 –

BB

|1010

|2020

|3030

|4040 xx11

xx22

CC

AA

ZZ = 70 = 70xx11 + 20 + 20xx22

Optimal point:Optimal point:



ZZ = $2,100 = $2,100

Example 1Example 1

Farmer’s HardwareFarmer’s Hardware

CHEMICAL CONTRIBUTIONCHEMICAL CONTRIBUTION

BrandBrand Nitrogen (lb/bag)Nitrogen (lb/bag) Phosphate (lb/bag)Phosphate (lb/bag)

Super-groSuper-gro 22 44

Crop-quikCrop-quik 44 33

Minimize Minimize ZZ = $6x = $6x11 + $3x + $3x22

subject tosubject to

22xx11 ++ 44xx22 16 lb of nitrogen 16 lb of nitrogen

44xx11 ++ 33xx22 24 lb of phosphate 24 lb of phosphate

xx11, , xx22 0 0Example 2Example 2

Farmer’s HardwareFarmer’s Hardware

14 14 –

12 12 –

10 10 –

8 8 –

6 6 –

4 4 –

2 2 –

0 0 –|22

|44

|66

|88

|1010

|1212

|1414 xx11

xx22

A

B

C

Z = 6x1 + 3x2

x1 = 0 bags of Super-grox2 = 8 bags of Crop-quikZ = $24

Example 2Example 2

Solving LP ProblemsSolving LP Problems

Exhibit S9.3Exhibit S9.3

Click on “Tools” to invoke “Solver.”

Objective function

Decision variables – bowls(x1)=B10; mugs (x2)=B11


Exhibit S9.4Exhibit S9.4

After all parameters and constraints have been input, click on “Solve.”

Objective function

Decision variables

C6*B10+D6*B11≤40

C7*B10+D7*B11≤120

Click on “Add” to insert contraints.


Exhibit 1Exhibit 1

Example 2: Olympic Bike CoExample 2: Olympic Bike Co..

• Model Formulation

Max 10x1 + 15x2 (Total Weekly Profit)

s.t. 2x1 + 4x2 < 100 (Aluminum Available)

3x1 + 2x2 < 80 (Steel Available)

x1, x2 > 0


• Partial Spreadsheet Showing SolutionA B C D67 Deluxe Professional8 Bikes Made 15 17.5009

10 412.5001112 Constraints Amount Used Amount Avail.13 Aluminum 100 <= 10014 Steel 80 <= 80

Decision Variables

Maximized Total Profit

A B C D67 Deluxe Professional8 Bikes Made 15 17.5009

10 412.5001112 Constraints Amount Used Amount Avail.13 Aluminum 100 <= 10014 Steel 80 <= 80

Decision Variables

Maximized Total Profit

Example 2: Olympic Bike Co.Example 2: Olympic Bike Co.

• Optimal Solution

According to the output:

x1 (Deluxe frames) = 15

x2 (Professional frames) = 17.5

Objective function value = $412.50


• Range of Optimality

Question

Suppose the profit on deluxe frames is increased to $20. Is the above solution still optimal? What is the value of the objective function when this unit profit is increased to $20?


• Sensitivity Report Adjustable Cells

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$B$8 Deluxe 15 0 10 12.5 2.5$C$8 Profess. 17.500 0.000 15 5 8.333333333

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$13 Aluminum 100 3.125 100 60 46.66666667$B$14 Steel 80 1.25 80 70 30



Answer

The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 7.5 and 22.5. Since 20

is within this range, the optimal solution will not change. The optimal profit will change: 20x1 +

15x2 = 20(15) + 15(17.5) = $562.50.



Question

If the unit profit on deluxe frames were $6 instead of $10, would the optimal solution change?


• Range of Optimality Adjustable Cells

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$B$8 Deluxe 15 0 10 12.5 2.5$C$8 Profess. 17.500 0.000 15 5 8.33333333

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$13 Aluminum 100 3.125 100 60 46.66666667$B$14 Steel 80 1.25 80 70 30

• Range of OptimalityAnswer

The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 7.5 and 22.5. Since 6

is outside this range, the optimal solution would change.


The Minimum Cost Network The Minimum Cost Network Flow Problem (MCNFP)Flow Problem (MCNFP)

• Extremely Useful Model in OR & EM• Important Special Cases of the MCNFP

– Transportation and Assignment Problems– Maximum Flow Problem– Minimum Cut Problem– Shortest Path Problem

• Network Structure– Some MCNFP LP’s have integer values !!!– Problems can be formulated graphically

Formulation of Shortest Path Formulation of Shortest Path ProblemsProblems

• Source node s has a supply of 1• Sink node t has a demand of 1• All other nodes are transshipment nodes• Each arc has capacity 1 • Tracing the unit of flow from s to t gives a

path from s to t

Shortest PathShortest Path

1 -11 2 3

4

5 10

11 2 3

4

1 0

10

0

0

77

1

Shortest Path ExampleShortest Path Example• In a rural area of Texas, there are six farms

connected my small roads. The distances in miles between the farms are given in the following table.

• What is the minimum distance to get from Farm 1 to Farm 6?

From Farm To Farm Distance 1 2 81 3 102 3 42 4 92 5 53 4 63 5 24 5 34 6 65 6 5

Formulation as Shortest PathFormulation as Shortest Path

s t

1

2 4

3

9

10

56

6

8 45

5

4

2

3

1 -1

0 0

00

LP FormulationLP Formulation

10

1

0

0

0

0

1st

min

5646

45352556

34244546

23133534

12252423

1312

5646453534

2524231312

54326594108

xxx

xxxxxxxxxxxxxxxx

xxxxxxxxxxxx

ij

Documents

Linear Programming OPIM 310-Lecture 2 Instructor: Jose Cruz