linear programming meeting multiple and diverse goals within the context of limited resources...

Linear ProgrammingLinear Programming

MEETING MULTIPLE ANDDIVERSE GOALS WITHIN

THE CONTEXT OF LIMITEDRESOURCES

Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions Philip A. Vaccaro , PhD

THE ALLOCATION PROBLEM

MGMT E-5050

HistoryHistory

Linear programming was conceptually developed

before World War II by the Soviet mathematician

Andrei Nikolaevich Kolmogorov ( 1903 – 1987 )

АндрейАндрей НиколаевичНиколаевич ΚΚолмогороволмогоров

Leonid Vitalyevich Kantorovich, another Soviet mathematician,

won the Nobel Prize in Economics for advancing the concepts of optimal planning.

ЛєонидЛєонид КанторовичКанторович

( 1912 – 1986 )

HistoryHistory

In 1947, George Bernard Dantzig developed the solution

procedure known as the simplex algorithm, while

working on Air Force logistics problems.

( 1914 – 2005 )

HistoryHistory

ALL DEVELOP ANOPTIMAL SOLUTION

Linear Programming Models

Application Examples

PRODUCT MIX PROBLEMPRODUCT MIX PROBLEM determining the optimal number of several products to make in order to maximize total profit or minimize total cost within the context of limited resources.

BLENDING PROBLEMBLENDING PROBLEM determining the lowest-cost mixture of food groups that will meet a set of minimum nutritional requirements.

Application ExamplesApplication Examples

FINANCIAL PROBLEMFINANCIAL PROBLEM determining the allocation of funds among various investment classes such as stocks,

bonds, real estate, and commodities so as to maximize total returns over time.

PROMOTION PROBLEMPROMOTION PROBLEM determining the allocation of marketing budgets

over various media such as billboards, radio, television, newspapers, and magazines

so as to maximize total exposure.

Problem StatementProblem Statement

A firm makes two kinds of clocks: regular and alarm..

3 resources are required to produce these clocks:

Available daily labor hours - 1,600

Available daily processing hours - 1,800

Available daily alarm assemblies - 350

Problem StatementProblem StatementContinued

Model DevelopmentModel Development

Let X1 = the number of regular clocks produced

Let X2 = the number of alarm clocks produced IF THERE WERE A 3rd and 4th TYPE CLOCK,

THEY WOULD BE DESIGNATED AS X3 AND X4RESPECTIVELY

The The Objective FunctionObjective Function

Maximize Z = 3X1 + 8X2

total daily profit

controllable, decision, or real variables

contribution marginsor

objective function coefficients

Labor Resource Constraint

2X1 + 4X2 =< 1,600 hours

real variables

usage coefficients right-hand side (RHS)

Processing Resource ConstraintProcessing Resource Constraint

6X1 + 2X2 =< 1,800 hours

real variables


Alarm Assemblies Resource ConstraintAlarm Assemblies Resource Constraint

0X1 + 1X2 =< 350 units

real variables


Non-Negativity ConstraintsNon-Negativity Constraints

X1 => 0

X2 => 0or X1 , X2 => 0

THE FIRM MUST PRODUCE NOTHINGOR SOMETHING OF EACH PRODUCT.NEGATIVE VALUES FOR PRODUCT

DO NOT EXIST !

The ModelThe Model


Subject to:

2X1 + 4X2 =< 1,600 ( labor hours )6X1 + 2X2 =< 1,800 ( processing hours ) X2 =< 350 ( alarm assemblies ) X1 => 0 X2 => 0

THE MODEL WILL FIND VALUES FOR THE

CONTROLLABLE VARIABLES ( X1 and X2 ) THAT WILL MAXIMIZE THE

OBJECTIVE FUNCTION ( Z ) SUBJECT TO THE THREE

RESOURCE CONSTRAINTS AND NON-NEGATIVITY CONSTRAINTS

The Graphical Method of Linear The Graphical Method of Linear ProgrammingProgramming

Four Steps:Four Steps:

I. Graph the resource constraints.

II. Identify the feasible solution region.

III. Compute the values of X1 and X2 at each

corner point of the feasible region.

IV. Select the corner point with the maximum profit. ( the optimal solution )

Plotting the 1Plotting the 1stst Resource Constraint Resource Constraint

2X1 + 4X2 =< 1,600 labor hours

assume that labor hours are the only resource needed to produce X1 and X2 clocks.

if we only made X1 s, we could make 800 units.

If we only made X2 s , we could make 400 units.

therefore, the coordinates for plotting the labor hour constraint are X1 = 800 and X2 = 400.

The Labor Constraint PlotThe Labor Constraint Plot

1000 900 800 700 600 500 400400 300 200 100 0

0 100 200 300 400 500 600 700 800800 900 1000X1

X2

2X1 + 4X2 = 1,600 labor hours

THE LINEAR INEQUALITYMUST BE CONVERTED TO

A LINEAR EQUALITY BEFOREIT CAN BE PLOTTED

Plotting the 2Plotting the 2ndnd Resource Constraint Resource Constraint

6X1 + 2X2 =< 1,800 process hours

assume that process hours are the only resource needed to produce X1 and X2 clocks.

if we only made X1 s , we could make 300 units.


therefore, the coordinates for plotting the process hour constraint are X1 = 300 and X2 = 900.

The Process Constraint PlotThe Process Constraint Plot

1000 900900 800 700 600 500 400 300 200 100 0

0 100 200 300300 400 500 600 700 800 900 1000X1

X2

6X1 + 2X2 = 1,800 process hours



Plotting the 3Plotting the 3rdrd Resource Constraint Resource Constraint

0X1 + 1X2 =< 350 alarm assemblies

assume that alarm assemblies are the only resource needed to produce X2 clocks.


therefore, the coordinates for plotting the alarm assemblies constraint are X1 = 0 and X2 = 350.

*

ALARM ASSEMBLIES

ARE NOT USED IN MAKING REGULARCLOCKS

The Alarm Assemblies Constraint PlotThe Alarm Assemblies Constraint Plot

1000 900 800 700 600 500 400400 300300 200 100 0

0 100 200 300 400 500 600 700 800 900 1000X1

X2

0X1 + 1X2 = 350 alarm assemblies



The Fully-Plotted GraphThe Fully-Plotted Graph

1000 900 800 700 600 500 400 300 200 100 0

0 100 200 300 400 500 600 700 800 900 1000XX11

XX22

CORNER POINTS A,B,C,D,E DEFINE THE FEASIBLE REGIONCORNER POINTS A,B,C,D,E DEFINE THE FEASIBLE REGION

FEASIBLEREGION

PROCESSING HOURSPROCESSING HOURS

LABOR HOURSLABOR HOURS

ALARM ASSEMBLIESALARM ASSEMBLIES

A

B C D

E

The The Corner PointCorner Point Evaluations Evaluations

Objective Function = Maximize Z = 3X1 + 8X2

Point A Z = 3 ( 0) + 8( 0) = $0.00

Point B Z = 3 ( 0) + 8(350) = $2,800.00

Point C Z = 3(100) + 8(350) = $3,100.00

Point D Z = 3(200) + 8(300) = $3,000.00

Point E Z = 3(300) + 8( 0) = $900.00

THE PROFITSAT EACHCORNER

POINTVIA THE

OBJECTIVEFUNCTION

Corner Point “C” CoordinatesCorner Point “C” Coordinates

2X1 + 4X2 = 1,6000X1 + 1X2 = 350

THE TWO CONSTRAINT LINES THAT INTERSECT TO FORM POINT “C” ARE:

AT THIS INTERSECTION, THE VALUES OF X1 AND X2

MUST BE IDENTICAL IN BOTH CONSTRAINTS.

SINCE X2 = 350 , WE SOLVE FOR X1:

2X1 + 4 (350) = 1,600 2X1 + 1,400 = 1,600 2X1 = 200 X1 = 100

Corner Point “D” CoordinatesCorner Point “D” Coordinates

2X1 + 4X2 = 1,600

6X1 + 2X2 = 1,800

THE TWO CONSTRAINT LINES THAT INTERSECT TO FORM POINT “D” ARE:

AT THIS INTERSECTION, THE VALUES OF X1 AND X2

MUST BE IDENTICAL IN BOTH CONSTRAINTS.

WE SOLVE FOR X2

BY CANCELING OUT X1 IN BOTH

EQUATIONS (CONSTRAINTS)

AS FOLLOWS


3 ( 2X1 + 4X2 = 1,600 ) 6X1 + 2X2 = 1,800

6X1 + 12X2 = 4,800 - 6X1 - 2X2 = - 1,800

10X2 = 3,000 X2 = 300


2X1 + 4X2 = 1,600

2X1 + 4(300) = 1,600

2X1 + 1,200 = 1,600

2X1 = 400

X1 = 200

WE SUBSTITUTE X2 = 300

INTO THE 1st EQUATION AND SOLVE

FOR X1:

TheThe Corner Point Corner Point EvaluationsEvaluations

The Corner Point Evaluations

THE OPTIMALSOLUTION

ISCORNER POINT

“C”

Optimal SolutionOptimal Solution

Slack Variables ( S )

Every resource has its own unique slack variable to represent it.

Its value is the difference between what was consumed and what was originally available for that particular resource.

In our example:

labor hours will be represented by “S1” process hours will be represented by “S2” alarm assemblies will be represented by “S3”

slackvariables

can becomputedonce theoptimalsolutionhas been

found

Slack Variable Slack Variable SS11 Computation ComputationGiven that X1 = 100 regular clocks

and X2 = 350

alarm clocks:

2X1 + 4X2 =< 1,600 labor hoursbecomes:

2X1 + 4X2 + 1S1 = 1,600(substituting)

2(100) + 4(350) + 1S1 = 1,600200 + 1,400 + 1S1 = 1,600

1,600 + 1S1 = 1,600therefore:

S1 = 0

total consumed hours originally available hours

Slack Variable Slack Variable SS11 Interpretation Interpretation

All labor hours are completely consumed in the optimal product mix solution.

The labor hour constraint is binding, that is , we are prevented from producing more clocks because labor hours are fully consumed.

Labor hours carry a positive shadow price , that is, we would be willing to buy additional labor hours if they were available.

Slack Variable Slack Variable SS22 Computation ComputationGiven that X1 = 100 regular clocks and

X2 = 350 alarm

clocks:

6X1 + 2X2 =< 1,800 process hoursbecomes:

6X1 + 2X2 + 1S2 = 1,800(substituting)

6(100) + 2(350) + 1S2 = 1,800600 + 700 + 1S2 = 1,800

1,300 + 1S2 = 1,800therefore:

S2 = 500

total consumed hours originally available hours


There are 500 process hours left over in the optimal product mix solution.

The process hour constraint is non-binding, because it does not prevent us from producing more clocks. Process hours carry a zero shadow price, meaning that we are not willing to buy additional hours since we still have an excess.

Given that X1 = 100 regular clocks and

X2 = 350 alarm clocks:

Slack Variable Slack Variable SS33 Computation Computation

0X1 + 1X2 =< 350 alarm assembliesbecomes:

0X1 + 1X2 + 1S3 = 350(substituting)

0(100) + 1(350) + 1S3 = 3500 + 350 + 1S3 = 350

350 + 1S3 = 350therefore:

S3 = 0

total consumed assemblies originally available assemblies


All alarm assemblies are completely consumed in the optimal product mix solution. The alarm assembly constraint is binding , that is, we are prevented from producing more clocks because alarm assemblies are fully consumed.

Alarm assemblies carry a positive shadow price , that is, we would be willing to buy additional alarm assemblies if they were available.

The Shadow PriceThe Shadow Price

Zero vs. Positive Shadow PricesZero vs. Positive Shadow Prices

Resources > 0 in theoptimal solution have zero shadow pricesbecause we have an

excess.

Resources = 0 in theoptimal solution have

positive shadow pricesbecause we would like to buy more of them.

If we pay less than the shadow price, total profit ( Z ) will increaseincrease

If we pay the shadow price exactly, total profit ( Z ) will not changenot change

If we pay more than the shadow price, total profit will decreasedecrease

Shadow Price Implications

Iso-Profit LinesIso-Profit Lines

From the Greek ( ίσώ ) meaning “equal”.

Any combination of the two products produced along this line will yield the same total profit *

They are identified by dashed lines.

Primarily used today to quickly determine if a certain level of profit (or cost) can be achieved before proceeding further.

COST IN A MINIMIZATION PROBLEM

The Clock Production GraphThe Clock Production Graph

10001000 900900 800800 700700 600600 500500 400400 300300 200200 100100 00

0 100 200 300 400 500 600 700 800 900 10000 100 200 300 400 500 600 700 800 900 1000X1X1

X2X2CORNER POINTS A,B,C,D,ECORNER POINTS A,B,C,D,E

DEFINE THE FEASIBLE REGIONDEFINE THE FEASIBLE REGION

FEASIBLEFEASIBLEREGIONREGION




A

B C D

E

UNFEASIBLEUNFEASIBLEREGIONREGION

$1,200.00 = 3X1 + 8X2


CONTINUED

If we wanted to determine if an arbitrary profit of$1,200.00 were achievable, we set the objectivefunction equal to $1,200.00 :

We find that we could make :

400 X1 clocks and 0 X2 clocks or

0 X1 clocks and 150 X2 clocks

in order to achieve a profit of $1,200.00


10001000 900900 800800 700700 600600 500500 400400 300300 200200 100100 00

0 100 200 300 400 500 600 700 800 900 10000 100 200 300 400 500 600 700 800 900 1000X1X1







A

B C D

E

$1,200.00 $1,200.00 Iso-ProfitIso-Profit Line Line


A PROFIT

OF $1,200.00

ISACHIEVABLE


CONTINUED


$2,400.00 = 3X1 + 8X2


800 X1 clocks and 0 X2 clocks or




10001000 900900 800800 700700 600600 500500 400400 300300 200200 100100 00

0 100 200 300 400 500 600 700 800 900 10000 100 200 300 400 500 600 700 800 900 1000X1X1







A

B C D

E


APROFIT

OF$2,400.00

ISACHIEVABLE



$3,000.00 = 3X1 + 8X2


1,000 X1 clocks and 0 X2 clocks or




10001000 900900 800800 700700 600600 500500 400400 300300 200200 100100 00

0 100 200 300 400 500 600 700 800 900 10000 100 200 300 400 500 600 700 800 900 1000X1X1

X2X2THE $3,000.00 ISO-PROFIT LINE IS JUST BELOW

CORNER POINTS “C” AND “D”





A

B C D

E



APROFIT

OF$3,000.00

ISACHIEVABLE


CONTINUED


$3,100.00 = 3X1 + 8X2


1,033 X1 clocks and 0 X2 clocks or




10001000 900900 800800 700700 600600 500500 400400 300300 200200 100100 00

0 100 200 300 400 500 600 700 800 900 10000 100 200 300 400 500 600 700 800 900 1000X1X1

X2X2THE $3,100.00 ISO-PROFIT LINE CROSSES THROUGH

CORNER POINT “C” ONLY





A

B C D

E



APROFIT

OF$3,100.00

ISACHIEVABLE


10001000 900900 800800 700700 600600 500500 400400 300300 200200 100100 00

0 100 200 300 400 500 600 700 800 900 10000 100 200 300 400 500 600 700 800 900 1000X1X1

X2X2THE $4,000.00 ISO-PROFIT LINE FALLS OUTSIDE THE FEASIBLE

REGION





A

B C D

E



A$4,000.00PROFITIS NOT

ACHIEVABLE

Units of product producedover and above the

minimum required in the optimal solution.

The only way it can bedifferentiated from aslack variable is by

placing a negative (-)sign in front of it.

The Surplus Variable ( S )

The Surplus VariableThe Surplus Variable

X1 => 60 regular clocks

X1 – S4 = 60 regular clocks

100 - 60 = 40 regular clocks

MEANS PRODUCE AT LEAST60 REGULAR CLOCKS( A QUOTA CONSTRAINT )

REWRITTEN AS A LINEAREQUALITY WHERE S4 IS THE

EXCESS OF REGULAR CLOCKS,IF ANY, PRODUCED IN THE

OPTIMAL SOLUTION

SINCE X1 = 100 REGULARCLOCKS IN THE OPTIMALSOLUTION, THE EXCESSPRODUCED IS 40 CLOCKS

EXAMPLE

Management / market imposed operating restrictions which, by their very nature, have a high probability of degrading the optimal solution, resulting in lower total profits or higher total costs.

QUOTA CONSTRAINTS ARE A NORMALOCCURRENCE IN THE LINEAR

PROGRAMMING PROBLEM BECAUSEMOST FIRMS DO NOT ENJOY THE

LUXURY OF MAKING WHAT THEY WANT

QUOTA CONSTRAINTS

X1 = regular clocks X2 = alarm clocks

X1 + X2 = 40

X1 + X2 => 40

X1 + X2 =< 40

PRODUCE EXACTLY 40 CLOCKS BETWEENREGULAR AND ALARM CLOCKS

PRODUCE AT LEAST 40 CLOCKS BETWEENREGULAR AND ALARM CLOCKS

PRODUCE NO MORE THAN 40 CLOCKSBETWEEN REGULAR AND ALARM CLOCKS

QUOTA CONSTRAINTS

1st

EXAMPLE

X1 + X2 = 40ALL COMBINATIONS OF X1 AND X2

ARE ON THE CONSTRAINT LINE ITSELF

X1 + X2 => 40ALL COMBINATIONS OF X1 AND X2

ARE ON THE CONSTRAINT LINE ANDABOVE IT

X1 + X2 =< 40ALL COMBINATIONS OF X1 AND X2

ARE ON THE CONSTRAINT LINE ANDBELOW IT

X1

X1

X1

X2

X2

X2

QUOTA CONSTRAINTSThe

ConstraintPlots


X1 = 2X2

X1 => 2X2

X1 =< 2X2

PRODUCE EXACTLY TWICE AS MANYREGULAR CLOCKS AS ALARM CLOCKS

PRODUCE AT LEAST TWICE AS MANYREGULAR CLOCKS AS ALARM CLOCKS

PRODUCE AT MOST, TWICE AS MANYREGULAR CLOCKS AS ALARM CLOCKS

QUOTA CONSTRAINTS

2nd EXAMPLE

X1 = 2X2 X1 => 2X2 X1 =< 2X2

X1 = 2X2

4 = 2 ( 2 ) 6 = 2 ( 3 ) 8 = 2 ( 4 )

etc.

ARBITRARILY SELECT VALUES FOR X1 AND X2

SUCH THAT THE LEFT SIDE OF THE LINEAR

EQUALITY EQUALS THE RIGHT SIDE.

0 1 2 3 4 5 6 7 8 9

543210

X1

X2

X1 = 2X2

ANY TWO PAIRS OFX1 AND X2 COORDINATESWILL ENABLE US TO PLOT

THE GENERAL CONSTRAINT

QUOTA CONSTRAINTS


X1 = 1.75X2

X1 => 1.75X2

X1 =< 1.75X2

PRODUCE EXACTLY 75% MOREREGULAR CLOCKS THAN ALARM CLOCKS

PRODUCE AT LEAST 75% MOREREGULAR CLOCKS THAN ALARM CLOCKS

PRODUCE AT MOST, 75% MOREREGULAR CLOCKS THAN ALARM CLOCKS

QUOTA CONSTRAINTS

3rd EXAMPLE

X1 = 1.75X2 X1 => 1.75X2 X1 =< 1.75X2

X1 = 1.75X2

7 = 1.75( 4)14 = 1.75( 8)21 = 1.75(12)

etc.

ARBITRARILY SELECT VALUES FOR X1 AND X2

SUCH THAT THE LEFT SIDE OF THE LINEAR

EQUALITY EQUALS THE RIGHT SIDE.

0 7 14 21 28

12

8

4

0 X1

X2

X1 = 1.75X2

ANY TWO PAIRS OFX1 AND X2 COORDINATESWILL ENABLE US TO PLOT

THE GENERAL CONSTRAINT

QUOTA CONSTRAINTS

Other ConstraintsOther Constraints

0 10 20 30 40 50 60

30

20

10

0

X2

X1

X1 = 20 on the line

X1 => 20 on and right of the line

X1 =< 20 on and left of the line

FEASIBLE REGION

4th Example

Other ConstraintsOther Constraints

0 10 20 30 40 50 60

30

20

10

0

X2

X1

X2 = 30 on the line

X2 => 30 on and above the line

X2 =< 30 on and below the line

FEASIBLE REGION

Graph Linear ProgrammingGraph Linear Programming

The objective function coefficients become costs.

The objective function becomes “Minimize Z”.

Usually more quota constraints are used.

The feasible region may

be unbounded.

“=“ and “=>”

X2

MINIMIZATION

UNBOUNDEDFEASIBLE

REGION

X1

Linear Programming ApplicationLinear Programming ApplicationHUMAN RESOURCESHUMAN RESOURCES

We have a choice of three different employee candidates: un-trained,semi-trained, and highly-trained for new positions in our department.The cost of training is $5.00 / hour, $8.50 / hour, and $10.50 / hour forun-trained, semi-trained, and highly-trained respectfully.

An un-trained worker requires 28 hours of training for the paintingprocess and 35 hours for the packing process.

A semi-trained worker requires 23 hours of training for the paintingprocess and 30 hours of training for the packing process.

A highly-trained worker requires only 15 and 20 hours for painting and packing process training.

We need at least 25 new employees. We only have 700 hours availablefor training in the painting process and 775 hours available for trainingin the packing process.

REQUIREMENTREQUIREMENT

Formulate a linear programming model that will minimize the total cost of training while hiring at least twenty-five new

employees.

Let X1 = the number of un-trained workers to hireLet X2 = the number of semi-trained workers to hireLet X3 = the number of highly-trained workers to hire

The ModelThe Model

Minimize Z = (63x$5.00)X1 + (53x$8.50)X2 + (35x$10.50)X3SUBJECT TO:

28X1 + 23X2 + 15X3 =< 700 painting training hrs.

35X1 + 30X2 + 20X3 =< 775 packing training hrs.

1X1 + 1X2 + 1X3 => 25 new employees

X1, X2, X3 => 0

$315.00 X1 + $450.50 X2 + $367.50 X3

Linear Programming ApplicationLinear Programming ApplicationADVERTISINGADVERTISING

The Salem Department of Tourism is developing a marketing strategy for next year. They would like to maximize the number of tourists thatchoose Salem for their vacation. They have narrowed their choices down to three types of television ads: regional, statewide, or local.

The research team has determined that a regional ad will reach 100,000people, a statewide ad will reach 70,000 people, and a local ad will reach 20,000 people.

The cost per ad is $8,000.00 for regional, $6,000.00 for statewide, and$800.00 for local.

There is a one million ($1,000,000.00) dollar budget for advertising. They would like to have at least twice as many regional ads as localads.

REQUIREMENTREQUIREMENT

Let X1 = the number of regional ads to placeLet X2 = the number of statewide ads to placeLet X3 = the number of local ads to place

Formulate a linear programming model that willmaximize the

number of peoplethey can reach

with their advertising while having at

least twice as manyregional ads as local

ads.

The ModelThe Model

Maximize Z = 100,000 X1 + 70,000 X2 + 20,000 X3

8000X1 + 6000X2 + 800X3 =< 1,000,000

X1 => 2X3

X1, X2, X3 => 0

SUBJECT TO:

BUDGETADVERTISING COSTS

EXPOSURE

AT LEAST TWICE AS MANY REGIONAL ADS

QM for WINDOWSQM for WINDOWSLinear Programming Linear Programming

WE SELECT THELINEAR PROGRAMMING

MODULE

WE WANT TO SOLVEA NEW PROBLEM

THE DIALOGUE BOX

THERE ARE THREERESOURCE

CONSTRAINTS

THERE ARE TWOPRODUCTS

( 2 REAL VARIABLES )WE WANT TO

MAXIMIZETOTAL PROFIT

THE DATA INPUTTABLE

THE MODEL

MAXIMIZE “Z” = $3.00 X1 + $8.00 X2

Subject to:

2X1 + 4X2 =< 1,600 labor hours

6X1 + 2X2 =< 1,800 process hours

0X1 + 1X2 =< 350 alarm assemblies

1st SOLUTION WINDOW( produce 100 regular clocks )( produce 350 alarm clocks )

( profit = $3,100.00 )

THE SHADOW PRICES

Labor hours - $1.50

Process hours - $0.00

Alarm assemblies - $2.00

BASIC VARIABLES ARE THOSEVARIABLES > 0

X1 = 100X2 = 350S2 = 500

NON-BASIC VARIABLES ( = 0 )

S1 = 0S3 = 0

THIS IS THESIMPLEX

ALGORITHMSOLUTION

TO OURPROBLEM

THE GRAPH METHODSOLUTION

*******THE FEASIBLE REGION

SHOWN IN PURPLE

THE OPTIMAL SOLUTIONIS AT THIS

CORNER POINTX1 = 100 , X2 = 350

Linear Programming Linear Programming

Template

Linear Programming

Solved ProblemsSolved Problems

Linear ProgrammingGraph Method COMPUTER-BASEDCOMPUTER-BASED

MANUALMANUAL

Solved ProblemSolved ProblemGRAPH METHOD OF LINEAR PROGRAMMINGGRAPH METHOD OF LINEAR PROGRAMMING

The Clothier ProblemThe Clothier Problem

A clothier makes coats and slacks. The two resources are wool clothand labor. The clothier has 150 square yards of wool and 200 hours oflabor available.Each coat requires three (3) square yards of wool and ten (10) hours oflabor, while each pair of slacks requires five (5) square yards of wool and four (4) hours of labor.The profit for a coat is $50.00 and the profit for a pair of slacks is $40.00.The clothier wants to determine the number of coats and slacks to makeso that profit will be maximized.

REQUIREMENT :

1. Formulate a linear programming model for this problem.2. Solve the model using the graph method.3. Solve the model using QM for WINDOWS.

Solved ProblemSolved ProblemTheThe

Clothier Clothier ProblemProblem

Let X1 = coats , Let X2 = slacks


subject to : 3X1 + 5X2 <= 150 square yards (wool) 10X1 + 4X2 <= 200 hours (labor)

X1 , X2 >= 0

The Model

Coordinates:

Wool: X1 = 5050 X1 = 0 Labor: X1 = 2020 X1 = 0 X2 = 0 X2 = 3030 X2 = 0 X2 = 5050

0 10 0 10 2020 30 40 30 40 50 50 60 70 60 70

5050

4040

3030

2020

1010

00 XX11 COATS COATS

XX22 SLACKS SLACKS

FEASIBLE

FEASIBLE

REGIO

N

REGIO

N

10X10X11 + 4X + 4X22 = 200 hours, labor= 200 hours, labor

3X3X11 + 5X + 5X22 = 150 sq. yards, wool = 150 sq. yards, wool

AA

BB

CC

DD

The Clothier ProblemSolved Problem

Corner Point “D” Coordinate Calculations

““1010” ( 3X” ( 3X1 1 + 5X+ 5X22 = 150) = 150) “ “33” (10X” (10X11 + 4X + 4X22 = 200) = 200)

30X30X11 + 50X + 50X22 = 1,500 = 1,500 30X30X11 + 12X + 12X22 = 600 = 600

38X38X22 = 900 = 900

XX22 = 23.68 = 23.68 ≈ 23≈ 23

Then: 3XThen: 3X11 + 5(23.68) = 150 + 5(23.68) = 150 3X3X11 + 118.4 = 150 + 118.4 = 150 3X3X11 = 31.6 = 31.6 XX11 = 10.53 = 10.53 ≈ 10≈ 10

X1 = 10 coats X2 = 23 slacks

Corner Point

Summary and

Solution

* * OPTIMAL SOLUTIONOPTIMAL SOLUTION

MAKE10 COATS

AND23 PAIRS

OF SLACKSWITH TOTALPROFIT OF$1,420.00

Make 10 Coats

Make 23 Slacks

Total Profit = $1,420.00

FEASIBLEREGION

Template

The Copperfield Mining Company owns two mines, each of which produces three grades of ore: high, medium, and low. The company has a contract to supply a smelting firm with at least twelve (12) tons of high-grade ore, eight (8) tons of medium-grade ore, and twenty four (24) tons of low-grade ore. Each mine produces a certain amount of each type of ore during each hour that it is in operation. Mine #1 produces 6, 2, and 4 tons respectively of high-, medium-, and low- grade ore per hour. Mine #2 produces 2, 2, and 12 tons respectively of high-, medium-, and low- grade ore per hour. It costs Copperfield $200.00 per hour to mine ore from mine #1, and $160.00 per hour to mine ore from mine #2. The company wants to determine the number of hours it needs to operate each mine so that its contractual obligations can be met at the lowest cost.

Solved ProblemSolved Problem

The Copperfield Mining Company

GRAPH METHOD OF LINEAR PROGRAMMINGGRAPH METHOD OF LINEAR PROGRAMMING


The Copperfield Mining CompanyThe Copperfield Mining Company

REQUIREMENT :

1. Formulate a linear programming model for this problem.2. Solve this model using the graph method.3. Solve this model using QM for WINDOWS.


The Copperfield Mining CompanyThe Copperfield Mining Company

The Model:

Let X1 = mine #1 , Let X2 = mine #2

Minimize Z = 200X1 + 160X2

subject to : 6X1 + 2X2 >= 12 tons , high-grade ore 2X1 + 2X2 >= 8 tons , medium-grade ore 4X1 + 12X2 >= 24 tons , low-grade ore

X1 , X2 >= 0

hourly operation cost

hourly production

rates

supply demands

Solved ProblemSolved ProblemThe Copperfield Mining CompanyThe Copperfield Mining Company

High-grade ore: X1 = 2 X1 = 0 X2 = 0 X2 = 6

Medium-grade ore: X1 = 4 X1 = 0 X2 = 0 X2 = 4

Low-grade ore: X1 = 6 X1 = 0 X2 = 0 X2 = 2

COORDINATES


0 1 2 3 4 5 6 7 80 1 2 3 4 5 6 7 8

88

66

44

22

00

XX22 ( Mine #2 ) ( Mine #2 )

( X1 Mine #1 )

6X6X11 + 2X + 2X22 = 12 tons , = 12 tons , high-grade orehigh-grade ore

2X2X11 + 2X + 2X22 = 8 tons , = 8 tons , medium-grade oremedium-grade ore

4X4X11 + 12X + 12X22 = 24 tons , = 24 tons , low-grade orelow-grade ore

AA

BB

CC

DD

FeasibleRegion

FeasibleFeasibleRegionRegion


Corner Point “B” Coordinate Calculations

6X6X11 + 2X + 2X22 = 12 = 122X2X11 + 2X + 2X22 = 8 = 8 4X4X11 = 4 = 4 XX11 = 1 = 1

Then: 6(1) + 2XThen: 6(1) + 2X22 = 12 = 12 2X2X22 = 6 = 6 XX22 = 3 = 3

Corner Point “C” Coordinate Calculations

4X4X11 + 4X + 4X22 = 16 = 16““22” (2X” (2X1 1 + 2X+ 2X2 2 = 8)= 8) 4X4X1 1 + 12X+ 12X22 = 24 = 24 -8X-8X22 = -8 = -8 XX22 = 1 = 1

Then: 2XThen: 2X11 + 2(1) = 8 + 2(1) = 8 2X2X1 1 = 6= 6 XX1 1 = 3= 3


Corner PointSummary

andSolution


solution

extrapolation

total number of tons

of each grade produced

from both mines

FEASIBLE REGION

Template

Solved ProblemsSolved Problems

Linear ProgrammingGraph Method

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linear programming meeting multiple and diverse goals within the context of limited resources...

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