linear programming meeting multiple and diverse goals within the context of limited resources...
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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
usage coefficients right-hand side (RHS)
Alarm Assemblies Resource ConstraintAlarm Assemblies Resource Constraint
0X1 + 1X2 =< 350 units
real variables
usage coefficients right-hand side (RHS)
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
Maximize Z = 3X1 + 8X2
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.
if we only made X2 s , we could make 900 units.
therefore, the coordinates for plotting the pro- cess 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
THE LINEAR INEQUALITYMUST BE CONVERTED TO
A LINEAR EQUALITY BEFOREIT CAN BE PLOTTED
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.
if we only made X2 s , we could make 350 units.
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 LINEAR INEQUALITYMUST BE CONVERTED TO
A LINEAR EQUALITY BEFOREIT CAN BE PLOTTED
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
Corner Point “D” CoordinatesCorner Point “D” Coordinates
3 ( 2X1 + 4X2 = 1,600 ) 6X1 + 2X2 = 1,800
6X1 + 12X2 = 4,800 - 6X1 - 2X2 = - 1,800
10X2 = 3,000 X2 = 300
Corner Point “D” CoordinatesCorner Point “D” Coordinates
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
Slack Variable Slack Variable SS22 Interpretation Interpretation
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, mean- ing 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
Slack Variable Slack Variable SS33 Interpretation Interpretation
All alarm assemblies are completely con- sumed 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
PROCESSING HOURSPROCESSING HOURS
LABOR HOURSLABOR HOURS
ALARM ASSEMBLIESALARM ASSEMBLIES
A
B C D
E
UNFEASIBLEUNFEASIBLEREGIONREGION
$1,200.00 = 3X1 + 8X2
Iso-Profit LinesIso-Profit Lines
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
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
PROCESSING HOURSPROCESSING HOURS
LABOR HOURSLABOR HOURS
ALARM ASSEMBLIESALARM ASSEMBLIES
A
B C D
E
$1,200.00 $1,200.00 Iso-ProfitIso-Profit Line Line
UNFEASIBLEUNFEASIBLEREGIONREGION
A PROFIT
OF $1,200.00
ISACHIEVABLE
Iso-Profit LinesIso-Profit Lines
CONTINUED
If we wanted to determine if an arbitrary profit of$2,400.00 were achievable, we set the objectivefunction equal to $2,400.00 :
$2,400.00 = 3X1 + 8X2
We find that we could make :
800 X1 clocks and 0 X2 clocks or
0 X1 clocks and 300 X2 clocks
in order to achieve a profit of $2,400.00
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
PROCESSING HOURSPROCESSING HOURS
LABOR HOURSLABOR HOURS
ALARM ASSEMBLIESALARM ASSEMBLIES
A
B C D
E
$2,400.00 $2,400.00 Iso-ProfitIso-Profit Line Line
APROFIT
OF$2,400.00
ISACHIEVABLE
Iso-Profit LinesIso-Profit Lines
If we wanted to determine if an arbitrary profit of$3,000.00 were achievable, we set the objectivefunction equal to $3,000.00 :
$3,000.00 = 3X1 + 8X2
We find that we could make :
1,000 X1 clocks and 0 X2 clocks or
0 X1 clocks and 375 X2 clocks
in order to achieve a profit of $3,000.00
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
X2X2THE $3,000.00 ISO-PROFIT LINE IS JUST BELOW
CORNER POINTS “C” AND “D”
FEASIBLEFEASIBLEREGIONREGION
PROCESSING HOURSPROCESSING HOURS
LABOR HOURSLABOR HOURS
ALARM ASSEMBLIESALARM ASSEMBLIES
A
B C D
E
$3,000.00 $3,000.00 Iso-ProfitIso-Profit Line Line
UNFEASIBLEUNFEASIBLEREGIONREGION
APROFIT
OF$3,000.00
ISACHIEVABLE
Iso-Profit LinesIso-Profit Lines
CONTINUED
If we wanted to determine if an arbitrary profit of$3,100.00 were achievable, we set the objectivefunction equal to $3,100.00 :
$3,100.00 = 3X1 + 8X2
We find that we could make :
1,033 X1 clocks and 0 X2 clocks or
0 X1 clocks and 387 X2 clocks
in order to achieve a profit of $3,100.00
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
X2X2THE $3,100.00 ISO-PROFIT LINE CROSSES THROUGH
CORNER POINT “C” ONLY
FEASIBLEFEASIBLEREGIONREGION
PROCESSING HOURSPROCESSING HOURS
LABOR HOURSLABOR HOURS
ALARM ASSEMBLIESALARM ASSEMBLIES
A
B C D
E
$3,100.00 $3,100.00 Iso-ProfitIso-Profit Line Line
UNFEASIBLEUNFEASIBLEREGIONREGION
APROFIT
OF$3,100.00
ISACHIEVABLE
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
X2X2THE $4,000.00 ISO-PROFIT LINE FALLS OUTSIDE THE FEASIBLE
REGION
FEASIBLEFEASIBLEREGIONREGION
PROCESSING HOURSPROCESSING HOURS
LABOR HOURSLABOR HOURS
ALARM ASSEMBLIESALARM ASSEMBLIES
A
B C D
E
$4,000.00 $4,000.00 Iso-ProfitIso-Profit Line Line
UNFEASIBLEUNFEASIBLEREGIONREGION
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 = regular clocks X2 = alarm clocks
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 = regular clocks X2 = alarm clocks
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 train- ing 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
Maximize Z = 50X1 + 40X2
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
Solved ProblemSolved ProblemGRAPH 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.
Solved ProblemSolved ProblemGRAPH METHOD OF LINEAR PROGRAMMINGGRAPH METHOD OF LINEAR PROGRAMMING
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
Solved ProblemSolved ProblemThe Copperfield Mining CompanyThe Copperfield Mining Company
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
Solved ProblemSolved ProblemThe Copperfield Mining CompanyThe Copperfield Mining Company
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
Solved ProblemSolved ProblemThe Copperfield Mining CompanyThe Copperfield Mining Company
Corner PointSummary
andSolution
Solved ProblemSolved ProblemThe Copperfield Mining CompanyThe Copperfield Mining Company
solution
extrapolation
total number of tons
of each grade produced
from both mines
FEASIBLE REGION
Template
Solved ProblemsSolved Problems
Linear ProgrammingGraph Method
COMPUTER-BASEDCOMPUTER-BASED
MANUALMANUAL