Linear Programming MGMT E-5050 THE ALLOCATION PROBLEM MEETING MULTIPLE AND DIVERSE GOALS WITHIN THE CONTEXT OF LIMITED RESOURCES Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions Philip A. Vaccaro , PhD

History Linear programming was conceptually developed before World War II by the Soviet mathematician Andrei Nikolaevich Kolmogorov ( 1903 – 1987 ) Андрей Николаевич Κолмогоров

History Leonid Vitalyevich Kantorovich, another Soviet mathematician, won the Nobel Prize in Economics for advancing the concepts of optimal planning. ( 1912 – 1986 ) Лєонид Канторович

History In 1947, George Bernard Dantzig developed the solution procedure known as the simplex algorithm, while working on Air Force logistics problems. ( 1914 – 2005 )

Linear Programming Models Integer Programming Goal Programming Dynamic Programming Nonlinear Programming The Graphical Method The Simplex Method The Transportation Algorithm The Assignment Algorithm ALL DEVELOP AN OPTIMAL SOLUTION

Application Examples PRODUCT MIX PROBLEM determining the optimal number of several products to make in order to maximize total profit or minimize total cost within the con- text of limited resources. BLENDING PROBLEM determining the lowest-cost mixture of food groups that will meet a set of minimum nutri- tional requirements.

Application Examples FINANCIAL 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 PROBLEM determining the allocation of marketing bud- gets over various media such as billboards, radio, television, newspapers, and magazines so as to maximize total exposure.

Problem 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 Statement Regular Clock 2 Labor Hours 6 Process Hours $3.00 Continued REQUIREMENT UNIT PROFIT Regular Clock 2 Labor Hours 6 Process Hours $3.00 Alarm Clock 4 Labor Hours 2 Process Hours $8.00 1 Alarm Assembly

Model Development Let X1 = the number of regular clocks produced Let X2 = the number of alarm clocks IF THERE WERE A 3rd and 4th TYPE CLOCK, THEY WOULD BE DESIGNATED AS X3 AND X4 RESPECTIVELY

The Objective Function controllable, decision, or real variables Maximize Z = 3X1 + 8X2 contribution margins or objective function coefficients total daily profit

Labor Resource Constraint real variables 2X1 + 4X2 =< 1,600 hours usage coefficients right-hand side (RHS)

Processing Resource Constraint real variables 6X1 + 2X2 =< 1,800 hours usage coefficients right-hand side (RHS)

Alarm Assemblies Resource Constraint real variables 0X1 + 1X2 =< 350 units usage coefficients right-hand side (RHS)

Non-Negativity Constraints X1 => 0 X2 => 0 or X1 , X2 => 0 THE FIRM MUST PRODUCE NOTHING OR SOMETHING OF EACH PRODUCT. NEGATIVE VALUES FOR PRODUCT DO NOT EXIST !

The 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 Programming Graph the resource constraints. Identify the feasible solution region. Compute the values of X1 and X2 at each corner point of the feasible region. Select the corner point with the maximum profit. ( the optimal solution ) Four Steps:

Plotting the 1st 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 Plot X2 THE LINEAR INEQUALITY MUST BE CONVERTED TO A LINEAR EQUALITY BEFORE IT CAN BE PLOTTED 1000 900 800 700 600 500 400 300 200 100 2X1 + 4X2 = 1,600 labor hours X1 0 100 200 300 400 500 600 700 800 900 1000

Plotting the 2nd 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 Plot X2 THE LINEAR INEQUALITY MUST BE CONVERTED TO A LINEAR EQUALITY BEFORE IT CAN BE PLOTTED 1000 900 800 700 600 500 400 300 200 100 6X1 + 2X2 = 1,800 process hours X1 0 100 200 300 400 500 600 700 800 900 1000

Plotting the 3rd 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 REGULAR CLOCKS

The Alarm Assemblies Constraint Plot X2 THE LINEAR INEQUALITY MUST BE CONVERTED TO A LINEAR EQUALITY BEFORE IT CAN BE PLOTTED 1000 900 800 700 600 500 400 300 200 100 0X1 + 1X2 = 350 alarm assemblies X1 0 100 200 300 400 500 600 700 800 900 1000

The Fully-Plotted Graph CORNER POINTS A,B,C,D,E DEFINE THE FEASIBLE REGION X2 1000 900 800 700 600 500 400 300 200 100 PROCESSING HOURS ALARM ASSEMBLIES B C D FEASIBLE REGION LABOR HOURS A E X1 0 100 200 300 400 500 600 700 800 900 1000

The Corner Point 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

Corner Point “C” Coordinates THE TWO CONSTRAINT LINES THAT INTERSECT TO FORM POINT “C” ARE: 2X1 + 4X2 = 1,600 0X1 + 1X2 = 350 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” Coordinates THE TWO CONSTRAINT LINES THAT INTERSECT TO FORM POINT “D” ARE: 2X1 + 4X2 = 1,600 6X1 + 2X2 = 1,800 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” 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” Coordinates WE SUBSTITUTE X2 = 300 INTO THE 1st EQUATION AND SOLVE FOR X1: 2X1 + 4X2 = 1,600 2X1 + 4(300) = 1,600 2X1 + 1,200 = 1,600 2X1 = 400 X1 = 200

The Corner Point Evaluations X1 X2 Z A $0.00 B 350 $2,800.00 C 100 $3,100.00 D 200 300 $3,000.00 E $900.00

The Corner Point Evaluations X1 X2 Z A $0.00 B 350 $2,800.00 C 100 $3,100.00 D 200 300 $3,000.00 E $900.00 THE OPTIMAL SOLUTION IS CORNER POINT “C”

Optimal Solution X1 = 100 X2 = 350 Z = $3,100.00 Produce 100 regular clocks Produce 350 alarm Total Profit $3,100.00 X1 = 100 X2 = 350 Z = $3,100.00

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”

Slack Variable S1 Computation Given that X1 = 100 regular clocks and X2 = 350 alarm clocks: 2X1 + 4X2 =< 1,600 labor hours becomes: 2X1 + 4X2 + 1S1 = 1,600 substituting: 2(100) + 4(350) + 1S1 = 1,600 200 + 1,400 + 1S1 = 1,600 1,600 + 1S1 = 1,600 therefore: S1 = 0 total consumed hours originally available hours

Slack Variable S1 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 S2 Computation Given that X1 = 100 regular clocks and X2 = 350 alarm clocks: 6X1 + 2X2 =< 1,800 process hours becomes: 6X1 + 2X2 + 1S2 = 1,800 substituting: 6(100) + 2(350) + 1S2 = 1,800 600 + 700 + 1S2 = 1,800 1,300 + 1S2 = 1,800 therefore: S2 = 500 total consumed hours originally available hours

Slack Variable S2 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.

Slack Variable S3 Computation Given that X1 = 100 regular clocks and X2 = 350 alarm clocks: 0X1 + 1X2 =< 350 alarm assemblies becomes: 0X1 + 1X2 + 1S3 = 350 substituting: 0(100) + 1(350) + 1S3 = 350 0 + 350 + 1S3 = 350 350 + 1S3 = 350 therefore: S3 = 0 total consumed assemblies originally available assemblies

Slack Variable S3 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.

including, for additional The Shadow Price The price we would be willing to pay (up to and including) for additional units of a resource if available The price we would be willing to pay, up to and including, for additional units of a resource if available OUNCES, POUNDS, TONS OF MATERIALS HOURS OF LABOR MACHINE PROCESS TIME.

Zero vs. Positive Shadow Prices Resources > 0 in the optimal solution have zero shadow prices because we have an excess. Resources = 0 in the optimal solution have positive shadow prices because we would like to buy more of them.

Shadow Price Implications If we pay less than the shadow price, total profit ( Z ) will increase If we pay the shadow price exactly, total profit ( Z ) will not change If we pay more than the will decrease

Iso-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 Graph CORNER POINTS A,B,C,D,E DEFINE THE FEASIBLE REGION X2 PROCESSING HOURS 1000 900 800 700 600 500 400 300 200 100 LABOR HOURS ALARM ASSEMBLIES UNFEASIBLE REGION B C D FEASIBLE REGION A E X1 0 100 200 300 400 500 600 700 800 900 1000

Iso-Profit Lines $1,200.00 = 3X1 + 8X2 400 X1 clocks and 0 X2 clocks CONTINUED If we wanted to determine if an arbitrary profit of $1,200.00 were achievable, we set the objective function equal to $1,200.00 : $1,200.00 = 3X1 + 8X2 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 Graph CORNER POINTS A,B,C,D,E DEFINE THE FEASIBLE REGION X2 PROCESSING HOURS 1000 900 800 700 600 500 400 300 200 100 LABOR HOURS ALARM ASSEMBLIES UNFEASIBLE REGION B C D $1,200.00 Iso-Profit Line FEASIBLE REGION A E X1 0 100 200 300 400 500 600 700 800 900 1000

Iso-Profit Lines $2,400.00 = 3X1 + 8X2 800 X1 clocks and 0 X2 clocks CONTINUED If we wanted to determine if an arbitrary profit of $2,400.00 were achievable, we set the objective function 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 Graph CORNER POINTS A,B,C,D,E DEFINE THE FEASIBLE REGION X2 PROCESSING HOURS 1000 900 800 700 600 500 400 300 200 100 LABOR HOURS ALARM ASSEMBLIES B C $2,400.00 Iso-Profit Line D FEASIBLE REGION A E X1 0 100 200 300 400 500 600 700 800 900 1000

Iso-Profit Lines $3,000.00 = 3X1 + 8X2 1,000 X1 clocks and 0 X2 clocks CONTINUED If we wanted to determine if an arbitrary profit of $3,000.00 were achievable, we set the objective function 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 Graph THE $3,000.00 ISO-PROFIT LINE IS JUST BELOW CORNER POINTS “C” AND “D” X2 PROCESSING HOURS 1000 900 800 700 600 500 400 300 200 100 LABOR HOURS ALARM ASSEMBLIES UNFEASIBLE REGION B C $3,000.00 Iso-Profit Line D FEASIBLE REGION A E X1 0 100 200 300 400 500 600 700 800 900 1000

Iso-Profit Lines $3,100.00 = 3X1 + 8X2 1,033 X1 clocks and 0 X2 clocks CONTINUED If we wanted to determine if an arbitrary profit of $3,100.00 were achievable, we set the objective function 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 Graph THE $3,100.00 ISO-PROFIT LINE CROSSES THROUGH CORNER POINT “C” ONLY X2 PROCESSING HOURS 1000 900 800 700 600 500 400 300 200 100 LABOR HOURS ALARM ASSEMBLIES UNFEASIBLE REGION B $3,100.00 Iso-Profit Line C D FEASIBLE REGION A E X1 0 100 200 300 400 500 600 700 800 900 1000

The Clock Production Graph THE $4,000.00 ISO-PROFIT LINE FALLS OUTSIDE THE FEASIBLE REGION AND THEREFORE SUCH A PROFIT IS UNACHIEVABLE X2 PROCESSING HOURS 1000 900 800 700 600 500 400 300 200 100 LABOR HOURS ALARM ASSEMBLIES UNFEASIBLE REGION B C D $4,000.00 Iso-Profit Line FEASIBLE REGION A E X1 0 100 200 300 400 500 600 700 800 900 1000

The Surplus Variable ( S ) Units of product produced over and above the minimum required in the optimal solution. The only way it can be differentiated from a slack variable is by placing a negative (-) sign in front of it.

EXCESS OF REGULAR CLOCKS, The Surplus Variable EXAMPLE X1 => 60 regular clocks MEANS PRODUCE AT LEAST 60 REGULAR CLOCKS ( A QUOTA CONSTRAINT ) REWRITTEN AS A LINEAR EQUALITY WHERE S4 IS THE EXCESS OF REGULAR CLOCKS, IF ANY, PRODUCED IN THE OPTIMAL SOLUTION X1 – S4 = 60 regular clocks SINCE X1 = 100 REGULAR CLOCKS IN THE OPTIMAL SOLUTION, THE EXCESS PRODUCED IS 40 CLOCKS S4 = 40 regular clocks

QUOTA CONSTRAINTS ARE A NORMAL LUXURY OF MAKING WHAT THEY WANT 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 NORMAL OCCURRENCE IN THE LINEAR PROGRAMMING PROBLEM BECAUSE MOST FIRMS DO NOT ENJOY THE LUXURY OF MAKING WHAT THEY WANT

1st EXAMPLE X1 = regular clocks X2 = alarm clocks X1 + X2 = 40 PRODUCE EXACTLY 40 CLOCKS BETWEEN REGULAR AND ALARM CLOCKS PRODUCE AT LEAST 40 CLOCKS BETWEEN REGULAR AND ALARM CLOCKS X1 + X2 => 40 X1 + X2 =< 40 PRODUCE NO MORE THAN 40 CLOCKS BETWEEN REGULAR AND ALARM CLOCKS

1st EXAMPLE PLOTS X2 X1 + X2 = 40 ALL COMBINATIONS OF X1 AND X2 ARE ON THE CONSTRAINT LINE ITSELF X1 X2 X1 + X2 => 40 ALL COMBINATIONS OF X1 AND X2 ARE ON THE CONSTRAINT LINE AND ABOVE IT X1 X2 X1 + X2 =< 40 ALL COMBINATIONS OF X1 AND X2 ARE ON THE CONSTRAINT LINE AND BELOW IT X1

2nd EXAMPLE X1 = regular clocks X2 = alarm clocks X1 = 2X2 PRODUCE EXACTLY TWICE AS MANY REGULAR CLOCKS AS ALARM CLOCKS X1 = 2X2 PRODUCE AT LEAST TWICE AS MANY REGULAR CLOCKS AS ALARM CLOCKS X1 => 2X2 PRODUCE AT MOST, TWICE AS MANY REGULAR CLOCKS AS ALARM CLOCKS X1 =< 2X2

THE GENERAL CONSTRAINT 2nd EXAMPLE PLOTS X1 = 2X2 X1 => 2X2 X1 =< 2X2 ANY TWO PAIRS OF X1 AND X2 COORDINATES WILL ENABLE US TO PLOT THE GENERAL CONSTRAINT ARBITRARILY SELECT VALUES FOR X1 AND X2 SUCH THAT THE LEFT SIDE OF THE LINEAR EQUALITY EQUALS THE RIGHT SIDE. X1 = 2X2 X2 X1 = 2X2 4 = 2 ( 2 ) 6 = 2 ( 3 ) 8 = 2 ( 4 ) etc. 5 4 3 2 1 X1 0 1 2 3 4 5 6 7 8 9

3rd EXAMPLE X1 = regular clocks X2 = alarm clocks X1 = 1.75X2 PRODUCE EXACTLY 75% MORE REGULAR CLOCKS THAN ALARM CLOCKS X1 = 1.75X2 PRODUCE AT LEAST 75% MORE REGULAR CLOCKS THAN ALARM CLOCKS X1 => 1.75X2 PRODUCE AT MOST, 75% MORE REGULAR CLOCKS THAN ALARM CLOCKS X1 =< 1.75X2

THE GENERAL CONSTRAINT 3rd EXAMPLE PLOTS X1 = 1.75X2 X1 => 1.75X2 X1 =< 1.75X2 ANY TWO PAIRS OF X1 AND X2 COORDINATES WILL ENABLE US TO PLOT THE GENERAL CONSTRAINT ARBITRARILY SELECT VALUES FOR X1 AND X2 SUCH THAT THE LEFT SIDE OF THE LINEAR EQUALITY EQUALS THE RIGHT SIDE. X1 = 1.75X2 X2 X1 = 1.75X2 7 = 1.75( 4) 14 = 1.75( 8) 21 = 1.75(12) etc. 12 8 4 X1 0 7 14 21 28

Other Constraint Plots X2 FEASIBLE REGION 30 20 10 X1 = 20 on the line X1 => 20 on and right of the line X1 =< 20 on and left of the line X1 0 10 20 30 40 50 60

Other Constraint Plots X2 FEASIBLE REGION 30 20 10 X2 = 30 on the line X2 => 30 on and above the line X2 =< 30 on and below the line X1 0 10 20 30 40 50 60

Graph Linear Programming MINIMIZATION The objective function coefficients become costs. The objective function becomes “Minimize Z”. Usually more quota constraints are used. The feasible region may be unbounded. X2 FEASIBLE REGION X1 “=“ and “=>”

Linear Programming Application HUMAN RESOURCES We have a choice of three different employee candidates: un-trained, semi-trained, and highly-trained for a new position in our department. The cost of training is $5.00 per hour, $8.50 per hour, and $10.50 per hour for un-trained, semi-trained, and highly-trained respectfully. An un-trained worker requires 28 hours of training for the painting process and 35 hours for the packing process. A semi-trained worker requires 23 hours of training for the painting process 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 available for training in the painting process and 775 hours available for training in the packing process.

REQUIREMENT 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 hire Let X2 = the number of semi-trained workers to hire Let X3 = the number of highly-trained workers to hire

The Model Minimize Z = (63x$5.00)X1 + (53x$8.50)X2 + (35x$10.50)X3 SUBJECT 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

Linear Programming Application ADVERTISING The Salem Department of Tourism is developing a marketing strategy for next year. They would like to maximize the number of tourists that choose 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 100,000 people, 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 local ads.

REQUIREMENT Formulate a linear programming model that will maximize the number of people they can reach with their advertising while having at least twice as many regional ads as local ads. Let X1 = the number of regional ads to place Let X2 = the number of statewide ads to place Let X3 = the number of local ads to place

The Model Maximize Z = 100,000 X1 + 70,000 X2 + 20,000 X3 ADVERTISING COSTS EXPOSURE 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: BUDGET AT LEAST TWICE AS MANY REGIONAL ADS

QM for WINDOWS Linear Programming

WE SELECT THE LINEAR PROGRAMMING MODULE

WE WANT TO SOLVE A NEW PROBLEM

THE DIALOGUE BOX

THERE ARE THREE RESOURCE CONSTRAINTS WE WANT TO MAXIMIZE TOTAL PROFIT THERE ARE TWO PRODUCTS ( 2 REAL VARIABLES )

THE DATA INPUT TABLE

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

( produce 100 regular clocks ) ( produce 350 alarm clocks ) THE SHADOW PRICES Labor hours - $1.50 Process hours - $0.00 Alarm assemblies - $2.00 1st SOLUTION WINDOW ( produce 100 regular clocks ) ( produce 350 alarm clocks ) ( profit = $3,100.00 )

BASIC VARIABLES ARE THOSE VARIABLES > 0 NON-BASIC VARIABLES ( = 0 ) X1 = 100 X2 = 350 S2 = 500 NON-BASIC VARIABLES ( = 0 ) S1 = 0 S3 = 0

THIS IS THE SIMPLEX ALGORITHM SOLUTION TO OUR PROBLEM

THE GRAPH METHOD SOLUTION THE OPTIMAL SOLUTION IS AT THIS CORNER POINT X1 = 100 , X2 = 350 THE GRAPH METHOD SOLUTION ******* THE FEASIBLE REGION SHOWN IN PURPLE

Linear Programming

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Linear Programming

Linear Programming Graph Method Solved Problems Linear Programming Graph Method COMPUTER-BASED MANUAL

Solved Problem GRAPH METHOD OF LINEAR PROGRAMMING The Clothier Problem A clothier makes coats and slacks. The two resources are wool cloth and labor. The clothier has 150 square yards of wool and 200 hours of labor available. Each coat requires three (3) square yards of wool and ten (10) hours of labor, 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 make so that profit will be maximized. REQUIREMENT: Formulate a linear programming model for this problem. Solve the model using the graph method. Solve the model using QM for WINDOWS.

Solved Problem The Clothier Problem The Model: Let X1 = coats , Let X2 = slacks Maximize Z = 50X1 + 40X2 s.t. 3X1 + 5X2 <= 150 square yards (wool) 10X1 + 4X2 <= 200 hours (labor) X1 , X2 >= 0

Solved Problem The Clothier Problem Coordinates: Wool: X1 = 50 X1 = 0 Labor: X1 = 20 X1 = 0 X2 = 0 X2 = 30 X2 = 0 X2 = 50 X2 SLACKS 50 40 30 20 10 10X1 + 4X2 = 200 hours, labor 3X1 + 5X2 = 150 sq. yards, wool B D FEASIBLE REGION A C X1 COATS 0 10 20 30 40 50 60 70

Solved Problem The Clothier Problem Corner Point “D” Coordinate Calculations “10” ( 3X1 + 5X2 = 150) “3” (10X1 + 4X2 = 200) 30X1 + 50X2 = 1,500 30X1 + 12X2 = 600 38X2 = 900 X2 = 23.68 ≈ 23 Then: 3X1 + 5(23.68) = 150 3X1 + 118.4 = 150 3X1 = 31.6 X1 = 10.53 ≈ 10 X1 = 10 coats X2 = 23 slacks

Solved Problem Corner Point X1 coats X2 slacks Profit ( Z ) A $0.00 B The Clothier Problem Corner Point Summary and Solution Corner Point X1 coats X2 slacks Profit ( Z ) A $0.00 B 30 $1,200.00 C 20 $1,000.00 D* 10 23 $1,420.00 * OPTIMAL SOLUTION

Make 10 Coats Make 23 Slacks Total Profit = $1,420.00

FEASIBLE REGION

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Solved Problem GRAPH METHOD OF LINEAR PROGRAMMING The Copperfield Mining Company 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- 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 Problem REQUIREMENT: GRAPH METHOD OF LINEAR PROGRAMMING The Copperfield Mining Company REQUIREMENT: Formulate a linear programming model for this problem. Solve this model using the graph method. Solve this model using QM for WINDOWS.

Solved Problem GRAPH METHOD OF LINEAR PROGRAMMING The Copperfield Mining Company The Model: Let X1 = mine #1 , Let X2 = mine #2 Minimize Z = 200X1 + 160X2 s.t. 6X1 + 2X2 >= 12 tons , high-grade ore 2X1 + 2X2 >= 8 tons , medium-grade ore 4X1 + 12X2 >= 24 tons , low-grade ore X1 , X2 >= 0

Solved Problem The Copperfield Mining Company Coordinates: 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

Solved Problem The Copperfield Mining Company 8 Feasible Region 6 X2 Mine #2 8 6 4 2 Feasible Region 6X1 + 2X2 = 12 tons , high-grade ore 2X1 + 2X2 = 8 tons , medium-grade ore 4X1 + 12X2 = 24 tons , low-grade ore A B Feasible Region C D X1 Mine #1 0 1 2 3 4 5 6 7 8

Solved Problem The Copperfield Mining Company Corner Point “B” Coordinate Calculations 6X1 + 2X2 = 12 2X1 + 2X2 = 8 4X1 = 4 X1 = 1 Then: 6(1) + 2X2 = 12 2X2 = 6 X2 = 3 Corner Point “C” Coordinate Calculations 4X1 + 4X2 = 16 “2” (2X1 + 2X2 = 8) 4X1 + 12X2 = 24 -8X2 = -8 X2 = 1 Then: 2X1 + 2(1) = 8 2X1 = 6 X1 = 3

Solved Problem Corner Point X1 Mine #1 X2 Mine #2 Cost ( Z ) A 6 The Copperfield Mining Company Corner Point Summary and Solution Corner Point X1 Mine #1 X2 Mine #2 Cost ( Z ) A 6 $960.00 B* 1 3 $680.00 C $760.00 D $1,200.00 * OPTIMAL SOLUTION

Solved Problem Operate Mine # 1 for one ( 1 ) hour The Copperfield Mining Company Corner Point Summary and Solution Operate Mine # 1 for one ( 1 ) hour Operate Mine # 2 for three ( 3 ) hours Total Cost = $680.00 ( Minimize Z = 200 ( 1 ) + 160 ( 3 ) = $680.00 )

Solved Problem Ore Grade (in tons) HIGH MEDIUM LOW MINE #1 6 2 4 The Copperfield Mining Company Ore Grade (in tons) HIGH MEDIUM LOW MINE #1 6 2 4 MINE #2 36 TOTAL 12 8 40 = > 12 tons , high-grade ore = > 8 tons , medium-grade ore = > 24 tons , low-grade ore

FEASIBLE REGION

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Linear Programming Graph Method Solved Problems Linear Programming Graph Method COMPUTER-BASED MANUAL