BY DR. IRENE P. SOLANO

INTRODUCTIONTHE COMPANY S PROFIT AND MINIMIZING ITS COST WITHOUT VIOLATING ANY LIMITATION OR RESTRICTION ON EXISTING VARIABLES, SUCH AS TIME AND QUANTITY OF AVAILABLE RAW MATERIALS IS ONE OF THE MANY GOALS OF A GOOD MANAGER y LINEAR PROGRAMMING IS A VITAL TOOL IN ACHIEVING THIS GOAL.y MAXIMIZING

DEFINITIONy LINEAR IMPLIES DIRECT PROPORTIONALITY OF RELATIONSHIP OF VARIABLES y PROGRAMMING MEANS MAKING SCHEDULES OR PLANS OF ACTIVITIES TO UNDERTAKE IN THE FUTURE y THEREFORE, LINEAR PROGRAMMING IS PLANNING BY THE USE OF LINEAR RELATIONSHIP OF VARIABLES INVOLVED. y LP MAKES USE OF CERTAIN MATHEMATICAL TECHNIQUES TO GET THE BEST POSSIBLE SOLUTION TO A PROBLEM INVOLVING LIMITED RESOURCES

A MINIMIZATION OR MAXIMIZATION PROBLEM CAN BE CLASSIFIED AS LINEAR PROGRAMMING IF IT HAS THE FOLLOWING PROPERTIES: 1) THE OBJECTIVE OF THE DECISION MAKER MUST BE EITHER TO MAXIMIZE OR MINIMIZE 2) THE ACTIONS OF THE DECISION MAKER MUST BE CONSTRAINED AND THE DECISION VARIABLES MUST NOT VIOLATE THE LIMITATIONS OR CONSTRAINTS

3) ALL VARIABLES HAVE A VALUE GREATER THAN OR EQUAL TO ZERO 4) THE RELATIONSHIP OF VARIABLES COULD BE EXPRESSED IN TERMS OF EQUATIONS OR INEQUALITIES

2 WAYS OF SOLVING A LINEAR PROGRAMMINGy BY GRAPHICAL METHOD (CAN ONLY BE USED IF

THE PROBLEM HAS 2 OR 3 VARIABLES)y BY

SIMPLEX METHOD (CAN HANDLE A PROBLEM HAVING ANY NUMBER OF VARIABLES)

THE GRAPHICAL METHODy MAKES USE OF GRAPHS TO ARRIVE AT THE

OPTIMUM SOLUTIONy OPTIMUM SOLUTION IS A SOLUTION THAT

MAKES THE OBJECTIVE FUNCTION AS LARGE AS POSSIBLE IN THE CASE OF MAXIMIZATION PROCESS AND AS SMALL AS POSSIBLE IN THE CASE OF MINIMIZATION PROCESS

y THE

SET OF ALL POINTS IN THE GRAPH SATISFYING THE CONSTRAINT IS CALLED THE FEASIBLE SOLUTION, AND THESE POINTS ARE LOCATED IN THE FEASIBLE REGION

2 MAIN PARTS OF A LINEAR PROGRAMy THE OBJECTIVE FUNCTION (INTRODUCED BY

THE WORD MAXIMIZE OR MINIMIZE )y THE

CONSTRAINTS OR LIMITATIONS (INTRODUCED BY THE WORD SUBJECT TO )

CONSTRAINTSy EXPRESSED IN EQUATIONS OR INEQUALITIES y 2 PARTS OF THE CONSTRAINTS

1) EXPLICIT CONSTRAINTS CONDITIONS OF THE PROBLEMS WHICH ARE TO BE EXPRESSED IN MATHEMATICAL SENTENCES 2) IMPLICIT ARE THOSE THAT ARE IMPLIED EX. THE QUANTITY OF TIME OR RAW MATERIAL AS A VARIABLE IN A GIVEN PROBLEM IS ALWAYS POSITIVE

STEPS TO FOLLOW IN USING THE GRAPHICAL METHOD1. 2. 3. 4. 5. 6. 7.

TABULATE THE GIVEN DATA REPRESENT THE UNKNOWN IN THE PROBLEM FORMULATE THE OBJECTIVE FUNCTION AND CONSTRAINTS GRAPH THE CONSTRAINTS SOLVE FOR THE COORDINATES AT THE POINT OF INTERSECTION OF LINES FIND THE OPTIMUM SOLUTION (SUBSTITUTE THE COORDINATES AT THE VERTICES OF THE FEASIBLE REGION IN THE OBJECTIVE FUNCTION) FORMULATE YOUR DECISION BY SELECTING THE HIGHEST VALUE OF THE OBJECTIVE IF IT IS MAXIMIZATION AND THE LOWEST VALUE IF IT IS MINIMIZATION

Example 1A FURNITURE COMPANY MAKES TWO PRODUCTS: TABLES AND CHAIRS WHICH MUST BE PROCESSED THROUGH ASSEMBLY & FINISHING DEPARTMENTS: ASSEMBLY DEPARTMENT IS AVAILABLE FOR 60 HOURS IN EVERY PRODUCTION PERIOD WHILE THE FINISHING DEPARTMENT IS AVAILABLE FOR 48 HOURS OF WORK. MANUFACTURING ONE TABLE REQUIRES 4 HOURS IN THE ASSEMBLY AND 2 HOURS IN THE FINISHING. EACH CHAIR REQUIRES 2 HOURS IN THE ASSEMBLY AND 4 HOURS IN THE FINISHING. ONE TABLE CONTRIBUTES P180 TO PROFIT WHILE A CHAIR CONTRIBUTES P100. THE PROBLEM IS TO DETERMINE THE NUMBER OF TABLES AND CHAIRS TO MAKE PER PRODUCTION PERIOD IN ORDER TO MAXIMIZE THE PROFIT.

1. Tabulate the given dataProduct Tables Chairs Availability Assembly Department 4 hrs 2 hrs 60 hrs Finishing Department 2 hrs 4 hrs 48 hrs Profit P180 100

2. Represent the unknown in the problemLet x be the number of tables to make per production period in order to maximize the profit y be the number of chairs to make per production period in order to maximize the profit

3. Formulate the objective function and constraintsObjective Function Maximize the profit 180x + 100y Constraints Subject to (Explicit Constraints) 4x + 2y 60 hrs (Availability of Assembly Dept) 2x + 4y 48 hrs (Availability of Finishing Dept)

(Implicit Constraints) x 0 ( value of x, number of tables ) y 0 ( value of y, number of chairs)

4. Graph the constraints4x + 2y 60 2x + 4y 48 x0 y0 1. Graph x 0 2. Graph y 0

3. Graph 4x + 2y 60 Replace inequality symbol with equal sign 4x + 2y = 60 Find the intercepts and test point (0, 0) Let x = 0 Let y = 0 Test Point (0, 0) 4(0) + 2y = 60 4x + 2(0) = 60 4x + 2y 60 2y = 60 4x = 60 4(0) + 2(0) 60 y = 30 x = 15 0 60 (0, 30) (15, 0) True

4. Graph 2x + 4y 48 Replace inequality symbol with equal sign 2x + 4y = 48 Find the intercepts and test point (0, 0) Let x = 0 Let y = 0 Test Point (0, 0) 2(0) + 4y = 48 2x + 4(0) = 48 2x + 4y 48 4y = 48 2x = 48 2(0) + 4(0) 48 y = 12 x = 24 0 48 (0, 12) (24, 0) True

feasible region is the intersection of the planes (shaded area)

5. Solve for the coordinates at the point of intersection of the lines4x + 2y 60 2x + 4y 48 4x + 2y = 60 2x + 4y = 48 equation 1 equation 2 Multiply equation 2 by 2 2 (2x + 4y = 48) 4x + 8y = 96 new equation 2 Equation 1 minus New Equation 2 4x + 2y = 60 4x + 8y = 96 (change the signs)(-) (-) (-)

-6y = -36 y=6

Solve for x (substitute y = 6 in equation 1) 4x + 2y = 60 equation 1 4x + 2(6) = 60 4x + 12 = 60 4x = 60 12 4x = 48 x = 12 Hence, the coordinates at the point of the intersection of the lines is (12, 6)

6. Find the optimum solutionCoordinates (0,12) (15,0) (12, 6) Maximize Profit 180x + 100y 180(0) + 100(12) = P1200 180(15) + 100(0) = P 2700 180(12) + 100(6) = P2760

7. Formulate the decisionSince (12, 6) produces the highest value for profit, we should make 12 tables and 6 chairs per production chair in order to maximize the profit at P2760.

Check whether the solution set is correct (optional)Explicit Constraints 4x + 2y 60 hrs (Availability of Assembly Dept) 4(12) + 2(6) 60 hrs 48 + 12 = 60 satisfied this constraint 2x + 4y 48 hrs (Availability of Finishing Dept) 2(12) + 4(6) 48 24 + 24 = 48 satisfied this constraint Implicit Constraints x0 x = 12 satisfied this constraint y 0 ( value of y, number of chairs) y = 6 satisfied this constraint

Example 2A SMALL GENERATOR BURNS TWO TYPES OF FUEL: LOW SULFUR AND HIGH SULFUR TO PRODUCE ELECTRICITY. FOR ONE HOUR, EACH GALLON OF LOW SULFUR EMITS 3 UNITS OF SULFUR DIOXIDE, GENERATES 4 KILOWATTS ELECTRICITY AND COSTS P160. EACH GALLON OF HIGH SULFUR EMITS 5 UNITS OF SULFUR DIOXIDE, GENERATES 4 KILOWATTS AND COSTS P150. THE ENVIRONMENTAL PROTECTION AGENCY INSISTS THAT THE MAXIMUM AMOUNT OF SULFUR DIOXIDE THAT CAN BE EMITTED PER HOUR IS 15 UNITS. SUPPOSE THAT AT LEAST 16 KILOWATTS MUST BE GENERATED PER HOUR, HOW MANY GALLONS OF LOW SULFUR AND HIGH SULFUR MUST BE UTILIZED PER HOUR IN ORDER TO MINIMIZE THE COST OF FUEL?

1. Tabulate the given dataTypes of Fuel Allowable Sulfur Dioxide Emission 3 units 5 units 15 units Required Electricity Generation 4 kilowatts 4 kilowatts 16 kilowatts Cost

Low Sulfur High Sulfur Constraints

P160 P150

2. Represent the unknown in the problemLet x = the no. of gallons of low sulfur that must be utilized y = the no. of gallons of high sulfur that must be utilized

3. Formulate the objective function and constraintsObjective Function Minimize the cost 160x + 150y Constraints Subject to (Explicit Constraints) 3x + 5y 15 units (Allowable Sulfur Dioxide Emission) 4x + 4y 16 kilowatts (Required Electricity Generation)

(Implicit Constraints) x 0 ( value of x or the number of gallons of low sulfur is zero or greater than zero meaning positive value) y 0 ( value of x or the number of gallons of high sulfur is zero or greater than zero meaning positive value)

4. Graph the constraints3x + 5y 15 4x + 4y 16 x0 y0 1. Graph x 0 2. Graph y 0

3. Graph 3x + 5y 15 Replace the inequality symbol with equal sign 3x + 5y = 15 Find the intercepts and test point (0, 0) Let x = 0 Let y = 0 Test point (0,0) 3(0) + 5y = 15 3x + 5(0) = 15 3x + 5y 15 5y = 15 3x = 15 3(0) + 5(0) 15 y=3 x=5 0 15 (0,3) (5,0) True

4. Graph 4x + 4y 16 Replace the inequality symbol with equal sign 4x + 4y =16 Find the intercepts and test point (0, 0) Let x = 0 Let y = 0 Test point (0,0) 4(0) + 4y = 16 4x + 4(0) = 16 4x + 4y 16 4y = 16 4x = 16 4(0) + 4(0) 16 y=4 x=4 0 16 (0,4) (4,0) Not True

5. Solve for the coordinates at the point of intersection of the lines3x + 5y = 15 4x + 4y = 16 4(3x + 5y = 15) 3(4x + 4y = 16) 12x + 20y = 60 12x + 12y = 48 (-) (-) (-) _____________ 8y = 12 y = 12/8 = 3/2 or 1.5

To find the value of x, substitute y = 3/2

3x + 5y = 15 3x + 5 (3/2) = 15 2[3x + 5(3/2) = 15] 6x + 15 = 30 6x = 30 15 6x = 15 x = 15/6 or 2.5 Therefore, the coordinates of the intersection of the lines are (2.5, 1.5)

in

6. Find the optimum solutionCoordinates (4,0) (5,0) (2.5, 1.5) Minimize Cost 160x + 150y 160 (4) + 150 (0) = P640 160 (5) + 150 (0) = P800 160 (2.5) + 150 (1.5) = P625

7. Formulate the decisionSince (2.5, 1.5) produces the lowest value for cost, we should utilize 2.5 gallons of low sulfur and 1.5 gallons of high sulfur in order to minimize the cost at P625

EvaluationA FIRM MANUFACTURES 2 PRODUCTS, A AND B. EACH PRODUCT IS PROCESSED BY 2 MACHINES, M1 AND M2. EACH UNIT OF TYPE A REQUIRES 1 HOUR OF PROCESSING BY M1 AND 2 HOURS IN M2 AND EACH UNIT OF TYPE B REQUIRES 3 HOURS ON M1 AND 1 HOUR ON M2. THE PROFIT ON PRODUCT A IS P20 PER UNIT AND ON PRODUCT B IS P30 PER UNIT. IF M1 IS AVAILABLE FOR 200 HOURS EACH MONTH AND M2 FOR 300 HOURS, HOW MANY UNITS OF EACH TYPE CAN BE MANUFACTURED IN ONE MONTH IN ORDER TO MAXIMIZE THE PROFITS?