Graphical Method of LP

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    LINEAR PROGRAMMING

    Boni Sena,S.T.,M.Eng.

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    WHATISLINEARPROGRAMMING ?

    How much fuelthat we need ?

    How many tirewhich can be carried on ?

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    LINEAR PROGRAMMING (STUDY CASE)

    Suppose a consumer has income $60 per week. Hewants to buy food A costs $2 per kilogram and food Bcosts $3 per kilogram. Arrange linear programming forthis problem !

    First step : Make table which describe the problem.You have to determine parameter or decision variable.Cost become parameter in this problem. See the table

    below :

    Parameter A B MaximumCapacity

    Cost $2 $3 $60

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    LINEAR PROGRAMMING (STUDY CASE)

    Second Step : State the table in the form ofmathematical expressions. Assume the consumerwants to use all of his budget.

    2 x + 3 y = 60 (where x 0 and y 0)

    Third step : Make graph from the math expressions.(Cartesian coordinates)

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    LINEAR PROGRAMMING (STUDY CASE)

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    LINEAR INEQUALITY

    Definition

    A linear inequality in the variables x and y is an inequality that canbe written in one of the form

    ax + by + c < 0 ,

    ax + by + c > 0,

    ax + by + c 0 ,

    ax + by + c 0

    a, b and c areconstant anda and b 0

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    See the graph below :

    y = mx + b

    y < mx + b

    y > mx + b

    Below the line,all points (x,y)

    satisfyy < mx+b

    Above the line,

    all points (x,y)satisfy

    y > mx+b

    the line itself,all points (x,y)

    satisfyy = mx+b

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    x > ax < a

    x = a

    See the graph below :

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    SYSTEMOF INEQUALITIES

    The solution of system of inequalities consists of allpoints whose coordinates

    Solve the system

    Draw the equation on the graph

    2 x + 3 y > 3

    x y

    2 y 1 > 0

    y > -2x + 3

    y x

    y > 1/2

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    y x

    Solution satisfy

    y > -2x +3

    y > 1/2

    THEANSWERIS..

    2 x + 3 y > 3

    x y

    2 y 1 > 0

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    LinearProgramming

    Constraints2 x + 3 y > 32 y 1 > 0

    Objective functionP (x) = ax + by

    Feasible points

    Solution

    Maximize / minimize

    Not only find the feasible region but alsoFind the way to maximize/minimize the

    Objective function

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    STUDY CASEA company produces two types of can openers : Manual

    and electric. Each require in its manufacture the use ofthree machines : A, B and C. Table 7.1 gives data relatingto the manufacturer of these can openers.

    Manual Electric HoursAvailable

    A 2 hr 1 hr 180B 1 hr 2 hr 160C 1 hr 1 hr 100Profit/Unit $4 $6

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    40

    80

    120

    160

    40 80 160120

    2x + y = 180

    FeasibleRegion

    x + y = 100

    x + 2y = 160

    Find the feasible region

    AB

    C

    D

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    40

    80

    120

    160

    40 80 160120

    FeasibleRegion

    Objective function (P) : 4 x + 6yy = - 2/3 x + P/6y = - 2/3 x + 100

    Try with P = 600Has no point in feasible region

    y = - 2/3 x + 50Try with P = 300

    AB

    Has infinitely many such points

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    40

    80

    120

    160

    40 80 160120

    FeasibleRegion

    Objective function (P) : 4 x + 6yy = - 2/3 x + P/6y = - 2/3 x + 100

    Try with P = 600Has no point in feasible region

    y = - 2/3 x + 50Try with P = 300

    AB

    Has infinitely many such points

    C

    DIsoprofit lines

    Maximum profit line

    x + 2y = 160x + y = 100

    The line whose y intercept is farthest from the origin and thathas at least one point in common with feasible region

    x = 40 and y = 60

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    40

    80

    120

    160

    40 80 160120

    FeasibleRegion

    Objective function (P) : 4 x + 6y

    AB

    C

    D(40,60)

    (90,0)(80,20)

    (0,80)

    P =4 x + 6y = 4(90)+6(0) = 360P =4 x + 6y = 4(80)+6(20) = 440P =4 x + 6y = 4(40)+6(60) = 520P =4 x + 6y = 4(0)+6(80) = 480

    Thus, P has maximum value at (40,60)