7/29/2019 Graphical Method of LP
1/16
LINEAR PROGRAMMING
Boni Sena,S.T.,M.Eng.
7/29/2019 Graphical Method of LP
2/16
WHATISLINEARPROGRAMMING ?
How much fuelthat we need ?
How many tirewhich can be carried on ?
7/29/2019 Graphical Method of LP
3/16
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
7/29/2019 Graphical Method of LP
4/16
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)
7/29/2019 Graphical Method of LP
5/16
LINEAR PROGRAMMING (STUDY CASE)
7/29/2019 Graphical Method of LP
6/16
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
7/29/2019 Graphical Method of LP
7/16
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
7/29/2019 Graphical Method of LP
8/16
x > ax < a
x = a
See the graph below :
7/29/2019 Graphical Method of LP
9/16
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
7/29/2019 Graphical Method of LP
10/16
y x
Solution satisfy
y > -2x +3
y > 1/2
THEANSWERIS..
2 x + 3 y > 3
x y
2 y 1 > 0
7/29/2019 Graphical Method of LP
11/16
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
7/29/2019 Graphical Method of LP
12/16
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
7/29/2019 Graphical Method of LP
13/16
40
80
120
160
40 80 160120
2x + y = 180
FeasibleRegion
x + y = 100
x + 2y = 160
Find the feasible region
AB
C
D
7/29/2019 Graphical Method of LP
14/16
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
7/29/2019 Graphical Method of LP
15/16
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
7/29/2019 Graphical Method of LP
16/16
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)
Top Related