lenier equation
DESCRIPTION
Linear Equation and EqualitiesTRANSCRIPT
Linear Programming Problem(LPP)
Describe the type of Linear Programming Problem (LPP)
Formulate a linear programming model from a description of a problem
Solve linear programming problems using the
1. Graphical method 2. Simplex method
Learning Objectives
Definition: LPP is a minimization (or maximization) problem where we are asked to minimize (or maximize) a given linear function
subject to one or more linear inequality constraints.
For example, minimized z=ax+by Where ax+by<0 x>0 or y>0
Linear Program Problem(LPP)
Objective Function: Z=c1x1+c2x2+………+ cnxn
Which is to be minimized or maximized is called the objective function of the General L.P.P.
Constraints: The inequalities are called the constraints of General L.P.P.
Non- negative restrictions: The set of inequalities is usually known as the set of non- negative restrictions of General L.P.P.
Solution: Values of unknowns x1, x2, ......, xn which satisfy the constraints of a General L.P.P. is called a solution to the General L.P.P.
Feasible Solution: Any solution to a General L.P.P. which satisfies the non-negative restrictions of the problem, called feasible solution to the General L.P.P.
Optimum Solution: Any feasible solution which optimizes (Minimize or Maximizes) the objective function of a General L.P.P. is called an optimum solution to the General L.P.P.
Graphical methodProblem 1:
Solve a linear programming problem Find the minimum value of Z=3x+8y objective function
4x+3y≥60 3x+5y≤75 constraints x, y≥ o
From 4x+3y≥60 we can find the value of X and Y as:
If X 0 15Then Y 20 0
From 3x+5y≤75 we can find the value of X and Y as:
If X 0 25then Y 15 0
Now we put the values of x and y in an x-y graph
0 5 10 15 20 25 300
5
10
15
20
25
Y-Values
x axis
y a
xis
Corner Point Objective function ValueZ=3x+8y
Remarks
A(15,0) 3*15+8*0=45
B(25,0) 3*25+8*0=75
C(6.8, 10.9)
So the optimum value of x=6.8 and y=10.9
3*6.8+8*10.9=107.66 Maximum
Evaluation of objective function
12
Example 2: A Simple Maximization Problem
Maximum z= 5x1 + 7x2
s.t. x1 < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1 > 0 and x2 > 0
Objective
Function
“Regular”Constraint
sNon-negativity Constraints
13
Example 2: Graphical Solution First Constraint Graphed
x2
x1
x1 = 6
(6, 0)
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10
Shaded regioncontains all
feasible pointsfor this constraint
14
Example 2: Graphical Solution Second Constraint Graphed
2x1 + 3x2 = 19
x2
x1
(0, 6.33)
(9 .5, 0)
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10
Shadedregion containsall feasible pointsfor this constraint
15
Example 2: Graphical Solution
Third Constraint Graphed x2
x1
x1 + x2 = 8
(0, 8)
(8, 0)
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10
Shadedregion containsall feasible pointsfor this constraint
16
Example 2: Graphical Solution
x1
x2
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10
2x1 + 3x2 = 19
x1 + x2 = 8
x1 = 6
Combined-Constraint Graph Showing Feasible Region
Feasible Region
Corner point Objective functionZ= 5x1 + 7x2 Remark
A(0,0) 5*0+7*0=0
B(0, 6.33) 5*0+7*6.33=44.31
C(5,3) 5*5+7*3=46 Maximum
D(6,2) 5*6+7*2=44
E(6,0) 5*6+7*0=30
Evaluation of objective function
Problem 3: A factory manufactures two articles A and B. To manufacture the article A, a certain machine has to be
worked for 1.5 hours and in addition a craftsman has to work for 2 hours.
To manufacture the article B, the machine has to be worked for 2.5 hours and in addition a craftsman has to work for 1.5 hours.
In a week the factory can avail of 80 hours of machine time and 70 hours of craftsman time.
The profit on each article A is Tk.5 an that on each article B is Tk.4.
If all the articles produced can be sold away, Find how many of each kind should be produced to earn
the maximum profit per week?
We need to find Data Summary
Step 1:
Decision variables
Article Hours on Profit per unitMachine Craftsman
X A 1.5 2 5
Y B 2.5 1.5 4
Hours Available(per week)
80 70
Where, x = Number of units of article A y = Number of units of article B Thus, the given problem is formulated as a
L.P.P as follows: Maximize Z= 5x+4y (Objective Function) Subject to the constraint: 1.5x+2.5y ≤80 2x+1.5y ≤70 x ≥0 , y≥0
Then for 1.5x+2.5y ≤80,
If, x 0 55.3
Then,y 32 0
& for 2x+1.5y ≤70
IF,X 0 35
Then,y 46.7 0
0 10 20 30 40 50 600
5
10
15
20
25
30
35
40
45
50
32
0
46.7
0
Y-Values
Article A
Art
icle
B
Corner Point (x, y)
Objective functionZ= 5x+4y
Value
A=(0,32) 5*0+4*32 Z(A)=128
B=(20,20) 5*20+4*20 Z(B)=180
O=(0,0) 5*0+4*0 Z(O)=0
C=(0,35) 5*35+4*0 Z(C)=175
For point B (20, 20) the objective value is maximum. So the solution is for
article A: 20 units and and article B: 20 units to earn the maximum profit per week
Problem 4: A firm makes two types of furniture: chairs and tables.
The profit for each product as calculated by the accounting department is Tk. 20 per chair and Tk. 30 per table.
Both products are processed on three machines M1,M2, M3. The time required in hours by each product and total time is available in hours per week on each machine as follows:
How should the manufacturer schedule his production in order to maximize profit?
Machine Chair Table Available time
M1 3 3 36
M2 5 2 50
M3 2 6 60
Let x= numbers of chairs y= numbers of tables In this case, the objective function is Maximize Z=20x+30y
And the constraint equations are
3x+3y≤365x+2y≤502x+6y≤60
Solution:
For 3x+3y≤36 the graph point will be
For 5x+2y≤50 the graph will be
X 0 12
Y 12 0
X 0 10
Y 25 0
For 2x+6y≤60 the graph will be
X 0 30
Y 10 0
Corner Point Objective functionZ=20x+30y
Values of z
A(0,10) 20*0+30*10 =300
B(3,9) 20*3+30*9 =330
C(8.6,3.3) 20*8.6+30*3.3 =273.3
D(10,0) 20*10+30*0 =200
Evaluation of objective function
So the maximum value of z is 330, and this occurs when x=3 and y=9.
so the manufacturer should use 330 units to maximize the production.
Problem 5. A company produces two articles X and Y.There are two departments through which the articles are processed. That is assembling and finishing.
The potential capacity of the assembling department is 60hours a week and that of the finishing department is 48 hours a week.
Production of one unit of X requires 4 hours in assembling and 2 hours in finishing.
Each of the unit Y requires 2 hours in assembly and 4 hours in finishing.
If profit is tk.8 for each unit of X and Tk.6 for each unit of Y,
Find out the number of units of X and Y to be prepared each week to give maximum profit??
Products Time required(one unit)
Total hours available
X Y
Assembly department
4 2 60
Finishing Department
2 4 48
Profits per unit
8 6
Solution:
Objective Function: Z= 8a+6b Subject to constraints: 4a+2b ≤ 60 2a+4b ≤ 48 Non-negativity requirement: X≥0,Y≥0
Then for 4a+2b ≤60,,
X 0 15
y 30 0
& for 2a+4b ≤48
X 0 24Y 12 0
0 5 10 15 20 25 300
5
10
15
20
25
30
35
Y-Values
X
Y
CornerPoint(x, y)
Objective functionZ= 8a+6b
Value
A=(0,0) 8*0+6*0 Z(A)=0
B=(15,0) 8*15+6*0 Z(B)=120
C=(12,6) 8*12+6*6 Z(C)=132
D=(0,12) 8*0+6*12 Z(D)=72
For point C (12, 6) the objective function z is maximum. So the solution is to get the maximum produced article, article x must produce 12 units and article y must produce 6 unit
1. A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day.
Because of limitations on production capacity, no more than 200 scientific and 170 graphing calculators can be made daily.
To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day.
If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits?
Real Life problems
You need to buy some file cabinets. You know that Cabinet X costs $10 per unit, requires 6 square feet of floor space, and holds 8 cubic feet of files.
Cabinet Y costs $20 per unit, requires 8 square feet of floor space, and holds 12 cubic feet of files.
You have been given $140 for this purchase, though you don't have to spend that much.
The office has room for no more than 72 square feet of cabinets.
How many of which model should you buy, in order to maximize storage volume?
What is the maximum storage?
Quiz test 3