lpp
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Abstract
A linear programming problem is simply an optimization problem. (A maximization or
minimization problem). A large number of optimization problems fall into the subcategory
linear programming problems. In order to solve a real world mathematical problem using
linear programming, first of all the wording or problem statement in words should be
translated into a set of inequalities or equations. It should have certain constraints for a
solution to exist. Also some assumptions will have to be made when formulating the
mathematical model for a LPP. It is possible to represent a linear programming problem
geometrically and solve it while also using a method named Simplex method. In some cases
even MatLAB is used to find solutions for LPP. What needed to be known of a LPP are
• What the components of the problem are-objective and variables affecting the
problem.
• How to formulate the problem-is it a minimization or maximization problem?
• What assumptions are underlying
• How to find a solution to a 2-dimensional problem graphically.
• How to find a solution using simplex method
The prerequisites for solving a LPP are knowledge of geometrical representation and the
knowledge in solving linear equations and linear inequalities.
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Problem description
A company manufactures and sells two models of lamps L1, L2 the profit being $15 and $10,
respectively. The process involves two workers W1 and W2 who are available for this kind of
work 100 and 80 hours per month, respectively. W1 assembles L1 in 20 minutes and L2 in 30
minutes. W2 paints L1 in 20 minutes and L2 in 10 minutes. It is assumed for this problem that
all lamps made can be sold without difficulty. The objective is to find the figures of L 1 and L2
for which the maximum profit is obtained.
This problem is a common problem for production departments. Managers need to determine
how much products they need to manufacture for a month in order to get the maximum profit,
while utilizing the presently available resources (both material and human resources).
The questions we need to answer for this problem is how many L1 and L2 need to be produced
for the maximum profit.
Simplification
Take that the quantity of L1 for maximum profit as also L1 and similarly quantity of L2 for
maximum profit is L2. Taking the profit as f (L1, L2) the equation for profit can be written as
f ( L1,L2) = ( unit profit of L1* L1) + (unit profit of L2* L2) PROFIT
Assumption: Here it is assumed that all manufactured number of lamps is sold by the end of
the month.
This assumption is made in order to further simplify this mathematical model.
f (L1, L2) = $ (15* L1 + 10*L2) -----------------------------------------(A)
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Considering the work done by worker W1 and the time constraints for that worker we can
identify that the time taken for his work possibly won’t exceed 100 hours per month.
(20 minutes * L1)+ (30 minutes*L2) ≤ (100* 60 minutes) PROCESSING TIME
20*L1 + 30*L2 ≤ 6000 -------------------------------------------------- (B)
Next considering the W2 worker and his time constraints,
(20 minutes * L1) + (10 minutes * L2) ≤ (80 * 60 minutes) PROCESSING TIME
20*L1 + 10*L2 ≤ 4800---------------------------------------------------- (C)
Also as the number of L1 and L2 cannot be lesser than zero anyway the constraints for these
conditions are,
L1 ≥0--------------------------------------------------------------------------- (D)
L2 ≥0--------------------------------------------------------------------------- (E)
General Statement of the Linear Programming Problem:
Objective Function:
Maximize f (L1, L2) = $ (15* L1 + 10*L2) ----------------------------------------- (A)
Constraint Equations:
Subject to:
20*L1 + 30*L2 ≤ 6000 -------------------------------------------------- (B)
20*L1 + 10*L2 ≤ 4800---------------------------------------------------- (C)
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L1 ≥0--------------------------------------------------------------------------- (D)
L2 ≥0--------------------------------------------------------------------------- (E)
The number of lamps sold can be different in the practical world. However by assuming that
all the lamps manufactured are sold there is no need to calculate the percentage of lamps sold.
There is no record about that. So this assumption is justified. Also since the number of lamps
sold of each type can never be lesser than zero the constraints (D) and (E) are justified.
Mathematical Model
The standard representation of a linear programming problem is
Maximize/minimize OBJECTIVE FUNCTION
Subject to CONSTRAINTS
Since in our problem we have only two variables number of L1 lamps = x and number of L2
lamps= y, rewriting the simplified problem in the form of a Linear Programming Problem
Model we get
Maximize f (x, y) = $ (15* x + 10*y) ----------------------------------------- (A)
Subject to:
20*x + 30*y ≤ 6000 -------------------------------------------------------------- (B)
20*x + 10*y ≤ 4800--------------------------------------------------------------- (C)
x ≥0---------------------------------------------------------------------------------- (D)
y≥0----------------------------------------------------------------------------------- (E)
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Solution of the Mathematical Problem
Graphical Method
In the graphical method following steps are followed.
Convert the constraint inequalities into equations. (Simply replace the unequal sign with equal
sign)
a. 20*x + 30*y =6000 -------------------------------------------------------------- (p)
b. 20*x + 10*y = 4800--------------------------------------------------------------- (q)
c. x =0---------------------------------------------------------------------------------- (r)
d. y=0----------------------------------------------------------------------------------- (s)
Draw the graphs for these linear equations on the x, y coordinate system.
Figure 1:- Graphs consisting the constraint equations
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Feasible region
2. Identify the feasible area for the solution. In this case the shaded region contains the
solution for the possible values for objective function. However in order to determine
the maximum value for the objective function, the corner points should be identified.
Corner points are points at which each of above graphs intersect each other.
Corner points:
a. The corner point at which equation p and q intersect, A is obtained by solving
the two graph equations parallel.
20*x + 30*y =6000 -------------------------------------------------------------- (p)
20*x + 10*y = 4800--------------------------------------------------------------- (q)
In matrix form the parallel equation can be written as,
=
By solving above equation we get values for x and y as,
x=210
y=60
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20 30
20 10
6000
4800
x
y
Figure 2: Corner point A found
Similarly the other four corner points at which the rest of the three graphs intersect can be
calculated.
Corner pint D at which p and r equations intersect y=0, x=200 D= (200, 0)
Corner point at C= (0, 0) and Corner point at B= (0,240)
Next by calculating the value of the objective function at the corner points the maximum and
minimum values for the object function can be found.
Results
Table 1:- Value of objective function at the corner points
Corner Points
Objective Function value
($)
(0,0) 0
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A= (210, 60)B
C D
(210,60) 3750
(200,0) 3000
(0,240) 2400
According to Table 1 It can be seen that the maximum value for the objective function is at
Corner point (210, 60), at which the objective function value i.e. Maximum profit of company
is $3750.
Therefore from this graphical method the figures of L1 and L2 for which the maximum profit
can be obtained is calculated as
Number of L1(x) = 210
Number of L2(y) = 60
Simplex method
In simplex method first of all the constraint inequalities of the problem statement should be
rewritten as equations by introducing a slack variables. Slack variables are the variables that
“pick up” the extra value needed to convert the inequalities into equalities. So by taking the
slack variables as u and v we rewrite the problem statement as,
Objective Function:
Maximize z = 15* x + 10*y ----------------------------------------- (A)
Number of L1=x, number of L2=y
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Subject to:
20*x + 30*y + u = 6000 -------------------------------------------------- (B)
20*x + 10*y + v = 4800---------------------------------------------------- (C)
x ,y, u, v ≥0--------------------------------------------------------------------------- (D)
Next we have to form the simplex tableau. For that we have to write above system of linear
equations in 3 rows as following.
Z - 15* x + 10*y = 0 Row 0
20*x + 30*y + u = 6000 Row 1
20*x + 10*y + v = 4800 Row 2
These equations have to be solved in terms of basic variables and non-basic variables. The
variables (other than the special variable z) which appear in only one equation are the basic
variables. Other variables except z are non-basic variables.
Basic variables= u, v
Non-Basic variables= x, y
A basic solution is obtained from the system of equations by setting the non-basic variables to
zero. This gives
x=0
y=0
u=6000
v=4800
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The simplex tableau
Next according to the first rule of Simplex method the row 0 has to be observed carefully. If If
all variables have a nonnegative coefficient in Row 0, the current basic solution is optimal. If
not, for speedy convergence the most negative coefficient is chosen. In this example as the
row zero contains negative coefficients. Therefore this basic solution IS NOT optimal. So we
pick the most negative coefficient -15. We call the variable containing this coefficient as the
‘entering variable’. In this case it is x. The idea is to pivot in order to make the non-basic
variable x become a basic variable. When choosing the pivot element first the values in the
right most column should be divided by the relevant pivot column element and the pivot
column element which gives the smallest value has to be chosen. This is the second rule of the
simplex method.
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x y u v z |
20 30 1 0 0 | 6000
20 10 0 1 0 | 4800
-15 -10 0 0 1 | 0 Row 0
Row 1
Row2
x y u v z |
20 30 1 0 0 | 6000
20 10 0 1 0 | 4800
-15 -10 0 0 1 | 0
6000/20= 300
4800/20= 240
Hence the coefficient 20 in the 2nd row is chosen as the pivoting element. And pivoting around
that value gives us the following solution.
Next we again check the row 0 for any negative coefficients according to first rule. In this
case again row 0 has a -2.5 negative coefficient. So the column y is our new pivot column.
Again applying the second rule we divide the right most column values by the pivot column
values and select the pivot element for which the above division is smallest.
Thus we can see that the value 20 in row 1 is our new pivot element. Therefore pivoting
around that element we obtain the following solution.
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x y u v z |
0 20 1 -1 0 | 1200
1 0.5 0 0.05 0 | 240
0 -2.5 0 0.75 1 | 3600
x y u v z |
0 20 1 -1 0 | 1200
1 0.5 0 0.05 0 | 240
0 -2.5 0 0.75 1 | 3600
1200/20= 60
240/0.5= 480
x y u v z |
0 1 0.05 -0.05 0 | 60
1 0 -0.025 0.075 0 | 210
0 0 0.125 0.625 1 | 3750
Now we can see that the last row (row 0) doesn’t have any negative coefficients.
Hence this solution is optimal according to the first rule of simplex method. Therefore as we
can see the variables x and y are now basic. Whereas, the variables u and v are now non-basic
variables.
All the above computations can be represented very compactly in tableau form .
Table 2: compact solution tableau
z x y u v RHS Basic solution
1 -15 -10 0 0 0 Basic u=6000,v=4800
0 20 30 1 0 6000 Non basic x=0 y=0
0 20 10 0 1 4800 Z=0
1 0 -2.5 0 0.75 3600 Basic x=240,u=1200
0 0 20 1 -1 1200 Non basic y=0 v=0
0 1 0.5 0 0.05 240 Z=3600
1 0 0 0.125 0.625 3750 Basic x=210,y=60
0 0 1 0.05 -0.05 60 Non basic u=0 v=0
0 1 0 -0.025 0.075 210 Z=3750
Results
From the simplex method by two pivoting steps we obtain the optimum solution for the
objective function which is the maximum monthly profit of the lamp manufacturing company.
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Also the production figures for L1 and L2 is also finally obtained as the basic variables of the
simplex method as x=210 and y=60.
It means the number of L1 lamps and number of L2 lamps needed to get the maximum profit
is 210 and 60 respectively.
In both methods we followed this result was obtained without any discrepancies.
The problem we solved has a feasible solution. It had a feasible region for solutions even
when the graphs of constraints were plotted. However there are cases in which a solution
cannot be founded for linear programming problems. The reasons for this are
1. The linear program is unbounded: this happens when there is no positive entry in the
entry variable column. So there can be no ratio calculated at the RHS column. Hence a
optimum solution is not possible to find either graphically or by simplex method.
2. The program is degenerate: In the Simplex Method there is an improvement in the
objective function in each step as the algorithm converges to the optimum. However, a
situation can arise where there is no improvement in the objective function from an
application of the algorithm, and this is referred to as degeneracy. Also there is a
possibility that cycling could occur, and the optimum would not be reached. This
occurs when the basic solution has a basic variable with the coefficient of zero. By
pivoting the same basic solution will be obtained with interchanged basic variable
values.
In order to a solution to exist the following conditions should be satisfied
1. The LPP must have a unique optimum solution : the basic solution itself
2. The LPP must have an alternative optimum solution: by the above two methods the
alternative optimum solution can be obtained
Improvements
These two methods are the frequently used methods for solving linear programming
problems. Apart from these two methods MatLAB can be used. By using MatLAB the
calculations can done using the optimization toolbox in several specifies iteration.
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In MatLAB the syntax linprog solves linear programming problems.
x = linprog(f,A,b) solves min/max f'*x such that A*x ≤ b. where A is the inequality matrix, b is a vector.
Conclusions
In this report we have presented how to solve a real world optimization problem using linear
programming model. Linear programming problems are either a maximization problem or
minimization problem. In this case the problem we chose is a maximization problem. To
obtain the maximum value of the objective function of our Model we followed two well
known mathematical methods. Namely, the graphical Method and the Simplex tableau
method. Both methods yielded the same values for the variables in order to get the optimum
solution. We followed both methods step by step by describing each step simply so that
anyone can follow these to solve any linear programming problem. Further we have
mentioned that by referring to the graphs of the constraints or by referring to the basic
solution of the simplex tableau without calculating we can determine whether any LPP has a
solution, if the solution is Feasible, or if the problem’s solution is unbound or degenerate.
The conditions for a LPP to have a solution is also mentioned in this report, after doing some
research on it.
References
© 1984-2011- The Math Works, Inc, Product Documentation,linprog [online]
http://www.mathworks.in/help/toolbox/optim/ug/f10534.html
[Accessed 09 November 2011]
Linear Programming. Chapter 4 [online]
http://www.mpri.lsu.edu/textbook/Chapter5.htm
[Accessed 09 November 2011]
Thomas S. Ferguson,2008, LINEAR PROGRAMMING,A Concise Introduction,
[pdf]
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Walter J. Mayer,1984, Concepts of Mathematical Modeling,[pdf] Dover
Publications,Inc,Mineola, NY
Jennifer Whitfield,Linear Programming, Sample Simplex Problem,[pdf],College Of
Science IT LAB.
Robert J. Vanderbei ,October 17, 2007, Linear Programming: Chapter 2 ,The
Simplex Method, Operations Research and Financial Engineering ,Princeton University,Princeton, NJ 08544http://www.princeton.edu/_rvdb
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