Transcript Slide 1

1
2
A linear program in which all the variables are non-negative and all the
constraints are equalities is said to be in standard form (or augmented
form)
LP standard form properties:
1) All the constraints (with the exception of the nonnegativity
restrictions on the variables) are equations with
nonnegative.
2) All the variables are nonnegative.
3) The objective function may be of the maximization or the
minimization type.
3
1- Conversion of inequalities into equation.
a) < constraints: Add slack variable to constraint (coefficient of 0 in
obj function)
Example:
x
1
 2x2  3
x
1
s
1
 2x2  s1  3
0
b) > constraints: Subtract surplus from constraint (coefficient of 0 in
obj function)
3x
1

x
2
5
3x
1
s
1
0

x
2
 S1  5
:
1- The right hand side of an equation can always be made
nonnegative by multiplying the equation by -1 , if necessary.
2- < inequality can be converted to a > by multiplying both side of the
inequality by -1
4
2- Conversion of unrestricted variable (xj) into nonnegative
variables.
x
j

x

j

 xj
,


xj , xj  0
3- Conversion of maximization to minimization
An objective function to be minimized instead of maximized
5
Example:
Express the following LP model in standard form
Maxim ize
Subjectto :
z  2 x1  3x2  5x3
x1  x2  x3  5
 6 x1  7 x2  9 x3  4
x1  x2  4 x3  10
x1 , x2  0
x3 unrestricted
6
Solution:
1- Subtracts the surplus S1 from L.H.S of first constraint and then multiply both sides
by -1 to obtain nonnegative R.H.S.
2- Add the slack s2 to L.H.S of second constraint.
3- Because the third constraint is already in equation form, no slack or surplus
needed.
4- Substitute unrestricted x3
Maximize
Subjectto :

x

j


x in the objective and all constraints
3

z  2x1  3x2  5x3  5x3

 x1  x2  x3  x3  s1  5

 6 x1  7 x2  9 x3  9 x3  s2  4

x1  x2  4 x3  4 x3  10
x1 , x2 , x3 , x3 , s1 , s2  0
7
It is a procedure for solving linear programming problems.
When decision variables are more than 2, it is always advisable to
use Simplex Method to avoid lengthy graphical procedure
The simplex method is not used to examine all the feasible solutions.
It is proven to be the most efficient method for solving LP problems.
The simplex algorithm is designed to locate the optimum basic
solution in an efficient manner.
It is an algebraic procedure. However, its underlying concepts are
geometric.
8
Feasible Region and Extreme Points
Feasible region is the collection of points that satisfy all constraints.
Corner points of the feasible region are known as extreme points.
LP in Standard Form
An LP is in standard form, when all variables are non-negative and all
constraints are equalities.
Basic Variables
A basic variable for an equation is a variable whose coefficient in the
equation is +1 and in all other equations is 0.
Non-basic Variable (NBV)
A non-basic variable does not appear in the basic solution. Its value is
0.
Non-basic variables in a problem = (n-m) = (# of variables - # of
equations)
Basic Feasible Solution (bfs) and Adjacent Feasible Solution
9

Steps involved (Graphical):
1. Locate an extreme point of the feasible region.
2. Examine each boundary edge intersecting at this point to see
whether movement along any edge increases the value of the
objective function.
3. If the value of the objective function increases along any edge,
move along this edge to the adjacent extreme point. If several
edges indicate improvement, the edge providing the greatest rate
of increase is selected.
4. Repeat steps 2 and 3 until movement along any edge no longer
increases the value of the objective function.
10
Example: Product Mix Problem
The N. Dustrious Company produces two products: I and II. The raw
material requirements, space needed for storage, production rates, and
selling prices for these products are given below:
The total amount of raw material available per day for both products is
15751b. The total storage space for all products is 1500 ft2, and a
maximum of 7 hours per day can be used for production. The company
wants to determine how many units of each product to produce per
day to maximize its total income.
Solution
 Step 0: Convert all the inequality constraints into equalities by
the use of slack variables. Let:
s1 = unused storage space
s2 = unused raw material
s3 = unused production time
As already developed, the LP model is:
Maximize:
Subject to:
Z  13x1  11x2
4 x1  5 x2  1500
5 x1  3x2  1575
x1  2 x2  420
x1  0
x2  0
…..Eq (1)
 Introducing these slack variables into the inequality constraints
and rewriting the objective function such that all variables are on
the left-hand side of the equation.
Equation (1) can be expressed as:
Z  13x1  11x2
0
4 x1  5 x2  s1
5 x1  3 x2
x1  2 x2
xi  0,
( A1)
 1500 ( B1)
 s2
 1575 (C1)
 s3  420 ( D1)
i  1,2
…..Eq (2)
The steps of the simplex method are:
1. Determine a starting basic feasible solution.
2. Select an entering variable using the optimally condition.
Stop if there is no entering variables.
3. Select a leaving variable using feasibility condition.
4. Determine the new basic solution by using appropriate
Gauss- Jordan computations. Go to step 2
 Step I: Set up the initial tableau using Eq. (2).
Z  13x1  11x2
0
4 x1  5 x2  s1
5 x1  3 x2
x1  2 x2
 1500 ( B1)
 s2
 1575 (C1)
 s3  420 ( D1)
xi  0, i  1,2
In any
iteration, a
variable that
has a
nonzero
value in the
solution is
called a
basic
variable.
( A1)
…..Eq (2)
 Step II: . Identify the variable that will be assigned a nonzero
value in the next iteration so as to increase the value of the
objective function. This variable is called the.entering variable
 It is that nonbasic variable which is associated with the
smallest negative coefficient in the objective function.
 If two or more nonbasic variables are tied with the smallest
coefficients, select one of these arbitrarily and continue.
 Step III: Identify the variable, called the leaving variable, which
will be changed from a nonzero to a zero value in the next
solution.
Basic
X1
Solution
Ratio
(Intercept)
s1
4
1500
1500/4=375
s2
5
1575
1575/5=315
s3
1
420
420/1=420
Because s2 is associated with the smallest (nonnegative) ratio,Thus s2
is the leaving variable, which automatically makes x1 basic at 5.
Replacing the leaving variable s2 with the entering variable x1 produces
the new basic solution (x1, s1, s3 )
* The computation of the new basic solution is based on the GaussJordan row operations.
Pivot row
Pivot column
Pivot column: the column associated with entering variable .
Pivot row: the row associated with leaving variable.
Pivot element: The intersection of pivot column and the pivot row.
The Gauss – Jordan computations needed to produce the new basic solution
include two types:
1. Pivot row
New pivot row = Current pivot row / Pivot element
2. All other rows, including z
New Row= Current row – (its pivot column coefficient)* (New pivot row)
Tableau
 The row operations proceed as follows:
 The coefficients in row C2 are obtained by dividing the
corresponding coefficients in row C1 by 5.
 The coefficients in row A2 are obtained by multiplying the
coefficients of row C2 by 13 and adding the products to the
corresponding coefficients in row Al.
 The coefficients in row B2 are obtained by multiplying the
coefficients of row C2 by -4 and adding the products to the
corresponding coefficients in row Bl.
 The coefficients in row D2 are obtained by multiplying the
coefficients of row C2 by -1 and adding the products to the
corresponding coefficients in row Dl.
 Step IV: . Enter the basic variables for the second tableau. The
row sequence of the previous tableau should be maintained,
with the leaving variable being replaced by the entering variable.
 Step V: Compute the coefficients for the second tableau. A
sequence of operations will be performed so that at the end the
x1 column in the second tableau will have the following
coefficients:
The second tableau yields the following feasible solution:
x1 = 315, x2 = 0, SI = 240, S2 = 0, S3 = 105, and Z = 4095
 Step VI: Check for optimality. The second feasible solution is
also not optimal, because the objective function (row A2)
contains a negative coefficient. Another iteration beginning with
step 2 is necessary.
 In the third tableau (next slide), all the coefficients in the
objective function (row A3) are positive. Thus an optimal solution
has been reached and it is as follows:
x1 = 270, x2 = 75, SI = 45, S2 = 0, S3 = 0, and Z = 4335
The rules for selecting the entering and leaving
variables are referred to as the optimally and feasibility
conditions
Optimality condition: The entering variable in a
maximization (minimization) problem is the non basic
variable having the most negative (positive) coefficient in
the z-row. Ties are broken arbitrarily. The optimum is
reached at the iteration where all the z-row coefficient of the
non basic variables are nonnegative (non positive).
Feasibility condition: For both maximization and
minimization problems, the leaving variables is the basic
variable associated with the smallest non-negative ratio
26