Post on 30-Apr-2015
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Algebraic Solution of LPPs - Simplex Method
To solve an LPP algebraically, we first put it in the standard form. This means all decision variables are nonnegative and all constraints (other than the nonnegativity restrictions) are equations with nonnegative RHS.
Converting inequalities into equations
21 32 xxz
Subject to
0,
623
63
21
21
21
xx
xx
xx
Maximize
Consider the LPP
We make the ≤ inequalities into equations by adding to each inequality a “slack” variable (which is nonnegative). Thus the given LPP can be written in the equivalent form
21 32 xxz
Subject to
0,,,
623
63
2121
221
121
ssxx
sxx
sxx
are slack variables.21, ss
Maximize
Thus we seem to have complicated the problem by introducing two more variables; but then we shall see that this is easier to solve. This is one of the “beauties” in mathematical problem solving.
The ≥ inequalities are made into equations by subtracting from each such inequality a “surplus” (non-negative) variable.
Thus the LPP
21 32 xxz Subject to
0,
223
63
21
21
21
xx
xx
xx
Maximize
is equivalent to the LPP
21 32 xxz
Subject to
0,,,
223
63
2121
221
121
ssxx
sxx
sxx
is a slack variable; is a surplus variable.2s1s
Maximize
If in a constraint, the RHS constant is negative, we make it positive by multiplying the constraint by -1.
Thus the LPP
21 32 xxz
Subject to
0,
223
63
21
21
21
xx
xx
xx
Maximize
is equivalent to the LPP
21 32 xxz
Subject to 1 2
1 2
1 2
3 6
3 2 2
, 0
x x
x x
x x
Maximize
Its standard form is the LPP
21 32 xxz
Subject to
0,,,
223
63
2121
221
121
ssxx
sxx
sxx
Maximize
are slack variables.21, ss
Dealing with unrestricted variables
If, in an LPP, a decision variable xi is unrestricted (in sign) i.e. it can take positive as well as negative values, then we can, by writing i i ix x x ,i ix x
are (defined below and are) nonnegative, make the LPP into an equivalent LPP where all the decision variables are ≥ 0.
Note:| |
;2i i
i
x xx
if 0 and otherwisei i i ix x x x
where
| |
2i i
i
x xx
Thus the LPP
Maximize 21 3xxz Subject to
1 2
1 2
1 2
2
4
unrestricted, 0
x x
x x
x x
is equivalent to the LPP
211 3xxxz
Subject to
1 1 2 1
1 1 2 2
1 1 2 1 2
2
4
, , , , 0
x x x s
x x x s
x x x s s
Maximize
Basic variables, Basic feasible Solutions
Consider an LPP (in standard form) with m constraints and n decision variables. We assume m ≤ n. We choose n –m variables and set them equal to zero. Thus we will be left with a system of m equations in m variables. If this mm square system has a unique solution, this solution is called a basic solution. If further if it is feasible, it is called a Basic Feasible Solution (BFS).
The n-m variables set to zero are called nonbasic and the m variables which we are solving for are known as basic variables. Thus a basic solution is of the form x = (x1, x2, …, xn) where n-m “components” are zero and the remaining m components form the unique solution of the square system (formed by the m constraint equations). Note that we may have a maximum of basic solutions.
n
m
Consider the LPP:
Maximize 21 32 xxz
Subject to
0,
623
63
21
21
21
xx
xx
xx
CThis is equivalent to the LPP (in standard form)
Maximize 21 32 xxz
Subject to
0,,,
623
63
2121
221
121
ssxx
sxx
sxx
are slack variables.21, ss
Nonbasic Nonbasic (zero) (zero) variablesvariables
Basic Basic variablesvariables
Basic Basic solutionsolution
AssocAssoc-iated -iated corner corner pointpoint
Feasible?Feasible?Object-Object-
ive value, ive value, z z
(0,0,6,6)(0,0,6,6) OO YesYes 00
(0,2,0,2)(0,2,0,2) BB YesYes 66
(0,3,-3,0)(0,3,-3,0) EE NoNo --
(6,0,0, -12)(6,0,0, -12) DD NoNo --
(2,0,4,0)(2,0,4,0) AA YesYes 44
CC YesYes48/748/7
OptimalOptimal)0,0,
7
12,
7
6(
),,,( 2121 ssxx
),( 21 xx
),( 11 sx
),( 21 sx
),( 12 sx
),( 22 sx
),( 21 ss ),( 21 xx
),( 11 sx
),( 21 sx
),( 12 sx
),( 22 sx
),( 21 ss
Graphical solution of the above LPP
x1
x2
O A D
B
E
C
(2,0) (6,0)
(0,2)
(0,3) (6/7, 12/7)Optimal point
(0,0)
Thus every Basic Feasible Solution corresponds to a corner(=vertex) of the set SF of all feasible solutions.
SF
Consider the LPP:
Maximize 21 3xxz
Subject to 1 2
1 2
1 2unrestricted
2
4
, 0
x x
x x
x x
Question 6 (Problem set 3.2A – Page 79)
This is equivalent to the LPP(in standard form)
Maximize 211 3xxxz
Subject to
0,,,,
4
2
21211
2211
1211
ssxxx
sxxx
sxxx
Nonbasic Nonbasic (zero) (zero) variablesvariables
Basic Basic variablesvariables
Basic Basic solutionsolution
Assoc-Assoc-iated iated corner corner pointpoint
Feasible?Feasible? Objective Objective value, zvalue, z
(0,0,0,2,4)(0,0,0,2,4) OO YesYes 00
(0,0,2,0,2)(0,0,2,0,2) BB YesYes 66
(0,0,4,-2,0)(0,0,4,-2,0) EE NoNo --
(0,-2,0,0,6)(0,-2,0,0,6) -- NoNo --
(0,4,0,6,0)(0,4,0,6,0) DD YesYes -4-4
(0,1,3,0,0)(0,1,3,0,0) CC YesYes 88
),,,,( 21211 ssxxx
),,( 111 sxx
),,( 121 sxx
),,( 211 xxx
),,( 211 sxx
),,( 221 sxx
),,( 211 ssx
),( 21 ss
),( 21 xx
),( 11 sx
),( 21 sx
),( 12 sx
),( 22 sx
Optimal value
Nonbasic Nonbasic (zero) (zero) variablesvariables
Basic Basic variablesvariables
Basic Basic solutionsolution
Associ-Associ-ated ated corner corner pointpoint
FeasibleFeasible??
Objective Objective value, zvalue, z
(2,0,0,0,6)(2,0,0,0,6) AA YesYes 22
(-4,0,0,6,0)(-4,0,0,6,0) -- NoNo --
(-1,0,3,0,0)(-1,0,3,0,0) -- NoNo --
N0 N0 SolutionSolution -- -- --
),,,,( 21211 ssxxx
),,( 221 sxx
),,( 212 ssx
),,( 211 ssx
),,( 121 sxx
),( 11 xx
),( 21 xx
),( 11 sx
),( 21 sx
Hence note that the number of Basic Solutions can be less than n
m
(-1,3)z maximum 8 at
0
B
E
D A
C
Direction of increasing z
21 3xxz
Example: Convert the following optimization problem into a LPP:
Maximize
1 2 1 2max{| 2 3 |, | 3 7 |}z x x x x
Subject to1 2
1 2
1 2
2
4
, 0
x x
x x
x x
Note that here the objective function is NOT linear. Let us put
1 2 1 2max{| 2 3 |, | 3 7 |}y x x x x
Hence 1 2 1 2| 2 3 | and | 3 7 |y x x y x x
Which is equivalent to
1 2 1 22 3 , (2 3 )y x x y x x
1 2 1 23 7 , 3 7y x x y x x
Hence the given optimization problem is equivalent to the LPP:
Maximize z ySubject to
1 2
1 2
2
4
x x
x x
1 2 1 22 3 0, 2 3 0x x y x x y
1 2 1 2
1 2
3 7 0, 3 7 0.
, , 0
x x y x x y
x x y