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A detailed explanation of the revised simplex method, an efficient approach to solving linear programming problems. The basic ideas, formulas, and steps involved in the method, including the initialization, pivot selection, and optimality test. It also discusses the importance of updating the b-1 matrix and the role of the method in large sparse lp problems.
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Revised Simplex Procedure
Basic Ideas :
Update B -1^ matrix by pivoting (as usual) Compute any other column that you need on the fly, i.e. new column =B-1^ initial column Update the reduced costs by the formula r = c (^) BB -1^ D - c
The Formulas Revisited
rD c (^) B B D (^) I cD D
= −^1. −
D .'^ j = B −^1 D. j
b ' = B −^1 b
xB = B −^1 b
x (^) D = (0,...,0)
Z = cB x (^) B
c = Original cost vector
ID = (1,2) ; IB =(3,4,5) ; B -1=I
BV Eq. # Z (^) x 1 x 2 x 3 x 4 x 5 RHS
x 3 1 0 2 1 1 0 0 40 x 4 2 0 1 1 0 1 0 30 x 5 3 0 1 0 0 0 1 15 Z Z (^1) − 4 − 3 0 0 0 0
Step 2:(stop?)
The are negative reduced costs so we
continue
Step 3:(Variable in)
Using the Greedy Rule, we select x 1 as the
new basic variable.
Step 4: (Variable out)
Applying the Ratio Test we take x 5 out of the
basis.
Step 5:(Update the indices I (^) B and I (^) D ,etc)
In contrast to the Primal Simplex Method,
here we do not update the entire simplex tableau. The main update is the B -1^ section of the tableau.
We use the usual pivot operation for this
purpose.
BV Eq. # (^) x 1 x 2 x 3 x 4 x 5 RHS
x 3 1 1 0 - 2 x 4 2 0 1 - 1 x 1 3 0 0 1 Z Z
In practice, when we update B -1^ we also update the RHS and the z-row elements under the B -1^ matrix (why?)
The first non-basic variable is selected by the
Greedy Rule
Since I (^) D = (2,5), this is x 2.
' = B
− 1 D ⋅ 2 =
b ' = B
− 1 b =
IB = (2,4,1) ; ID = (3,5), etc
BV Eq. #
x 1 x 2 x 3 x 4 x 5 x 3 RHS x 4 1 1 0 −^2 x 1 2 0 1 −^1 3 0 0 1 15 Z Z
cB = c (^) I (^) B = (3,0, 4)
Correction:
BV Eq. # (^) x 1 x 2 x 3 x 4 x 5 RHS
x (^2 1 1 0) - 2 x 4 2 - 1 1 1 x 1 3 0 0 1 Z Z
New B -
There is a negative reduced cost so we continue
rD = c (^) B B −^1 D − c (^) D = (3,0, 4)
1 0 − 2 − 1 1 1 0 0 1
1 0 0
0 0 1
− (0,0) = (3, −2)
We have to perform the Ratio Test on column
b ' = B
− 1 b =
D = S ⋅ I (^) D =
1
0
0
0
0
1
D ⋅' 5 =
− 2
1
1
