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Unformatted text preview: MATRIX FORMULATION OF THE LPps In this lecture we shall look at the matrix formulation of the LPPs. We see that the Basic feasible solutions are got by solving the matrix equation BX = b where B is a m× m nonsingular submatrix of the contraint matrix of the LPP. Look at the LPP (in standard form) Maximize z = c1 x1 + c2 x2 +... + cn xn Subject to the constraints a11 x1 + a12 x2 +... + a1n xn = b1 a21 x1 + a22 x2 +... + a2 n xn = b2
. . am1 x1 + am 2 x2 +... + amn xn = bm x1 , x2 ,..., xn ≥ 0, b1 , b2 ,..., bm ≥ 0 Using the notations x1 x 2 X = . , . xn b1 b 2 b= . . bm c = [ c1 c2 . . cn ] , a11 a 21 A = . . am1 a12 a22 . . . . am 2 . . a1n a2 n amn We can write the LPP in matrix form as: Maximize Subject to Here z=cX
AX = b X ≥ 0, b ≥ 0 c denotes the row vector [ c1 c2 . . cn ] Basic Feasible Solutions We assume that the rank of the matrix A is m. (This means that all the constraints are LI. We also assume m ≤ n.) After possibly rearranging the columns of A, let A = [B, N] where B is a mxm invertible submatrix of A and N is the mx(nm) submatrix formed by the remaining columns of A. The solution XB X = XN to the equations AX = b where X B = B −1b and X N = 0 is called a basic solution of the system. If X B ≥ 0 then X is called a Basic feasible solution (BFS). We give an example illustrating these. Consider the LPP Maximize Subject to z = 2 x1 + x2
x1 + x2 ≤ 6 x1 + 2 x2 ≤ 10 x1 , x2 ≥ 0 Maximize z = 2 x1 + x2 + 0 x3 + 0 x4 Subject to x1 + x2 + x3
x1 + 2 x2 x1 , x2 , x3 , x4 ≥ 0 =6 + x4 = 10 In matrix form this can be written as: Maximize z = c X Subject to AX = b
X ≥ 0, b ≥ 0 6 1 1 1 0 where A = , b = 10 , 1 2 0 1 c =[ 2 1 0 0] x1 x 2 X= x3 x4 There are 6 basic solutions of which 4 are feasible. We give them below. B
1 1 1 2 B1 2 −1 −1 1 B1b
2 4 10 −4 XT
[2
4 0 0] Feasible? z 8 Maximum Y N Y Y N Y 1 1 1 1 1 0 0 1 0 1 1 −1 1 − 1 0 1 [ 10 0 − 4 0] 6 [ 6 0 0 4] 4 5 [ 0 5 1 0] 1 6 − 2 12 5 0 1 2 1 2 1 0 0 1 0 1/ 2 1 − 1/ 2 1 −2 0 1 [0 6 0 − 2] 1 0 0 1 1 0 0 1 6 10 [ 0 0 6 10] Problem 2 Problem Set 7.1 C Page 296 Consider the following LPP Maximize z = 5 x1 + 12 x2 + 4 x3 Subject to x1 + 2 x2 + x3 + x4 = 10
2 x1 − 2 x2 − x3 x1 , x2 , x3 , x4 ≥ 0 =2 In matrix form this can be written as: Maximize z = c X Subject to AX = b
X ≥ 0, b ≥ 0 10 1 2 1 1 where A = , b = 2 2 −2 −1 0 c = [ 5 12 4 0] There are only 5 4 < basic solutions of 2 x1 x 2 X= x3 x4 which 3 are feasible. We give them below. B
1 2 2 −2 B 1 Bb
1 X
[4 T Feasible? z
56 cB B −1b 2/ 6 2/ 6 2/ 6 − 1/ 6 1/ 3 1/ 3 2 / 3 − 1/ 3 0 1 1/ 2 −1/ 2 4 3 4 6 3 0 0] Y Y Y  Maximum 1 2 1 2 1 −1 1 0 [4
[1 0 6 0]
0 0 9] 44 5  1 9 2 1 − 2 − 1 2 1 −2 0 1 1 −1 0 Does Not exist 0 − 1/ 2 1 1 0 1 −1 1 −1 12 [0 − 1 0 12] N N −2 12 [0 0 − 2 12] ...
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 Spring '11
 ritadubey

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