# 310-2slides7 - 6.1 Simplex Method for LP Katta G. Murty...

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6.1 Simplex Method for LP Katta G. Murty Lecture slides First: Put problem in Standard Form which is: Min z = cx subject to Ax = b and x 0. All variables nonnegative variables. Only equality constraints. Every LP can be put in standard form by following simple steps. 1. Convert obj. to min. form. 2. Convert ineq. constraints involving 2 or more variables into eqs. with appropriate slacks (slack vars. are always nonnegative vars.). 2 x 1 x 4 + x 7 ≤− 13 becomes 2 x 1 x 4 + x 7 + s 1 = 13, s 1 0. x 1 2 x 5 ≥− 8 becomes x 1 2 x 5 s 2 = 8, s 2 0. 3. A variable with only one bound (lower or upper). Convert into eq. with slack. Use eq. to eliminate variable. 79

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x 1 6 becomes x 1 + s 3 =6 or x 1 =6 s 3 where s 3 0. x 2 4 becomes x 2 s 4 =4 or x 2 =4+ s 4 where s 4 0. 4. A Variable with both lower & upper bound restrictions. If lower bound 0, leave it as nonegativity restriction; & treat upper bound converted into an eq. as a constraint. If lower bound 6 =0 ,say 6 x 3 10, make x 3 =6+ s 5 where 0 s 5 4. Now treat bounds on s 5 as above. 5. Put all equality constraints in detached coeF. tableau form. If there are any unrestricted variables, eliminate them by pivoting. Example: max z 0 = x 1 2 x 2 + x 3 x 4 subject to x 1 + x 2 + x 3 + x 4 6 x 1 x 2 x 3 x 4 ≤− 7 2 x 1 + x 2 x 3 =12 2 x 1 10, x 2 5, x 3 0, x 4 unrestricted. Second: Transform all RHS constants in constraint rows 80
into nonnegative numbers. Example: min z = 3 x 4 +2 x 2 2 x 1 x 3 s. to x 4 2 x 2+ x 1 x 3 = 12 x 2 x 3 + x 5 2 x 4 + x 1 = 2 3 x 5 2 x 1 + x 4 2 x 3 =6 and x j 0 j . The resulting tableau, called Original Tableau is of following form: x 1 ... x j n z Original a 11 ... a 1 j 1 n 0 b 1 0 constraint . . . . . . . . . . . . . . . rows a m 1 mj mn 0 b m 0 Original c 1 ... c j n 1 z 0 obj. row Third: Look for variables whose col. vecs. (among con- straint rows only, ignoring obj. row) are unit vecs. 1. If all unit vectors found, for i =1to m , select a variable 81

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associated with i th unit vec. as i th basic variable or basic variable in i th row . Leads to a feasible basic vector . Price out all basic columns: For i =1to m , convert cost coef. o± i th basic var. to 0, by subtracting suitable multiple o± i th row ±rom obj. row. Now select z as basic variable in obj. row. The matrix con- sisting o± basic cols. in proper order in present tableau is unit ma- trix, so present tableau is canonical tableau WRT present basic vector . Go to Phase II: Simplex algo. to solve original LP with it. 2. I± one or more unit vectors are missing in original tableau, we don’t have ±esible basic vector to start simplex algo. Now we construct a Phase I problem to ²nd ±easible basic vector ±or original problem ²rst. For i m ,i ± i± original tableau has i th unit vector, select a variable associated with it as i th basic variable; i± original tableau does not have i th unit vector, introduce an artiFcial variable associated with i th unit vector 82
& 0 cost coef. in original obj. row, and select it as i th basic var. All artiFcials are required to be

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## This note was uploaded on 11/06/2011 for the course ISE 421 taught by Professor Km during the Spring '11 term at King Fahd University of Petroleum & Minerals.

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310-2slides7 - 6.1 Simplex Method for LP Katta G. Murty...

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