Lecture5

# Degeneracy may be temporary degenerate point is

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Unformatted text preview: effkients of primal constraints form the left side coeffkients of the dual constraint The objective coefficient of the same variable becomes the RHS of the dual constraint Given a primal problem in standard form, the dual can be constructed according to the Table 5-2 Example 5.1-l l Example 5.1-I Table 5-2 Standard Objective Objective Constraints Minimize > Maximize Maximize i Minimize The Primal in standard form Primal Problem Maximize 2 = subject to Variables Unrestricted Unrestricted 6, +?.r, j.K, + 1 I.K, + d.K, Maximize z = 5, + 12s: +4x, +0x, +.K, < 10 s.t. ?.K, - .K, +3.K, = .K, + ‘.K? + .K, + .K, = 10 8 2.K, - .K2 + 3.K, + o.K, = 8 .K,..K, .I, > 0 .K, .I, ..K,..K, 2 0 EM-602 I QM-710 (NJ) Lecture 5 Page 5-9 Example 5.1-l Example 5.1-l Dual Problem l Mmunlze w = IOy, + 8>r2 subject to x, : \, + 2~: >- 5 I, 2l’,-v,>I2 _ .r,: J’, +3v. 2 1 _ Observe from the last constraint that yc unrestricted Is dominated by yl 10, giving Mimmlze w = IOv, +8v, subject to ~,+ZV,_ 25 _ 2y,-v,212 _ v, + 3-V? 2 4 x,: F, +Ov? 2 0 (implies that v, 2 0) ~3, ._vI unrestricted v, 2 0 _vI unrestncted Example 5.1-2 Example 5.1-2 Primal Problem The Primal In standard form Minimize z = 15x, + 12~~ subject to s, fl. 2 I.5 subject to Minimize z = 15x, +12x, s, +x2 -xl = I.5 2.y +4x, +I, = 5 X,.X2 ..Kl..K4 t 0 ?K, +?S, 5 j .K,._K, > 0 Example 5.1-2 Dual Problem Maximize w = I\$; + 5\,, subject to y, +?.I!, S 15 .v, +1y, 5 I? -J’, 2 0 (or J’, 2 0) y, .,vl unrestncted EM-602 I QM-710 (NJ) Lecture 5 Page 5-l 0...
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