Linear Programming Lecture Notes - Penn State Personal Web Server

More documents

Recommendations

Info

Since bi ≥ 0, we will require xai ≥ 0. If xai = 0, then this is simply the original constraint. Thus if we can find values for the ordinary decision variables x so that xai = 0, then constraint i is satisfied. If we can identify values for x so that all the artificial variables are zero and m variables of x are non-zero, then the modified constraints described by Equation 6.2 are satisfied and we have identified an initial basic feasible solution. Obviously, we would like to penalize non-zero artificial variables. This can be done by writing a new linear programming problem: ⎧ ⎪⎨ (6.3) P1 ⎪⎩ min e T xa s.t. Ax + Imxa = b x, xa ≥ 0 Remark 6.1. We can see that the artificial variables are similar to slack variables, but they should have zero value because they have no true meaning in the original problem P . They are introduced artificially to help identify an initial basic feasible solution to Problem P . Lemma 6.2. The optimal objective function value in Problem P1 is bounded below by 0. Furthermore, if the optimal solution to problem P1 has xa = 0, then the values of x form a feasible solution to Problem P . Proof. Clearly, setting xa = 0 will produce an objective function value of zero. Since e > 0, we cannot obtain a smaller objective function value. If at optimality we have xa = 0, then we know that m of the variables in x are in the basis and the remaining variables (in both x and xa) are not in the basis and hence at zero. Thus we have found a basic feasible solution to Problem P . Example 6.3. Consider the following problem: min x1 + 2x2 s.t. x1 + 2x2 ≥ 12 (6.4) 2x1 + 3x2 ≥ 20 x1, x2 ≥ 0 We can convert the problem to standard form by adding two surplus variables: min x1 + 2x2 s.t. x1 + 2x2 − s1 = 12 (6.5) 2x1 + 3x2 − s2 = 20 x1, x2, s1, s2 ≥ 0 It’s not clear what a good basic feasible solution would be for this. Clearly, we cannot set x1 = x2 = 0 because we would have s1 = −12 and s2 = −20, which is not feasible. We can introduce two artificial variables (xa1 and xa2) and create a new problem P1. min xa1 + xa2 s.t. x1 + 2x2 − s1 + xa1 = 12 (6.6) 2x1 + 3x2 − s2 + xa2 = 20 x1, x2, s1, s2, xa1, xa2 ≥ 0 92
A basic feasible solution for our artificial problem would let xa1 = 12 and xa2 = 20. The pertinent matrices in this case are: 1 2 −1 0 1 0 12 A = b = 2 3 0 −1 0 1 20 1 0 1 2 −1 0 B = N = B 0 1 2 3 0 −1 −1 12 b = 20 ⎡ ⎤ 0 1 ⎢ cB = cN = ⎢0 ⎥ 1 ⎣0⎦ 0 c T BB −1 b = 32 c T BB −1 N − c T N = 3 5 −1 −1 We can construct an initial tableau for this problem as: ⎡ ⎤ z x1 x2 s1 s2 xa1 xa2 RHS z ⎢ (6.7) ⎢ 1 3 5 −1 −1 0 0 32 ⎥ xa1 ⎣ 0 1 2 −1 0 1 0 12 ⎦ 0 2 3 0 −1 0 1 20 xa2 This is a minimization problem, so if zj −cj > 0, then entering xj will improve (decrease) the objective value because ∂z/∂xj < 0. In this case, we could enter either x1 or x2 to improve the objective value. Let’s assume we enter variable x1. Performing the minimum ratio test we see: (6.8) z xa1 xa2 x1 ⎡ ⎢ ⎣ z x1 x2 s1 s2 xa1 xa2 RHS 1 3 5 −1 −1 0 0 32 0 1 2 −1 0 1 0 12 0 2 3 0 −1 0 1 20 ⎤ ⎥ ⎦ MRT(x1) 12 20/2 = 10 Thus xa2 leaves the basis and x1 enters. The new tableau becomes: ⎡ ⎤ z x1 x2 s1 s2 xa1 xa2 RHS z ⎢ (6.9) ⎢ 1 0 1/2 −1 1/2 0 −3/2 2 ⎥ xa1 ⎣ 0 0 1/2 −1 1/2 1 −1/2 2 ⎦ 0 1 3/2 0 −1/2 0 1/2 10 In this case, we see that x2 should enter the basis. Performing the minimum ratio test, we obtain: (6.10) z xa1 x1 ⎡ z ⎢ 1 ⎢ ⎣ 0 0 x1 0 0 1 x2 1/2 1/2 3/2 s1 −1 −1 0 s2 1/2 1/2 −1/2 xa1 0 1 0 xa2 −3/2 −1/2 1/2 ⎤ RHS 2 ⎥ 2 ⎦ 10 MRT(x2) 4 20/3 Thus we see that xa2 leaves the basis and we obtain: ⎡ ⎤ z x1 x2 s1 s2 xa1 xa2 RHS z ⎢ (6.11) ⎢ 1 0 0 0 0 −1 −1 0 ⎥ x2 ⎣ 0 0 1 −2 1 2 −1 4 ⎦ 0 1 0 3 −2 −3 2 4 x1 93
Page 1:
Linear Programming: Penn State Math
Page 4 and 5:
8. Caratheodory Characterization Th
Page 7 and 8:
List of Figures 1.1 Goat pen with u
Page 9 and 10:
4.6 Convex Direction: Clearly every
Page 11:
Preface Stop! Stop right now! This
Page 14 and 15:
x Goat Pen Figure 1.1. Goat pen wit
Page 16 and 17:
We have formulated the general maxi
Page 18 and 19:
Figure 1.3. Contour Plot of z = x 2
Page 20 and 21:
Figure 1.5. A Level Curve Plot with
Page 22 and 23:
and thus our vector is v = (x1 −
Page 24 and 25:
Figure 1.7. Gradients of the Bindin
Page 26 and 27:
finishing toys and 120 hours per we
Page 28 and 29:
2. Graphically Solving Linear Progr
Page 30 and 31:
S r x 0 Br( x0) Figure 2.2. A Bound
Page 32 and 33:
S Every point on this line is an al
Page 34:
6. Problems with Unbounded Feasible
Page 37:
to “prove” your example works.
Page 40 and 41:
Definition 3.6 (Matrix Transpose).
Page 42 and 43:
3. Matrices and Linear Programming
Page 44 and 45:
Remark 3.16. We can deal with const
Page 46 and 47:
Theorem 3.20. Each elementary row o
Page 48 and 49:
Gauss-Jordan Elimination Computing
Page 50 and 51:
and right hand side vector b = [7,
Page 52 and 53:
Gauss-Jordan elimination to solve t
Page 54 and 55: which clearly has a solution for al
Page 56 and 57: which is a contradiction. Case 2: S
Page 58 and 59: Definition 3.46 (Basic Variables).
Page 60 and 61: This leads to the following linear
Page 62 and 63: This is not in a form Matlab likes,
Page 64 and 65: Example 4.5. Figure 4.1 illustrates
Page 66 and 67: Figure 4.3. A hyperplane in 3 dimen
Page 68 and 69: 4. Rays and Directions Recall the d
Page 70 and 71: for all λ > 0. This can only be tr
Page 72 and 73: Theorem 4.31. Let P ⊆ R n be a po
Page 74 and 75: Figure 4.9. A Polyhedral Set: This
Page 76 and 77: Figure 4.10. Visualization of the s
Page 78 and 79: It is clear that P is bounded. In f
Page 80 and 81: x1 x5 x2 x λx2 +(1− λ)x3 x4 x3
Page 82 and 83: If there is some i such that c T di
Page 84 and 85: Consider now the fact xj = 0 for al
Page 86 and 87: Example 5.9. Consider the Toy Maker
Page 88 and 89: using this information, we can comp
Page 90 and 91: Figure 5.1. The Simplex Algorithm:
Page 92 and 93: Using this information, we can see
Page 94 and 95: Based on this information, we can c
Page 96 and 97: Using the rule you developed in Exe
Page 98 and 99: Exercise 52. Consider the diet prob
Page 100 and 101: Suppose we choose s4 as the leaving
Page 103: CHAPTER 6 Simplex Initialization In
Page 107 and 108: variables. This is because we start
Page 109 and 110: To remove the artificial variables
Page 111 and 112: Remark 6.6. In the case of a minimi
Page 113 and 114: Example 6.13. Suppose we solve the
Page 115 and 116: That is, s1 = −12 and s2 = −20
Page 117 and 118: The complete problem describing McL
Page 119: * data section */ data; set DAY :=
Page 122 and 123: This leads to a vector of reduced c
Page 124 and 125: 2.1. Lexicographic Minimum Ratio Te
Page 126 and 127: Proof. The initial basis Im yieds a
Page 128 and 129: for i = 1, . . . , n. Then we have:
Page 130 and 131: Revised Simplex Algorithm (1) Ident
Page 132 and 133: simplex tableau). We obtain: z1 −
Page 134 and 135: Proof. We can prove Farkas’ Lemma
Page 136 and 137: A1· Half-space cy > 0 A2· c Am·
Page 138 and 139: 3. The Karush-Kuhn-Tucker Condition
Page 140 and 141: Remark 8.9. The expressions: ∗ A
Page 142 and 143: (x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
Page 144 and 145: Since w1, w2 ≥ 0, we know that w
Page 146 and 147: Example 8.13. Consider the followin
Page 148 and 149: Figure 8.5. This figure illustrates
Page 150 and 151: (9.4) Proof. Rewrite Problem D as:
Page 152 and 153: This vector will be related to c in
Page 154 and 155:
3. Strong Duality Lemma 9.7. Proble
Page 156 and 157:
Likewise beginning with the KKT con
Page 158 and 159:
egions are polyhedral sets 1 . Cert
Page 160 and 161:
Figure 9.2. The simplex algorithm b
Page 162 and 163:
If a selfish profit maximizer wishe
Page 164 and 165:
Recall the optimal full tableau for
Page 166 and 167:
For the sake of space, we will prov
Page 168 and 169:
We can choose either s1 or s2 as a
show all

Linear Programming Lecture Notes - Penn State Personal Web Server

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?