Linear Programming Lecture Notes - Penn State Personal Web Server

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For the sake of space, we will provide the dual simplex algorithm for a maximization problem: ⎧ ⎪⎨ max cx P ⎪⎩ s.t. Ax = b x ≥ 0 We will then shown how to adjust the dual simplex algorithm for minimization problems. Dual Simplex Algorithm in Algebraic Form (1) Choose an initial basic solution xB and corresponding basis matrix B so that wA·j−cj ≥ 0 for all j ∈ J , where J is the set of non-basic variables and w = cBB −1 . (2) Construct a simplex tableau using this initial solution. (3) If b = B −1 b ≥ 0, then an optimal solution has been achieved; STOP. Otherwise, the dual problem is feasible (since zj − cj ≥ 0). GOTO STEP 4. (4) Choose a leaving variable (row) xBi = bi so that bi < 0. (5) Choose the index of the entering variable (column) xj (j ∈ J ) using the following minimum ratio test: zj − cj aji = min zk − ck |aki | : k ∈ J , aki < 0 (6) If no entering variable can be selected (aki ≥ 0 for all k ∈ K) then the dual problem is unbounded and the primal problem is infeasible. STOP. (7) Using a standard simplex pivot, pivot on element aji , thus causing xBi to become 0 (and thus feasible) and causing xj to enter the basis. GOTO STEP 3. Algorithm 9. The Matrix form of the Dual Simplex Algorithm The pivoting step works because we choose the entering variable specifically so that the reduced costs will remain positive. Just as we chose the leaving variable in the standard simplex algorithm using a minimum ratio test to ensure that B −1 b remains positive, here we use it to ensure that zj − cj remains non-negative for all j ∈ J and thus we assure dual feasibility is maintained. The convergence of the dual simplex algorithm is outside of the scope of this course. However, it suffices to understand that we are essentially solving the dual problem in the primal simplex tableau using the simplex algorithm applied to the dual problem. Therefore under appropriate cycle prevention rules, the dual simplex does in fact converge to the optimal (primal) solution. Theorem 9.16. In the absence of degeneracy, or when using an appropriate cycling prevention rule, the dual simplex algorithm converges and is correct. Example 9.17. Consider the following linear programming problem: max − x1 − x2 s.t. 2x1 + x2 ≥ 4 x1 + 2x2 ≥ 2 x1, x2 ≥ 0 154
Then the standard form problem is given as: max − x1 − x2 s.t. 2x1 + x2 − s1 = 4 x1 + 2x2 − s2 = 2 x1, x2 ≥ 0 The coefficient matrix for this problem is: 2 A = 1 1 2 −1 0 0 −1 In standard form, there is no clearly good choice for a starting basic feasible solution. However, since this is a maximization problem and we know that x1, x2 ≥ 0, we know that the objective function −x1 − x2 must be bounded above by 0. A basic solution that yields this objective function value occurs when s1 and s2 are both non-basic and x1 and x2 are both non-basic. If we let B = −1 0 0 −1 Then we obtain the infeasible solution: b = B −1 −4 b = −2 Likewise we have: w = cBB −1 = 0 0 since both s1 and s2 do not appear in the objective function. We can compute the reduced costs in this case to obtain: z1 − c1 = wA·1 − c1 = 1 ≥ 0 z2 − c2 = wA·2 − c2 = 1 ≥ 0 z3 − c3 = wA·3 − c3 = 0 ≥ 0 z4 − c4 = wA·4 − c4 = 0 ≥ 0 Thus, the fact that w ≥ 0 and the fact that zj − cj ≥ 0 for all j, shows us that we have a dual feasible solution and based on our use of a basic solution, we know that complementary slackness is ensured. We can now set up our initial simplex tableau for the dual simplex algorithm. This is given by: z s1 s2 ⎡ ⎢ ⎣ z x1 x2 s1 s2 RHS 1 1 1 0 0 0 0 −2 −1 1 0 −4 0 −1 -2 0 1 −2 ⎤ ⎥ ⎦ 155
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Linear Programming: Penn State Math
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8. Caratheodory Characterization Th
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List of Figures 1.1 Goat pen with u
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4.6 Convex Direction: Clearly every
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Preface Stop! Stop right now! This
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x Goat Pen Figure 1.1. Goat pen wit
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We have formulated the general maxi
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Figure 1.3. Contour Plot of z = x 2
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Figure 1.5. A Level Curve Plot with
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and thus our vector is v = (x1 −
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Figure 1.7. Gradients of the Bindin
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finishing toys and 120 hours per we
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2. Graphically Solving Linear Progr
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S r x 0 Br( x0) Figure 2.2. A Bound
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S Every point on this line is an al
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6. Problems with Unbounded Feasible
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to “prove” your example works.
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Definition 3.6 (Matrix Transpose).
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3. Matrices and Linear Programming
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Remark 3.16. We can deal with const
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Theorem 3.20. Each elementary row o
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Gauss-Jordan Elimination Computing
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and right hand side vector b = [7,
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Gauss-Jordan elimination to solve t
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which clearly has a solution for al
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which is a contradiction. Case 2: S
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Definition 3.46 (Basic Variables).
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This leads to the following linear
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This is not in a form Matlab likes,
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Example 4.5. Figure 4.1 illustrates
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Figure 4.3. A hyperplane in 3 dimen
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4. Rays and Directions Recall the d
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for all λ > 0. This can only be tr
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Theorem 4.31. Let P ⊆ R n be a po
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Figure 4.9. A Polyhedral Set: This
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Figure 4.10. Visualization of the s
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It is clear that P is bounded. In f
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x1 x5 x2 x λx2 +(1− λ)x3 x4 x3
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If there is some i such that c T di
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Consider now the fact xj = 0 for al
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Example 5.9. Consider the Toy Maker
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using this information, we can comp
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Figure 5.1. The Simplex Algorithm:
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Using this information, we can see
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Based on this information, we can c
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Using the rule you developed in Exe
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Exercise 52. Consider the diet prob
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Suppose we choose s4 as the leaving
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CHAPTER 6 Simplex Initialization In
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A basic feasible solution for our a
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variables. This is because we start
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To remove the artificial variables
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Remark 6.6. In the case of a minimi
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Example 6.13. Suppose we solve the
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Page 126 and 127: Proof. The initial basis Im yieds a
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Page 132 and 133: simplex tableau). We obtain: z1 −
Page 134 and 135: Proof. We can prove Farkas’ Lemma
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Page 140 and 141: Remark 8.9. The expressions: ∗ A
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