Linear Programming Lecture Notes - Penn State Personal Web Server

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combination of (x1, x2) on the line 3x1 + x2 = 120 for x1 ∈ [16, 35] will provide the largest possible value z(x1, x2) can take in the feasible region S. 20 2.4 A Linear Programming Problem with no solution. The feasible region of the linear programming problem is empty; that is, there are no values for x1 and x2 that can simultaneously satisfy all the constraints. Thus, no solution exists. 21 2.5 A Linear Programming Problem with Unbounded Feasible Region: Note that we can continue to make level curves of z(x1, x2) corresponding to larger and larger values as we move down and to the right. These curves will continue to intersect the feasible region for any value of v = z(x1, x2) we choose. Thus, we can make z(x1, x2) as large as we want and still find a point in the feasible region that will provide this value. Hence, the optimal value of z(x1, x2) subject to the constraints +∞. That is, the problem is unbounded. 22 2.6 A Linear Programming Problem with Unbounded Feasible Region and Finite Solution: In this problem, the level curves of z(x1, x2) increase in a more “southernly” direction that in Example 2.10–that is, away from the direction in which the feasible region increases without bound. The point in the feasible region with largest z(x1, x2) value is (7/3, 4/3). Note again, this is a vertex. 23 3.1 The feasible region for the diet problem is unbounded and there are alternative optimal solutions, since we are seeking a minimum, we travel in the opposite direction of the gradient, so toward the origin to reduce the objective function value. Notice that the level curves hit one side of the boundary of the feasible region. 49 3.2 Matlab input for solving the diet problem. Note that we are solving a minimization problem. Matlab assumes all problems are mnimization problems, so we don’t need to multiply the objective by −1 like we would if we started with a maximization problem. 50 4.1 Examples of Convex Sets: The set on the left (an ellipse and its interior) is a convex set; every pair of points inside the ellipse can be connected by a line contained entirely in the ellipse. The set on the right is clearly not convex as we’ve illustrated two points whose connecting line is not contained inside the set. 52 4.2 A convex function: A convex function satisfies the expression f(λx1+(1−λ)x2) ≤ λf(x1) + (1 − λ)f(x2) for all x1 and x2 and λ ∈ [0, 1]. 53 4.3 A hyperplane in 3 dimensional space: A hyperplane is the set of points satisfying an equation a T x = b, where k is a constant in R and a is a constant vector in R n and x is a variable vector in R n . The equation is written as a matrix multiplication using our assumption that all vectors are column vectors. 54 4.4 Two half-spaces defined by a hyper-plane: A half-space is so named because any hyper-plane divides R n (the space in which it resides) into two halves, the side “on top” and the side “on the bottom.” 54 4.5 A Ray: The points in the graph shown in this figure are in the set produced using the expression x0 + dλ where x0 = [2, 1] T and d = [2, 2] T and λ ≥ 0. 56 viii
4.6 Convex Direction: Clearly every point in the convex set (shown in blue) can be the vertex for a ray with direction [1, 0] T contained entirely in the convex set. Thus [1, 0] T is a direction of this convex set. 57 4.7 An Unbounded Polyhedral Set: This unbounded polyhedral set has many directions. One direction is [0, 1] T . 58 4.8 Boundary Point: A boundary point of a (convex) set C is a point in the set so that for every ball of any radius centered at the point contains some points inside C and some points outside C. 59 4.9 A Polyhedral Set: This polyhedral set is defined by five half-spaces and has a single degenerate extreme point located at the intersection of the binding constraints 3x1 + x2 ≤ 120, x1 + 2x2 ≤ 160 and 28 16x1 + x2
Page 1: Linear Programming: Penn State Math
Page 4 and 5: 8. Caratheodory Characterization Th
Page 7: List of Figures 1.1 Goat pen with u
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
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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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That is, s1 = −12 and s2 = −20
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The complete problem describing McL
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* data section */ data; set DAY :=
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This leads to a vector of reduced c
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2.1. Lexicographic Minimum Ratio Te
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Proof. The initial basis Im yieds a
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for i = 1, . . . , n. Then we have:
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Revised Simplex Algorithm (1) Ident
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simplex tableau). We obtain: z1 −
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Proof. We can prove Farkas’ Lemma
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A1· Half-space cy > 0 A2· c Am·
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3. The Karush-Kuhn-Tucker Condition
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Remark 8.9. The expressions: ∗ A
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(x ∗ 1,x ∗ 2) = (16, 72) 3x1 +
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Since w1, w2 ≥ 0, we know that w
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Example 8.13. Consider the followin
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Figure 8.5. This figure illustrates
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(9.4) Proof. Rewrite Problem D as:
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This vector will be related to c in
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3. Strong Duality Lemma 9.7. Proble
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Likewise beginning with the KKT con
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egions are polyhedral sets 1 . Cert
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Figure 9.2. The simplex algorithm b
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If a selfish profit maximizer wishe
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Recall the optimal full tableau for
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For the sake of space, we will prov
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We can choose either s1 or s2 as a
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