Linear Programming Lecture Notes - Penn State Personal Web Server

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he wants to know how many shamrocks he should make and store on each day to minimize his cost. To determine an answer to this problem, note that we have time as a parameter: time runs over three days. Let xt be the number of shamrocks McLearey makes on day t (t = 1, 2, 3) and let yt be the number of shamrocks McLearey stores on day t. There is also a parameter y0 = 10, the number of shamrocks left over from last year. McLearey’s total cost (in cents) can the be written as: (6.35) z = 200x1 + 200x2 + 200x3 + y1 + y2 + 5y3 Additionally, there are some constraints linking production, storage and demand. These constraints are depicted graphically in Figure 6.2. Multiperiod inventory models operate on a principle of conservation of flow. Manufactured goods and previous period inventories flow into the box representing each period. Demand and next period inventories flow out of the box representing each period. This inflow and outflow must be equal to account for all shamrocks produced. This is depicted below: Initial Inventory Enters y0 x1 x2 x3 Day 1 Manufactured Causes Shamrocks to Enter Boxes y1 y2 y3 Remaining Shamrocks goto Inventory Day 2 Day 3 d1 d2 d3 Demand Causes Shamrocks to Leave Boxes Final Inventory Leaves Figure 6.2. Multiperiod inventory models operate on a principle of conservation of flow. Manufactured goods and previous period inventories flow into the box representing each period. Demand and next period inventories flow out of the box representing each period. This inflow and outflow must be equal to account for all production. This means that: (6.36) yt−1 + xt = yt + dt ∀t This equation says that at period t the amount of inventory carried over from period t − 1 plus amount of shamrocks produced in period t must be equal to the total demand in period t plus any left over shamrocks at the end of period t. Clearly we also know that xt ≥ 0 for all t, since you cannot make a negative number of shamrocks. However, by also requiring yt ≥ 0 for all t, then we assert that our inventory can never be negative. A negative inventory is a backorder. Thus by saying that yt ≥ 0 we are also satisfying the requirement that McLearey satisfy his demand in each period. Note, when t = 1, then yt−1 = y0, which is the parameter we defined above. 104
The complete problem describing McLearey’s situation is: ⎧ min 200x1 + 200x2 + 200x3 + y1 + y2 + 5y3 ⎪⎨ s.t. yt−1 + xt = yt + dt ∀t ∈ {1, 2, 3} (6.37) xt ≤ 150 ∀t ∈ {1, 2, 3} ⎪⎩ xt, yt ≥ 0 ∀t ∈ {1, 2, 3} Constraints of the form xt ≤ 150 for all t come from the fact that McLearey can produce at most 150 shamrocks per day. This simple problem now has 6 variables and 6 constraints plus 6 non-negativity constraints and it is non-trivial to determine a good initial basic feasible solution, especially since the problem contains both equality and inequality constraints. A problem like this can be solved in Matlab (see Chapter 3.11), or on a commercial or open source solver like the GNU <strong>Linear</strong> <strong>Programming</strong> Kit (GLPK, http://www.gnu.org/ software/glpk/). In Figure 6.3 we show an example model file that describes the problem. In Figure 6.4, we show the data section of the GLPK model file describing McLearey’s problem. Finally, figure 6.5 shows a portion of the output generated by the GLPK solver glpsol using this model. Note that there is no inventory in Year 3 (because it is too expensive) even though it might be beneficial to McLearey to hold inventory for next year. This is because the problem has no information about any other time periods and so, in a sense, the end of the world occurs immediately after period 3. This type of end of the world phenomenon is common in multi-period problems. 105
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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
Page 66 and 67: Figure 4.3. A hyperplane in 3 dimen
Page 68 and 69: 4. Rays and Directions Recall the d
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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
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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 and 104: CHAPTER 6 Simplex Initialization In
Page 105 and 106: A basic feasible solution for our a
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: That is, s1 = −12 and s2 = −20
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 +
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Page 148 and 149: Figure 8.5. This figure illustrates
Page 150 and 151: (9.4) Proof. Rewrite Problem D as:
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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
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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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