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EMIS 8374 LP Review: The Ratio Test

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1 Main Steps of the Simplex Method 1.Put the problem in row-0 form. 2.Construct the simplex tableau. 3.Obtain an initial BFS. 4.If the current BFS is optimal then goto step 9. 5.Choose a non-basic variable to enter the basis. 6.Use the ratio test to determine which basic variable must leave the basis. 7.Perform the pivot operation on the appropriate element of the tableau. 8.Goto Step 4. 9.Step 9. Stop.

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2 Example 1: Step 1 LP in Row-0 Form Maximize z s.t. z - 4.5 x1 - 4 x2 = 0 30 x1 + 12 x2 + x3 = 6000 10 x1 + 8 x2 + x4 = 2600 4 x1 + 8 x2 + x5 = 2000 x1, x2, x3, x4, x5 0 Original LP Maximize 4.5 x1 + 4 x2 s.t. 30 x1 + 12 x2 6000 10 x1 + 8 x2 2600 4 x1 + 8 x2 2000 x1, x2 0

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3 Example 1: Steps 2 and 3 Initial BFS: BV = {z, x3, x4, x5}, NBV = {x1, x2} z = 0, x3 = 6,000, x4 = 2,600, x5 = 2,000 x1 = x2 = 0

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4 Example 1: Steps 4 and 5 x1 and x2 are eligible to enter the basis. Select x1 to become a basic variable

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5 Example 1: Step 6 How much can we increase x1? Constraint in Row 1: 30 x1 + 12 x2 + x3 = 6000 => x3 = 6000 - 30 x1 - 12 x2. x2 = 0 (it will stay non-basic) x3 0 Thus x1 200.

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6 Example 1: Step 6 How much can we increase x1? Constraint in Row 2: 10 x1 + 8 x2 + x4 = 2600 => x4 = 2600 - 10 x1 - 8 x2 x2 = 0 (it will stay non-basic) x4 0 Thus x1 260.

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7 Example 1: Step 6 How much can we increase x1? Constraint in Row 3: 4 x1 + 8 x2 + x5= 2000 => x5 = 2000 - 4 x1 - 8 x2 x2 = 0 (it will stay non-basic) x5 0 Thus x1 500.

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8 Example 1: Step 6 From constraint 1, we see that we can increase x1 up to 200, but we must reduce x3 to zero to satisfy the constraints. From constraint 2, we see that we can increase x1 up to 260, but we must also reduce x4 to zero to satisfy the constraints. From constraint 3, we see that we can increase x1 up to 500, but we must reduce x5 to zero to satisfy the constraints. Since x3 is the limiting variable, we make it non- basic as x1 becomes basic.

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9 Step : The Ratio Test Row 1: 30 x1 + 12 x2 + x3 = 6000 => 30 x1 + x3 = 6000 => x1 6000/30 = 200. Row 2: 10 x1 + 8 x2 + x4 = 2600 => 10 x1 + x4 = 2600 => x1 2600/10 = 260. Row 3: 4 x1 + 8 x2 + x5= 2000 => 4 x1 + x5 = 2000 => x1 2000/4 = 500.

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10 Example 1: Ratio Test The minimum ratio occurs in row 1. Thus, x3 leaves the basis when x1 enters.

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11 Example 1: Steps 7 (Pivot) Pivot on the x1 column of row 1 to make x1 basic and x3 non-basic.

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12 Example 1: Steps 7 (Pivot) First ERO: divide row 1 by 30

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13 Example 1: Steps 7 (Pivot) Second ERO: Add –10 times row 1 to row 2

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14 Example 1: Steps 7 (Pivot) Third ERO: Add –4 times row 1 to row 3

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15 Example 1: Steps 7 (Pivot) Fourth ERO: Add 4.5 times row 1 to row 0

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16 Example 1: Steps 4 BV = {z, x1, x4, x5}, NBV = {x2, x3} z = 900, x1 = 200, x4 = 600, x5 = 1200 Increasing x2 may lead to an increase in z.

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17 Example 1: Ratio Test The minimum ratio occurs in row 2. Thus, x4 leaves the basis when x2 enters.

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18 Example 1: Pivoting x2 into the basis BV = {z, x1, x2, x3}, NBV = {x4, x5} z = 1250, x1 = 100, x2 = 200, x3 = 600 This an optimal BFS.

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