RevisedSimplexTest

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22. def test(a: MatrixD, b: VectorD, c: VectorD, x_B: Array[Int] = null): Unit

    Test the Revised Simplex Algorithm for solving Linear Programming problems.

    Test the Revised Simplex Algorithm for solving Linear Programming problems.

    a

    the constraint matrix

    b

    the limit/RHS vector

    c

    the cost vector

    x_B

    the indices of the intial basis

23. def test1(): Unit

    Test case 1: Initialize matrix 'a', vectors 'b' and 'c', and optionally the basis 'x_B'.

    Test case 1: Initialize matrix 'a', vectors 'b' and 'c', and optionally the basis 'x_B'. For Revised Simplex, matrix 'a' must be augmented with an identity matrix and vector 'c' augmented with zeroes. ------------------------------------------------------------------------- Minimize z = -1x_0 - 2x_1 + 1x_2 - 1x_3 - 4x_4 + 2x_5 Subject to 1x_0 + 1x_1 + 1x_2 + 1y_3 + 1y_4 + 1x_5 <= 6 2x_0 - 1x_1 - 2x_2 + 1y_3 + 0y_4 + 0x_5 <= 4 0x_0 + 0x_1 + 1x_2 + 1y_3 + 2y_4 + 1x_5 <= 4 where z is the objective variable and x is the decision vector. ------------------------------------------------------------------------- Solution: primal x_1 = 4, x_7 = 8, x_4 = 2 dual y_1 = -2, y_2 = 0, y_3 = -1 objF f = -16 i.e., x = (4, 8, 2), x_B = (1, 7, 4), y = (-2, 0, -1), f = -16

    See also

    Linear Programming and Network Flows, Example 5.1

24. def test2(): Unit

    Test case 2: Solution: x = (2/3, 10/3, 0), x_B = (0, 1, 5), f = -22/3

    Test case 2: Solution: x = (2/3, 10/3, 0), x_B = (0, 1, 5), f = -22/3

    See also

    Linear Programming and Network Flows, Example 5.2

25. def test3(): Unit

    Test case 3: Solution: x = (1/3, 0, 13/3), x_B = (0, 2, 4), f = -17

    Test case 3: Solution: x = (1/3, 0, 13/3), x_B = (0, 2, 4), f = -17

    See also

    Linear Programming and Network Flows, Example 3.9

26. def test4(): Unit

    Test case 4: randomly generated LP problem.

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