# Math 407A: Linear Optimization

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1 Math 407A: Linear Optimization Lecture 4: LP Standard Form 1 1 Author: James Burke, University of Washington

2 LPs in Standard Form Minimization maximization Linear equations to linear inequalities Lower and upper bounded variables Interval variable bounds Free variable Two Step Process to Standard Form

3 LPs in Standard Form We say that an LP is in standard form if its matrix representation has the form

4 LPs in Standard Form We say that an LP is in standard form if its matrix representation has the form max s.t. c T x Ax b 0 x

5 LPs in Standard Form We say that an LP is in standard form if its matrix representation has the form max c T x It must be a maximization problem. s.t. Ax b 0 x

6 LPs in Standard Form We say that an LP is in standard form if its matrix representation has the form max c T x It must be a maximization problem. s.t. Ax b Only inequalities of the correct direction. 0 x

7 LPs in Standard Form We say that an LP is in standard form if its matrix representation has the form max c T x It must be a maximization problem. s.t. Ax b Only inequalities of the correct direction. 0 x All variables must be non-negative.

8 Every LP can be Transformed to Standard Form

9 Every LP can be Transformed to Standard Form minimization maximization

10 Every LP can be Transformed to Standard Form minimization maximization To transform a minimization problem to a maximization problem multiply the objective function by 1.

11 Every LP can be Transformed to Standard Form minimization maximization To transform a minimization problem to a maximization problem multiply the objective function by 1. linear inequalities

12 Every LP can be Transformed to Standard Form minimization maximization To transform a minimization problem to a maximization problem multiply the objective function by 1. linear inequalities If an LP has an inequality constraint of the form a i1 x 1 + a i2 x a in x n b i, it can be transformed to one in standard form by multiplying the inequality through by 1 to get a i1 x 1 a i2 x 2 a in x n b i.

13 Every LP can be Transformed to Standard Form linear equations

14 Every LP can be Transformed to Standard Form linear equations The linear equation a i1 x i + + a in x n = b i can be written as two linear inequalities a i1 x a in x n b i and a i1 x a in x n b i. or equivalently a i1 x a in x n b i a i1 x 1 a in x n b i

15 Every LP can be Transformed to Standard Form variables with lower bounds

16 Every LP can be Transformed to Standard Form variables with lower bounds If a variable x i has lower bound l i which is not zero (l i x i ), one obtains a non-negative variable w i with the substitution x i = w i + l i. In this case, the bound l i x i is equivalent to the bound 0 w i.

17 Every LP can be Transformed to Standard Form variables with lower bounds If a variable x i has lower bound l i which is not zero (l i x i ), one obtains a non-negative variable w i with the substitution x i = w i + l i. In this case, the bound l i x i is equivalent to the bound 0 w i. variables with upper bounds

18 Every LP can be Transformed to Standard Form variables with lower bounds If a variable x i has lower bound l i which is not zero (l i x i ), one obtains a non-negative variable w i with the substitution x i = w i + l i. In this case, the bound l i x i is equivalent to the bound 0 w i. variables with upper bounds If a variable x i has an upper bound u i (x i u i ) one obtains a non-negative variable w i with the substitution x i = u i w i. In this case, the bound x i u i is equivalent to the bound 0 w i.

19 Every LP can be Transformed to Standard Form variables with interval bounds

20 Every LP can be Transformed to Standard Form variables with interval bounds An interval bound of the form l i x i u i can be transformed into one non-negativity constraint and one linear inequality constraint in standard form by making the substitution x i = w i + l i.

21 Every LP can be Transformed to Standard Form variables with interval bounds An interval bound of the form l i x i u i can be transformed into one non-negativity constraint and one linear inequality constraint in standard form by making the substitution x i = w i + l i. In this case, the bounds l i x i u i are equivalent to the constraints 0 w i and w i u i l i.

22 Every LP can be Transformed to Standard Form free variables

23 Every LP can be Transformed to Standard Form free variables Sometimes a variable is given without any bounds. Such variables are called free variables. To obtain standard form every free variable must be replaced by the difference of two non-negative variables. That is, if x i is free, then we get with 0 u i and 0 v i. x i = u i v i

24 Transformation to Standard Form Put the following LP into standard form. minimize 3x 1 x 2 subject to x 1 + 6x 2 x 3 + x 4 3 7x 2 + x 4 = 5 x 3 + x x 2, x 3 5, 2 x 4 2.

25 Transformation to Standard Form The hardest part of the translation to standard form, or at least the part most susceptible to error, is the replacement of existing variables with non-negative variables.

26 Transformation to Standard Form The hardest part of the translation to standard form, or at least the part most susceptible to error, is the replacement of existing variables with non-negative variables. Reduce errors by doing the transformation in two steps.

27 Transformation to Standard Form The hardest part of the translation to standard form, or at least the part most susceptible to error, is the replacement of existing variables with non-negative variables. Reduce errors by doing the transformation in two steps. Step 1: Make all of the changes that do not involve a variable substitution.

28 Transformation to Standard Form The hardest part of the translation to standard form, or at least the part most susceptible to error, is the replacement of existing variables with non-negative variables. Reduce errors by doing the transformation in two steps. Step 1: Make all of the changes that do not involve a variable substitution. Step 2: Make all of the variable substitutions.

29 Must be a maximization problem minimize 3x 1 x 2 subject to x 1 + 6x 2 x 3 + x 4 3 7x 2 + x 4 = 5 x 3 + x x 2, x 3 5, 2 x 4 2.

30 Must be a maximization problem minimize 3x 1 x 2 subject to x 1 + 6x 2 x 3 + x 4 3 7x 2 + x 4 = 5 x 3 + x x 2, x 3 5, 2 x 4 2. Minimization = maximization

31 Must be a maximization problem minimize 3x 1 x 2 subject to x 1 + 6x 2 x 3 + x 4 3 7x 2 + x 4 = 5 x 3 + x x 2, x 3 5, 2 x 4 2. Minimization = maximization maximize 3x 1 + x 2.

32 Inequalities must go the right way.

33 Inequalities must go the right way. x 1 + 6x 2 x 3 + x 4 3

34 Inequalities must go the right way. x 1 + 6x 2 x 3 + x 4 3 becomes

35 Inequalities must go the right way. x 1 + 6x 2 x 3 + x 4 3 becomes x 1 6x 2 + x 3 x 4 3.

36 Equalities are replaced by 2 inequalities. 7x 2 + x 4 = 5

37 Equalities are replaced by 2 inequalities. becomes 7x 2 + x 4 = 5

38 Equalities are replaced by 2 inequalities. becomes 7x 2 + x 4 = 5 7x 2 + x 4 5

39 Equalities are replaced by 2 inequalities. becomes 7x 2 + x 4 = 5 7x 2 + x 4 5 and

40 Equalities are replaced by 2 inequalities. becomes 7x 2 + x 4 = 5 7x 2 + x 4 5 and 7x 2 x 4 5

41 Grouping upper bounds with the linear inequalities. The double bound 2 x 4 2 indicates that we should group the upper bound with the linear inequalities.

42 Grouping upper bounds with the linear inequalities. The double bound 2 x 4 2 indicates that we should group the upper bound with the linear inequalities. x 4 2

43 Step 1: Transformation to Standard Form Combining all of these changes gives the LP maximize 3x 1 + x 2 subject to x 1 6x 2 + x 3 x 4 3 7x 2 + x 4 5 7x 2 x 4 5 x 3 + x 4 2 x x 2, x 3 5, 2 x 4.

44 Step 2: Variable Replacement The variable x 1 is free, so we replace it by

45 Step 2: Variable Replacement The variable x 1 is free, so we replace it by x 1 = z + 1 z 1 with 0 z+ 1, 0 z 1.

46 Step 2: Variable Replacement The variable x 1 is free, so we replace it by x 1 = z + 1 z 1 with 0 z+ 1, 0 z 1. x 2 has a non-zero lower bound ( 1 x 2 ) so we replace it by

47 Step 2: Variable Replacement The variable x 1 is free, so we replace it by x 1 = z + 1 z 1 with 0 z+ 1, 0 z 1. x 2 has a non-zero lower bound ( 1 x 2 ) so we replace it by z 2 = x or x 2 = z 2 1 with 0 z 2.

48 Step 2: Variable Replacement The variable x 1 is free, so we replace it by x 1 = z + 1 z 1 with 0 z+ 1, 0 z 1. x 2 has a non-zero lower bound ( 1 x 2 ) so we replace it by z 2 = x or x 2 = z 2 1 with 0 z 2. x 3 is bounded above (x 3 5), so we replace it by

49 Step 2: Variable Replacement The variable x 1 is free, so we replace it by x 1 = z + 1 z 1 with 0 z+ 1, 0 z 1. x 2 has a non-zero lower bound ( 1 x 2 ) so we replace it by z 2 = x or x 2 = z 2 1 with 0 z 2. x 3 is bounded above (x 3 5), so we replace it by z 3 = 5 x 3 or x 3 = 5 z 3 with 0 z 3.

50 Step 2: Variable Replacement The variable x 1 is free, so we replace it by x 1 = z + 1 z 1 with 0 z+ 1, 0 z 1. x 2 has a non-zero lower bound ( 1 x 2 ) so we replace it by z 2 = x or x 2 = z 2 1 with 0 z 2. x 3 is bounded above (x 3 5), so we replace it by z 3 = 5 x 3 or x 3 = 5 z 3 with 0 z 3. x 4 is bounded below ( 2 x 4 ), so we replace it by

51 Step 2: Variable Replacement The variable x 1 is free, so we replace it by x 1 = z + 1 z 1 with 0 z+ 1, 0 z 1. x 2 has a non-zero lower bound ( 1 x 2 ) so we replace it by z 2 = x or x 2 = z 2 1 with 0 z 2. x 3 is bounded above (x 3 5), so we replace it by z 3 = 5 x 3 or x 3 = 5 z 3 with 0 z 3. x 4 is bounded below ( 2 x 4 ), so we replace it by z 4 = x or x 4 = z 4 2 with 0 z 4.

52 Step 2: Transformation to Standard Form Substituting x 1 = z + 1 z 1 into gives maximize 3x 1 + x 2 subject to x 1 6x 2 + x 3 x 4 3 7x 2 + x 4 5 7x 2 x 4 5 x 3 + x 4 2 x x 2, x 3 5, 2 x 4.

53 Step 2: Transformation to Standard Form maximize 3z z1 + x 2 subject to z 1 + z1 6x 2 + x 3 x 4 3 7x 2 + x 4 5 7x 2 x 4 5 x 3 + x 4 2 x z + 1, 0 z 1, 1 x 2, x 3 5, 2 x 4.

54 Step 2: Transformation to Standard Form Substituting x 2 = z 2 1 into maximize 3z z1 + x 2 subject to z 1 + z1 6x 2 + x 3 x 4 3 7x 2 + x 4 5 7x 2 x 4 5 x 3 + x 4 2 x 4 2 gives 0 z + 1, 0 z 1, 1 x 2, x 3 5, 2 x 4.

55 Step 2: Transformation to Standard Form maximize 3z z1 + z 2 1 subject to z 1 + z1 6z 2 + x 3 x = 3 7z 2 + x = 12 7z 2 x = 12 x 3 + x 4 2 x z + 1, 0 z 1, 0 z 2, x 3 5, 2 x 4.

56 Step 2: Transformation to Standard Form maximize 3z z1 + z 2 subject to z 1 + z1 6z 2 + x 3 x 4 3 7z 2 + x z 2 x 4 12 x 3 + x 4 2 x z + 1, 0 z 1, 0 z 2, x 3 5, 2 x 4.

57 Step 2: Transformation to Standard Form Substituting x 3 = 5 z 3 into maximize 3z z1 + z 2 subject to z 1 + z1 6z 2 + x 3 x 4 3 7z 2 + x z 2 x 4 12 x 3 + x 4 2 x 4 2 gives 0 z + 1, 0 z 1, 0 z 2, x 3 5, 2 x 4.

58 Step 2: Transformation to Standard Form maximize 3z z1 + z 2 subject to z 1 + z1 6z 2 z 3 x = 8 7z 2 + x z 2 x 4 12 z 3 + x = 3 x z + 1, 0 z 1, 0 z 2, 0 z 3, 2 x 4.

59 Step 2: Transformation to Standard Form Substituting x 4 = z 4 2 into maximize 3z z1 + z 2 subject to z 1 + z1 6z 2 z 3 x 4 8 7z 2 + x z 2 x 4 12 z 3 + x 4 3 x 4 2 gives 0 z + 1, 0 z 1, 0 z 2, 0 z 3, 2 x 4.

60 Step 2: Transformation to Standard Form maximize 3z z1 + z 2 subject to z 1 + z1 6z 2 z 3 z = 10 7z 2 + z = 14 7z 2 z = 14 z 3 + z = 1 z = 4 0 z + 1, 0 z 1, 0 z 2, 0 z 3, 0 z 4. which is in standard form.

61 Step 2: Transformation to Standard Form After making these substitutions, we get the following LP in standard form:

62 Step 2: Transformation to Standard Form After making these substitutions, we get the following LP in standard form: maximize 3z z1 + z 2 subject to z 1 + z1 6z 2 z 3 z z 2 + z z 2 z 4 14 z 3 + z 4 1 z z + 1, z 1, z 2, z 3, z 4.

63 Transformation to Standard Form: Practice Transform the following LP to an LP in standard form. minimize x 1 12x 2 + 2x 3 subject to 5x 1 x 2 + 3x 3 = 15 2x 1 + x 2 20x x 2, 1 x 3 4

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