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Chapter 7. The Revised Simplex Method

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1 Chapter 7. The Revised Simplex Method
Recall Theorem 3.1, same basis ๏ƒž same dictionary Entire dictionary can be constructed using original data as long as we know which variables are basic. Suppose we have the following form after adding slack variables to the standard LP. (or any LP with equality constraints and nonnegativity, will be discussed later in Chapter 8) maximize ๐‘ โ€ฒ ๐‘ฅ subject to ๐ด๐‘ฅ=๐‘ ๐‘ฅโ‰ฅ0 Where, ๐ด= ๐‘Ž 11 โ€ฆ ๐‘Ž 1๐‘› โ‹ฎ โ‹ฑ โ‹ฎ ๐‘Ž ๐‘š1 โ€ฆ ๐‘Ž ๐‘š๐‘› 1 โ€ฆ 0 โ‹ฎ โ‹ฑ โ‹ฎ 0 โ€ฆ 1 OR

2 This dictionary can be represented compactly using matrix notation.
ex) max 19 ๐‘ฅ ๐‘ฅ ๐‘ฅ ๐‘ฅ 4 s.t. 3 ๐‘ฅ 1 +2 ๐‘ฅ 2 + ๐‘ฅ 3 +2 ๐‘ฅ 4 โ‰ค225 ๐‘ฅ ๐‘ฅ ๐‘ฅ ๐‘ฅ 4 โ‰ค117 4 ๐‘ฅ 1 +3 ๐‘ฅ 2 +3 ๐‘ฅ 3 +4 ๐‘ฅ 4 โ‰ค320 ๐‘ฅ 1 , ๐‘ฅ 2 , ๐‘ฅ 3 , ๐‘ฅ 4 โ‰ฅ0 Add slack variables and, after two iterations of the simplex method, we obtain ๐‘ง=1782โˆ’2.5 ๐‘ฅ ๐‘ฅ 4 โˆ’3.5 ๐‘ฅ 5 โˆ’8.5 ๐‘ฅ 6 ๐‘ฅ 1 = 54 โˆ’0.5 ๐‘ฅ 2 โˆ’0.5 ๐‘ฅ 4 โˆ’0.5 ๐‘ฅ ๐‘ฅ 6 ๐‘ฅ 3 = 63 โˆ’0.5 ๐‘ฅ 2 โˆ’0.5 ๐‘ฅ ๐‘ฅ 5 โˆ’1.5 ๐‘ฅ 6 ๐‘ฅ 7 = ๐‘ฅ 2 โˆ’0.5 ๐‘ฅ ๐‘ฅ ๐‘ฅ 6 This dictionary can be represented compactly using matrix notation. OR

3 We write the constraints as ๐ด๐‘ฅ=๐‘ with
๐ด= , ๐‘= , ๐‘ฅ= ๐‘ฅ 1 ๐‘ฅ 2 ๐‘ฅ ๐‘ฅ 4 ๐‘ฅ 5 ๐‘ฅ 6 ๐‘ฅ 7 . We express ๐ด๐‘ฅ as ๐ต ๐‘ฅ ๐ต +๐‘ ๐‘ฅ ๐‘ with ๐ต= , ๐‘= , ๐‘ฅ ๐ต = ๐‘ฅ 1 ๐‘ฅ 3 ๐‘ฅ 7 , ๐‘ฅ ๐‘ = ๐‘ฅ 2 ๐‘ฅ ๐‘ฅ 5 ๐‘ฅ (In the text, ๐ด ๐ต , ๐ด ๐‘ are used instead of ๐ต, ๐‘.) Also ๐‘โ€ฒ= Express ๐‘ โ€ฒ ๐‘ฅ as ๐‘ ๐ต โ€ฒ ๐‘ฅ ๐ต + ๐‘ ๐‘ โ€ฒ ๐‘ฅ ๐‘ , with ๐‘ ๐ต โ€ฒ= , ๐‘ ๐‘ โ€ฒ= OR

4 Then ๐ด๐‘ฅ=๐‘ ๏ƒž ๐ต ๐‘ฅ ๐ต +๐‘ ๐‘ฅ ๐‘ =๐‘ ๏ƒž ๐ต ๐‘ฅ ๐ต =๐‘โˆ’๐‘ ๐‘ฅ ๐‘
If ๐ต is nonsingular (which is true for the example), we can premultiply ๐ต โˆ’1 on both sides. ๐‘ฅ ๐ต = ๐ต โˆ’1 ๐‘โˆ’ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ (matrix form of constraints in current dictionary with basis ๐ต) For objective row, ๐‘ง= ๐‘ โ€ฒ ๐‘ฅ= ๐‘ ๐ต โ€ฒ ๐‘ฅ ๐ต + ๐‘ ๐‘ โ€ฒ ๐‘ฅ ๐‘ = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘โˆ’ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ + ๐‘ ๐‘ โ€ฒ ๐‘ฅ ๐‘ = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘+ ๐‘ ๐‘ โ€ฒโˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘+ ๐‘—โˆˆ๐‘ ๐‘ ๐‘— โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐ด ๐‘— ๐‘ฅ ๐‘— = ๐‘ฆ โ€ฒ ๐‘+ ๐‘—โˆˆ๐‘ ๐‘ ๐‘— โˆ’๐‘ฆโ€ฒ ๐ด ๐‘— ๐‘ฅ ๐‘— ( if we let ๐‘ฆโ€ฒโ‰ก ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ) Matrix representation of a dictionary z= ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘+ ๐‘ ๐‘ โ€ฒโˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ OR

5 Note that matrix ๐ต corresponding to basic variables in a dictionary is nonsingular:
Pf) Can show that ๐ต is nonsingular by showing that ๐ต ๐‘ฅ ๐ต =๐‘ has a unique solution. The existence of a solution is evident: since the basic feasible solution ๐‘ฅ โˆ— satisfies ๐ด ๐‘ฅ โˆ— =๐‘ and ๐‘ฅ ๐‘ โˆ— =0, it satisfies ๐ต ๐‘ฅ ๐ต โˆ— =๐ด ๐‘ฅ โˆ— โˆ’๐‘ ๐‘ฅ ๐‘ โˆ— =๐‘. To verify that there are no other solutions, consider an arbitrary vector ๐‘ฅ ๐ต such that ๐ต ๐‘ฅ ๐ต =๐‘ and set ๐‘ฅ ๐‘ =0. Since the resulting vector ๐‘ฅ satisfies ๐ด ๐‘ฅ =๐ต ๐‘ฅ ๐ต +๐‘ ๐‘ฅ ๐‘ =๐‘, it must satisfy the bottom ๐‘š equations in the dictionary representing ๐‘ฅ โˆ— . But then ๐‘ฅ ๐‘ =0 implies ๐‘ฅ ๐ต = ๐‘ฅ ๐ต โˆ— . Hence the solution to ๐ต ๐‘ฅ ๐ต =๐‘ is unique and ๐ต is nonsingular. ๏ฟ OR

6 Matrix representation of dictionary
z= ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘+ ๐‘ ๐‘ โ€ฒโˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ ๐‘ฅ ๐ต = ๐ต โˆ’1 ๐‘โˆ’ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ In tableau form, โˆ’๐‘ง+ 0 โ€ฒ ๐‘ฅ ๐ต + ๐‘ ๐‘ โ€ฒโˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ =โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐ผ ๐‘ฅ ๐ต ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ = ๐ต โˆ’1 ๐‘ Note that ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘= ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐ด ๐‘ 1 , ๐ด ๐‘ 2 ,โ€ฆ, ๐ด ๐‘ ๐‘› =๐‘ฆโ€ฒ ๐ด ๐‘ 1 , ๐ด ๐‘ 2 ,โ€ฆ, ๐ด ๐‘ ๐‘› We frequently use ๐‘ฆ โ€ฒ = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 , i.e. ๐‘ฆ is the solution of ๐‘ฆ โ€ฒ ๐ต= ๐‘ ๐ต โ€ฒ. Hence ๐‘ ๐‘ โ€ฒโˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ = ๐‘—โˆˆ๐‘ ๐‘ ๐‘— โˆ’๐‘ฆโ€ฒ ๐ด ๐‘— ๐‘ฅ ๐‘— and โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘=โˆ’ ๐‘ฆ โ€ฒ ๐‘ OR

7 Interpretation with elementary row operations
Current tableau in matrix notation is: โˆ’๐‘ง+ 0 โ€ฒ ๐‘ฅ ๐ต + ๐‘ ๐‘ โ€ฒโˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ =โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ ๐ผ ๐‘ฅ ๐ต ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ = ๐ต โˆ’1 ๐‘ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘= ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐ด ๐‘ 1 , ๐ด ๐‘ 2 ,โ€ฆ, ๐ด ๐‘ ๐‘› =๐‘ฆโ€ฒ ๐ด ๐‘ 1 , ๐ด ๐‘ 2 ,โ€ฆ, ๐ด ๐‘ ๐‘› Here, โˆ’๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ can be viewed differently: โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘ = โˆ’๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐‘Ž 1 โ€ฒ โ‹ฎ ๐‘Ž ๐‘š โ€ฒ =(โˆ’ ๐‘ฆ โ€ฒ ) ๐‘Ž 1 โ€ฒ โ‹ฎ ๐‘Ž ๐‘š โ€ฒ i.e., we take linear combination of rows of ๐‘ using weights (โˆ’ ๐‘ฆ ๐‘– ) to ๐‘–โˆ’๐‘กโ„Ž constraint and add it to ๐‘ ๐‘ โ€ฒ in the zeroth eq. (or subtract ๐‘ฆ โ€ฒ ๐‘ from ๐‘ ๐‘ โ€ฒ) Similarly, for basic variables, we add โˆ’๐‘ฆ โ€ฒ ๐ต to ๐‘ ๐ต โ€ฒ to the zeroth equation. Then, ๐‘ ๐ต โ€ฒโˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐ต=0. Also, the right-hand side constant in the zeroth equation can be obtained by adding โˆ’๐‘ฆ โ€ฒ ๐‘ to 0 (initial value). OR

8 Hence given the initial tableau, we obtain zeroth equation of the updated tableau by taking a linear combination of rows of the initial tableau using weights โˆ’๐‘ฆ vector and add it to zeroth equation (or subtract it with weight vector ๐‘ฆ). It is the net effect of many elementary row operations to the zeroth equation performed on the tableau. The value ๐‘ ๐‘— โˆ’๐‘ฆโ€ฒ ๐ด ๐‘— is called reduced cost. It is 0 for every basic variable. OR

9 Initially, we have After reordering of columns OR

10 Updated tableau (cBโ€™B-1 = yโ€™ ) = cBโ€™ โ€“ cBโ€™B-1B = cNโ€™ โ€“ cBโ€™B-1N
OR

11 We take linear combination of rows of ๐ต:๐‘|๐‘ .
Also ๐ต โˆ’1 gives information on what elementary row operations have been performed on the constraints. Updated ๐‘–โˆ’๐‘กโ„Ž constraint = (๐‘–โˆ’๐‘กโ„Ž row of ๐ต โˆ’1 ) ๐ต:๐‘|๐‘ , so ๐‘—โˆ’๐‘กโ„Ž component of (๐‘–โˆ’๐‘กโ„Ž row of ๐ต โˆ’1 ) is the weight we multiply to the ๐‘—โˆ’๐‘กโ„Ž original constraint. We take linear combination of rows of ๐ต:๐‘|๐‘ . Hence updated tableau carries information on what elementary operations have been performed (the net effect) to obtain the current tableau. zeroth equation obtained from ๐‘โˆ’ ๐‘ฆ โ€ฒ ๐ด (negative of objective value: 0โˆ’ ๐‘ฆ โ€ฒ ๐‘) ๐‘–โˆ’๐‘กโ„Ž constraint : (๐‘–โˆ’๐‘กโ„Ž row of ๐ต โˆ’1 ) [B : N] (r.h.s: ( ๐‘–โˆ’๐‘กโ„Ž row of ๐ต โˆ’1 ) ๐‘ ) OR

12 Recall the proof of strong duality theorem, in which we claimed that, at the optimal tableau, the negative of the coefficients of slack variables in zeroth equation is an optimal dual solution. It is the ๐‘ฆโˆ’vector obtained from ๐‘ฆโ€ฒ= ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 . The updated tableau may be given using original sequence of variables. Then, OR

13 Therefore, ๐‘ โ€ฒ โˆ’ ๐‘ฆ โ€ฒ ๐ดโ‰ค0, โˆ’๐‘ฆโ€ฒโ‰ค0. Here, ๐‘ฆ โ€ฒ = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 .
In an optimal tableau, we have all coefficients in zeroth equation are โ‰ค0. Therefore, ๐‘ โ€ฒ โˆ’ ๐‘ฆ โ€ฒ ๐ดโ‰ค0, โˆ’๐‘ฆโ€ฒโ‰ค0. Here, ๐‘ฆ โ€ฒ = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 . With this ๐‘ฆ, we have ๐‘ฆ โ€ฒ ๐ดโ‰ฅ ๐‘ โ€ฒ , ๐‘ฆโ€ฒโ‰ฅ0, hence ๐‘ฆ is a dual feasible solution. Also, โˆ’ ๐‘ฆ โ€ฒ ๐‘ in the r.h.s of ๐‘งโˆ’row is the negative of the current primal objective value, which is also the negative of the objective value of the dual feasible solution ๐‘ฆ. So we have primal feasible solution ๐‘ฅ and dual feasible solution ๐‘ฆ which give the same objective value. Hence ๐‘ฅ is a primal optimal solution and ๐‘ฆ is a dual optimal solution. Since the dual constraint for the ๐‘—โˆ’๐‘กโ„Ž primal structural variable is ๐‘ฆโ€ฒ ๐ด ๐‘— โ‰ฅ ๐‘ ๐‘— , we obtain ๐‘ฆโ€ฒ ๐ด ๐‘— โˆ’ ๐‘ฆ ๐‘š+๐‘— = ๐‘ ๐‘— if we subtract nonnegative surplus variable to convert it into an equation. Then ๐‘ ๐‘— โˆ’๐‘ฆโ€ฒ ๐ด ๐‘— =โˆ’ ๐‘ฆ ๐‘š+๐‘— . So the coefficient of the ๐‘—โˆ’๐‘กโ„Ž primal structural variable in the zeroth equation can be viewed as the negative of ๐‘—โˆ’๐‘กโ„Ž surplus variable of the dual problem. At an optimal tableau, ๐‘ฆ ๐‘š+๐‘— โ‰ฅ0 for ๐‘—=1,โ€ฆ,๐‘›. OR

14 In summary, each tableau that appears in the simplex iterations gives primal basic feasible solution and, as a by-product, gives a dual solution (it is a dual basic solution although we do not prove it here). The dual solution is not necessarily dual feasible, but always gives an objective value of the dual problem which is the same as the primal objective value of the current primal solution. Also the primal and the dual solution pairs satisfy the complementary slackness condition (see the next slide) If we obtain a dual feasible solution, i.e. the coefficients in the zeroth equation are all nonpositive, we obtain primal, dual feasible solution with the same objective value. Hence optimality of the primal solution is established. Other types of algorithms can be designed using the complementary slackness optimality conditions. (e.g. dual simplex method, some algorithms for network flow problems, ...) OR

15 (recall CS conditions)
Complementary slackness between primal and dual solutions during simplex iterations. (recall CS conditions) ๐‘–=1 ๐‘š ๐‘Ž ๐‘–๐‘— ๐‘ฆ ๐‘– โˆ— = ๐‘ ๐‘— or ๐‘ฅ ๐‘— โˆ— =0 (or both), for every ๐‘—=1,2,โ€ฆ,๐‘› ๐‘—=1 ๐‘› ๐‘Ž ๐‘–๐‘— ๐‘ฅ ๐‘— โˆ— = ๐‘ ๐‘– or ๐‘ฆ ๐‘– โˆ— =0 (or both), for every ๐‘–=1,2,โ€ฆ,๐‘š (Suppose have primal b.f.s. ๐‘ฅ โˆ— , and dual solution ๐‘ฆ โˆ— = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 .) Structural variable ๐‘ฅ ๐‘— โˆ— >0 ๏ƒž ๐‘ฅ ๐‘— โˆ— basic ๏ƒž ๐‘ ๐‘— โˆ’ ๐‘ฆ โˆ— โ€ฒ๐ด ๐‘— =0 ๐‘ฆ โˆ— โ€ฒ๐ด ๐‘— โ‰  ๐‘ ๐‘— ๏ƒž ๐‘ ๐‘— โˆ’ ๐‘ฆ โˆ— โ€ฒ๐ด ๐‘— โ‰ 0 ๏ƒž ๐‘ฅ ๐‘— โˆ— nonbasic ๏ƒž ๐‘ฅ ๐‘— โˆ— =0 ๐‘Ž ๐‘– โ€ฒ ๐‘ฅ โˆ— โ‰ (<) ๐‘ ๐‘– ๏ƒž ๐‘ฅ ๐‘›+๐‘– โˆ— >0, hence basic ๏ƒž ๐‘ ๐‘›+๐‘– โˆ’๐‘ฆโ€ฒ ๐ด ๐‘›+๐‘– =0โˆ’ ๐‘ฆ ๐‘– โˆ— =0 ๐‘ฆ ๐‘– โˆ— โ‰ 0 ๏ƒž ๐‘ ๐‘›+๐‘– โˆ’ ๐‘ฆ โˆ— โ€ฒ ๐ด ๐‘›+๐‘– โ‰ 0 ๏ƒž ๐‘ฅ ๐‘›+๐‘– nonbasic, hence ๐‘ฅ ๐‘›+๐‘– โˆ— =0 ๏ƒž ๐‘Ž ๐‘– โ€ฒ ๐‘ฅ โˆ— = ๐‘ ๐‘– OR

16 (more in Chapter 10. Sensitivity Analysis later.)
A basis ๐ต is called an optimal basis if ๐‘ ๐‘— โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐ด ๐‘— โ‰ค0 for all ๐‘—โˆˆ๐‘ (we have ๐‘ ๐‘— โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐ด ๐‘— =0 for all ๐‘—โˆˆ๐ต) and ๐‘ฅ ๐ต โˆ— = ๐ต โˆ’1 ๐‘โ‰ฅ0 (we have ๐‘ฅ ๐‘ โˆ— =0). If ๐‘ ๐‘– โ†’ ๐‘ ๐‘– ยฑ1, then the objective value ๐‘ฆ โ€ฒ ๐‘โ†’ ๐‘ฆ โ€ฒ ๐‘ยฑ ๐‘ฆ ๐‘– (as long as we have the same basis). Hence ๐‘ฆ ๐‘– represent the amount of change in the objective value (with current basis ๐ต) when ๐‘ ๐‘– changes by 1 unit. Note that if we change the value of ๐‘ ๐‘– while ๐‘ฅ ๐ต โˆ— = ๐ต โˆ’1 ๐‘โ‰ฅ0, current basis is still the optimal basis, hence optimal objective value changes by ๐‘ฆ ๐‘– . But, if the nonnegativity of the basic variables is violated by changing ๐‘ ๐‘– , the current basis is no longer an optimal basis for the changed problem. (more in Chapter 10. Sensitivity Analysis later.) OR

17 Ex) OR

18 Current dictionary OR

19 ๐‘ ๐‘ โˆ’ ๐‘ ๐ต ๐ต โˆ’1 ๐‘ = 13, 17, 0, 0 โˆ’ 3.5, 8.5, = โˆ’2.5, 1.5, โˆ’3.5, โˆ’8.5 OR

20 Updated tableau OR

21 Apply same elementary row operations
OR

22 cj โ€“ yโ€™Aj = - ym+j cj โ€“ yโ€™ej = - yj OR

23 The Revised Simplex Method
Efficient implementation of the simplex method. Recall the needed operations for the simplex method: Find the entering and the leaving variable, update dictionary. (If we know the basis, entire tableau can be constructed. Hence we do not really need to update the dictionary.) Find the entering variable xj : ( ๐‘ ๐‘— โˆ’ ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 ๐ด ๐‘— >0, ๐‘—โˆˆ๐‘) 1) Find ๐‘ฆ โ€ฒ = ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 by solving ๐‘ฆ โ€ฒ ๐ต= ๐‘ ๐ต โ€ฒ 2) Evaluate ๐‘ ๐‘— โˆ’๐‘ฆโ€ฒ ๐ด ๐‘— for all ๐‘—โˆˆ๐‘ Choose a nonbasic variable with ๐‘ ๐‘— โˆ’๐‘ฆโ€ฒ ๐ด ๐‘— >0 as the entering nonbasic variable. Find the leaving variable: ๐‘ฅ ๐ต = ๐‘ฅ ๐ต โˆ— โˆ’ ๐ต โˆ’1 ๐‘ ๐‘ฅ ๐‘ = ๐‘ฅ ๐ต โˆ— โˆ’ ๐ต โˆ’1 ๐ด ๐‘ ๐ต โˆ’1 ๐ด ๐‘ 2 โ€ฆ ๐ต โˆ’1 ๐ด ๐‘ ๐‘› ๐‘ฅ ๐‘ = ๐‘ฅ ๐ต โˆ— โˆ’ ๐‘—โˆˆ๐‘ ๐ต โˆ’1 ๐ด ๐‘— ๐‘ฅ ๐‘— OR

24 ๐‘ฅ ๐ต โ† ๐‘ฅ ๐ต โˆ— โˆ’๐‘ก๐‘‘ (๐‘‘= ๐ต โˆ’1 ๐ด ๐‘— ) (Find ๐‘‘ by solving ๐ต๐‘‘= ๐ด ๐‘— )
Suppose ๐‘ฅ ๐‘— , ๐‘—โˆˆ๐‘ is chosen as the entering nonbasic variable. As we increase ๐‘ฅ ๐‘— from 0 to ๐‘กโ‰ฅ0 while keeping other nonbasic variables at 0, we get ๐‘ฅ ๐ต โ† ๐‘ฅ ๐ต โˆ— โˆ’๐‘ก๐‘‘ (๐‘‘= ๐ต โˆ’1 ๐ด ๐‘— ) (Find ๐‘‘ by solving ๐ต๐‘‘= ๐ด ๐‘— ) Then determine the largest value ๐‘ก which makes ๐‘ฅ ๐ต โˆ— โˆ’๐‘ก๐‘‘โ‰ฅ0, and find the basic variable which becomes 0, say ๐‘ฅ ๐‘™ . (minimum ratio test) i.e., ๐‘™= ๐‘Ž๐‘Ÿ๐‘”๐‘š๐‘–๐‘› ๐‘–โˆˆ๐ต ๐‘ฅ ๐‘– โˆ— ๐‘‘ ๐‘– | ๐‘‘ ๐‘– >0 and set ๐‘ก โˆ— = ๐‘š๐‘–๐‘› ๐‘–โˆˆ๐ต ๐‘ฅ ๐‘– โˆ— ๐‘‘ ๐‘– | ๐‘‘ ๐‘– >0 Update: Set ๐‘ฅ ๐‘— = ๐‘ก โˆ— ๐‘ฅ ๐ต โˆ— โ† ๐‘ฅ ๐ต โˆ— โˆ’ ๐‘ก โˆ— ๐‘‘ Replace the leaving column ๐ด ๐‘™ of ๐ต by entering column ๐ด ๐‘— , and ๐‘ฅ ๐‘— enters the basis in place of ๐‘ฅ ๐‘™ . (position in the basis is important) OR

25 ex) Current dictionary OR

26 Find entering nonbasic variable ๐‘ฅ ๐‘— such that ๐‘ ๐‘— โˆ’๐‘ฆโ€ฒ ๐ด ๐‘— >0 (๐‘ฆ= ๐‘ ๐ต โ€ฒ ๐ต โˆ’1 )
First solve ๐‘ฆโ€ฒ๐ต= ๐‘ ๐ต โ€ฒ. OR

27 From ๐‘ฅ ๐ต = ๐‘ฅ ๐ต โˆ— โˆ’ ๐ต โˆ’1 ๐ด ๐‘ ๐‘ฅ ๐‘ , consider ๐‘ฅ ๐ต โˆ— โˆ’๐‘ก ๐ต โˆ’1 ๐ด 4
Solve ๐ต๐‘‘= ๐ด 4 to find ๐ต โˆ’1 ๐ด 4 . OR

28 Remarks Recall the economic significance of dual variables discussed in Chapter 5. In the revised simplex method, the vector ๐‘ฆ obtained from ๐‘ฆ โ€ฒ ๐ต= ๐‘ ๐ต โ€ฒ (๐‘ฆโ€ฒ ๐ด ๐‘— = ๐‘ ๐‘— , ๐‘—โˆˆ๐ต) may be interpreted as a dual solution (may not be dual feasible) and it gives tentative values of the resources. Let ๐ด ๐‘— be the column of ๐ด corresponding to the variable ๐‘ฅ ๐‘— . It represents the amounts of resources consumed when we perform one unit of activity ๐‘— (๐ด๐‘ฅ= ๐ด ๐‘— ๐‘ฅ ๐‘— =๐‘). Then, for a nonbasic variable ๐‘ฅ ๐‘— , it compares the profit obtained from performing one unit of activity ๐‘— ( ๐‘ ๐‘— ) with the value of the resources consumed when we do one unit of activity ๐‘— (๐‘ฆโ€ฒ ๐ด ๐‘— ). If ๐‘ ๐‘— โˆ’ ๐‘ฆ โ€ฒ ๐ด ๐‘— >0, it means performing activity ๐‘— gives more profit than the value of the resources consumed. Hence it is desirable to increase the level of activity ๐‘— (make ๐‘ฅ ๐‘— basic). If all reduced costs are โ‰ค0, there are no profitable activities remaining under current values of resources, hence current solution is optimal. OR


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