1 5.6 No-Standard Formulations  What do you do if your problem formulation doeshave the Standard Form?  What do you do if your problem formulation does not have the Standard Form?  This is an important issue because the simplex procedure we described on the standard form, eg  This is an important issue because the simplex procedure we described relies very much on the standard form, eg – the RHS coefficients are non-negative – availability of m slack variables – opt=max – and so on.

2 Standard Form opt=max ~  b i ≥ 0, for all i.

3 Violation # 1 Minimization  Observe that the problem z*:= min x Œ X f(x) z*:= min x Œ X f(x) is equivalent to the problem z’:= max x Œ X  f(x) z’:= max x Œ X  f(x) in that both have the. Also, z’=  z*. in that both have the same set of optimal solutions. Also, z’=  z*.

4 RemediesRemedies  1. Multiply the coefficients of the objective function by -1 and maximize the new objective function.  2. Change the simplex algorithm a bit.  :  Remark:  Australia is a country so in principle you can use either of these two approaches.  Australia is a free country so in principle you can use either of these two approaches.  In his notes, Moshe prefers the approach. I prefer the first.  In his notes, Moshe prefers the second approach. I prefer the first.

Example

6 The coefficients of the objective function are c 1 =1 ; c 2 =-1 ; c 3 = -2 c 1 =1 ; c 2 =-1 ; c 3 = -2 so using the first appraoch, we multiply them by -1 to obtain c’ 1 = -1 ; c’ 2 = 1 ; c’ 3 = 2 c’ 1 = -1 ; c’ 2 = 1 ; c’ 3 = 2 We now use these new coefficients with opt=max. Thus the form is: We now use these new coefficients with opt=max. Thus the equivalent form is:

7

8 After setting the simplex tableau and conducting two pivot operations we obtain the above final tableau. The optimal solution is thus: x*=(10/3,0,70/3,30,0,0) x*=(10/3,0,70/3,30,0,0) The optimal value of the modified objective function is: Z’ = 130/3 Thus the optimal value of the original objective function is equal to Z* = - Z’ = - 130/3. Z* = - Z’ = - 130/3.

9  Thus the optimal value of the original objective function is equal to Z* = - Z’ = - 130/3. Z* = - Z’ = - 130/3.

10 The Second Approach  Modification in the Simplex Algorithm for opt=min:  1.:  1. Optimality Test: – Stop if all the reduced costs are. – Stop if all the reduced costs are nonpositive.  2. :  2. Greedy Rule: – Select the variable with the most reduced cost. – Select the variable with the most positive reduced cost.

11 Example (Continued)

12  We do not change the data, only modify theand. Thus, the initial tableau is the same for opt=max and opt=min.  We do not change the data, only modify the optimality test and greedy rule. Thus, the initial tableau is the same for opt=max and opt=min.  Step 2 (Optimality Test): Since there are reduced costs, we continue. Since there are positive reduced costs, we continue.  Step 3 (Variable in): We select x 3 as the new basic variable because its reduced costs is the. We select x 3 as the new basic variable because its reduced costs is the largest.

13  Step 4 (Variable out):  The (usual) Ratio Test identifies x 5 as the leaving basic variable.  Step 5 (Pivoting):  As usual  New tableau:

14  All the reduced costs are, so we stop.  All the reduced costs are nonpositive, so we stop.  Optimal solution: x=(10/3,0,70/3,30,0,0)  Optimal value of the objective function: Z* = -130/3.

15 Violation #2 Varaibles allowed to be negative  Use the following fact:  Any number (positive or negative) can be expressed as the  Any number (positive or negative) can be expressed as the difference of two positive numbers.  Thus, if say x j is not required to be nonnegative, we can introduce two additional variables, say x’ j and x” j and set  x j = x’ j -x” j  with x’ j,x” j >= 0.

16  Clearly, if x’ j > x” j then x j >0, whereas if x’ j x” j then x j >0, whereas if x’ j <x” j then x j <0. And if x’ j =x” j then x j =0.  Thus, x j is indeed unrestricted in sign ()  Thus, x j is indeed unrestricted in sign (URS)

17 Example 5.6.3

18 where

19 Violation #3 Negative RHS  This is handled by multiplying the respective constraint by and taking care of the inequality sign if necessary (changing = and >= to = and >= to <=).

20 Example  Observe that in fixing this violation we created another one!

21 Violation # 4 <= constraint  We convert a “>=“ constraint to a “=“ constraint by introducing a.  We convert a “>=“ constraint to a “=“ constraint by introducing a surplus variable.  In many respects surplus variables are similar to.  In many respects surplus variables are similar to slack variables.

22 Example  Convince yourself that (a) is equivalent to (b). (a) } (b) = Correction (old notes)

23 Violation #5 = constraint  There are two strategies to handle this violation:  1. Use the “=“ constraint to one of the variables (Assignment # 2 !!!!)  1. Use the “=“ constraint to eliminate one of the variables (Assignment # 2 !!!!)  2. Use.  2. Use artificial variables.  In we prefer the.  In we prefer the second approach.

24  These variables are called “artificial” because they are used temporarily, and ultimately will disappear from the model (be set to zero).  Their sole purpose in life is to facilitate the construction of a.  Their sole purpose in life is to facilitate the construction of a basic feasible solution.

25 Example  Convince yourself that the first is equivalent to the second provided the artificial variable (x 3 ) is equal to zero!!!!!  If you are not convinced see your ophthalmologist!  How do we force x 3 to be equal to ??  How do we force x 3 to be equal to zero ??  This is a that will be addressed shortly.  This is a good question that will be addressed shortly.

26 An Overall Example 5.6.7

27  There are a number of violations here!  Opt = min  An equality constraint  A >= constraint  A negative RHS

28  To fix the “min” violation, we consider a modified objective function, namely  Z’ = - Z = -3x x 2 so the objective now is: so the objective now is: max -3x x 2 max -3x x 2 : Constraints: The first is in standard form (x 1 <= 4). To bring it to the canonical form we simply use a, say x 3, and write The first is in standard form (x 1 <= 4). To bring it to the canonical form we simply use a slack variable, say x 3, and write x 1 + x 3 = 4 x 1 + x 3 = 4

29 The second constraints (-2x 2 = -12) has two violations. The negative RHS is handled by a multiplication by -1 to produce 2x 2 = 12. So we just have to deal with the “=“ violation.

30  The “=“ violation is handled by an artificial variable, say x 4. The resulting canonical form of the constraint is thus 2x 2 + x 4 = 12 2x 2 + x 4 = 12 The third constraint (3x 1 + 2x 2 >= 18) violates the “ = 18) violates the “<=“ condition. It is handled by a surplus variable and an artificial variable. The surplus variable, say x 5, will produce an “=“ constraint: 3x 1 + 2x 2 - x 5 = 18 3x 1 + 2x 2 - x 5 = 18

31 The “=“ violation” is then handled by an artificial variable, say x 6. This yields the following canonical form: 3x 1 + 2x 2 - x 5 + x 6 = 18 3x 1 + 2x 2 - x 5 + x 6 = 18

32 Complete Reformulation

33 Important Remark  Moshe wrote “We (all of us!) shall adopt the convention that the auxiliary variables (slack, surplus, artificial) are “named” in such a way that the initial basis appears in the last m columns of the “LHS”.”  I don’t like this and would prefer that the variables kept their original names.

34 _ In any case, observe that the initial basis consists of: Slack variables Artificial variables Artificial variables