-z c = c T - c T B B-1 A 1 - c T B B-1 b. x B B -1 A 0 B -1 b. (a) (b) Figure 1. Simplex Tableau in Matrix Form

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1 3. he Revised Simple Method he LP min, s.t. A = b ( ),, an be represented by Figure (a) below. At any Simple step, with known and -, the Simple tableau an be represented by Figure (b) below. he minimum is found if. -z RHS -z A b -z RHS -z = - - A - - b - A - b (a) (b) Figure. Simple ableau in Matri Form Remark. ased on our onvention, the z-row of the tableau is - - b, negative of the atual objetive funtion value. his is the onsequene of defining redued osts by - - A as in the tetbook. Figure suggests that knowing and - an generate the whole Simple tableau, and hene an eeute the Simple method. Moreover, if there are simple rules to determine the new basi variables and to generate the new -, then the Simple iterations an be arried out without keeping trak of the whole Simple tableau. he Revised Simple method is suh a proedure. asially, it eeutes the eat Simple method by keeping trak of the hange of the urrent basi variables,ur to the new basi variables,new, and of the inverse of the urrent basis ( ur ) - to the inverse of the new basis ( new ) -. Other than these steps, it detets the optimality and the unboundedness of an LP as the Simple method does. In the following, we indiate how to eeute the Simple steps by the Revised Simple method. Cheking the Optimality Condition and Piking the Entering Variable: he redued ost oeffiient of any variable j is given by j = j ( ur) - A j. In partiular, hek one by one j for non-basi variable j. If j ( ) for all non-basi variable j for a minimization (maimization) problem, the urrent FS is minimal (maimal). If not, pik any in most ases the first non-basi variable j with j < (>) as the

2 entering variable for a minimization (maimization) problem. y doing so, it is unneessary to alulate all non-basi j and apply the steepest asending or desending rule. In any ase, the rule does not guarantee to give the least number of iterations. Cheking the Unboundedness Condition and Piking the Leaving Variable: Let enter be the entering variable and A,enter be its original olumn. With known inverse of the urrent basis ( ur ) -, the olumn of enter in the Simple tableau, A, enter = ( ur ) - A,enter. If all the entries of the olumn vetor A, enter are non-positive, the LP is unbounded and the Simple iteration stops. Suppose the LP is bounded. Let the urrent RHS b = ( ur ) - b >. hen A, enter and b provide inputs for the minimum ratio test to determine the entering variable. Updating and : he new basi variable,new an be formed from the original basi variable,ur by replaing the leaving variable with the entering variable, and the new vetor of the ost oeffiients of the basi variables,new an be formed aordingly. Updating - : A new - is formed by pivoting, whih is equivalent to pre-multiplying by an elementary matri. Suppose that A, enter = a a and the minimal ratio test says that, enter m, enter a leave, enter (> ) is the element to be pivoted on. he required pre-multiplying elementary matri E = [e ij ] is found by hanging the (leave)th olumn of the identity matri I suh that e i,leave =, aleave,enter if i leave, ai,enter, if i leave. aleave,enter Note that oneptually ( ur ) - is given by the olumns in the tableau of the slak variables. he pivoting operations is equivalent to pre-multiplying A with E, whih turns the olumns in the tableau of the slak variables into ( new ) -. hus,

3 ( new ) - = E( ur ) -. Eample 3.. Solve the LP ma +, s.t. +,, +,,, by the revised Simple method. Sol. For benhmarking, we first solve the LP by the Simple method, whih we have turned the objetive funtion into min - -. Note that in this ase the RHS is - - b, the negative value of the atual objetive funtion. Refer to the tableaus below. z =, with =, =, 3 = = 5 = =. Note that the variables 3,, 5, and are slak variables added to the first to the fourth strutural onstraints, respetively z RHS ratio -z z RHS ratio -z z RHS ratio -z he following are the steps to solve the problem by the Revised Simple method. First Iteration: = [ 3,, 5, ] and ur = I 3

4 = [,,, ]. is non-basi. ur ( ) A = = -. he FS is not minimal and an be the entering variable., A = ( ur ) - A, = I - A, = [-,,, ]. he RHS is ( ur ) - b = [,,, ]. he minimal ratio test gives as the leaving variable. With, A = [-,,, ] and the last basi variable leaving,,new = [ 3,, 5, ], (,new ) = [,,, -], and E =. ( new ) - = E ( ur ) - = E (I) - = E =. Seond Iteration: = [ 3,, 5, ] and ur = = [,,, -]. is non-basi., A = ur A = =. ur ( ) A = -, A = - - [,,, -] = - (< ). is the entering variable. o find the leaving variable, the urrent RHS b = ( ur ) - b = =. From, A and b, the minimal ratio test gives 5 as the leaving variable. E = and ( new ) - = E ( ur ) - =

5 5 =. hird Iteration: = [ 3,,, ] and ur = = [,, -, -]. 5 and are the non-basi variables. ur A 5 = and ur A =. (Note. We an guess the values without multiplying out ur A 5 and ur A. Why? Hint. 5 and are two of the slak variables that form the initial FS.) 5 = 5-5, A = -[,, -, -] = and = -, A = -[,, -, -] =. Sine the redued ost oeffiients of all the non-basi variables are non-negative (in fat, positive), the minimal solution is got. = 3 ur b = =,, 5 z = ur ur b = [,,, ] =. Chek that the intermediate and final results of the Revised Simple method are eatly the same as those of the Simple method.