Summary
Highlights
The video starts by continuing from Lesson 5, where the initial simplex tableau was prepared. The initial solution had decision variables 'a' and 'b' at 0, slack variable 's1' at 45, 's2' at 70, and the objective function value at 0.
To improve the objective function, the first step is to determine the 'entering variable'. This is found by selecting the most positive entry in the 'Cj - Zj' row (the last row). In this case, '7' is the highest positive entry, which corresponds to decision variable 'b'. Thus, 'b' is the entering variable.
Next, the 'leaving variable' (or outgoing variable) must be determined. This is done by dividing the entries in the 'Solution' column by the corresponding entries in the 'entering variable' column ('b' column). The smallest positive quotient among these results identifies the leaving variable. Comparing the quotients (9 and 7), '7' is the smallest positive value, corresponding to slack variable 's2'. Therefore, 's2' is the leaving variable and will be replaced by 'b' in the basis column.
To construct the new tableau, the pivot element ('10') must be changed to '1', and all other entries in the pivot column (except the first two and last two entries) must be changed to '0'. This is achieved using elementary row operations, specifically multiplying the pivot row by the reciprocal of the pivot element (1/10) to make the pivot element '1', and then using another operation to make the other entry in the pivot column '0'.
The new pivot row (replacing 's2' with 'b') is calculated by multiplying each entry of the old pivot row by the reciprocal of the pivot element (1/10). This results in new entries: 7/10, 1, 0, 1/10, and 7.
The next step is to change the '5' in the pivot column to '0'. This involves another elementary row operation: New Row = (-A * New Pivot Row) + Old Row, where 'A' is the element to be changed to zero (in this case, 5). This calculation leads to new entries for the 's1' row: 11/2, 0, 1, -1/2, and 10.
The 'Cj' row (sum of products of basic variable coefficients and column entries) and 'Cj - Zj' row (Cj values minus the calculated Zj values) are then calculated. The objective function value for this second solution is 49. The 'Cj - Zj' row has entries: 11/10, 0, 0, -7/10.
The second solution indicates s1 = 10, b = 7, a = 0, s2 = 0, and the objective function value (P) = 49. Since there is still a positive entry (11/10) in the 'Cj - Zj' row, the solution is not yet optimal, indicating the need for a third simplex tableau. The variable 'a' (corresponding to 11/10) will be the entering variable for the next tableau, and 's1' will be the leaving variable (as 20/11 is the smallest positive quotient).
The video concludes by assigning the development of the third simplex tableau as practice. The student is asked to identify the new entering and leaving variables, the pivot element, and confirm an increased objective function value. The solution is optimal when all entries in the 'Cj - Zj' row are zero or negative.
The 'pivot element' is the entry at the intersection of the 'pivot column' (the column of the entering variable, 'b') and the 'pivot row' (the row of the leaving variable, 's2'). In this tableau, the pivot element is '10'.