Math 414 Lecture 30. The greedy algorithm provides the initial transportation matrix.

Transcription

1 Math Lecture The greedy algorithm provides the initial transportation matrix. matrix P P Demand W ª «2 ª2 «W ª «W ª «ª «ª «Supply The circled x ij s are the initial basic variables. Erase all other values to get a transportation skeleton.

2 skeleton W W W Supply P ª «ª «ª «ª «P ª2 «ª «Demand Converting a skeleton to a transportation tableau: The empty squares of the skeleton must be filled with the dual variables and slacks. The s go on the right border; the s on the bottom border, and the the o ij s in the squares not occupied by the x ij s. The ij th dual constraint with slacks is x ij : + +o ij = c ij. For a basic variable x ij, the dual slack is o ij = (Complementary Slackness Theorem). Thus + + o ij = c ij becomes + = c ij. For each of the circled basic variables we have an equation: + = c ij.

3 skeleton W W W Supply P ª «ª «ª «ª «P ª2 «ª «Demand For each of the circled basic variables we have an equation: + = c ij. There are dual variables but only independent equations, hence only are basic variables and one, let s pick p, is a parameter. Thus p =. Use the equations + = c ij to solve for the remaining dual variables p 2, p, w,..., w. Write the s on the right border; the s on the bottom border.

4 P P W ª «ª2 «W ª «W For the parameters, x ij =. So instead of writing a circled x ij in the square, write the dual slack o ij : + +o ij = c ij. Hence o ij = c ij - -. Use this equation to fill in the uncircled squares for o ij. These would be objective row entries in a simplex tableau. The result is a transportation tableau. ª «ª «ª «

5 tableau P P W ª «9 - ª2 «W 2 ª «W Now we successively improve these tableau to an optimal solution (none of the interior o ij s are negative). The circled x ij are > and the and are unrestricted. Hence the dual solution will be feasible (and the primal will be optimal) iff the uncircled o ij are >. In a simplex tableau, these would be objective row coefficients which must be made positive. Loop: If no o ij is negative, stop, you have an optimal solution. Otherwise Choose the next entering basic variable as usual. How ª «ª «ª «

6 Pick the most negative (uncircled) entry o ij. Erase o ij ; shade its square; add an empty circle. This will contain the new entering basic variable x ij. tableau W W W P ª «ª «- 2 ª «9 ª «ª «P ª2 «ª «In row i, the circled x ij values must sum to s i, i.e., S j x ij = s i. In column j, the circled values must sum to d j. Thus if one circled x ij value in a row or column increases (decreases), some other value must decrease (increase). Suppose we increase the green square by one, what are the consequences

7 Chase these increases and decreases around until you find a loop. This loop begins and ends at the shaded entering variable; its other corners are the circled basic variables. tableau P P W ª «+ ª «- ª2 «W ª «- ª «ª «+ ª «Alternately mark the corners of the loop + or - starting with a + on the entering variable. We will increase the + values; decrease the - ones. How much can be added and subtracted from the loop variables W

8 The - values cannot decrease below since x ij >. The maximum amount of change = the minimum of the values labeled -. Add this to the + variables; subtract it from the - variables. tableau P P W ª «+ ª «- ª «W ª «- W ª «ª «+ ª «The - variable with the minumum value goes to ; it is the departing basic variable (pick one if two or more go to ). Remove the square s circled basic variable. Picking the most negative o ij equals selecting a simplex column with the most negative objective coefficient. Picking the first basic variable which goes to equals picking the variable with the minimum q ratio to be the departing variable.

9 Erase the old s, s, and o ij s (getting a skeleton) recalculate the new ones (converting the skeleton to a tableau as before). Repeat the loop. tableau P P W ª «ª «ª «W ª «W ª «ª «

10 tableau W W W P ª «ª «ª «ª «P ª «ª «

11 tableau W W W P ª «ª «ª «ª «P 2 ª «ª «

12 tableau W W W P ª «ª «ª «ª «P 2 ª «ª «

13 tableau W W W P ª «ª «ª «ª «P 2 ª «ª «ª «

14 tableau W W W P ª «ª «ª «ª «P ª «ª «

15 tableau W W W P ª «ª «- ª «ª «P 2 ª «ª «2

16 tableau P P W ª «ª «ª «ª «ª «- Optimal. Solution: min z = = 2 at x =, x 2 =, x 2 =, x 2 =, x =, x =, the rest = W W ª «2

17 TRANSPORTATION ALGORITHM Input: an initial transportation tableau. Output: an optimal transportation matrix. Repeat until an optimal solution is found: v Set p =. v For the remaining,, calculate, using + = c ij for each circled basic x ij. calculate o ij using o ij = c ij - - for each parameter x ij. v If no objective coefficient (uncircled o ij ) is negative, stop. You have an optimal solution. Otherwise: v Pick the variable with the most negative o ij as the entering variable. Erase o ij ; shade the square; add a circle. v v Find a loop, starting with the entering variable and whose other corners are basic (circled). Alternately mark the corners + and -, starting with a + on the entering variable. Increase the + and decrease the - corner by the minimum of the - values. v Remove the circled departing variable (a minimum - value) which went to.