Transportation Tableau Solution Methodology: Location Planning and Analysis | Study notes Production and Operations Management

MGMT250 Notes October 21, 2004

1. Schedule and Announcements

a. .Return Exams – Average 69 out of 84 (add 4 points to your grade on blue book, Boston Won),

high 98%, low 65%. If you have questions, please see me, at office, will answer more next class if

more details needed.

b. Location Analysis Transportation Tableau.

c. Look at Computer Software to solve Transportation Tableau from CD ROM on Book.

d. Homework due next time (May Begin Forecasting).

2. Location Planning and Analysis, The Transportation Tableau.

Transportation Solution Methodology:

Each node is a supply or destination node, each arc is a shipping route with a cost per unit shipped.

Let us make up an example. We have 3 plants that assemble computers located in Auburn, Boston and

Chicago, we have retail outlets for these computers located in Exeter, Fort Lauderdale, and Grafton.

The capacities of the assembly plants are 200, 100, and 150 respectively. The expected demands at

each of the retail outlets is 150, 100 and 120, respectively. The costs for shipping in dollars per unit

from Auburn to each of the outlets in order are 3, 7, and 7, respectively. The costs for shipping in

dollars per unit from Boston to each of the outlets is 12, 4, and 8. Finally the costs for shipping in

dollars per unit from Chicago to each of the locations is 8, 10 and 16, respectively. What is the best

way to ship among the various locations?

Graphically the problem requires finding the minimum cost shipping or delivery routes between a

source and its destination or a supplier and its demander. (see board for graphic of problem).

We can also put this problem in “Tableau form”.

The steps to solve the problem are to:

1. Set up the Tableau: Make certain supply and demand are equal. If they are not, determine the

difference among the two amounts. Create a dummy source or destination with a supply (or demand)

equal to the difference, so that demand and supply are equal. The costs for shipping over cells that

appear in the dummy column or row are set to zero.

2. Develop an initial solution using one of many techniques (let us use the inituitive

approach).

Steps: a. find the cell with the lowest cost, allocate to this cell the minimum of the row or column

amount for that cell.

b. decrease both the “rim” amounts by the allocation amount, put a line through the one that has been

fully allocated (if both have been fully allocated, select one line to put through, arbitrarily).

c. find the “uncovered” cell with the lowest cost, and repeat the allocation and crossing out of columns

and rows. (if a rim requirement has only 0 left, then allocate an “e” value (which is actually assumed to

be zero).

d. repeat steps b and c,until all units have been allocated.

Let’s make the initial allocation.

What is the cost of this initial “solution?”

3. Is this Solution the Best One? (Evaluating Empty Cells with the MODI approach).

The Modified Distribution Method determines if there is a way to reduce the costs by determining if

there is a negative value in the unoccupied cell evaluations.

After you have determined a feasible, non-degenerate solution, you evaluate the tableau by:

(Feasible Solution is when the sum of the rows and columns equal their rim requirements)

(Non-Degenerate is when the number of occupied cells is equal to the number of rows + number of

columns -1)

1. Determining row and column “indices” by:

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MGMT250 Notes October 21, 2004

1. Schedule and Announcements a. .Return Exams – Average 69 out of 84 (add 4 points to your grade on blue book, Boston Won), high 98%, low 65%. If you have questions, please see me, at office, will answer more next class if more details needed. b. Location Analysis Transportation Tableau. c. Look at Computer Software to solve Transportation Tableau from CD ROM on Book. d. Homework due next time (May Begin Forecasting). 2. Location Planning and Analysis, The Transportation Tableau. Transportation Solution Methodology: Each node is a supply or destination node, each arc is a shipping route with a cost per unit shipped. Let us make up an example. We have 3 plants that assemble computers located in Auburn, Boston and Chicago, we have retail outlets for these computers located in Exeter, Fort Lauderdale, and Grafton. The capacities of the assembly plants are 200, 100, and 150 respectively. The expected demands at each of the retail outlets is 150, 100 and 120, respectively. The costs for shipping in dollars per unit from Auburn to each of the outlets in order are 3, 7, and 7, respectively. The costs for shipping in dollars per unit from Boston to each of the outlets is 12, 4, and 8. Finally the costs for shipping in dollars per unit from Chicago to each of the locations is 8, 10 and 16, respectively. What is the best way to ship among the various locations? Graphically the problem requires finding the minimum cost shipping or delivery routes between a source and its destination or a supplier and its demander. (see board for graphic of problem). We can also put this problem in “Tableau form”. **The steps to solve the problem are to:

Set up the Tableau** : Make certain supply and demand are equal. If they are not, determine the difference among the two amounts. Create a dummy source or destination with a supply (or demand) equal to the difference, so that demand and supply are equal. The costs for shipping over cells that appear in the dummy column or row are set to zero. 2. Develop an initial solution using one of many techniques (let us use the inituitive approach). Steps: a. find the cell with the lowest cost, allocate to this cell the minimum of the row or column amount for that cell. b. decrease both the “rim” amounts by the allocation amount, put a line through the one that has been fully allocated (if both have been fully allocated, select one line to put through, arbitrarily). c. find the “uncovered” cell with the lowest cost, and repeat the allocation and crossing out of columns and rows. (if a rim requirement has only 0 left, then allocate an “e” value (which is actually assumed to be zero). d. repeat steps b and c,until all units have been allocated. Let’s make the initial allocation. What is the cost of this initial “solution?” 3. Is this Solution the Best One? (Evaluating Empty Cells with the MODI approach). The Modified Distribution Method determines if there is a way to reduce the costs by determining if there is a negative value in the unoccupied cell evaluations. After you have determined a feasible, non-degenerate solution, you evaluate the tableau by: (Feasible Solution is when the sum of the rows and columns equal their rim requirements) (Non-Degenerate is when the number of occupied cells is equal to the number of rows + number of columns -1)
Determining row and column “indices” by:

a. assign a value of zero to the outside of row 1. b. Determine the column index for each occupied cell in row 1 using the following equation: Column Index = Cell Cost Value - Row Index. c. Once you have a column index you may be able to calculate the row index for a new row (using the equation from b). d. Continue until all row and column indices have been calculated.

Obtain cell evaluations for the empty cells using the equation: Cell Evaluation = Cell Cost - (Row Index + Column Index). The results of the cell evaluations will tell you if it is an optimal solution or not. What the rules for this evaluation? Is the initial solution to our tableau optimal? 4. Obtaining an Improved Solution Does a negative cell evaluation exist in any of the unoccupied cells? If one does not exist then, stop, solution is optimal, otherwise:
Identify the largest negative cell evaluation. This is the cell that will improve the solution the most. Once the cell for the greatest improvement has been identified use the reallocation method to improve the solution. To improve solution:
Develop a “stepping stone” path from the unoccupied cell back to itself. The stepping stones are occupied cells where you can “turn” to get back to unoccupied evaluation cell.
Label any “odd numbered stepping stone cell” in our path (beginning with the unoccupied evaluation cell) a “+” cell or a recipient cell. Label any “even numbered stepping stone cell” in our path a “-” cell, or donor cell.
The amount that will be reallocated is the minimum value from among all the donor cells.
This reallocation is completed by adding the value identified in step 3 to all the recipient cells, and subtracting this value from all the donor cells. The cell with the smallest value identified in step 3 becomes “unoccupied”. What is the new cost of the improved solution? How else could you have calculated it? 5. Repeat steps 3 and 4 until solution is optimal (e.g. no more negative evaluation cells). How to use as Location Problem?

Transportation Tableau Solution Methodology: Location Planning and Analysis, Study notes of Production and Operations Management

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