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196 The Transportation Problem Recall the following example of a Transportation Problem: CCC has 1000 computers at each of three plants this month. Three customers have requested 1100, 800, and 1100 computers. These data are summarized in the table below, along with the cost of shipping one computer from each plant to each customer. As Distribution Manager, you have been asked to develop a least-cost shipping plan for CCC. Customer 1 Customer 2 Customer 3 Supply Plant 1 5 3 2 1000 Plant 2 4 7 8 1000 Plant 3 6 7 4 1000 Demand 1100 800 1100

197 Formulating the Trans. Problem A FIA is proposed now for solving a general transportation problem. To do so, however, it is assumed that the problem is balanced , meaning that total supply equals total demand. If not, then... Using vertices to represent the plants and customers, and edges to connect each plant to each customer to indicate the possibility of shipping computers between the two associated locations, problem formulation resulted in the desire to find a least-cost shipping plan in the following complete bipartite graph: P3 P2 C2 C3 C1 P1 800 1100 1100 1000 1000 1000 4 5

198 Handling Too Much Demand (Supplies) Plants Customers (Demands) 2 1 2 3 3 1 4 3 7 6 8 7 4 2 5 1000 1000 1000 1100 800 1600 The algorithm for solving the transportation problem requires that total supply = total demand. Question: What do you do if total supply < total demand? Dum 3000 3500 0 0 0 500 3500 Note: Shipping one unit from the Dummy Plant to Customer j means that Customer j will not receive one unit of their demand. Means that some customers will not receive all of their demand.

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199 Handling Too Much Supply (Supplies) Plants Customers (Demands) 2 1 2 3 3 1 4 3 7 6 8 7 4 2 5 1500 1000 1000 1100 800 1100 The algorithm for solving the transportation problem requires that total supply = total demand. Question: What do you do if total supply > total demand? Dum 3500 3000 0 0 0 500 3500 Note: Shipping one unit from Plant i to the Dummy Customer means that Plant i will have one unit of supply not shipped. Means that some plants will not ship all of their supplies.

200 Solving The Trans. Problem Based on the steps of the simplex algorithm , a FIA is now developed for solving a balanced Transportation Problem and works with the following transportation tableau : C1 C2 C3 Sup. P1 5 3 2 1000 P2 4 7 8 1000 P3 6 7 4 1000 Dem. 1100 800 1100 Step 0 ( Initialization ). Find an initial feasible solution f 0 ∈ F and set k = 0. Step 1 ( Move ). Use a movement mechanism in an attempt to find a new feasible solution f k +1 ∈ F with Z ( f k +1 ) strictly “better” than Z ( f k ). If successful, move to f k +1 , set k = k + 1, and return to Step 1. Otherwise, stop. Recall how a general Finite Improvement Algorithm works: Cells

201 Finding an Initial Shipping Plan C1 C2 C3 Sup. P1 5 3 2 1000 P2 4 7 8 1000 P3 6 7 4 1000 Dem. 1100 800 1100 = cost 1000 100 1000 0 100 0 900 0 100 100 0 800 0 800 0 12600 Step 0 ( Initialization ). Find an initial shipping plan (tableau). One approach for doing so is a greedy algorithm that works with the cells (building blocks) of the transportation tableau, one at a time.

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202 Finding an Initial Shipping Plan C1 C2 C3 Sup. P1 5 3 2 1000 1000 P2 4 1000 7 8 1000 P3 6 100 7 800 4 100 1000 Dem. 1100 800 1100 12600 = cost nonbasic cells basic cells Note: Cells in which nothing is being shipped (cells 11, 12, 22, and 23) are called nonbasic cells . The others are called basic cells (cells 13, 21, 31, 32, and 33). Likewise for the variables.

203 Finding an Initial Shipping Plan The Matrix Minimum Method Repeat the following steps until there are no more cells: Step 1. Identify a remaining cell ij whose unit shipping cost is the least. Step 2. Set x ij = min{remaining supply at Plant i , remaining demand for Customer j }. Step 3. Reduce the supply at Plant i and the demand for Customer j by x ij . Step 4. If the supply at Plant i has decreased to 0, then cross off row i . Otherwise, the demand for Customer j is now 0, so cross off column j.

204 The Movement Mechanism C1 C2 C3 Sup. P1 5 3 2 1000 1000 P2 4 1000 7 8 1000 P3 6 100 7 800 4 100 1000 Dem. 1100 800 1100 12600 = cost Movement Mechanism: See if there is a nonbasic (“empty”) cell in which you can ship 1 unit and reduce the total cost. To do so, for each nonbasic cell, compute a reduced cost whose value indicates the change in the current transportation costs if one unit is shipped in that nonbasic cell. +1

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205 Computing Reduced Costs = ) rc( 11 x C1 C2 C3 Sup. P1 5 3 2 1000 1000 P2 4 1000 7 8 1000 P3 6 100 7 800 4 100 1000 Dem. 1100 800 1100 12600 = cost +1 − 1 +1 − 1 +5 − 2 − 6 = + 4 +1 ≥ 0 Note: To compute rc( x ij ), start in cell ij and, using only the basic cells , alternate between rows and columns to find the unique “cycle” that returns to cell ij. Then, alternately add and subtract the unit shipping costs around this cycle.

206 Test for Optimality = ) rc( 12 x 1 ) rc( 11 = x = ) rc( 22 x = ) rc( 23 x C1 C2 C3 Sup. P1 5 3 2 1000 1000 P2 4 1000 7 8 1000 P3 6 100 7 800 4 100 1000 Dem. 1100 800 1100 12600 = cost +1 − 1 +1 − 1 +3 − 2 + 4 − 7 = − 2 < 0 +7 +1 − 1 +1 − 4 + 6 − 7 = +2 > 0 +1 − 1 +8 − 4 + 6 − 4 = +6 > 0 Not optimal

207 Find the Direction of Movement The nonbasic cell whose reduced cost is most negative is used to create the direction of movement (the rule of steepest descent). 2 ) rc( 12 − = X C1 C2 C3 Sup. P1 5 3 2 1000 1000 P2 4 1000 7 8 1000 P3 6 100 7 800 4 100 1000 Dem. 1100 800 1100 12600 = cost +1 − 1 +1 − 1 The direction of movement is the balancing cycle.

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208 Find the Amount of Movement C1 C2 C3 Sup. P1 5 3 + 1 2 1000 − 1 1000 P2 4 1000 7 8 1000 P3 6 100 7 800 − 1 4 100 + 1 1000 Dem. 1100 800 1100 12600 = cost Question: From the fact that rc( x 12 ) = –2, you can improve the current plan by shipping 1 unit in cell 12. In fact, each unit shipped in cell 12 reduces the costs by 2. So why not ship 1000 units there? Answer: Because each unit shipped in cell 12 requires reducing the shipment in some of the basic cells by the same amount...and the values of these basic variables cannot be reduced below 0, so decreases} value whose cycle the in cell basic a is : { min ij x ij = t * = 800

209 Move to the New Shipping Plan Having determined that t * = 800 is the amount to ship in the chosen nonbasic cell, create the new shipping plan by adjusting the amounts shipped in each basic cell in the associated cycle. Note: The nonbasic cell whose value increased to t * becomes basic (cell 12, in this example) and the basic cell in the cycle whose value decreases to 0 becomes nonbasic (cell 32, in this example). C1 C2 C3 Sup. P1 5 3 2 1000 1000 P2 4 1000 7 8 1000 P3 6 100 7 800 4 100 1000 Dem. 1100 800 1100 12600 = cost +1 − 1 +1 − 1 +800 − 800 +800 − 800 200 900 0 11000

210 Computing Reduced Costs Again C1 C2 C3 Sup. P1 5 3 800 2 200 1000 P2 4 1000 7 8 1000 P3 6 100 7 4 900 1000 Dem. 1100 800 1100 11000 = cost = ) ( rc 22 x 0 1 6 4 2 5 ) ( rc 11 > = − + − = x 0 6 4 6 4 8 ) ( rc 23 > = − + − = x 0 2 3 2 4 7 ) ( rc 32 > = − + − = x +1 − 1 +1 − 1 +1 − 1 +7 − 4 + 6 − 4 +4 > 0 + 2 − 3 = stop Question: Is the final shipping plan x* obtained by the algorithm the best, that is, is x* optimal? Yes!

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211 The Stepping Stone Algorithm Step 0: ( Initialization ) Find an initial shipping plan x (for example, by using the Matrix Minimum Method). Step 1: ( Move ) Compute a reduced cost for each nonbasic cell. If all reduced costs are ≥ 0, then stop. Otherwise, choose any nonbasic cell ij whose reduced cost is < 0, create a new shipping plan as follows, and then return to Step 1: (a) Find the cycle of cells that starts and ends at cell ij. (b) Set t* = min{ x ij : x ij is being reduced in the cycle}. (c) Starting in cell ij , alternately increase and decrease the values of the cells in the cycle by t *. The cell whose value reduces to 0 becomes nonbasic.

212 Interpreting STORM Output TRANSPORTATION - OPTIMAL SOLUTION - SUMMARY REPORT ------- Cell ------ Unit Cell Row Column Amount COST COST PLANT 1 CUSTOMER 2 800 3.0000 2400.0000 PLANT 1 CUSTOMER 3 200 2.0000 400.0000 PLANT 1 Subtotal = 2800.0000 PLANT 2 CUSTOMER 1 1000 4.0000 4000.0000 PLANT 2 Subtotal = 4000.0000 PLANT 3 CUSTOMER 1 100 6.0000 600.0000 PLANT 3 CUSTOMER 3 900 4.0000 3600.0000 PLANT 3 Subtotal = 4200.0000 Total Cost = 11000.0000 Number of iterations = 2

213 Interpreting STORM Output TRANSPORTATION - OPTIMAL SOLUTION - DETAILED REPORT ------- Cell ------ Unit Cell Reduced Row Column Amount COST COST COST PLANT 1 CUSTOMER 1 0 5.0000 0.0000 1.0000 PLANT 1 CUSTOMER 2 800 3.0000 2400.0000 0.0000* PLANT 1 CUSTOMER 3 200 2.0000 400.0000 0.0000* PLANT 1 Subtotal = 2800.0000 PLANT 2 CUSTOMER 1 1000 4.0000 4000.0000 0.0000* PLANT 2 CUSTOMER 2 0 7.0000 0.0000 4.0000 PLANT 2 CUSTOMER 3 0 8.0000 0.0000 6.0000 PLANT 2 Subtotal = 4000.0000 PLANT 3 CUSTOMER 1 100 6.0000 600.0000 0.0000* PLANT 3 CUSTOMER 2 0 7.0000 0.0000 2.0000 PLANT 3 CUSTOMER 3 900 4.0000 3600.0000 0.0000* PLANT 3 Subtotal = 4200.0000 Total Cost = 11000.0000 * Basic cells Number of iterations = 2

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214 Variations of the Trans. Prob. •Capacitated/Uncapacitated Problems •Prohibited Routes •Transshipment Nodes •Incorporating Unequal Production Costs •Incorporating Unequal Revenue •Lower Bounds on Supplies and Demands

215 Summary The transportation problem is a combinatorial optimization problem. For a balanced problem, in which the total supply equals the total demand, the Stepping Stone Algorithm is a finite improvement algorithm that generates a sequence of shipping plans, each of which has less total cost that the preceding one. The Matrix Minimum Method is one algorithm for finding an initial feasible solution, that is, an initial shipping plan.