Distribution Model Meaning Types Transportation Model Assignment Model
Distribution Model • Meaning • Types • Transportation Model • Assignment Model
Distribution Models • Distribution problems are a special type of linear programming problem. • Two types of Model. They are: i. Transportation Problems: Deals with shipment from number of sources to number of destinations. – ii. Typically each source is supply limited and each destination has a known demand Assignment Problems: Deals with finding the best one to one match for each of a given no of candidates. As for example: – Assigning worker to machine – Assigning teacher to classes • Objective is to get maximum reward or minimum cost.
Transportation Model • Wheat is harvested in Midwest and stored in grain elevators in three diff cities-Kansas Omaha and des moines. Three grain elevators supply three flour mills located in Chicago, St. Louis and Cincinnati. Each grain elevator is able to supply the Each following no of of wheat Mill demand thetons following no of tons to mills. Mill Demand Elevators Supply 1 Kansas 150 2 Omaha 175 3 Des Moines 275 Total 600 tons The cost of transporting one ton of wheat from each grain elevator to each mill are Grain elevators A. Chicago B. St. Louis C. Cincinnati 1. Kansas 6 8 10 2. Omaha 7 11 11 3. Des 4 5 12 A Chicago 200 B St. Louis 100 C Cincinnati 300 Total 600 tons The problem is to how many tons of wheat to transport from each grain elevators to each mill on a monthly basis in order to minimize total cost of transportation
Solution of Transportation Model • Several methods are: i. The northwest corner methods ii. Minimum cell cost method iii. Vogel’s approximation method(VAM) iv. The stepping- stone solution method
Solution of Transportation Model • Transportation models are solved manually with in the context of a tableau as in the simplex method. From To A B C Supply 1 6 8 10 150 2 7 11 11 175 3 4 5 12 275 Demand 200 100 300 600
Solution of Transportation Model • The northwest corner method: Here the largest possible allocation is made to the cell in the upper left hand corner of the tableau, followed by allocation to adjacent feasible cells. From To A B C Supply 1 6 8 10 150 2 7 11 11 175 3 4 5 12 275 Demand 200 100 300 600
Solution of Transportation Model • The northwest corner method: Here the largest possible allocation is made to the cell in the upper left hand corner of the tableau, followed by allocation to adjacent feasible cells. i. Start with the northwest corner, allocate the smaller amount of either the row supply or column demand ii. Subtract from the row supply and the column demand the amount allocated iii. If the column demand is zero move to the cell next on the right if the row supply is zero move down to the cell in the next row. If both are zero move first to the next cell on the right then down on cell
Solution of Transportation Model • The northwest corner method: Here the largest possible allocation is made to the cell in the upper left hand corner of the tableau, followed by allocation to adjacent feasible cells. From To A 1 150 2 50 3 Demand B 6 Supply 8 7 100 4 200 C 100 10 150 11 25 11 175 5 275 12 275 300 600
Solution of Transportation Model • The northwest corner method: Here the largest possible allocation is made to the cell in the upper left hand corner of the tableau, followed by allocation to adjacent feasible cells. From To A 1 150 B 6 50 7 3 4 200 Supply 8 2 Demand C 100 10 150 11 25 11 175 5 275 12 275 300 600 • Total cost =150*6+50*7+100*11+25*11+275*12=5925
Solution of Transportation Problems (least cost Method) • Logic is to allocate to the cells with the lowest cost • The next allocation is made the cell that has the minimum cost • Repeat this way until all rim requirements have been made.
Solution of Transportation Model (least cost method) Total cost =4550 From To A B 1 6 2 7 3 Demand 200 4 C 25 8 11 75 100 Supply 125 175 5 300 10 150 11 175 12 275 600
Solution of Transportation Problems (least cost Method) • Logic is to allocate to the cells with the lowest cost • The next allocation is made the cell that has the minimum cost • Repeat this way until all rim requirements have been made.
Solution of Transportation Problems (least cost Method) • Logic is to allocate to the cells with the lowest cost • The next allocation is made the cell that has the minimum cost • Repeat this way until all rim requirements have been made.
Solution of Transportation Problems (least cost Method) • Logic is to allocate to the cells with the lowest cost • The next allocation is made the cell that has the minimum cost • Repeat this way until all rim requirements have been made.
Solution of Transportation Problems (least cost Method) • Logic is to allocate to the cells with the lowest cost • The next allocation is made the cell that has the minimum cost • Repeat this way until all rim requirements have been made.
Solution of Transportation Model Balance the Table • Transportation solution technique requires that the problem be balanced ie total supply must equal total demand. Two causes of imbalance are excess supply and excess demand. i. ii. Excess supply: If it is present in any table first balance it by adding a dummy column (dummy destination). Excess demand: When total demand exceeds total supply a dummy source row is added to balance the table. • In both case the shipment costs are set to zero
Solution of Transportation Model Balance the Table i. Excess supply: If it is present in any table first balance it by adding a dummy column (dummy destination). Source/ destination R S T Supply A 1 2 3 200 B 4 1 5 100 Demand 80 120 60 260/300 Dummy Supply Source/desti nation R S T A 1 2 3 0 200 B 4 1 5 0 100 Demand 80 120 60 40(excess supply) 300/300
Solution of Transportation Model Balance the Table i. Excess demand: When total demand exceeds total supply a dummy source row is added to balance the table. Source/ destination R S T Supply A 1 2 3 100 B 4 1 5 110 Demand 80 120 60 260/210 Source/destination R S T Supply A 1 2 3 100 B 4 1 5 110 Dummy row 0 0 0 50 80 120 Demand 60 260/260
Solution of Transportation Model Balance the Table: Northwest corner solution i. Excess demand: When total demand exceeds total supply a dummy source row is added to balance the table. Source/ destination R S T Supply A 1 2 3 100 B 4 1 5 110 Demand 80 120 60 260/210 Source/destination R A 80 1 B 4 Dummy row 0 0 80 120 Demand S T Supply 20 2 3 100 1 10 5 110 50 60 260/260
Assignment Model The Hungarian Method: The following algorithm applies the above theorem to a given n × n cost matrix to find an optimal assignment. • Step 1. Subtract the smallest entry in each row from all the entries of its row. • Step 2. Subtract the smallest entry in each column from all the entries of its column. • Step 3. Draw lines through appropriate rows and columns so that all the zero entries of the cost matrix are covered and the minimum number of such lines is used. • Step 4. Test for Optimality: (i) If the minimum number of covering lines is n, an optimal assignment of zeros is possible and we are finished. (ii) If the minimum number of covering lines is less than n, an optimal assignment of zeros is not yet possible. In that case, proceed to Step 5. • Step 5. Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to Step 3.
Example 1 • • • You work as a sales manager for a toy manufacturer, and you currently have three salespeople on the road meeting buyers. Your salespeople are in Austin, TX; Boston, MA; and Chicago, IL. You want them to fly to three other cities: Denver, ; Edmonton, and Fargo, ND. The table below shows the cost of airplane tickets in dollars between these cities. From/To Denver Edmonton Fargo Austin 250 400 350 Boston 400 600 350 Chicago 200 400 250 Where should you send each of your salespeople in order to minimize airfare?
Example 1 • Step 1. Subtract 250 from Row 1, 350 from Row 2, and 200 from Row 3. From/To Denver Edmonton Fargo Austin 250 400 350 Boston 400 600 350 Chicago 200 400 250 From/To Denver Edmonton Fargo Austin 0 150 100 Boston 50 250 0 Chicago 0 200 50
Example 1 • Step 2. Subtract 0 from Column 1, 150 from Column 2, and 0 from Column 3. From/To Denver Edmonton Fargo Austin 0 150 100 Boston 50 250 0 Chicago 0 200 50 From/To Denver Edmonton Fargo Austin 0 0 100 Boston 50 100 0 Chicago 0 50 50
Example 1 • Step 3. Cover all the zeros of the matrix with the minimum number of horizontal or vertical lines. From/To Denver Edmonton Fargo Austin 0 0 100 Boston 50 100 0 Chicago 0 50 50 Step 4. Since the minimal number of lines is 3, an optimal assignment of zeros is possible and we are finished. Assignment Cost Austin -Edmonton 400 Boston-Fargo 350 Chicago-Denver 200
Example 2 Machine/Job A B C D 1 8 5 2 4 2 6 7 11 10 3 3 5 7 6 4 5 10 12 9 Determine the optimum assignment of jobs to machine for the Data
Example 3 Machine/Job A B C D 1 12 16 14 10 2 9 8 13 7 3 15 12 9 11 Table contains information on cost to run 3 jobs on four machines. Determine an assignment plan that will minimize cost
Solution of Transportation Model
Solution of Transportation Model
Solution of Transportation Model
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