management science – mng221
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Management Science – MNG221. Lecture 1: Transportation, Transshipment, and Assignment. Transportation, Transshipment , and Assignment. T ransportation , transshipment, and assignment problems are special types of linear programming model formulations. - PowerPoint PPT PresentationTRANSCRIPT
Management Science – MNG221
Lecture 1: Transportation, Transshipment, and Assignment
Transportation, Transshipment, and Assignment
• Transportation, transshipment, and assignment problems are special types of linear programming model formulations.• They are part of a larger class of linear
programming problems known as network flow problems.• There solution approaches are variations of
the traditional simplex solution procedure.
Transportation ModelTransportation, Transshipment, and Assignment
Transportation Model
• Transportation models are formulated to solve problems where:A product is transported from a number of
sources to a number of destinations at the minimum possible cost.
Each source is able to supply a fixed number of units of the product, and each destination has a fixed demand for the product.
Transportation Model
• In a transportation problem, items are allocated from sources to destinations at a minimum cost.• The linear programming model for a
transportation problem has constraints for supply at each source and demand at each destination.
Transportation Model• A Transportation can be a:
1. Balanced transportation model in which supply equals demand, all constraints are equalities. Supply = Demand
2. Unbalanced transportation model in which supply exceeds demand or demand exceeds supply:Supply < Demand
Supply > DemandTransportation problems are usually solved manually within a tableau format.
The Transshipment ModelTransportation, Transshipment, and Assignment
Transshipment Model
• The transshipment model is an extension of the transportation model in which intermediate transshipment points are added between the sources and destinations.
• An example of a transshipment point is a distribution center or warehouse located between plants and stores.
Transshipment Model
• In a transshipment problem, items may be transported:1. From sources through transshipment
points on to destinations, 2. From one source to another,3. From one transshipment point to
another,
Transshipment Model
4. From one destination to another, 5. Or directly from sources to
destinations, 6. Or some combination of these
alternatives.• The transshipment model includes
intermediate points between sources and destinations.
The Assignment ModelTransportation, Transshipment, and Assignment
Assignment Model
• The Assignment Model is a special form of a linear programming model that is similar to the transportation model.
• The differences, however, is that the supply at each source and the demand at each destination are each limited to one unit.
Assignment Model
• The linear programming formulation of the assignment model is similar to the formulation of the transportation model, except all the supply values for each source equal one, and all the demand values at each destination equal one.
Transportation: Worked ExampleTransportation, Transshipment, and Assignment
Transportation: Worked Example
• Wheat is harvested in the Midwest and stored in grain elevators in three different cities - Kansas City, Omaha, and Des Moines.
• These grain elevators supply three flour mills, located in Chicago, St. Louis, and Cincinnati.
• The cost of transporting one ton of wheat from each grain elevator (source) to each mill (destination) differs according to the distance and rail system.
• The problem is to determine how many tons of wheat to transport from each grain elevator to each mill on a monthly basis in order to minimize the total cost of transportation.
Transportation: Worked ExampleGrain Elevator Supply
1. Kansas City 150
2. Omaha 175
3. Des Moines 275
Total 600
Mill Demand
A. Chicago 200
B. St. Louis 100
C. Cincinnati 300
Total 600
Mills
Grain Elevator A. Chicago B. St. Louis C. Cincinnati
1. Kansas City $6 $8 $10
2. Omaha $7 $11 $11
3. Des Moines $4 $5 $12
Transportation: Worked ExampleTransportation, Transshipment, and Assignment
Transportation
Transportation: Worked ExampleGrain Elevator Supply
1. Kansas City 150
2. Omaha 175
3. Des Moines 275
Total 600
Mill Demand
A. Chicago 200
B. St. Louis 100
C. Cincinnati 300
Total 600
Mills
Grain Elevator A. Chicago B. St. Louis C. Cincinnati
1. Kansas City $6 $8 $10
2. Omaha $7 $11 $11
3. Des Moines $4 $5 $12
Transportation: Worked ExampleMinimize Z = $61A + $81B + $101C + $72A + $112B + $112C + $43A + $53B + $123C
Subject to:Supply Constraints• x1A + x1B + x1C = 150
• x2A + x2B + x2C = 175
• x3A + x3B + x3C = 275
Demand Constraints• x1A + x2A + x3A = 200
• x1B + x2B + x3B = 100
• x1C + x2C + x3C = 300
Non-negativity• xij ≥ 0
Xij - the number of tons of wheat transported from each grain elevator, i - (where i = 1, 2, 3), to each mill, j -(where j = A, B, C).
Example Supply Constraint, x1A + x1B + x1C = 150 Kansas City to all three mills: Chicago (x1A), St. Louis (x1B), and Cincinnati (x1C).
Constraints are an equation ( = ) and not a inequality (≤), because all of wheat available is needed to meet demand of 600 tons.
TransportationMinimize Z = $61A + $81B + $101C + $72A + $112B + $112B + $43A + $53B + $123C
Subject to:x1A + x1B + x1C = 150 x1A + x2A + x3A = 200 xij ≥ 0x2A + x2B + x2C = 175 x1B + x2B + x3B = 100 x3A + x3B + x3C = 275 x1C + x2C + x3C = 300
•Each cell in a transportation tableau is analogous to a decision variable that indicates the amount allocated from a source to a destination.
•The supply and demand values along the outside of the rim of the tableau are called rim requirements.
FromTo
Transportation: Worked Example
• 2 -Methods for solving a Transportation Model –1. The Stepping-Stone Method 2. The Modified Distribution Method (also known
as MODI)
BUT
• Must be given an Initial Solution1. The Northwest Corner Method2. The Minimum Cell Cost Method3. Vogel’s Approximation Method
TransportationFrom
To Northwest Corner Method the largest possible allocation is made in the cell in the upper left hand corner of the tableau, followed by allocations to adjacent feasible cells.
The initial solution is complete when all rim requirements are satisfied.
150
50 100 25
275
The cost of the initial cost is obtained by substituting the allocations in the objective functionZ = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(150) + 8(0) + 10(0) + 7(50) + 11(100) + 11(25) + 4(0) + 5(0) + 12(275) = $5,925
TransportationFrom
To The Minimum Cost Method as much as possible is allocated to the cell with the minimum cost.
The initial solution is complete when all rim requirements are satisfied.
200 75
25 125
175
The cost of the initial cost is obtained by substituting the allocations in the objective functionZ = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(0) + 8(25) + 10(125) + 7(0) + 11(0) + 11(175) + 4(200) + 5(75) + 12(0) = $4,550
Transportation
• Vogel's Approximation Model is based on the concept of penalty cost or regret.
• A Penalty Cost is the difference between the largest and next largest cell cost in a row (or column).
• VAM allocates as much as possible to the minimum cost cell in the row or column with the largest penalty cost.
Transportation• Vogel's Approximation Model Steps1. Determine the penalty cost for each row and
column by subtracting the lowest cell cost in the row or column from the next lowest cell cost in the same row or column.
• Column A: L C -4, Next LC -6. As such 6-4 = 2
2. Select the row or column with the highest penalty cost (breaking ties arbitrarily or choosing the lowest-cost cell).
Transportation
• Vogel's Approximation Model Steps3. Allocate as much as possible to the
feasible cell with the lowest transportation cost in the row or column with the highest penalty cost.
4. Repeat steps 1, 2, and 3 until all rim requirements have been met.
TransportationFrom
To
2 3
2
1
4
1
175
Transportation
FromTo
2 3
2
After each VAM cell allocation, all row and column penalty costs are recomputed
1
1
175
100 8
Transportation
FromTo
2
2
After each VAM cell allocation, all row and column penalty costs are recomputed
1
175
100 825 150
150
Transportation
FromTo
The cost of the initial cost is obtained by substituting the allocations in the objective functionZ = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(0) + 8(0) + 10(150) + 7(175) + 11(0) + 11(0) + 4(25) + 5(100) + 12(150) = $5,125
175
10025 150
150
Transportation
• After obtaining the initial solution, the problem may be solved using:
1. Stepping-Stone Method2. Modified Distribution Method (MODI)
Stepping Stone Method – determines if there is a cell with no allocation that would reduce cost if used.
Modified Distribution Method (MODI)– a modified version of the stepping-stone method. However, the individual cell cost changes are determined mathematically.
Transportation
• Stepping Stone Method
• Step 1: Determine the stepping-stone path and cost changes for each empty cell in the tableau.
• Step 2: Allocate as much as possible to the empty cell with the greatest net decrease in cost.
• Step 3: Repeat steps 1 and 2 until an empty cell have positive cost changes that indicates an optimal solution.
TransportationFrom
To Step 1: Determine the stepping-stone path and cost changes for each empty cell in the tableau.
But First Identify Empty Cells :
1A2A2B3C
200 75
25 125
175
X
X X
X
TransportationFrom
To Evaluation of Cell1A
In evaluating the empty cells the constraint of the problems cannot be violated, and feasibility must be maintained.
Review of the cost increase/reduction of the process.
$6 - $8 + $5 - $4 = -$1
200 75
25 125
175
+1
-1 +1
-1
1A 1B 3B 3A
TransportationFrom
To Evaluation of Cell2A
Review of the cost increase/reduction of the process.
200 75
25 125
175+1
-1 +1
-1
2A 2C 1C 1B
-1 +1
3B 3A
+$7 - $11 + $10 - $8 +$5 - $4 = -$1
TransportationFrom
To Evaluation of Cell1A2A2B3C
Review of the cost increase/reduction of the process.
+$11 - $11 + $10 - $8 = +$2
200 75
25 125
175+1
-1 +1
-1
2B 2C 1C 1B
TransportationFrom
To Evaluation of Cell1A2A2B3C
Review of the cost increase/reduction of the process.
+$12 - $10 + $8 - $5 = +$5
200 75
25 125
175
+1
-1+1
-1
2B 2C 1C 1B
TransportationFrom
To Summary
1A – reduction $12A – reduction $12B – increase $23C – increase $5
200 75
25 125
175
TransportationFrom
To Evaluation of Cell1A2A2B3C
Identify the minimum in the stepping stone path.
200 75
25 125
175
+1
-1 +1
-1-25= 025
-25 = 175 + 25 = 100
The cost of the initial cost is obtained by substituting the allocations in the objective functionZ = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(25) + 8(0) + 10(125) + 7(0) + 11(0) + 11(175) + 4(175) + 5(100) + 12(0) = $4,525
The Assignment ModelTransportation, Transshipment, and Assignment
Assignment ModelAn assignment problem is a special form of transportation problem where all supply and demand values equal one.
Steps of the assignment solution method are:1. Perform row reductions by subtracting the
minimum value in each row from all row values.
Assignment ModelSteps of the assignment solution method (Continued)2. Perform column reductions by subtracting
the minimum value in each column from all column value.
3. In the completed Opportunity Cost Table, cross out all zeros, using the minimum number of horizontal or vertical lines.
Assignment ModelSteps of the assignment solution method (Continued)4. If fewer than m lines are required (where m
= the number of rows or column), subtract the minimum uncrossed value from all the other uncrossed values, and add this same minimum uncrossed value where two lines intersect. Leave all other values unchanged.
Assignment ModelSteps of the assignment solution method (Continued)5. If m lines are required, the tableau contains
the optimal solution and m unique assignments are made. If fewer than m lines are required, repeat step 4.
The Assignment Model- A Worked Example
Transportation, Transshipment, and Assignment
Assignment Model - Example• The Atlantic Coast Conference has four
basketball games on a particular night.
• The conference office wants to assign four teams of officials to the four games in a way that will minimize the total distance traveled by the officials.
• The distances in miles for each team of officials to each game location are shown below.
Assignment Model
Step 1: Perform row reductions by subtracting the minimum value in each row from all row values.
Assignment Model
Step 1: Perform column reductions by subtracting the minimum value in each column from all column value.
Assignment ModelStep 3: In the completed opportunity cost table, cross out all zeros, using the minimum number of horizontal or vertical lines.
Assignment ModelStep 4: If fewer than m lines are required (where m = the number of rows or column), subtract the minimum uncrossed value from all the other uncrossed values, and add this same minimum uncrossed value where two lines intersect. Leave all other values unchanged.
Assignment ModelStep 4: If m lines are required, the tableau contains the optimal solution and m unique assignments are made. If fewer than m lines are required, repeat step 4.
Assignment DistanceTeam A-Atlanta 90Team B-Raleigh 100Team C-Durham 140Team D-Clemson 120Total 420
Assignment ModelUnbalanced Assignment Tableau with a dummy column
•When demand exceeds supply, a dummy row is added to the assignment tableau.•When supply exceeds demands, a dummy column is added to the assignment tableau.