assignment and transportation

C h a p t e r 6 S u p p l e m e n t

Transportation andAssignment SolutionProceduresChapter Outline

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Chapter Outline

Learning Objectives

After completing this chaptersupplement, you should be able to:1. Use the transportation method to solve

problems manually.2. Deal with special cases in solving

transportation problems.

3. Use the assignment (Hungarian) method to solve problems manually.

4. Deal with special cases in solvingassignment problems.

Transportation Problems, 6S-4Formulating the Model, 6S-4

Solution of the TransportationModel, 6S-4Overview of the Solution

Technique, 6S-6Finding an Initial Feasible

Solution, 6S-6

Finding an Initial FeasibleSolution: The Northwest-Corner Method, 6S-7Finding an Initial Feasible

Solution: Intuitive Approach, 6S-8

Finding an Initial FeasibleSolution: Vogel’sApproximation Method, 6S-11

Evaluating a Solution forOptimality, 6S-13

Evaluation Using the Stepping-Stone Method, 6S-13

Evaluation Using the MODIMethod, 6S-16

Developing an ImprovedSolution, 6S-18

Special Issues, 6S-20

Assignment Problems, 6S-24The Hungarian Method, 6S-25Summary of Procedure, 6S-29Requirements for Use of the

Hungarian Method, 6S-29Special Situations, 6S-29

Summary, 6S-32

Glossary, 6S-32

Solved Problems, 6S-33

Discussion and ReviewQuestions, 6S-42

Problems, 6S-43


6S-4 Part 2 Deterministic Decision Models

Transportation Problems

Solution of the Transportation Model

Certain types of linear programming problems can be solved using special-purpose algo-rithms instead of using the simplex method. Prior to the widespread use of computers tosolve LP problems, special-purpose algorithms were particularly useful because they en-abled practitioners to obtain solutions to these special cases with much less computationalburden than simplex would have required. Now, these special-purpose algorithms providefurther insight into LP problems and their solutions.

This chapter supplement describes two special-purpose algorithms: the transporta-tion model and the assignment model. Model formulation, manual solution of models, andformatting models for computer solution are covered for each of these classes of prob-lems. The discussion begins with the transportation method.

The transportation model is usually applied to distribution-type problems, in which suppliesof goods that are held at various locations are to be distributed to other receiving locations.For example, a company may have 10 warehouses that are used to supply 50 retail stores.Obviously, there are many different combinations of warehouse-store supply lines thatcould be used. Generally, some of these combinations will involve transportation costs thatare higher than others. The purpose of using an LP model would be to identify a distribu-tion plan that would minimize the cost of transporting the goods from the warehouses tothe retail stores, taking into account warehouse supplies and store demands as well as trans-portation costs. Other examples of transportation problems include shipments from factoriesto warehouses, shipments between departments within a company, and production sched-uling. Moreover, some companies use the transportation method to compare location alter-natives (i.e., to decide where to locate factories and warehouses in order to achieve theminimum-cost distribution configuration).

Formulating the ModelA transportation problem typically involves a set of sending locations, which are referred to asorigins, and a set of receiving locations, which are referred to as destinations. In order to de-velop a model of a transportation problem, it is necessary to have the following information:

1. Supply quantity (capacity) of each origin.2. Demand quantity of each destination.3. Unit transportation cost for each origin-destination route.

The transportation algorithm requires the assumption that all goods be homogeneous,so that any origin is capable of supplying any destination, and the assumption that trans-portation costs are a direct linear function of the quantity shipped over any route.

In this supplement we will demonstrate how transportation and assignment problemscovered in Chapter 6 of the textbook can be manually solved.

In Chapter 6 of the textbook, the following example was used to illustrate the formulationand solution of the transportation model. In this supplement, we will show how this modelcan be solved using the MODI (Modified Distribution) and stepping-stone methods.


Let’s again consider this example. Harley’s Sand and Gravel Pit has contracted to pro-vide topsoil for three residential housing developments. Topsoil can be supplied from threedifferent “farms” as follows:

Weekly CapacityFarm (cubic yards)

A 100B 200C 200

Demand for the topsoil generated by the construction projects is

Weekly Demand Project (cubic yards)

1 502 1503 300

The manager of the sand and gravel pit has estimated the cost per cubic yard to ship overeach of the possible routes:

Cost per Cubic Yard to

From Project #1 Project #2 Project #3

Farm A $4 $2 $8Farm B 5 1 9Farm C 7 6 3

This constitutes the information needed to solve the problem. The next step is toarrange the information into a transportation table. This is shown in Table 6S-1. The ori-gins (farms) are listed down the left side of the table, and their respective supply quantitiesare listed down the right side of the table. The destinations (projects) are listed across thetop of the table, and their respective demands are listed across the bottom of the table. Theunit shipping costs are shown in the upper right-hand corner of each cell, which representsa shipping route. Hence, the cost per cubic yard to ship topsoil from farm A to project #1 is$4. (For convenience, dollar signs are not shown.)

Chapter 6 Supplement Transportation and Assignment Solution Procedures 6S-5

Table 6S-1 Transportation Table for Harley’s Sand and Gravel

To: Project Project Project SupplyFrom: #1 #2 #3

Farm A 100

Farm B 200

Farm C 200

Demand 50 150 300

4 2 8

9

3

1

6

5

7


Overview of the Solution TechniqueThe transportation solution technique is similar in certain respects to the simplex techniquebecause both involve an initial feasible solution that is evaluated to determine if it can beimproved. Moreover, both involve displaying initial and improved solutions in a series oftableaus or tables. However, as noted earlier, the transportation method requires consider-ably less computational effort. Moreover, it is not unusual to discover that the initial feasi-ble solution in a transportation problem is the optimum solution. Figure 6S-1 gives anoverview of the transportation method.

A solution to a transportation problem consists of quantities that are assigned to thevarious routes (i.e., cells in the table). These values can range from zero, which implies thatno units will be shipped over that route, to a maximum that equals the smaller of two quan-tities: the row (supply) and column (demand) totals. The logic of the maximum quantity issimple: The quantity shipped cannot exceed the available supply in a row, and it should notexceed the amount demanded (column total). Such a solution is a feasible solution. Thestarting point of the transportation method is a feasible solution.

Finding an Initial Feasible SolutionA feasible solution is one in which assignments are made in such a way that all supply anddemand requirements are satisfied. In general, the number of nonzero (occupied) cellsshould equal one less than the sum of the number of rows and the number of columns in a


Figure 6S-1 Overview of the Transportation Method

Develop animproved solution

by reallocation

Determine aninitial feasible

solution

Is thesolutionoptimal?

Start

Yes

No

Use NW-corner rule or intuitive method or Vogel’s approximation method

Evaluate using stepping-stone or MODI

Stop


transportation table. In the case of a table with 3 rows and 3 columns, the number of occu-pied cells should be 3 � 3 � 1 � 5 in order to be able to use the transportation algorithm.Sometimes, fewer occupied or completed cells appear in a solution. When that happens, thesolution is referred to as a degenerate solution: such a solution requires modification in orderto be able to determine if it is optimal. This topic is covered later in the supplement.

Aside from a degenerate case, the transportation method will generate solutions thathave the number of completed cells equal to the number of rows plus the number ofcolumns minus 1. Other feasible solutions that use more of the cells are imaginable but notreally desirable. For instance, it may be possible to imagine a solution that uses all of thecells. However, the cell costs are marginal costs to transport single units. Often, there arefixed costs associated with the number of cells (routes) used. For example, in the Harleyproblem, consider that, undoubtedly, each route requires a separate truck, which represents,perhaps, a high fixed cost. Hence, the more routes, the more trucks, and the higher the fixedcost. In addition, a solution that uses completed cells equal to the number of rows plusnumber of columns minus 1 will provide the minimum cost transportation schedule.Therefore, it would be desirable to devise a solution that used as few routes as possible. Ingeneral, such a solution will require the above-mentioned number of completed cells.

A number of different approaches can be used to find an initial feasible solution. Threeof these are described here:

1. The northwest-corner method.2. An intuitive approach.3. Vogel’s approximation method (VAM).

The northwest-corner method is a systematic approach for developing an initial feasiblesolution. Its chief advantages are that it is simple to use and easy to understand. Its chiefdrawback is that it does not take transportation costs into account. Consequently, such a so-lution may require much additional effort to obtain the optimal solution.

The northwest-corner method gets its name because the starting point for the alloca-tion process is the upper-left-hand (northwest) corner of the transportation table. For theHarley problem, this would be the cell that represents the route from farm A to project # 1.The following set of principles guides the allocation:

1. Begin with the upper-left-hand cell and allocate as many units as possible to that cell.This will be the smaller of the row supply and the column demand. Adjust the row andcolumn quantities to reflect the allocation.

2. Remain in a row or column until its supply or demand is completely exhausted or sat-isfied, allocating the maximum number of units to each cell in turn, until all supply hasbeen allocated (and all demand has been satisfied because we assume total supply anddemand are equal).

For the Harley problem, the sequence would be as follows (see Table 6S-2):

1. Beginning in cell A–1, allocate 50 units, exhausting demand in column 1 and leaving50 units of supply in row A.


Finding an Initial Feasible Solution: The Northwest-Corner Method

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2. Staying in row A, move to cell A–2, where supply is now 50 and demand is 150 units.Allocate 50 units to this cell, exhausting the supply of row A and leaving 100 units incolumn 2.

3. Staying in column 2, move down to cell B–2, where supply is 200 units. Allocate100 units to this cell, exhausting demand in column 2 and leaving 100 units of supplyin row B.

4. Staying in row B, move to cell B–3 and allocate 100 units, exhausting that row’s supplyand leaving 200 units in column 3.

5. Staying in column 3, move down to cell C–3 and allocate 200 units, exhausting both therow and column quantities.

In terms of minimizing total transportation cost, this solution may or may not be op-timal. We shall make that determination very shortly. At this point, let’s simply compute thetotal cost this solution would generate if it was implemented.

The total cost is found by multiplying the quantities in “completed” (i.e., nonempty)cells by the cell’s unit cost, then summing those amounts. Thus:

Total cost � 50(4) � $ 20050(2) � 100100(1) � 100100(9) � 900200(3) � 600

$1,900

As noted earlier, the main drawback of the northwest-corner method is that it does notconsider cell (route) costs in making the allocation. Consequently, if this allocation is opti-mal, that can be attributed to chance rather than the method used.

Finding an Initial Feasible Solution: The Intuitive ApproachThis approach, also known as the minimum-cost method, uses lowest cell cost as the basis forselecting routes. The procedure is as follows:

1. Identify the cell that has the lowest unit cost. If there is a tie, select one arbitrarily. Al-locate a quantity to this cell that is equal to the lower of the available supply for the rowand the demand for the column.

2. Cross out the cells in the row or column that has been exhausted (or both, if both havebeen exhausted), and adjust the remaining row or column total accordingly.



Farm A 50 (first) 50 (second) 100

Farm B 100 (third) 100 (fourth) 200

Farm C 200 (last) 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

Table 6S-2 Initial Feasible Solution for Harley Using Northwest-CornerMethod


3. Identify the cell with the lowest cost from the remaining cells. Allocate a quantity to thiscell that is equal to the lower of the available supply of the row and the demand for thecolumn.

4. Repeat steps 2 and 3 until all supply and demand have been allocated.

Let’s see how this method can be applied to the Harley problem. The cell with the low-est cost is B–2, in which the cost is $1 (see Table 6S-3a). Farm B has a supply of 200 cubicyards of gravel, whereas project # 2 has a demand for only 150 cubic yards. Consequently,we allocate 150 cubic yards (the lesser of the two quantities) to cell B–2. Because this ex-hausts the demand of project # 2, we cross out the cells in that column and the demand, andwe change the row total to 50 (see Table 6S-3b).

Of the cells that remain, C–3 has the lowest cost (A–2 cannot be used because it hasbeen crossed out). The supply of farm C is 200 cubic yards; the demand of project # 3 is 300cubic yards. The lesser of these is 200 cubic yards, so that quantity is placed in cell C–3.Since the supply of farm C has been completely used, the cells in row 3 are crossed out alongwith the row total. The remaining demand for project # 3 is 100 units, so that column totalmust be adjusted accordingly (see Table 6S-4).

Of the remaining cells, A–1 has the lowest cell cost. The column total is 50 and the rowtotal is 100; hence, the quantity 50 is assigned to cell A–1. Because the demand of project # 1has been satisfied, the cells in column A must be crossed out, and 50 units must be sub-tracted from the supply of farm A, learning 50 units of supply (see Table 6S-4).

At this point, only two remaining cells have not been crossed out: cells A–3 and B–3.Cell A–3 has the lower cost, so it is next in line for allocation. The remaining supply is 50units, and the remaining demand is 100 units; consequently, the quantity 50 is placed in A–3.



Farm A 100

Farm B 200

Farm C 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

Lowest cell cost

Table 6S-3a Find the Cell That Has the Lowest Unit Cost


Farm A 100

Farm B 150 200 50

Farm C 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

Table 6S-3b Allocate 150 Units to Cell B–2


This completes the use of supply for farm A, and it leaves a demand of 50 units for project 3.The last remaining cell, then, receives a quantity of 50 units, canceling the remaining sup-ply and demand for both its row and column, (see Table 6S-5).

We can easily verify that this is a feasible solution by checking to see that the row andcolumn totals of the assigned cell quantities equal the supply and demand totals for the rowsand columns. For example, 50 � 50, or 100 units have been allocated in the first row, whichequals the supply of farm A. Likewise, 50 units have been assigned in the first column, whichequals the demand of project 1. In addition, the number of occupied cells equals five, whichis equal to the number of rows plus number of columns minus 1 � (3 � 3 � 1 � 5).

This solution may or may not be optimal. In the next section, the procedure for testingfor optimality will be described. For now, let us simply compute the total cost of this solu-tion and compare it to that of the northwest-corner solution. Here we have

Cell Cost per Unit Number of Units Transportation Cost

A–1 $4 50 $200A–3 8 50 400B–2 1 150 150B–3 9 50 450C–3 3 200 600

500 $1,800



Farm A 50 100 50

Farm B 150 200 50

Farm C 200 200

Demand 50 150 300 500100

4 2 8

9

3

1

6

5

7

Table 6S-4 200 Units Are Assigned to Cell C–3 and 50 Units AreAssigned to cell A–1


Farm A 50 50 100

Farm B 150 50 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

Table 6S-5 Completion of the Initial Feasible Solution for theHarley Problem Using the Intuitive Approach


Compared to the plan generated using the northwest-corner method, this one has atotal cost that is $100 less. The fact that this plan is less costly than the previous one wasexpected: The northwest-corner method did not involve the use of cost information in al-locating units. Whether this plan is optimal, or can be improved on, remains to be seen.

Finding an Initial Feasible Solution: Vogel’s Approximation MethodIn addition to the intuitive approach and the northwest-corner method, we can use Vogel’sapproximation to obtain an initial solution.

Steps of the Vogel’s Approximation Method1. Determine the penalty cost for each row and each column of the transportation

tableau. Penalty cost is obtained by subtracting the smallest cost from the next smallestcost in a given row or column.

2. Select the row or column with the highest penalty cost. In that row/column, allocate asmany units as possible to the cell with the smallest cost.

3. Reduce the row supply and column demand by the amount allocated in step 2. If therow supply is now zero, eliminate the row from further consideration. If the columndemand is now zero, eliminate the column from further consideration. Go to step 1.

Now we are going to use the Vogel’s approximation method with the Harley problem.Step 1. In establishing the penalty cost for row 1 (farm A), we subtract the lowest cost in

row one from the second lowest cost in row 1. For farm A, the lowest cost is 2 (unit shippingcost from farm A to project 2). The second lowest cost is 4 (unit shipping cost from farm A toproject 1). The penalty cost for row 1 is 2, since $4 � $2 � $2. For column 1, farm A to project1 has the lowest per-unit transportation cost ($4) and farm B to project 1 has the second low-est transportation cost. Therefore, the penalty cost for column 1 (project 1) is $5 � $4 � $1.

Proceeding in this fashion for the rest of the rows and columns, we obtain the follow-ing penalty costs:

Farm A � 2 Project 1 � 1Farm B � 4 Project 2 � 1Farm C � 3 Project 3 � 5

Step 2. Since column (project) 3 has the largest penalty cost, it is selected. In column3, the shipping route from farm C to project 3 has the lowest shipping cost per unit ($3).Thus, we allocate as many units as possible (200 units) to it. Since the demand for farm 3 isreduced to zero, we eliminate farm C from further consideration and return to step 1.

Step 1. After eliminating farm C from further consideration, we recalculate thepenalty costs. Results are given below.

Farm A � 2 Farm B � 4 Project 1 � 1 Project 2 � 1 Project 3 � 1

Note that the penalty cost for column (project) 3 was reduced from 5 to 1 due to the elim-ination of row 3. Since the smallest value in column 3 is 8 and the next smallest value is 9,the penalty cost for column 3 is 1.

Step 2. Since farm B has the largest penalty cost ($4), it is selected. In row 2, the ship-ping route from farm B to project 2 has the lowest shipping cost per unit ($1). At this stage,the maximum units that can be allocated to farm B/project 2 shipping route is 150 units.

Step 3. Allocation of 150 units results in elimination of project 2 from further consid-eration because the demand at project 2 is exhausted.



Step 1. After eliminating project 2 from further consideration, we recalculate thepenalty costs. Results are given below.

Farm A � 4 Farm B � 4 Project 1 � 1 Project 3 � 1

Step 2. Since there is a tie between row 1 and row 2, we select the row that gives thesmallest unit cost after allocation. Therefore, we select row 1 and allocate as much as possi-ble to a feasible cell with the smallest per-unit shipping cost. Since farm A to project 1 has thesmallest per-unit shipping cost ($4) in row 1, we allocate 50 units from farm A to project 1.

Step 3. Allocation of 50 units results in elimination of project 1 from further consid-eration because the demand at project 1 is exhausted. We continue in this fashion until weobtain the transportation tableau given in Table 6S-6. Note that in this table, we not onlyshow the units allocated to particular shipping routes, but we also include the penalty costsat each iteration for each row and column.

Based on Table 6S-6, we have the following shipment quantities and transportationcosts for each cell:

Cell Cost per Unit Number of Units Transportation Cost

A–1 $4 50 $200

A–3 8 50 400

B–2 1 150 150

B–3 9 50 450

C–3 3 200 600

Total 500 $1,800

Total transportation cost of the above initial allocation tableau is $1,800, which is equal tothe total cost of the intuitive method.

In the previous section, we discussed three different methods for obtaining an initialfeasible solution. In the next section, we will investigate whether we can improve the initialfeasible solution.


Project Project Project Supply PC 1 PC 2 PC 3#1 #2 #3

Farm A 50 50 100 2 2 4

Farm B 150 50 200 4 4 4

Farm C 200 200 3 — —

Demand 50 150 300 500

*PC 1 1 1 5

PC 2 1 1 1

PC 3 1 — 1

4 2 8

9

3

1

6

5

7

*Penalty costs at iteration 1.

Table 6S-6 Vogel’s Approximation Initial Allocation Tableau with Penalty Costs


Evaluating a Solution for OptimalityThe test for optimality for a feasible solution involves a cost evaluation of empty cells (i.e.,routes to which no units have been allocated) to see if an improved solution is possible. Weshall consider two methods for cell evaluation:

1. The stepping-stone method.2. The MODI method.

The stepping-stone method involves a good deal more effort than the MODI method, as youwill note. However, it is discussed here because it provides an intuitive understanding of theevaluation process. When a solution is not optimal, the distribution plan must be revised byreallocating units into and out of various cells, and the stepping-stone method will be usedfor the reallocation.

Evaluation Using the Stepping-Stone MethodThe stepping-stone method involves tracing a series of paths in the transportation table,using one such path for each empty cell. The path represents a shift of one unit into anempty cell, and it enables the manager or analyst to answer a “what if …?” question: Whatimpact on total cost would there be if one unit were shifted into an unused route? The re-sult is a cost change per unit shifted into a cell. If the shift results in a cost savings, the step-ping-stone path also can be used to determine the maximum number of units that can beshifted into the empty cell, as well as modifications to other completed cells needed to com-pensate for the shift into the previously unused cell.

The name stepping-stone relates to an analogy of crossing a pond or stream by movingfrom stone to stone; in the case of a transportation solution, the “stones” are the occupied cells.

The initial feasible solution we found using the northwest-corner method is reproducedin Table 6S-7. We know it is not optimal because the intuitive method and Vogel’s approxi-mation method generated an initial solution that has a lower total cost. However, it will beinstructive to analyze the northwest-corner solution in order to see how the stepping-stonemethod works and to gain some insight into the process of developing an improved solution.

Only the unoccupied cells need to be evaluated because the question at this point is nothow many units to allocate to a particular route but only if converting a cell from zero unitsto nonzero (a positive value) would decrease or increase total costs. The unoccupied cells areA–3, B–1, C–1, and C–2. They must be evaluated one at a time, but in no particular order.

Let’s begin with cell B–1. We start by placing a � in the cell being evaluated, whichstands for the addition of one unit (e.g., a cubic yard of topsoil) to the cell. This will be the


Table 6S-7 Initial Feasible Solution Obtained Using theNorthwest-Corner Method


Farm A 50 50 100

Farm B 100 100 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7


only empty cell on the stepping path. In order to maintain the column total of 50, we mustsubtract one unit from an occupied cell; cell A � 2 is the only option because cell A � 1 isthe only occupied cell in column 1. This is designated by placing a � in cell A � 1 (see Table6S-8). Because we subtracted one unit from row A, we must compensate for this, which wecan do by adding a unit (i.e., placing a � sign) in cell A � 2. Similarly, we compensate for theaddition of one unit to column 2 by subtracting a unit from cell B � 2, and place a � in thatcell to reflect this. Because we initially added one unit to row B in cell B � 1, this last subtrac-tion also compensates for that, and we have traced a completed path, which we can use toevaluate B � 1. Before doing that, let’s consider rules that will guide tracing these paths.

Rules for Tracing Stepping-Stone Paths

1. All unoccupied cells must be evaluated. Evaluate cells one at a time.2. Except for the cell being evaluated, only add or subtract in occupied cells. (It is permissible

to skip over occupied cells to find an occupied cell from which the path can continue.)3. A path will consist of only horizontal and vertical moves, starting and ending with the

empty cell that is being evaluated.4. Alternate � and � signs, beginning with a � sign in the cell being evaluated.

Note that it is not necessary to actually alter the quantities in the various cells to reflect theone-unit change; the � and � signs suffice.

The general implication of the plus and minus signs is that cells with a � sign meanone unit would be added; cells with a � sign indicate one unit would be subtracted. The netimpact of such a one-unit shift can be determined by adding the cell costs with signsattached and noting the resulting value. Thus, for cell B–1, we have a net change of � 2:

Cells B–1

� cells � cells5 42 1

�7 �5

�7 � 5 � �2

This means that for each unit shifted into cell B–1 in this way (which is the only way a shiftcould be made), the total cost would increase by $2. Consequently, such a shift would not bedesirable.

Turning our attention to cell C–1, we begin its evaluation by placing � sign in that cell(see Table 6S-9). We can move horizontally or vertically to an occupied cell. Suppose wemove to cell C–3 and place a � sign there. Next, we move vertically to cell B–3, place a �sign in it, then horizontally to B–2, place a � sign there, move up to cell A–2, place a � sign



Farm A 50 50 100

Farm B 100 100 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7<

<< <

Table 6S-8 Evaluation Path for Cell B–1


Table 6S-10 Evaluation Paths for Cells A–3 and C–2

there, and, then, move horizontally to A–1 and place a � sign there. Coincidentally, some ofthe cells have � or � signs that are the same as in the previous evaluation path. However,this is not necessary; each evaluation path is independent of others in terms of assigning � and � signs. The evaluation of this path is �10.

Cell C–1

� cells � cells7 39 12 4

�18 �8

�10

Hence, using this path for reallocation would increase total cost by $10 per unit. Again, thiswould be undesirable.

Evaluation paths for empty cells A–3 and C–2 are shown in Table 6S-10. Their netchanges are

Cell A–3 Cell C–2

� cells � cells � cells � cells8 2 6 11 9 9 3

�9 �11 �15 �4

�2 �11

The negative value for cell A–3 indicates an improved solution is possible: For each unit wecan shift into that cell, the total cost will decrease by $2. The next question is, how manyunits can be reallocated into that cell?



Farm A 50 50 100

Farm B 100 100 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

<

<

<

< <

<


Farm A 50 50 100

Farm B 100 100 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

<

<

<

<

<<

<

<

Table 6S-9 Evaluation Path for Cell C–1


Before addressing that question, let’s consider an alternative method for cell evaluationthat avoids having to trace all of the evaluation paths.

Evaluation Using the MODI MethodThe MODI (MOdified DIstribution) method of evaluating a transportation solution foroptimality involves the use of index numbers that are established for the rows and columns.These are based on the unit costs of the occupied cells. The index numbers can be used toobtain the cell evaluations for empty cells without the use of stepping-stone paths.

There is one index number for each column and one for each row. These can be conve-niently displayed along the left and upper edges of a matrix. The index numbers are deter-mined using occupied cells so that the sum of the row index and the column index equals thecell’s unit transportation cost:

Row index � Column index � Cell costri � kj � cij (6S-1)

The index numbers are determined sequentially in a manner dictated by the position of oc-cupied cells. The number of equations is equal to the number of occupied cells(s), and thenumber of unknown variables is the number of rows plus the number of columns (6). Sincethe number of equations is less than the number of unknown variables, there is no uniquesolution to this, system of equations. However, a solution can be found by assigning a valueof zero as the index number of one of the rows or columns, reducing the number of un-known variables by one and making it equal to the number of equations.

The method will be illustrated by developing index numbers for the initial feasible so-lution for the Harley problem generated by the northwest-corner method, which is repeatedin Table 6S-11.

We begin by arbitrarily assigning a value of zero as the index for row 1. Once a row indexhas been established, it will enable us to compute column index numbers for all occupiedcells in that row. Similarly, once a column index number has been determined, index num-bers for all rows corresponding to occupied cells in that column can be determined.

The index number for column 1 is based on the fact that the sum of its value and therow index number must equal the cell cost of $4 for cell A–1. Thus, 0 � k1 � 4, so k1 � �4.Similarly, using occupied cell A–2, the index number for column 2 can be determined:0 � k2 � 2, so k2 � �2.



Farm A 50 50 100

Farm B 100 100 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

k1 k2 k3

Index numbers to be computed

r2

r3

r 1� 0

Table 6S-11 Initial Feasible Solution Obtained Using the Northwest-Corner Method


Knowledge of the index number for column 2 enables us to compute the index num-ber for row B using the unit cost for occupied cell B–2: r2 � 2 � 1, so r2 � �1. The value of r2

then enables us to compute the index number for column 3: �1 � k3 � 9, so k3 � �10. Theremaining index number, that of row 3, can be determined using the unit cost of occupiedcell C–3 and the column 3 index number, k3: r3 � 10 � 3, so r3 � �7.

The complete set of row and column index numbers is shown in Table 6S-12. Gener-ally, it is advisable to do a quick check of the values by confirming that for all occupied cells,the sum of the row and the column index number equals the unit cell cost.

We can now readily determine the cell evaluations (improvement potentials) for eachof the unoccupied cells using the relationship:

Cell evaluation � Cell cost � Row index � Column index

reduced costseij � cij � ri � kj (6S-2)

These determinations can be made in any order. For example, the cell evaluation for A–3is e13 � c13 � r1 � k3, or 8 � 0 � 10 � �2. This implies an improvement (decrease in totalcost) of $2 per unit for units that can be shifted into cell A–3. Similarly, for empty cell B–1,the improvement potential is 5 � (�1) �4 � �2, which indicates any units shifted intothis cell would increase total cost by $2 each. For unoccupied cell C–1, the evaluation is7 � (�7) � 4 � �10, and for cell C–2, the cell evaluation is 6 � (�7) � 2 � �11. Thesevalues are summarized in Table 6S-13. Note that they agree with the values we computedearlier using the stepping-stone method.



Farm A 50 50 �2 100

Farm B �2 100 100 200

Farm C �10 �11 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

k1 ��4 k2 � �2 k3 � �10

r1 � 0

r2 � �1

r3 � �7

Table 6S-12 Index Numbers for Initial Northwest-Corner Solution to the Harley Problem

Table 6S-13 Cell Evaluations for Northwest-Corner Solution for the Harley Problem


Farm A 50 50 �2 100

Farm B �2 100 100 200

Farm C �10 �11 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

k1 � �4 k2 � �2 k3 � �10

r1 � 0

r2 � �1

r3 � �7

*

* The numbers in circles represent reduced cost (cell evaluation).


When all evaluations are positive or zero, an optimal solution has been found. If one ormore is negative, the cell with the largest negative should be brought into solution becausethat route has the largest potential for improvement per unit. In this case, we found that cellA–3 had an evaluation of –2, which represented an improvement potential of $2 per unit.Hence, an improved solution is possible.

You can use the following summary of the MODI procedure to guide you through theprocess.

Rules for Using MODI

(As you determine each index number, write it in the appropriate location along the edge ofthe table.)

1. Begin by assigning a value of zero to the index number for any row or column of the table(arbitrarily selected).

2. For occupied cells in the row or column selected in step 1, assign a value to the row orcolumn index that is equal to the unit cost of the occupied cell. Ignore empty cells.

3. Use the index numbers generated by the first two steps and this step to obtain theremaining index numbers. In doing so, work only with unit costs of occupied cells. Use thisrelationship to compute a row index number: ri � cij � kj. For column index numbers, usethis relationship: kj � cij � ri.

4. Use the row and column index numbers to obtain cell evaluation values for eachunoccupied cell. Compute cell evaluations using this relationship: eij � cij � ri � kj.

5. Select the cell with the most negative eij value and allocate as many units as possible tothat cell and go to step 2. If all eij � 0, stop; further improvements to the solution are notpossible. Therefore, an optimal solution has been determined.

Developing an Improved SolutionDeveloping an improved solution to a transportation problem requires focusing on the un-occupied cell that has the largest negative cell evaluation. In the Harley problem, the onlynegative evaluation was for cell A–3.

Improving the solution involves reallocating quantities in the transportation table.More specifically, we want to take advantage of the improvement potential of cell A–3 bytransferring as many units as possible into that cell. The stepping-stone path for that cell isnecessary for determining how many units can be reallocated while retaining the balance ofsupply and demand for the table. The stepping-stone path also reveals which cells must havequantity changes and both the magnitude and the direction of the changes. The stepping-stone path for cell A–3 is reproduced in Table 6S-14. The � signs in the path indicate unitsto be added; the � signs indicate units to be subtracted. The limit on subtraction is the


Table 6S-14 Stepping-Stone Path for Cell A–3


Farm A 50 100

Farm B 100 100 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

<

<

<

<


smallest quantity in a cell with a negative sign along the cell path. The rationale for using thesmallest quantity is that using a larger quantity will result in a negative shipment quantityfor at least one route. There are two quantities in negative positions, 50 and 100. Because50 is the smaller quantity, that amount will be shifted in the following manner: Subtract50 units from each cell on the path with a � sign, and add 50 units to the quantity of eachcell with a � sign in it. The result is shown in Table 6S-15. A quick check reveals that thesums of quantities in each row and in each column are equal to original row and columntotals.

With each iteration (new solution), it is necessary to evaluate the empty cells to see iffurther improvement is possible. This requires use of either the MODI or the stepping-stone method. Both will yield the same values. Suppose we use the MODI method.

We begin by setting the index number for row 1 equal to zero. The column 1 indexnumber is found using the equality 0 � k1 � 4. Solving, we find k1 � �4. Similarly, for col-umn 3, 0 � k3 � 8, so k3 � �8. Using cell B–3 as a “pivot,” the index number for row 2can be found from the equality r2 � k3 � 9. Because k3 was found to equal �8, this meansthat r2 � �1. This value now allows us to compute the index number for column 2 becauser2 � k2 � 1: 1 � k2 � 1, so k2 � 0. Lastly, the index number for row 3 can be determined onthe basis of the unit cost of cell C–3 and our finding that k3 � �8. Thus, r3 � 8 � 3, so r3 � �5.These values provide the basis for computing evaluations for the empty cells using the rela-tionship eij � cij � ri � kj:

Cell A–2: e12 � c12 � r1 � k2 2 � 0 �0 � �2Cell B–1: e21 � c21 � r2 � k1 5 � 1 �4 � 0Cell C–1: e31 � c31 � r3 � k1 7 � (�5) �4 � �8Cell C–2: e32 � c32 � r3 � k2 6 � (�5) �0 � �11

Because none of these numbers is negative, this is an optimal solution.The index numbers and the cell evaluations are summarized in Table 6S-16. You may

recall that this was the same solution obtained using the intuitive method for the initial fea-sible solution (see Table 6S-4). At that point, it was determined that the total cost for thedistribution plan was $1,800.

The transportation method is summarized below.

Summary of the Transportation Method1. Obtain an initial feasible solution. Use either the northwest-corner method, the intuitive

method, or the Vogel’s approximation method. Generally, the intuitive method and Vogel’sapproximation are the preferred approaches.

2. Evaluate the solution to determine if it is optimal. Use either the stepping-stone methodor MODI. The solution is not optimal if any unoccupied cell has a negative cell evaluation.


Table 6S-15 Distribution Plan after Reallocation of 50 Units


Farm A 50 50 100

Farm B 150 50 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7


3. If the solution is not optimal, select the cell that has the most negative cell evaluation.Obtain an improved solution using the stepping-stone method.

4. Repeat steps 2 and 3 until no cell evaluations (reduced costs) are negative. Once you haveidentified the optimal solution, compute its total cost.

Special IssuesA number of special issues are discussed in this section in order to round out your under-standing of the transportation model. They are

1. Determining if there are alternate optimal solutions.2. Recognizing and handling degeneracy (too few occupied cells to permit evaluation of

a solution).3. Avoiding unacceptable or prohibited route assignments.4. Dealing with problems in which supply and demand are not equal.5. Solving maximization problems.

Alternate Optimal Solutions Sometimes transportation problems have multiple optimalsolutions. In such instances, it can be useful for a manager to be aware of alternate solutions,because this gives the manager an option of bringing nonquantitative considerations intothe decision.

In the case of the transportation problem, the existence of an alternate solution is sig-naled by an empty cell’s evaluation equal to zero. In fact, you may have noted that cell B–1had an evaluation equal to zero in the final solution of the Harley problem (see Table 6S-16).We can find out what that alternate solution is by reallocating the maximum number of unitspossible around the stepping-stone path for that cell. That path is shown in Table 6S-17a.The smallest quantity in a negative position on that path is 50 units. Shifting those 50 unitsresults in the distribution plan shown in Table 6S-17b.

As a check, note that the total cost is the same as before, $1,800:

100(8) � 50(5) � 150(1) � 200(3) � $1,800

Degeneracy In a transportation problem, degeneracy occurs when there are too few occu-pied cells to enable all the empty cells to be evaluated. In the case of the stepping-stonemethod, this means that there will be at least one empty cell for which an evaluation path


Table 6S-16 Index Numbers and Cell Evaluations


Farm A 50 �2 50 100

Farm B 0 150 50 200

Farm C �8 �11 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

�4 0 �8

0

�1

�5

* The numbers in circles represent reduced cost of cell evaluation.** Alternate optimal solution because cell evaluation � 0.

**

*


cannot be constructed. For the MODI method, it means that it will be impossible to deter-mine all of the row and column index numbers.

It is relatively simple to determine if a solution is degenerate: A solution is degenerate if thenumber of occupied cells is less than the number of rows plus the number of columns minus 1.

Test for Degeneracy

If the number of occupied cells is equal to R � C � 1, where

R � number of rows and C � number of columns, then there is no degeneracy

A quick check of the alternate solution to the Harley problem developed in the preced-ing section will reveal that the solution was degenerate: In Table 6S-17b, the number ofcompleted cells is 4, while the required number of completed cells is 3 � 3 � 1 � 5. Thispresented no difficulty because we were not concerned with evaluating the empty cellssince the transportation table represented an optimal solution. We simply were interested incomparing the total cost of that solution with the total cost of the original optimal solutionto verify that the two yielded the same total cost (alternate optimal solutions).

However, try to trace a stepping-stone path for any of the empty cells and you willunderstand the nature of the problem. It should be mentioned that this particular case issomewhat atypical in that usually some paths can be traced, but not all of them. Similarly, ifyou attempt to compute index numbers for the rows and columns, you will be unable tocompute them for row B, column 1, or column 2.

Obviously, some modification has to be made in order to determine if a given solutionis optimal. The modification is to treat some of the empty cells as occupied cells. This is


Table 6S-17a Index Numbers and Cell Evaluations


Farm A 50 50 100

Farm B 150 50 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

<

< <

<

Table 6S-17b Alternate Optimal Solution


Farm A 100 100

Farm B 50 150 200

Farm C 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7


accomplished by placing a delta (�) in one of the empty cells.1 The delta represents an ex-tremely small quantity (e.g., .001 unit); it is so small that supply and demand for the rowand column involved will be unaffected even without modifying other quantities in the rowor column, and so small that total cost will not change.

The purpose of the delta is to enable evaluation of the remaining empty cells. Thechoice of location for the delta can be somewhat tricky. Some empty cells may be unsuitableif they do not enable evaluations of the remaining empty cells. Moreover, the delta cannotbe placed in a cell that later turns out to be in a negative position of a cell path involved inreallocation because delta will be the “smallest quantity in a negative position” and shiftingthat minute quantity around the cell path will leave the solution virtually unchanged.Consequently, a certain amount of trial and error may be necessary before a satisfactorylocation can be identified for delta.

The technique can be demonstrated for the degenerate alternate solution of the Harleyproblem. Suppose that after some experimentation, cell A–1 has been selected for the locationof delta. (Not all choices would be acceptable. For example, try placing the delta in cell C–2and compute the improvement potentials for the empty cells. Remember that delta cannot bein a cell with a negative sign). The resulting index numbers generated using MODI and theimprovement potential for empty cells based on delta in cell A–1 are shown in Table 6S-18.Since all of the cell evaluations are nonnegative, this confirms that the solution is optimal.

Unacceptable Routes In some cases, certain origin-destination combinations may be un-acceptable. This may be due to weather factors, equipment breakdowns, labor problems, orskill requirements that either prohibit or make undesirable certain combinations (routes).

Suppose that in the Harley problem route A–3 was suddenly unavailable because of recentflooding. In order to prevent that route from appearing in the final solution (as it originallydid), the manager could assign a unit cost to that cell that was large enough to make that routeuneconomical and, hence, prohibit its occurrence. One rule of thumb would be to assign a costthat is 10 times the largest cost in the table. Thus, because the largest cost is $9, a unit cost of $90could be assigned instead of the original cost of $8 per unit. Then this revised problem couldbe solved using the northwest-corner or intuitive method for an initial solution and either step-ping-stone or MODI to evaluate the initial solution and any possible reallocations.

The optimal solution is shown in Table 6S-19. The prohibited route may appear in anonoptimal solution, but it will be eliminated by the time the optimal solution is reached.


1Actually, the number of deltas needed will equal the difference between the number of completed cells and R � C �1.However, you will only be exposed to the most common case in which one more completed cell is needed.

Table 6S-18 Harley Alternate Solution Modified for Degeneracy


Farm A � �2 100 100

Farm B 50 150 0 200

Farm C �8 �11 200 200

Demand 50 150 300 500

4 2 8

9

3

1

6

5

7

k1 ��4 k2 � 0 k3 � �8

r1 � 0

r2 � �1

r3 � �5

* The numbers in circles represent cell evaluation (reduced cost).

*


Unequal Supply and Demand Up to this point, examples have involved cases in which supplyand demand were equal. As you might guess, in most situations, the two are not equal. Whensuch a situation is encountered, it is necessary to modify the original problem so that supply anddemand are equal. This is accomplished by adding either a dummy column or a dummy row; adummy row is added if supply is less than demand and a dummy column is added if demand isless than supply. The dummy is assigned unit costs of zero for each cell, and it is given a supply(if the row total is less than the column total) or a demand (if the column total is less than therow total) equal to the difference between supply and demand. Quantities in dummy routes inthe optimal solution are not shipped. Rather, they serve to indicate which supplier will hold theexcess supply, and how much, or which destination will have a shortage, and how much it willbe short.

Let’s consider an example. Suppose that farm C in the Harley problem has experiencedan equipment breakdown, and it will be able to supply only 120 cubic yards of topsoil for aperiod of time. Therefore, total supply will be 80 units less than total demand. This willrequire adding a dummy origin with a supply of 80 units. The modified problem is shownin Table 6S-20, and the final solution is shown in Table 6S-21. We interpret the solution indi-cating that project 3 will be short 80 units per week until the equipment is repaired. Note,though, that this analysis has considered only transportation costs, and that other factors, suchas shortage costs or schedules of the projects, may dictate some other course of action.

If the intuitive approach is used to obtain the initial feasible solution when a dummy isinvolved, make assignments to the dummy last. Hence, begin by assigning units to the cellwith the lowest nonzero cost, then the next lowest nonzero cost, and so on. For the Harleyproblem, this would mean that units would be assigned first to cell B–2 because its cost of


Table 6S-19 Solution to Harley Problem with a Prohibited Route


Farm A 50 50 �80 100

Farm B �2 100 100 200

Farm C �10 �11 200 200

Demand 50 150 300 500

4 2 90

9

3

1

6

5

7

r1 � 0

r2 � �1

r3 � �7

Prohibitedroute

Table 6S-20 A Dummy Origin Is Added to Make Up 80 Units


Farm A 100

Farm B 200

Farm C 120

Dummy 80

50 150 300 300

4 2 8

9

3

1

6

5

7

0 0 0

k1 � �4 k2 � �2 k3 � �10


$1 is the lowest nonzero cell cost. In the next section, we will illustrate the solution proce-dure for the assignment problem.

Maximization Problems Some transportation-type problems concern profits or revenuesrather than costs. In such cases, the objective is to maximize rather than to minimize. Suchproblems can be handled by adding one additional step at the start: Identify the cell with thelargest profit and subtract all the other cell profits from that value. Then replace the cell prof-its with the resulting values. These values reflect the opportunity costs that would be incurredby using routes with unit profits that are less than the largest unit profit. Replace the originalunit profits with these opportunity costs and solve in the usual way for the minimum oppor-tunity cost solution. This will be identical to maximizing the total profit. For example, sup-pose in the Harley problem, the cell values had been unit profits instead of unit costs. CellB-3 had the largest dollar value: $9. Hence, each cell’s dollar amount would be subtractedfrom 9. For cell A-1, the resulting opportunity cost would have been 9 � 4 � 5, and so on.Cell B-3 would have an opportunity cost of 0, making it the most desirable route.

The remainder of the steps for developing an initial feasible solution, evaluation ofempty cells, and reallocation are identical to those used for cost minimization. When theoptimal distribution plan has been identified, use the original cell values (i. e., profits) tocompute the total profit for that plan (see Solved Problem 3).

The assignment problem is a special case of the linear programming problem that is similarto the transportation problem. It differs from the transportation problem such that the de-mand at each destination and supply at each source is equal to one. The general explanation,linear programming formulation, and Excel solution of the assignment problem was coveredin Chapter 6. In this chapter supplement, we will demonstrate the specialized solutionmethod for the assignment problem called the Hungarian method. The following example isused to illustrate the Hungarian method.

Example 6S-1

A manager has prepared a table that shows the cost of performing each of four jobsby each of four employees (see Table 6S-22). According to this table, Job 1 will cost $15if done by employee A, $20 if it is done by employee B, and so on. The manager has


Table 6S-21 Solution Using the Dummy Origin


Farm A 50 50 100

Farm B 150 50 200

Farm C 120 120

Dummy 80 80

50 150 300 500

4 2 8

9

3

1

6

5

7

0 0 0

Assignment Problems


stated that his goal is to develop a set of job assignments that will minimize the totalcost of getting all four jobs done. It is further required that the jobs be performedsimultaneously, thus requiring one job being assigned to each employee.

Although the manager recognizes that this problem can be solved using thesimplex routine, he also knows that he can solve the problem by hand using theHungarian method.


Table 6S-22 Job Costs for Each Possible Pairing

Employee

A B C D

1 $15 20 18 24

Job2 12 17 16 153 14 15 19 174 11 14 12 13

The Hungarian MethodThe Hungarian method provides a simple heuristic that can be used to find the optimal setof assignments. It is easy to use, even for fairly large problems. It is based on minimizationof opportunity costs that would result from potential pairings. These are additional coststhat would be incurred if the lowest-cost assignment is not made, in terms of either jobs(i.e., rows) or employees (i.e., columns). For example, we can see in Table 6S-22 that thelowest processing cost for job 1 is $15 when done by employee A. Therefore, if the jobwere assigned for some reason to employee B, the additional (i.e., opportunity) costwould be $20 � $15 � $5. Similarly, if job 1 were assigned to employee C, the opportunitycost would be $18 � $15 � $3, and if it were assigned to employee D, the opportunity costwould be $24 � $15 � $9. We can perform similar calculations for the other rows by iden-tifying the lowest cost for each row, then subtracting that value from each of the other costsin that row to obtain job opportunity costs for all job assignments. This is usually referredto as a row reduction. The results of the row reduction for the costs are shown in Table 6S-23.The procedure for a row reduction is summarized in the following.

Procedure for Row Reduction1. Identify the minimum value for each row.2. Subtract the minimum value in each row from all the values in that row.3. Use the resulting values to develop a new table.

Now, the same logic can be applied from the perspective of the employees; this is calledcolumn reduction. That is, because each employee will have a job, and because there arecost differences among employees, there can be opportunity costs in that respect. These willbe in addition to the job opportunity costs because the minimum-cost (column) assign-ments will not necessarily be the same as the minimum-cost row assignments. For instance,the lowest-cost assignment for job 1 was employee A. However, the lowest-cost assignmentfor employee A would be job 4, which has a cost of $11. The opportunity costs for employ-ees can be determined using the values obtained from the row reductions because they arein addition to those opportunity costs.


The procedure is similar to that for a row reduction:

Procedure for Column Reduction

1. Use the row-reduced matrix and identify the lowest-cost value in each column.2. Subtract the lowest-cost value in each column for each of the values in that column.3. Use the resulting values to form a new table.

The column reductions for this example are illustrated in Table 6S-24. Notice that becausethe minimum cost in column A is zero, there is no change in costs with a column reduction.This will always be the case for a row or column with a minimum cost equal to zero. Sinceboth the second and third columns have minimum costs of $1, the values in the new table areall one less in those columns than in the previous table. Similarly, the new values in the lastcolumn are $2 less than in the previous table.

Once both the row and column opportunity costs have been determined, we can attemptto make the minimum-cost assignments. Recognizing that assignments with opportunity costsof zero reflect minimum costs, it would be desirable to attempt to make assignments only withmatches that have costs of zero. For instance, assigning any job to employee A would have azero cost. However, once a job is assigned to A, no other job can be assigned to A (i.e., theother zero costs in that column become irrelevant). Consequently, even though it may appearon the surface that there are enough zeros to make zero-cost assignments, the fact that eachassignment eliminates an entire row and column from further consideration (i.e., assignments)means that a complete set of zero-cost assignments might not be possible at this juncture.

A quick method of determining if a set of zero opportunity cost assignments can bemade is to find the minimum number of lines needed to “cover” all the zero costs. That is, ifwe draw a line through the zero costs in column A, this will account for four of the sevenzeros. How many such lines will be needed at minimum to cross out all zeros? The answer istwo more (a line through row 3 and one through row 4), for a total of three (see Table 6S-25).Now, if the minimum number of covering lines equals the number of rows (or columns, becausethis is a square table), an optimal assignment is possible. In this case, apparently, an optimalassignment is not possible because only three lines were necessary. (Note that there is an-other way of covering the second zero in row 3; a vertical line could have been drawnthrough column B. The point is that only three lines, however drawn, would be needed.)

We must, therefore, make further reductions. No further row or column reductionsare possible because there is a zero in every row and column. To get around this, we do the


Original costs Cost after the row reduction

EmployeeRow

Employee

A B C D minimum A B C D

1 15 20 18 24 15 1 0 5 3 9

Job2 12 17 16 15 12

Job2 0 5 4 3

3 14 15 19 17 14 3 0 1 5 3

4 11 14 12 13 11 4 0 3 1 2

Row

reduction

Table 6S-23 Row Reduction


following: Subtract the smallest uncovered value ($1 in this case), from every other uncoveredvalue and adding that same amount to values that lie at an intersection of two coveringlines. This is done in Table 6S-26. The rationale for this is: Subtracting the smallest uncov-ered value reveals the next smallest increment in opportunity costs, whereas adding thatamount to intersections removes those assignments from consideration. Because both areat an intersection means that another assignment already exists in both that row and thatcolumn; hence, we can ignore those intersection possibilities.

We now can repeat the process of finding the minimum number of covering lines. In therevised table, a minimum of four lines is needed. Hence, an optimal assignment is now possible.


Table 6S-24 Column Reduction of Opportunity (Row Reduction) Costs

Revised (row reduction) costs

Employee

A B C D

1 0 5 3 9

Job2 0 5 4 3

3 0 1 5 3

4 0 3 1 2

Columnminimum 0 1 1 2

Column reduction

Employee

A B C D

1 0 4 2 7

Job2 0 4 3 1

3 0 0 4 1

4 0 2 0 0

A B C D

1 0 4 2 7

2 0 4 3 1

3 0 0 4 1

4 0 2 0 0

Table 6S-25 Determine the Minimum Number of Lines Needed to Coverthe Zeros


To avoid confusion in making the assignments, begin by identifying a row or columnthat has only a single zero. Candidates are column B and C and rows 1 and 3 because eachhas only one zero. Suppose we take row 1. We can denote the assignment of job 1 to em-ployee A by boxing in the 0 at that intersection, and because this assignment eliminates job1 and employee A from further consideration, we can cross out both column A and row 1.Next, we might select row 3 because it has a single zero, and box that one. This eliminates row3 and employee B, and they can be crossed out. Similarly, we can box the zero in column Cand row 4, and cross out that column and row 4. Finally, we can box the remaining zero incolumn D, row 2 to complete our assignments. These assignments are shown in Table 6S-27.

The total cost of the assignments can be determined by referring to the original costtable (Table 6S-22). The total cost is

Assignment Cost

1–A $152–D 153–B 154–C 12

Total cost � $57

It is instructive to note that this cost is equal to the sum of the row and column reduction costsplus the reduction amount for the revised (final) cost table. That is, the row reduction amounts’total was $15 � $12 � $14 � $11 � $52 (see Table 6S-23); and the column reduction amounts’total was $0 � $1 � $1 � $2 � $4 (see Table 6S-24). The revised cost table involved an addi-tional reduction of $1 for all uncovered numbers. Hence, the total reduction was $52 � ($1 �$1 � $2) � $1 � $57, which agrees with the total cost amount just determined above.


Employee

A B C D

1 0 3 1 6

Job2 0 3 2 0

3 1 0 4 1

4 1 2 0 0

Table 6S-27 Optimal Assignments

A B C D Revised costs

1 0 4 2 7 A B C D

2 0 4 3 1 1 0 3 1 6

3 0 0 4 1 2 0 3 2 0

4 0 2 0 0 3 1 0 4 1

4 1 2 0 0

Smallest uncovered cost

Table 6S-26 Further Revision of the Cost Table


Summary of Procedure1. Perform a row reduction on the cost table by subtracting the least cost in each row from

all costs in that row.2. Perform a column reduction on the cost table that results from the row reduction by

subtracting the least cost in each column from all costs in that column.3. Determine if an optimal assignment can be made by drawing the minimum number of

horizontal and/or vertical lines necessary to cover all zero costs. If the number of linesequals the number of rows, go to step 5.

4. If the minimum number of lines is less than the number of rows, identify the small-est uncovered opportunity cost. Subtract that amount from all uncovered costs andadd that amount to the covered costs that lie at line intersections. Go to step 3.

5. Make the assignments. Begin with a row or column that has a single zero. Box that zeroto indicate the assignment and eliminate that row and column from further considera-tion by drawing a line through the row and another line through the column. Continueassigning rows or columns with single zero-cost elements, then choose arbitrarily for as-signments where multiple zero-cost elements exist. Box that zero in that location andeliminate the row and column from further consideration by drawing a line through therespective row and column. Stop when all the zeros in the assignment table are covered.

Requirements for Use of the Hungarian MethodSituations in which the Hungarian method can be used are characterized by the following:

1. There needs to be a one-for-one matching of two sets of items.2. The goal is to minimize costs (or to maximize profits) or a similar objective (e.g., time,

distance, etc.).3. The costs or profits (etc.) are known or can be closely estimated.

Special SituationsCertain situations can arise in which the model deviates slightly from that previously described.Among those situations are the following:

1. The number of rows does not equal the number of columns.2. The problem involves maximization rather than minimization.3. Certain matches are undesirable or not allowed.4. Multiple optimal solutions exist.

The procedure outlined previously for assignment problems requires an equal numberof rows and columns. However, certain problems may not satisfy that requirement. For ex-ample, a situation might involve four jobs that need processing but there are only three ma-chines available for processing. Consequently, one job will not be processed immediately. Inorder to perform the analysis, and to learn which job will not be processed, an extra “ma-chine” must be added to the table. In analyzing the problem, one job will be assigned to thenonexistent machine. Hence, that will be the job that is not immediately processed.

Example 6S-2

Prepare this assignment table so that the optimal set of assignments can be madeusing the previously described procedure.



Job

1 2 3 4

A 15 19 12 16Machine B 23 21 18 17

C 20 16 11 19

SolutionCompensate for too few machines by adding another machine (row) to the table.Because no such machine exists, three will be no cost for the assignment. Hence, usecosts of 0 for this dummy row:

Job

1 2 3 4

A 15 19 12 16

MachineB 23 21 18 17C 20 16 11 19D 0 0 0 0

Then, proceed as before. Note that the column reduction step will have no effectbecause each column already has a zero. Simply skip that step and go on to the next step.

If the goal is to maximize rather than to minimize, one extra step must be added to thestart of the process: Identify the largest value in each column and, then, subtract all num-bers in each column from the column maximum. Having done that, perform the same stepsthat would be required if the problem were minimization because the modified values rep-resent opportunity costs. The set of assignments that minimizes the opportunity costs willalso maximize the original values.

Example 6S-3

The following table contains profits that would be realized from various possiblepairings. Prepare the table so that the optimal solution can be obtained using theHungarian method for minimization.

1 2 3

A 14 22 30B 20 18 40C 11 12 50

SolutionIdentify the maximum value in each column and then subtract every value in a givencolumn from the column maximum, as shown below:

Original values1 2 3

A 14 22 30B 20 18 40C 11 12 50

Column maximum 20 22 50



Opportunity costs1 2 3

A 6 0 20B 0 4 10C 9 10 0

In certain instances, a particular match or pairing may be either undesirable or other-wise unacceptable. For example, an employee may not have the skills necessary to performa particular job or a machine may not be equipped to handle a particular operation. For amanual solution when such a restriction is present a capital M, representing a large cost, isoften placed in the table in the position for that pairing. The M is equivalent to an artificialvariable (infeasible solution using LP simplex procedure). Analysis is performed as usual ex-cept that the M is ignored throughout the analysis. That is, the M is not used in any reduc-tions, nor is any value added to it or subtracted from it during the course of the analysis.

Example 6S-4

Determine the optimal set of pairings (minimum cost) given the following cost table.Note that assignment B–3 is undesirable, as denoted by the M in that position.

1 2 3

A 8 7 2B 1 4 MC 7 9 3

SolutionNotice how the M does not change throughout the analysis:

Original values After row reduction

1 2 3 1 2 3

A 8 7 2 A 6 5 0B 1 4 M B 0 3 MC 7 9 3 C 4 6 0

After column reduction

1 2 3

A 6 2 0B 0 0 MC 4 3 0

After further reduction, optimal solution

1 2 3

A 4 0 0B 0 0 MC 2 1 0

Total min. cost � 1 � 7 � 3 � 11



In some cases, there are multiple optimal solutions to a problem. This condition can beeasily recognized when making the optimal assignments: No unique 0 will exist at somepoint, resulting in more than one choice for assignment and, hence, more than one optimalsolution. It should be noted that all optimal solutions will yield the same value of the objec-tive function (e.g., the same minimum cost).

Example 6S-5

Given the final assignment table, identify two optimal solutions.

1 2 3

A 4 0 0B 0 3 2C 1 0 0

SolutionThe first assignment must be B–1, because B–1 is the only 0 that appears in a singlerow or column. Having made that assignment, there are two choices for the remain-ing two rows, and two choices for the remaining two columns. This results in twopossible solutions, as shown:

Optimal Solution I Optimal Solution II

1 2 3 1 2 3

A 4 0 0 A 4 0 0B 0 3 2 B 0 3 2C 1 0 0 C 1 0 0

This chapter supplement covers the transportation solution procedures. In the first part ofthe supplement, how to obtain an initial feasible solution is explained. Three differentmethods of obtaining an initial feasible solution are northwest-corner rule, intuitivemethod, and Vogel’s approximation. After obtaining an initial feasible solution, two special-ized methods of solving the transportation problem are demonstrated: (1) stepping-stonemethod and (2) MODI method. The supplement continues with an explanation of how todeal with special cases, including prohibited routes, maximization problems, degeneratesolutions, and alternate optimal solutions. The last part of the supplement describes thesolution procedure for the assignment problem. The specialized solution procedure calledthe Hungarian method is described with an example. The chapter supplement concludeswith the coverage of special situations for the assignment model.

Assignment Problem A problem that requires pairing two sets of items given a set ofpaired costs or profits in such a way that the total cost (profit) of the pairings is minimized(maximized).Column Reduction A step in the Hungarian method, whereby the smallest number in eachcolumn in subtracted from all of the numbers in that column.


Summary

Glossary


Degeneracy A condition that can occur in a transportation model. The number of occupiedcells is too small to permit evaluation for optimality without first modifying the solution.Hungarian Method A specialized solution procedure to solve the assignment problem.Provides a faster solution to the assignment problem than the simplex method.MODI MOdified DIstribution method, used to evaluate a transportation solution foroptimality. Involves the use of row and column index numbers.Northwest-Corner Method A Procedure for obtaining an initial feasible solution to a transportation problem that begins by allocating units to a upper-left-hand corner ofa transportation table and proceeds to make subsequent allocations moving in a southeastdirection on the transportation table.Row Reduction A step in the assignment method, whereby the smallest number in eachrow of a table is subtracted from all the numbers in its row.Stepping-Stone Method A procedure for determining if a solution to a transportation prob-lem is optimal that involves tracing closed paths from each empty cell through occupied cells.

Solved Problems

Solve this transportation problem for the minimum-cost solution. If there is an alternateoptimal solution, identify it.

First, check to see if supply and demand are equal. Demand is 50 � 55 � 17 � 122;supply is 61 � 61 � 122. Hence, they are equal.

Next, determine an initial feasible solution. This can be done using either thenorthwest-corner method, the intuitive method or the Vogel’s approximation method.Because the question did not specify which one, either can be used. Suppose we use theintuitive method. The lowest cell cost is $2 in A–1, so we begin by allocating as many unitsas possible to that cell. With a supply of 61 and a demand of 50, the most we can allocateis 50 (which is the smaller of the two quantities). This exhausts the demand of column 1and leaves 11 units still to be allocated in the first row. Because column 1 demand isexhausted, we draw a line through the costs in column 1 to remove them from furtherconsideration.


To:1 2 3 SupplyFrom:

A 61

B 61

Demand 50 55 17

93

6

2

5 4

Problem 1

Solution


A 50 61

B 61

Demand 50 55 17 122

93

6

2

5 411


The next lowest cost is the $3 in cell A–2, where demand is 55 and the remainingsupply is 11. Because 11 is the smaller of the two, allocate 11 units, and adjust the rowand column totals accordingly. Cross out the costs in row 1, since that supply has beenexhausted. This results in the following:

The remaining two cells are filled in the same manner, with the result:

Because the row and column quantities allocated, add to the row and column totals, this isa feasible solution.

Next, we must ensure that the minimum number of occupied cells exists, or else wewill have to adjust for degeneracy. That number is one less than the sum of the number ofrows and the number of columns. With two rows and three columns, the desired numberof occupied cells is: 2 � 3 � 1 � 4. Since this agrees with the number we have, there is nodegeneracy and we can proceed to the next step, which is evaluation for optimality.

We can use either the stepping-stone method or MODI because no method wasspecified. Since MODI is not as messy as stepping-stone, let’s use MODI. We begin with anindex number of 0 for the first row, which we write at the left of the first row. Then, for anycompleted cells in the first row, the column index number is the sum of the cell cost � 0.Hence, for column 1 we have an index number of 2, and for column 2, the index number is3. Once we have determined the index number for column 2, we can compute the indexnumber for the second row because cell B–2 is occupied. It is equal to the cell cost for B–2minus the index for column 2. Hence, it is 6 � 3 � 3. And with that row index, we candetermine the column index for column 3. It is equal to the cell cost for B–3 minus theindex for the second row: 4 � 3 � 1. The following table shows these index numbers.

Now, we can compute the cell evaluations for the empty cells, using the relationship: Cellevaluation � Cell cost � Row index � Column index. For cell B–1, this is 5 � 3 � 2 � 0; forcell A–3, it is 9 � 0 � 1 � �8. These are shown in the following table as circled values.



A 50 11 61

B 61

Demand 50 5544 17 122

93

6

2

5 4


A 50 11 61

B 44 17 61

Demand 50 55 17 122

93

6

2

5 4


A 50 11 61

B 44 17 61

Demand 50 55 17 122

93

6

2

5 4

k1 � 2 k2 � 3 k3 � 1

r1 � 0

r2 � 3


Because there are no negative cell evaluations, this is the optimal solution. Thus, tominimize transportation cost, ship 50 units from A to 1, 11 units from A to 2, 44 unitsfrom B to 2, and 17 units from B. The total cost will be

Cost � Quantity � Route Route per Unit Cost

A–1 $2 50 $100A–2 3 11 33B–2 6 44 264B–3 4 17 68

Total cost � $465

The appearance of a cell evaluation of zero for cell B–1 indicates that an alternatesolution with the same total cost exists. To determine what that other solution is, we musttrace the stepping-stone path for the cell that has the zero evaluation. The stepping stonefor cell B–1 is given in the following table.

The negative positions in the path are cells A–1 and B–2. The smallest quantity in anegative position is the 44 units in cell B–2. 44 is this amount that can be reallocatedaround the path, subtracting it from the quantity at each negative position and addingit at each positive position. After doing this, the alternate solution becomes apparent.It is

If you compute the total cost, you will see that it is the same as the total cost of theoriginal optimal solution. Hence, the two solutions are equivalent in that regard.



A 50 11 �8 61

B 0 44 17 61

Demand 50 55 17 122

93

6

2

5 4

k1 � 2 k2 � 3 k3 � 1

r1 � 0

r2 � 3


A 50 11 61

B 44 17 61

Demand 50 55 17 122

93

6

2

5 4


A 6 55 61

B 44 17 61

Demand 50 55 17 122

93

6

2

5 4

<<

<<


For the following transportation table, determine the initial feasible solution using theVogel’s approximation method.

In establishing the penalty cost for row 1 (Cleveland), we subtract the lowest cost in row 1from the second lowest cost in row 1. For Cleveland, the lowest cost is $50 (unit shippingcost from Cleveland to Ft. Wayne), and the second lowest cost is $60 (unit shipping costfrom Cleveland to South Bend); therefore, $60 � $50 � $10 is the initial penalty cost forrow 1 (Chicago). For column 1, Bowling Green to Chicago has the lowest per-unittransportation cost ($70) and Columbus to Chicago has the next lowest cost ($75).Therefore, the penalty cost for column 1 (Chicago) is $75 � $70 � $5

Proceeding in this fashion for the rest of the rows and columns, we obtain thefollowing penalty costs:

Cleveland � 10 Columbus � 15 Bowling Green � 20 Cincinnati � 7Chicago � 5 South Bend � 5 Indianapolis � 0 Ft. Wayne � 5

Since row 3 (Bowling Green) has the largest penalty cost, it is selected. In row 3, theshipping route from Bowling Green to Ft. Wayne has the lowest shipping cost per unit($35). Thus, we allocate as many units as possible (100 units) to it. Since the demand inFt. Wayne is reduced to zero, we eliminate Ft. Wayne from further consideration.

After eliminating Ft Wayne from further consideration, penalty costs are recalculated:

Cleveland � 10 Columbus � 10 Bowling Green � 15 Cincinnati � 7Chicago � 5 South Bend � 5 Indianapolis � 0

Since Bowling Green has largest penalty cost, it is selected again. Since Bowling Green toIndianapolis has the smallest cost for the Bowling Green supply point, allocate 175 units fromBowling Green to Indianapolis. This then eliminates Bowling Green from furtherconsideration.

The updated penalty costs are

Cleveland � 10 Columbus � 10 Cincinnati � 7Chicago � 5 South Bend � 5 Indianapolis � 0

Since Cleveland and Columbus have the largest penalty cost, Columbus is arbitrarlyselected. Since Columbus to Indianapolis has the smallest cost for the Columbus supplypoint, allocate 125 units from Columbus to Indianapolis. This eliminates Indianapolisfrom further consideration.

Continuing in this fashion gives the following completed transportation table withthe penalty costs. Note that the solution is degenerate because the number of allocations� 6 � (m � n � 1 � 4 � 4 � 1 � 7).


To:From: Chicago South Bend Indianapolis Fort Wayne Total

Cleveland 150

Columbus 175

Bowling 275Green

Cincinnati 100

Total 200 100 300 100 700

80 60 70 50

40

35

60485590

55

55

65

85

75

70

Problem 2

Solution


Solve this transportation problem for the maximum profit. The values in the upper-right-hand corner of each cell represent profit per unit for that cell. Use the intuitive method forthe initial solution and the MODI method to evaluate the empty cells.

a. Because this is a maximization problem, our first step is to convert it to a minimizationproblem so that we can solve it in the same way we solve all transportation problems.To do this, we note that the largest unit profit in the table is 9. Subtract all of the unitprofits in the table from 9, and enter the resulting opportunity costs in a new table:

Transportation Table with Opportunity Costs


Problem 3

To:A B C CapacityFrom:

#1 40

#2 50

#3 60

Demand 60 30 60 150

68

7

2

9

4 4 3

5

Solution


#1 40

#2 50

#3 60

Demand 60 30 60 150

31

2

7

0

5 5 6

4

To:From: Chicago South Bend Indianapolis Fort Wayne Total PC 1 PC 2 PC 3 PC 4 PC 5

Cleveland 150 150 10 10 10 20 —

Columbus 50 125 175 15 10 10 10 10

Bowling 175 100 275 20 15 — — —Green

Cincinnati 100 100 7 7 7 35 —

Total 200 100 300 100 700

Penalty 5 5 0 5Cost 1

Penalty 5 5 0 —Cost 2

Penalty 5 5 0 —Cost 3

Penalty 5 5 — —Cost 4

Penalty 5 — — —Cost 5

80 60 70 50

40

35

60485590

55

55

65

85

75

70


b. Now solve this revised problem as a regular transportation problem where the objectiveis to minimize the (opportunity) costs. The resulting solution will maximize profits.

(1) Make the initial set of allocations using the intuitive method: Begin with cell A–2because it has the lowest cell cost (0). Allocate the smaller of the row capacity (50)and the column demand (60). Thus, allocate 50 units, and adjust the row andcolumn totals accordingly. Cross out the costs in row 2 because its capacity is nowexhausted. The next lowest cost is for cell B–1. Allocate 30 units and cross out thecosts in column B and the column total of 30. Continue in this manner until allunits have been allocated to cells. The resulting shipping plan is

Intuitive Method Solution

(2) Determine the index numbers of MODI: Begin by assigning a row index numberof zero to the first row. Then, you can determine column index numbers using theoccupied cells in the first row. For column B, the index number is 1 and forcolumn C, it is 3. (Recall that the sum of the row and column index numbers foran occupied cell must equal the cell cost.) Next, use the index number for column Cand the cell cost for C–3 to obtain the index number for row 3. We can see that thesum of the row 3 index number (unknown at this point) and the column C indexnumber (3) must equal 6. (6 � r3 � 3 and r3 � 3) Hence, the row index numbermust equal 3. Continuing in this manner, we obtain index numbers for theremaining rows and columns. These are shown in the next table.

(3) Use the index numbers to evaluate the empty cells: For each empty cell, subtractthe sum of the row and column index numbers for that cell from the cell’s unitcost. The result is the cell evaluation. For cell A–1, we have 7 � (0 � 2) � �5. Forcell B–2, we have 2 � (�2 � 1) � �3. For cell C–2 we have 4 � (�2 � 3) � �3.For cell B–3, we have 5 �(3 � 1) � �1. These are shown in the following table.

(4) Because none of the cell evaluations are negative, this solution is optimal.(5) Find the total profit by using the unit profits for the original transportation

table, not the table containing the opportunity costs. This yields

30(8) � 10(6) � 50(9) � 10(4) � 50(3) � $940



#1 30 10 40

#2 50 50

#3 10 50 60

Demand 60 30 60 150

31

2

7

0

5 5 6

4

A B C Capacity

#1 �5 30 10 40

#2 50 �3 �3 0

#3 10 �1 50 60

Demand 60 30 60 150

31

2

7

0

5 5 6

4r1 � 0

r2 � �2

r3 � �3

k1 � �2 k2 � �1 k3 � �3


Given this table of job costs determine the optimal set of assignments of employees tojobs.

EmployeeA B C

1 6 4 7Job 2 9 5 2

3 5 7 1

First, note that the figures in the table represent costs, so this is a minimization problem.

Step 1. First, write the minimum value in each row down the right side of the table. Then sub-tract each row minimum from every value in the row. This will produce at least onezero in each row (See Step1 in the following table.)

Step 2. In the resulting table, write the minimum value in each column across the bottom,and then subtract the column minimum from each value in the column. Note thatwhen the minimum is zero, the numbers in the column remain the same. (See Step 2in the following table.)

Step 3. Now draw horizontal and/or vertical crossout lines such that all zeros are crossedout using as few lines as possible. If the number of crossout lines equals the num-ber of rows in a square table, the optimal solution can be identified in that table.Here there are only two crossout lines, so at least one more table will be needed.(See Step 3 in the following table.)

Step 4. Identify the smallest number that is not crossed out. It is 2. Subtract that numberfrom every number that is not crossed out. Also, add it to every intersection ofcrossout lines. If a number is crossed out but not at an intersection of crossoutlines, don’t change its value. (See Step 4 in the following table.)

Step 5. Redraw the crossout lines using as few as possible. If the number of lines equals thenumbers of rows (it does), an optimal solution can be identified. To identify opti-mal assignments, begin by finding a row (or column) that has only one zero. Placea box around that zero and cross out every other number in that row and column.Find another row or column that has only one zero, place a box around it, and crossout every other value in that row and column. Repeat this procedure until no fur-ther zeros exist. (Note: In some instances, there may not be a row or column withonly one zero. This means there will be at least one alternate solution. Refer to theexample in the text for the procedure.) (See Step 5 in the following table.)

Original Table

Employee Row A B C minimum

1 6 4 7 4Job 2 9 5 2 2

3 5 7 1 1

Step 1. Subtract row minimum.

A B C

1 2 0 32 7 3 03 4 6 0

2 0 0 Column minimum


Problem 4

Solution



Problem 5

Solution

Step 2. Subtract column minimum.

A B C

1 0 0 32 5 3 03 2 6 0

2 0 0 Column minimum

Step 3. Cross out zeros.

Step 4. Subtract 2 from unlined numbers.

Step 5.

A B C

1 0 0 5Job 2 3 1 0

3 0 4 0

Since for column B and job (row) 2, there is a single zero, the optimal assignments areemployee B to job 1, employee C to job 2, and employee A to job 3. Relating theassignments to the original table reveals that the minimum cost is 4 � 2 � 5 � 11.

EmployeeA B C

1 6 4 7Job 2 9 5 2

3 5 7 1

The following table contains processing costs for two jobs on various machines. Each jobmust be processed on one machine. Determine a set of pairings that will minimize totalprocessing costs.

Job1 2

A 8 7Machine B 5 4

C 3 1

The problem requires a pairing, or matching, of jobs and machines. Thus, it is anassignment problem. Assignment techniques require square tables (i.e., in this case, anequal number of jobs and machines). Because there are three machines and only two jobs,a third (dummy) job must be added. Processing costs of zero are assigned to each of thepossible pairings of the dummy job and the machines, resulting in this table:

Job1 2 3

A 8 7 0Machine B 5 4 0

C 3 1 0


Ordinarily, the next step would be to perform a row reduction. Note, however, that thesmallest value in each row is a zero. Consequently, a row reduction (i.e., subtracting thesmallest value in each row from the other values in the row) will result in the same values.Therefore, we move to the next step, which is a column reduction. The smallest value inthe first column is 3, the smallest in the second column is 1, and the smallest in the thirdcolumn is 0. The column reduction is illustrated in the next set of tables.

Job1 2 3

A 8 7 0Machine B 5 4 0 (Original values)

C 3 1 0Minimum column value: 3 1 0

Job (Values after the1 2 3 column reduction)


C 0 0 0

Now, we test to determine if the optimal solution can be found at this point. This involvescrossing out the zeros in the table with as few (horizontal or vertical) lines as possible. Asshown in the preceding table, only two lines are required. If the lines had equaled the numberof rows (or columns, because we had a square table after adding the dummy job), the optimalset of assignments could be made. Because that is not the case, we move on to the next step.

Identify the smallest value that is not crossed out. It is the 2 in location B–1. Circleit, as in the preceding table, and then subtract it from any value that has not beencrossed out. Also, add it to the intersection of the crossing lines. The results are shownin the next table.

Job1 2 3


C 0 0 0

At this point, three lines (not shown) would be needed to cross out all of the zero values.Therefore, we can identify the optimal solution. To do this, begin with a row or columnthat has only one zero. In this case, the first row has only one zero; so does the secondcolumn. Hence, either of those could be chosen for the initial assignment. Suppose wechoose A–3. We indicate this by drawing a box around the 0 at a A–3, and then cross outall of the values in row A and all of the values in column 3.

Job1 2 3


C 0 0 2

We must continue to make assignments to rows or columns that have only one zero,until none with unique zeros remain. Zeros that have been crossed out should be ignored.

Chapter 6S Transportation and Assignment Solution Procedures 6S-41



Discussion and Review Questions

At this point, row B has only one remaining zero, and the second column also has onezero. Therefore, either of these could be chosen. Suppose we choose B–1. Drawing a boxaround that zero and crossing out all of the other values in row B and column 1 leavesonly the 0 at C–2.

Job1 2 3


C 0 0 2

Thus, the last assignment is C–2, and our optimal set of assignments is A–3, B–1, and C–2.The (minimum) cost associated with this set of assignments can be determined byreferring to the optimal table of values and noting the cost of each assignment. Thus:

Assignment Cost

A–3 0B–1 5C–2 1

6

Because job 3 (the dummy) has been assigned to machine A, this tells us that, in reality,machine A will not have a job assigned to it.

1. In solving a problem using the MODI method, how do we know if a solution is optimal?2. In solving a problem using the stepping-stone method, how do we know if a solution

is optimal?3. Would it ever make sense to have a situation that required both a dummy source and

a dummy destination? Why or why not? Explain briefly.4. In using the stepping-stone method for a minimization problem, all of the reduced

costs for each empty cell are positive, except one reduced cost that is zero. What doesthis situation indicate? Briefly explain.

5. For a minimization transportation problem, if a solution is not optimal, how do youdecide which empty cell to ship unit to?

6. For a minimization transportation problem, if a solution is not optimal, once the cellis selected to ship units to, how do you decide how many units to ship?

7. How would you interpret a shipment of 100 units from a dummy source to an existingdestination?

8. How would you interpret a shipment of 200 units from an existing source to a dummydestination?

9. In using the MODI method, how many stepping stones would need to be evaluated ateach iteration? In using the stepping-stone method, how many stepping stones wouldneed to be evaluated at each iteration?

10. What is meant by the term degeneracy? How can it be overcome? Briefly explain.11. We have studied three initial starting solution methods. Which one of the initial start-

ing solution methods is least likely to result in the smallest cost? Why?12. Will the northwest-corner solution rule always result in a higher total cost than the

Vogel’s approximation rule? Why, or why not?


1. Consider the transportation problem of the California Dishwasher Company:

a. Develop an initial feasible solution using the northwest-corner method. Computethe total cost for this solution.

b. Evaluate the solution using the stepping-stone method. Is the solution optimal?Explain.

c. Repeat the evaluation using MODI and compare your cell evaluations to those ob-tained using the stepping-stone method.

d. Obtain an improved solution and evaluate it using MODI. Is it optimal?e. What is the total cost for your optimal solution?

2. Consider the transportation problem of Sunbelt Heat Pumps:

a. Develop an intuitive solution to this problem; evaluate it, and revise if necessary toobtain the minimum-cost solution. Then, compute the total cost.

b. Solve the problem using the northwest-corner and stepping-stone methods. Arethere alternate optimal solutions to this problem? How do you know?

3. Rod Steele, superintendent of Rochester Forging, has developed the following transporta-tion model. Solve it for the minimum-cost solution using the intuitive method for the ini-tial solution and MODI for evaluation. Is there an alternate optimal solution? What is it?


Problems

To:Store 1 Store 2 Store 3 SupplyFrom:

Warehouse A 50

Warehouse B 15

Warehouse C 55

Demand 25 50 45 120

12 20 15

4

8

11

14

9

20

To: Cincinnati Miami Memphis Tampa SupplyFrom: (C) (M) (Me) (T)

Houston 98(H)

Atlanta 80(A)

Demand 35 50 25 68

12 5 7 2

16311

To: Albany Buffalo Cleveland SupplyFrom: (A) (B) (C) (tons per day)

Pittsburgh 40(Pi)

Toronto 80(T)

Philadelphia 60(Ph)

Demand 70 60 50(tons per day)

14 8 3

7

4

5

12

9

6


5. The manager of Home Office Supplies, Gigi Staples, has just received demand forecastsand capacity (supply) figures for next month. These are summarized along with unitshipping costs in the following transportation table.

a. Using the intuitive method and MODI, find the minimum-cost solution.b. Which destinations will not receive their entire demand? How many units short

will each be?

6. Consider the transportation problem of Doors Plus, a producer of steel doors for publicschools:


To: D1 D2 D3 D4 D5 SupplyFrom:

O1 220

O2 260

O3 200

Demand 140 180 150 140 195

8

7

12

4

6

13

12

10

9

11

5

16

9

6

9

4. a. Find the initial solution to the following transportation problem using thenorthwest-corner method and optimum solution using the stepping-stonemethod. What is the total cost?

b. Find the minimum-cost solution using the intuitive method and MODI.

To: Detroit Danyer Kansas SupplyFrom: (De) (Dn) City (K)

Chicago 64(C)

St. Louis 44(S)

Omaha 36(O)

Demand 48 48 48

2 17 12

6

10

12

5

9

7

a. Using the intuitive method and MODI, find the minimum-cost solution.b. Now suppose that the Rochester–Chicago route is temporarily unavailable. Begin-

ning with the solution from (a), determine a distribution plan that will avoid thisroute. How much extra does this plan cost compared to the plan when all routesare possible? Note: Costs are in $100s.

To: Charlotte Columbus Chicago SupplyFrom: (Ca) (Co) (Ch)

Rochester 160(R)

Dallas 125(D)

Bulfalo 30(B)

Demand 75 90 150

26 15 22

17

20

12

24

18

28


7. The Future Furniture company recently began construction of a new warehouse. Dur-ing the construction period, several changes have occurred that require development ofa new distribution plan. The current figures for supply are

Capacity Plant (pieces per week)

East L.A. (ELA) 600West L.A. (WLA) 800El Toro (ET) 400

Current figures for demand are

Demand (pieces Warehouse per week)

Orange County #1 (OC1) 300Orange County #2 (OC2) 400Long Beach (LB) 200Los Angeles (LA) 900

Average shipping costs per unit are

To: Orange County #1 Orange County #2 Long Beach Los Angeles From: (OC1) (OC2) (LB) (LA)

East L.A. (ELA) $8 $9 $6 $4West L.A. (WLA) 10 7 2 3El Toro (ET) 5 7 9 12

a. Develop an initial feasible solution using the intuitive method.b. Find the optimal solution. What are the total cost?c. If the El Toro–Orange County #1 route becomes temporarily unavailable, what

distribution plan would minimize total cost? How much more would it cost thanif the El Toro–Orange County #1 route could be used?

8. Rework the Harley problem from the first section of this supplement, treating the cellvalues as unit profits instead of unit costs. Determine the optimal distribution plan andthe total profit for that plan.

9. Assume in the preceding problem that route A–3 is temporarily unavailable.a. Develop a solution that will maximize total profit without involving A–3.b. What are the total profits for this plan?

10. Consider the information given in Problem 6. Assume that the route from Dallas toColumbus is prohibited. Use Vogel’s approximation method and MODI to obtain theminimum cost solution.

11. Solve Problem 1 treating the cell values as revenue per unit instead of cost per unit.Develop the initial solution using the intuitive approach. What is the total revenue as-sociated with your optimal solution?

Chapter 6Supplement Transportation and Assignment Solution Procedures 6S-45


12. Consider the following transportation table:

a. Develop a distribution plan that minimizes total shipping cost.b. If route A–3 is unacceptable for some reason, what distribution plan would be

optimal? What would its total cost be?

13. Suppose the cell values in the previous problem represent unit profits and that allroutes are acceptable. What distribution plan would be best in terms of maximizingprofits? What would be the total profit for that plan?

14. Reconsider the situation presented in problem 6. Suppose that Chicago demand is tem-porarily reduced to 80 units.a. Using Vogel’s approximation method MODI, find the optimal solution.b. What is the total cost of the distribution plan?c. Which location will have excess capacity? How much?

15. a. For the following transportation tableau (cost minimization problem), determinethe initial feasible solution using Vogel’s approximation.

b. Based on the solution found in part a, determine the optimal solution usingMODI.

16. a. For the following cost minimization transportation tableau, determine the initialfeasible solution using Vogel’s approximation method.


To: Warehouse Warehouse Warehouse SupplyFrom: (W#1) (W#2) (W#3)

Plant A 700(PA)

Plant B 200(PB)

Plant C 200(PC)

Demand 300 400 400

50 32 40

20

42

30

28

16

35

To: Milwaukee, St. Louis, (S) Dayton, (D) SupplyFrom: WI (M) MO OH

Wichita, KS (W) 150

Omaha, NE (O) 175

Ames, IA (A) 275

Demand 200 100 300

6 9 10

11

12

7

5

8

4

To:A B C D SupplyFrom:

1 40

2 80

3 130

Demand 90 80 30 50

1612

24

18

23

42

14

27

3134 26

333


b. Use the initial feasible solution found in part a and determine the minimum-costtransportation schedule using MODI.

17. A soft drink manufacturer, Sara Soda, Ltd., has recently begun negotiations with bro-kers in the areas where it intends to distribute its products. Before finalizing the agree-ments, however, manager Dave Pepper wants to determine shipping routes and costs.The firm has three plants with capacities as shown below:

Capacity Plant (cases per week)

Metro 40,000Ridge 30,000Colby 25,000

Estimated demands in each of the warehouse localities are

Demand (cases Warehouse per week)

RS1 24,000RS2 22,000RS3 23,000RS4 16,000RS5 10,000

The estimated shipping costs per case for the various routes are

To:From: RS1 RS2 RS3 RS4 RS5

Metro .80 .75 .60 .70 .90Ridge .75 .80 .85 .70 .85Colby .70 .75 .70 .80 .80

a. Determine the optimal shipping plan that will minimize total shipping cost underthese conditions:

b. Route Ridge–RS4 is unacceptable.c. All routes are acceptable.d. What is the additional cost of the Ridge–RS4 route not being acceptable?

18. A shop foreman has prepared the following table, which shows the costs for variouscombinations of job-machine assignments:

MachineA B C

1 20 35 22Job 2 42 18 25

3 6 23 15

a. Perform a row reduction on the cost table.b. Perform a column reduction on the costs that resulted from the row reduction.c. Can an optimal assignment be made? Why or why not?d. What is the optimal (minimum-cost) assignment for this problem.e. What is the total cost for the optimum assignment?



19. The foreman of a machine shop wants to determine a minimum-cost matching for op-erators and machines. The foreman has determined the hourly cost for each of four op-erators for the four machines, as shown in the following cost table.

MachineA B C D

1 70 80 75 64

Operator2 55 52 58 543 58 56 64 684 62 60 67 70

a. Determine the minimum-cost assignment for this problem.b. What is the total cost for the optimal assignment?c. Is there an alternate optimal assignment? What is it? Calculate the total cost for the

alternate optimal assignment.

20. The approximate travel times for officiating crews for college basketball for four gamesscheduled for a weekend are shown in the following table:

Game siteSyracuse Buffalo Rochester Ithaca

A 1.2* 1.4 0.2 1.5Crew B 1.0 2.0 0.5 1.0

C 1.2 3.4 2.4 0.5D 2.1 3.1 1.1 0.8

*Time in hours

a. What set of crew assignments will minimize travel time?b. What is the total travel time required for the optimal assignment?

21. Suppose in the preceding problem that because of a previous conflict between crew Dand one of the teams playing at Ithaca, crew D cannot be assigned to that location.With this in mind, determine an optimal set of assignments. What effect, if any, doesthis restriction have on the total travel time for all crews?

22. (Refer to Problem 20.) Suppose that a basketball game in Niagara Falls has been can-celed, freeing up another crew (crew E) to select from. Crew E’s travel times to Syra-cuse, Buffalo, Rochester, and Ithaca are 1.8, .5, .6, and 2.0 hours, respectively. Assume,too, that the D–Ithaca assignment is undesirable.a. Determine the set of assignments that will minimize total travel time.b. Which crew will not be assigned?c. Is there an alternate optimal solution? Explain.

23. An industrial engineer has prepared a table that shows the costs for each possible com-bination of job and machine for four jobs, as shown:

MachineA B C D

1 45 75 80 35Job 2 55 60 60 65

3 70 65 50 454 60 75 70 65



a. Determine the set of assignments that will minimize total processing cost.b. What is total processing cost?c. Is there an alternate optimal solution? What is it?

24. If the numbers in the previous problem reflected profits rather than costs, what set ofassignments would have been optimal?

25. (Refer to Problem 23.) Assuming cost minimization objective. Suppose that job 1 couldnot be assigned to machine D because of a technical problem. What set of assignmentswould now minimize total cost? What additional cost is incurred because of the tech-nical problem?

26. An analyst has kept track of the number of defectives produced by five workers on fivedifferent machines. The results are shown below for a run of 400 units per machine:

MachineA B C D E

1 9 7 4 6 22 7 4 5 2 1

Worker 3 3 4 3 2 34 9 7 8 6 55 0 3 2 4 3

a. Determine a set of assignments that will minimize the total number of errors fora given run size.

b. Is there an alternate optimal solution? If so, what is it?c. What is the total number of defectives expected for the optimal assignment?

27. In the previous problem, if another machine (machine F) was available and the num-ber of defectives produced by the five workers for a run of 400 units was 4, 4, 2, 3,and 1, respectively, determine the following:a. A set of assignments that will minimize the total number of expected defectives.b. Which machine will not be used?c. Is there an alternate optimal solution? If so, what is it?

28. A manager has four jobs that must be assigned. Estimated processing times for eachemployee are shown in the accompanying table.

EmployeeSmith Jones Green Mehl

1 6.2* 8.0 5.4 4.8

Job2 6.0 7.2 5.8 4.43 5.5 6.0 6.6 6.84 6.3 6.6 7.0 7.3

*Time in hours.

a. Determine a set of assignments that will minimize total processing time.b. What is the total processing time for the optimal assignments?

29. A company has four manufacturing plants and four warehouses. Each of the plants willsupply one of the warehouses. Each warehouse has a demand that is equal to a factory’soutput for a given period. Assume that all capacities and all demands are 300 units perperiod. The shipping costs for all routes are shown in the table below. The costs are inthousands.



Warehouse1 2 3 4

A 2.4 1.8 2.6 1.8Plant B 2.0 1.9 2.2 2.2

C 1.4 1.7 2.0 1.4D 1.6 2.1 2.1 1.6

a. Determine a set of assignments that will minimize total shipping cost for a period.b. What is the total shipping cost for your plan?c. Is there an alternate optimal solution? If so, what is it?d. Would the optimal solution change if the values represented shipping time instead

of shipping cost? Explain.

30. On Monday morning, the manager of a small print shop, Carri Fonts, finds that fourjobs must be handled on a “rush” basis. Fortunately, there are four employees availableto work on these jobs, and each will handle one of the jobs. Each employee has aslightly different estimated completion time for each job, as shown in the table below:

Completion time (hours)Job

A B C D

Tom 4.2 4.1 5.4 5.0

EmployeeDick 4.4 4.0 5.2 4.8Harry 4.3 4.2 5.0 4.9Jane 4.0 4.1 5.4 5.0

Carri wants to determine how to assign the employees to jobs so that the total comple-tion time is as low as possible. Determine the set of assignments that will minimize thecompletion time of all jobs.



assignment and transportation

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