operations management by stevenson, chapter 8s

Rev.Confirming Pages

8S-1

8

The transportation problem involves finding the lowest-cost plan for distributing stocks of

goods or supplies from multiple origins to multiple destinations that demand the goods. For

instance, a firm might have three factories, all of which are capable of producing identi-

cal units of the same product, and four warehouses that stock or demand those products, as

depicted in Figure 8S.1 . The transportation model can be used to determine how to allocate

the supplies available from the various factories to the warehouses that stock or demand those

goods, in such a way that total shipping cost is minimized (i.e., the optimal shipping plan).

Usually, analysis of the problem will produce a shipping plan that pertains to a certain period

of time (day, week), although once the plan is established, it will generally not change unless

one or more of the parameters of the problem (supply, demand, unit shipping cost) changes.

8S-1 INTRODUCTION Although Figure 8S.1 illustrates the nature of the transportation problem, in real life manag-

ers must often deal with allocation problems that are considerably larger in scope. A beer

maker may have four or five breweries and hundreds or even thousands of distributors, and

an automobile manufacturer may have eight assembly plants scattered throughout the United

States and Canada and thousands of dealers that must be supplied with those cars. In such

cases, the ability to identify the optimal distribution plan makes the transportation model very

important.

LO8S.1. Describe the nature of a transportation problem.

SUPPLEMENT TO

C H A P T E R

SUPPLEMENT OUTLINE

8S.1 Introduction 8S-1

8S.2 Obtaining an Initial Solution 8S-3

The Intuitive Lowest-Cost Approach 8S-3

8S.3 Testing for Optimality 8S-4

Evaluating Empty Cells: The Stepping-

Stone Method 8S-5

Evaluating Empty Cells: The MODI

Method 8S-8

8S.4 Obtaining an Improved Solution 8S-9

8S.5 Special Problems 8S-12

Unequal Supply and Demand 8S-12

Degeneracy 8S-12

Summary of Procedure 8S-14

8S.6 Location Decisions 8S-14

8S.7 Other applications 8S-14

8S.8 Computer Solutions 8S-15

Key Terms 8S-16

Solved Problems 8S-16

Problems 8S-21

Discussion and Review

Questions 8S-25

The Transportation Model LEARNING OBJECTIVES

After completing this supplement you

should be able to:

LO8S.1 Describe the nature of a

transportation problem.

LO8S.2 Solve transportation problems

manually and interpret the

results.

LO8S.3 Set up transportation problems in

the general linear programming

format.

LO8S.4 Interpret computer solutions.

SCREENCAM TUTORIAL

ste24102_ch08s_08S1-08S25.indd 8S-1ste24102_ch08s_08S1-08S25.indd 8S-1 1/24/14 7:20 PM1/24/14 7:20 PM


8S-2 Supplement to Chapter Eight The Transportation Model

The shipping (supply) points can be factories, warehouses, departments, or any other place

from which goods are sent. Destinations can be factories, warehouses, departments, or other

points that receive goods. The information needed to use the model consists of the following:

1. A list of the origins and each one’s capacity or supply quantity per period.

2. A list of the destinations and each one’s demand per period.

3. The unit cost of shipping items from each origin to each destination.

This information is arranged into a transportation table (see Table 8S.1 ).

The transportation model is one of a class of linear programming models, so named

because of the linear relationships among variables. In the transportation model, transporta-

tion costs are treated as a direct linear function of the number of units shipped.

Use of the transportation model implies that certain assumptions are satisfied. The follow-

ing are the major ones:

1. The items to be shipped are homogeneous (i.e., they are the same regardless of their

source or destination).

2. Shipping cost per unit is the same regardless of the number of units shipped.

3. There is only one route or mode of transportation being used between each origin and

each destination.

e celxmhhe.com/stevenson12e

FIGURE 8S.1 The Transportation problem

involves determining a minimum-

cost plan for shipping from

multiple sources to multiple

destinations D

(demand)

S

(supply)S

(supply)

S

(supply)

D

(demand)

D

(demand)

D

(demand)

TABLE 8S.1 A transportation table

Cost to ship

one unit from

factory 1 to

warehouse A

Factory 1

can supply 100

units per period

Total supply

capacity per

period

Total demand

per period

Warehouse B

can use 90 units per period

Warehouse

Factory

11774

88312

516108

A B C D

2

3

80 90 120 160 450

450

150

200

100

Demand

Supply



Supplement to Chapter Eight The Transportation Model 8S-3

The solution to the problem shown in Figure 8S.1 would look like the one shown in

Figure 8S.2 .

The transportation model starts with the development of a feasible solution, which is

then sequentially tested and improved until an optimal solution is obtained. The description

of the technique on the following pages focuses on each of the major steps in the process in

this order:

1. Obtaining an initial solution.

2. Testing the solution for optimality.

3. Improving sub optimal solutions.

8S-2 OBTAINING AN INITIAL SOLUTION To begin the process, it is necessary to develop a feasible distribution plan. A number of

different methods are available for obtaining such a plan. The discussion here will focus

on the intuitive approach , a heuristic approach that yields an initial solution that is often

optimal or near optimal.

The Intuitive Lowest-Cost Approach With the intuitive approach, cell allocations are made according to cell cost, beginning with

the lowest cost. The procedure involves these steps:

1. Identify the cell with the lowest cost.

2. Allocate as many units as possible to that cell, and cross out the row or column (or both)

that is exhausted by this.

3. Find the cells with the next lowest cost from among the feasible cells.

4. Repeat steps (2) and (3) until all units have been allocated.

Cell 1-D has the lowest cost ($1) (see Table 8S.2 ). The factory 1 supply is 100, and the ware-

house D demand is 160. Therefore, the most we can allocate to this cell is 100 units. Since

LO8S.2 Solve transportation problems manually

Intuitive approach Heuristic

approach to obtaining an initial

solution.

FIGURE 8S.2 Solution for the problem shown

in Figure 8S.1

Warehouse

Factory

11774

88312

516108

A B C D

2

3

80

80

90 120

90 110

10 90

70

160 450

450

150

200

100

Demand

Supply

TABLE 8S.2 Warehouse

Factory

11774

88312

516108

A B C D

2

3

80 90 120

100

60 450

450

150

200

1000

Demand

Supply

160




8S-3 TESTING FOR OPTIMALITY Testing for optimality and revising sub optimal solutions involves analysis of each unused

(empty) cell to determine the potential for reducing the total cost of the solution, This is

accomplished by transferring one unit into an empty cell and noting its impact on costs. If

the supply of 100 is exhausted, we cross out the costs in the first row along with the supply

of 100. In addition, we must adjust the column total to reflect the allocation, which leaves

60 units of unallocated demand in column D.

The next lowest cost is $3 in cell 2-B. Allocating 90 units to this cell exhausts the column

total and leaves 110 units for the supply of factory 2 (see Table 8S.3 ). Also, we cross out the

costs for column B.

The next lowest cost (that is not crossed out) is the $5 in cell 3-D. Allocating 60 units to

this cell exhausts the column total and leaves 90 units for row 3. We now cross out the costs

in column D (see Table 8S.4 ).

At this point, there is a tie for the next lowest cell cost: cell 3-A and cell 2-C each has a

cost of $8. Break such a tie arbitrarily. Suppose we choose cell 3-A. The demand is 80 units,

and the remaining supply is 90 units in the row. The smaller of these is 80, so that amount is

placed in cell 3-A (see Table 8S.5 ). This exhausts the column, so we cross out the cell costs in

column A. Also, the remaining supply in factory 3 is now 10.

The other cell in the tie, cell 2-C, still remains. The remaining supply in the row is 110 and

demand in the column is 120. The smaller of these two amounts, 110, is assigned to the cell,

and the row costs are crossed out. The row total is changed to zero, and the column total is

changed to 10 units. The only remaining cell with a cost that has not been crossed out is cell

3-C. Both supply and demand are 10 units, so 10 units are assigned to the cell. This completes

the initial allocation. (See Table 8S.6 .)

Now let’s determine if this initial solution obtained using the intuitive method is optimum.

TABLE 8S.3

Warehouse

Factory

11774

88312

516108

A B C D

2

3

80 90

90

120

100

60 450

450

150

200

100

110

Demand

Supply

160

TABLE 8S.4

Warehouse

Factory

11774

88312

516108

A B C D

2

3

80 90

90

120

100

60

60

450

450

150

200

100

110

90

Demand

Supply

160

TABLE 8S.5

Warehouse

Factory

11774

88312

516108

A B C D

2

3

80

80

90

90

120

100

60

60

450

450

150

200

100

110

10

90

Demand

Supply

160

TABLE 8S.6 Initial allocation using intuitive approach

Warehouse

Factory

11774

88312

516108

A B C D

2

3

80

80

90

90 110

10

120

100

60

450

450

150

200

100

Demand

Supply

160




costs are increased, that implies that using the cell would increase total costs. If costs remain

the same, that implies the existence of an alternative option with the same total cost as the

current plan. However, if analysis reveals a cost decrease, the implication is that an improved

solution is possible. The test for optimality requires that every unused cell be evaluated for

potential improvement. Either of two methods can be used: stepping-stone or MODI.

Evaluating Empty Cells: The Stepping-Stone Method In the stepping-stone method , cell evaluation proceeds by borrowing one unit from a full cell

and using it to assess the impact of shifting units into the empty cell. For instance, if a shift

of one unit causes an increase of $5, total costs would be increased by $5 times the number

of units shifted into the cell. Obviously, such a move would be unwise since the objective is

to decrease costs.

The name stepping-stone derives from early descriptions of the method that likened the

procedures to crossing a shallow pond by stepping from stone to stone. Here, the occupied

cells are the “stones”; shifting units into empty cells requires borrowing units from occupied

cells. To maintain the balance of supply and demand for every row and column, a shift of one

unit into an empty cell requires a series of shifts from other occupied cells.

The best way to understand the evaluation process is to consider a simple example.

Suppose we want to evaluate cell 1-A (see Table 8S.7 ). We must shift one unit into that cell,

which implies that one unit will be shipped from factory 1 to warehouse A. However, factory

1 has only 100 units it can ship, and all are allocated to warehouse D. Therefore, to ship one

unit to warehouse A, we borrow a unit from cell 1-D. This creates two problems: Column A

now has one extra unit, and Column D is short one unit (i.e., A has 1 � 80 � 81 units but

needs only 80 units, and D has 99 � 60 � 159 but needs 160 units). We remedy these prob-

lems by subtracting one unit from cell 3-A and adding it to cell 3-D; this gives column A a

total of 1 � 79 � 80 and column D a total of 99 � 61 � 160. (Note: Instead of making each

addition or subtraction of the single unit, a � or � sign is simply inserted in the appropriate

cell to signify the operation.)

Let’s see what effect such a shift would have on costs. For each cell to which a unit was

added, the cost will increase by the transportation cost for that cell; for cells that gave up a

unit, costs will decrease by the cell’s transportation cost. We can summarize this with a T

account as follows.

Stepping-stone method: Cell

evaluation by borrowing one

unit from a full cell to assess the

impact of shifting units into the

empty cell.

TABLE 8S.7 Evaluation path for cell 1-A

Warehouse

Factory

1

1774

(+)

(+) (−)

88312

516108

A B C D

2

3

80

80

(−)

90

90 110

10

120

100

60

450

450

150

200

100

Demand

Supply

160

Cell 1-A � �

(1–A) �4 (1–D) �1

(3–D) �5 (3–A) �8

�9 �9

0

Thus, shifting units around this path would have no impact whatsoever on transportation

costs: use of cell 1-A would be an equivalent-cost alternative. Management may prefer to use

it for other reasons, or management may prefer to stay with the original solution. In any case,

knowledge of such alternatives adds a certain degree of flexibility to decision making.




This means that for every unit shifted into cell 2-A, the total cost would increase by $12.

Thus we should avoid shipping units from factory 2 to warehouse A.

At this point, let’s consider some helpful rules for obtaining evaluation paths, since you

may still be unsure of how to do it.

1. Start by placing a � sign in the cell you wish to evaluate.

2. Move horizontally (or vertically) to a completed cell (a cell that has units assigned to it).

It is OK to pass through an empty cell or a completed cell without stopping. Choose a

cell that will permit your next move to another completed cell. Assign a minus sign ( � )

to the cell.

3. Change direction and move to another completed cell. Again, choose one that will per-

mit your next move. Assign a plus sign ( � ) to the cell.

4. Continue this process of moving to completed cells and alternating � and � signs until

you can complete a closed path back to the original cell. Make only horizontal and ver-

tical moves.

5. You may find it helpful to keep track of cells that have been evaluated by placing the

cell evaluation value in the appropriate cell with a circle around it.

So far, we have not discovered any improvement potential, but five unused cells still must

be analyzed. Before we evaluate the remaining empty cells, it is important to mention that

constructing paths using the stepping-stone method requires a minimum number of occupied

cells, and unless that number exists, it will be impossible to evaluate all the empty cells in this

manner. The number of occupied cells must equal the sum of the number of rows and columns

minus 1, or R � C � 1, where R � Number of rows and C � Number of columns. In this

example, R � 3 and C � 4, so we must have 3 � 4 � 1 � 6 used or completed cells (which

we do have). If there are too few occupied cells, the matrix is said to be degenerate. A method

for overcoming this problem is explained later in this supplement.

For now, let’s continue evaluating the unused cells. Suppose we now consider cell 2-A.

Begin by adding a unit to the empty cell. Moving to the right in row 2, we have what seems

to be a choice: borrow a unit from 90 or from 110. However, borrowing from 90 will leave

column B one unit short, and since adding and borrowing must involve occupied cells and

there are no others in the B column, the 90 cannot be used. Instead, we must borrow from

the 110 and add one unit to the 10 in cell 3-C. We can complete our path back to the original

cell by subtracting a unit from the 80 units in cell 3-A. The � / � path is shown in Table 8S.8 .

The impact on total cost for the path associated with cell 2-A would be:

Cell 2-A � �

(2–A) �12 (2–C) � 8

(3–C) �16 (3–A) � 8

�28 �16

�12

TABLE 8S.8 Evaluation path for cell 2-A

Warehouse

Factory

1

1774

(+)

(+) (−)88312

516108

A B C D

2

3

80

80(−)

90

90 110

10

120

100

60

450

450

150

200

100

Demand

Supply

160




Let’s try another one, say, 3-B. Start by placing a � sign in cell 3-B. Move to the 90 in cell

2-B, and place a � sign there. Move to the 110 in cell 2-C, and place a � sign there. Move to

cell 3-C, and give it a � sign. The path is shown in Table 8S.9 .

This one was fairly easy. Let’s see what the impact on cost would be:

TABLE 8S.9 Evaluation path for cell 3-B

Warehouse

Factory

1

1774

(1)

(1) (2)

88312

516108

A B C D

2

3

80

80

(2)

90

90 110

10

120

100

60

450

450

150

200

100

Demand

Supply

160

0

+12

TABLE 8S.10 Evaluation path for cell 2-D

Warehouse

Factory

1

1774

(1)

(1) (2)

88312

516108

A B C D

2

3

80

80

(2)

90

90110

10

120

100

60

450

450

150

200

100

Demand

Supply

160

0

+12

21

Cell 3-B � �

(3–B) �10 (2–B) � 3

(2–C) � 8 (3–C) �16

�18 �19

�1

The � 1 indicates that for each unit we can shift into cell 3-B, our cost will decrease by $1. If

we could shift 100 units, we could save $100 over the present solution. However, at this point

we are not ready to make any changes since some other empty cell might yield even greater

savings, and we can only make one shift per table. Evaluation of cell 2-D is similar to the

evaluation of cell 3-B. The evaluation path is shown in Table 8S.10 .

The cost impact would be:

Cell 2-D � �

(2–D) � 8 (3–D) � 5

(3–C) �16 (2–C) � 8

�24 �13

�11

Cell 1-B � �

(1–B) �7 (1–D) � 1

(3–D) �5 (3–C) �16

(2–C) �8 (2–B) � 3

�20 �20

0

Thus, no improvement is possible.

There are two cells remaining to be evaluated: 1-B and 1-C. The evaluation of cell 1-B is a lit-

tle more involved than the previous paths. Begin by placing a � in cell 1-B. Move to cell 1-D

(100 units) and give it a � . Move vertically to 3-D (60 units) and give it a � . Move to the left

and put a � in cell 3-C (10 units). Move up to 110 and put a � in that cell. Move to the left to

90 units (cell 2-B) and give it a � sign, completing the path. The path is shown in Table 8S.11 .

The cost impact of the path would be zero.




Hence, for each unit we can transfer into this empty cell, we improve the cost by $5.

At this point, each of the unused cells has been evaluated. The resulting costs are

summarized below.

Thus, no improvement is possible using this cell path.

The last empty cell is 1-C. Its evaluation path is shown in Table 8S.12 (above). Begin with

1-C as � , 1-D as � , 3-D as � , and 3-C as � . The cost impact would be:

Cell 1-C � �

(1–C) � 7 (3–D) � 1

(3–D) � 5 (3–C) �16

�12 �17

�5

Cell . . . . . . . . 1-A 2-A 3-B 2-D 1-B 1-C

Cost . . . . . . . . 0 � 12 � 1 � 11 0 � 5

Cells that have a positive or zero evaluation present no opportunities for improvement. The

fact that 1-A and 1-B are both zero indicates that at least one other equivalent alternative

exists. However, they are of no interest because two cells have negative cell evaluations, indi-

cating that the present solution is not an optimum. In a later section, you will learn how to

develop an improved solution.

Evaluating Empty Cells: The MODI Method Another method for evaluating empty cells is the modified distribution method (MODI) . It involves computing row and column index numbers that can be used for cell evaluation.

In many respects, it is simpler than the stepping-stone method because it avoids having to

trace cell evaluation paths. Nevertheless, the cell evaluations it produces are identical to those

obtained using the stepping-stone method. Note, however, that if a solution is not optimal, one stepping-stone path must be traced to obtain an improved solution. This is explained in detail

in the next section.

The MODI procedure consists of these steps:

1. Make an initial allocation using the intuitive method.

2. Obtain an index number for each row and column. Do this using only completed cells.

Note that there will always be at least one completed cell in each row and in each column.

a. Begin by assigning a zero to the first row.

b. Determine the column index for any completed cells in row 1 using the relationship:

Column index � Cell cost � Row index. For example, if a cell cost is $8 per unit, the

column index will be 8 � 0 � 8.

Modified distribution method (MODI) Evaluating

empty cells by computing row

and column index numbers.

TABLE 8S.12 Evaluation path for cell 1-C

Warehouse

Factory

1

1774

(1)(2)

(2)

88312

516108

A B C D

2

3

80

80

(1)

90

90110

10

120

100

60

450

450

150

200

100

Demand

Supply

160

0 0

+12 +11

21

TABLE 8S.11 Evaluation path for cell 1-B

Warehouse

Factory

1

1774

(1)

(1)

(2)

(2)

88312

516108

A B C D

2

3

80

80

(1)(2)

90

90 110

10

120

100

60

450

450

150

200

100

Demand

Supply

160

0

+12

21




c. Each new column value will permit the calculation of at least one new row value,

and vice versa. Continue until all rows and columns have index numbers, using only

completed cells.

3. Obtain cell evaluations for empty cells using the relationship:

Cell evaluation � Cell cost � (Row index � Column index)

Comments:

1. Row or column values may be positive, negative, or zero.

2. A reallocation requires that new row and column cost values be calculated.

Let’s see how the MODI method can be used for cell evaluation.

We begin by assigning a zero to row 1 (see Table 8S.13 ). Since we can only work with

occupied cells and since cell 1-D is the only such cell in row 1, we focus on it. Since the sum

of the row and column index numbers must add to the cell cost (which is 1), we can see that

the index number for column D must be 1. There are no other occupied cells in row 1, so we

shift to Column D, cell 3-D. Again, the sum of the row and column index numbers must add

to the cell cost (which is 5). Since the column index for D is 1, the index number for row 3

must be 4, as 5 � 1 � 4.

We can use this row 3 index number and the other two occupied cells in row 3 to obtain

index numbers for columns A and C. For column A, the A index number plus 4 must equal the

3-A cell cost of 8. Hence, the index for A is 4: 8 � 4 � 4. For C, the index plus 4 must equal

16; hence, the index number is 12.

Next, for row 2, using cell 2-C and a C index of 12, the row index is � 4. Then, using cell

2-B, the column index is 7:3 � ( � 4) � 7. This completes the row and column index numbers.

The cell evaluation for each empty cell is determined by subtracting from the cost of the

empty cell the sum of the row and column index numbers:

Cell evaluation � Cell cost � (Row index � Column index)

Table 8S.14 shows all of the evaluations. Note that these agree with the evaluations obtained

using the stepping-stone method. Because some evaluations are negative, the solution is

not optimal.

8S-4 OBTAINING AN IMPROVED SOLUTION The presence of negative cell evaluations is evidence that an improved solution is possible. By

the same token, if no negatives appear, an optimum has been achieved.

In this case, there are two cells with negative evaluations: 3-B, with a value of � 1, and

1-C, with a value of � 5. Select the value that implies more improvement (i.e., � 5), and

ignore the other.

The implication of the � 5 is that a savings of $5 per unit will be realized for every unit

that can be shifted into cell 1-C. Recall the path used to evaluate the cell; it is reproduced in

Table 8S.15A.

TABLE 8S.13 MODI index numbers

Warehouse

Factory

1

1

1

7

12

7

7

4

4

88312

516108

A B C D

2

34

0

24

80

80

90

90 110

10

120

100

60

450

450

150

200

100

Demand

Supply

160

TABLE 8S.14 MODI cell evaluation

Cell Evaluation

1-A 4 � (0 � 4) � 0 1-B 7 � (0 � 7) � 0 1-C 7 � (0 � 12) � � 5 2-A 12 � ( � 4 � 4) � 12 2-D 8 � ( � 4 � 1) � 11 3-B 10 � (4 � 7) � � 1




Note that the revised solution has a total cost that is $50 less than the previous solution

($2,300 versus $2,350), which is exactly equal to the cell evaluation value times the number

of units shifted: � 5 � 10 units � � $50.

Is this the optimum? Once again, we must evaluate each empty cell to determine if further

improvements are possible. Recall that we must check to see if there are enough completed

cells to allow all empty cells to have evaluation paths. Thus,

R � C � 1 � 3 � 4 � 1 � 6

Since there are six completed cells, the second round of cell evaluations can proceed.

The cells paths and their values are shown in Tables 8S-17 to 8S-22 . Since all are either

zero or positive, the solution is the optimum. (Note: MODI could have been used instead of

stepping-stone.)

In the course of the evaluation, a single unit was added or subtracted to certain cells, as

denoted by the � and � signs. We want to repeat that procedure, but instead of shifting one

unit, we want to shift as many as possible. The quantities in the cells that have � signs (i.e.,

10 in cell 3-C and 100 in cell 1-D) are potential candidates for shifting, since the � sign

implies units are subtracted from those cells. It is the smaller of these quantities that is the

limiting factor. That is, we can subtract 10 from both numbers (10 and 100), but if we try to

subtract 100 from both, we end up with 0 and � 90. The negative value would imply shipping

from the warehouse back to the factory! Hence, 10 units can be shifted. Always shift the mini-

mum quantity that occupies a negative position in the � path.

To accomplish the shift, add 10 units to each cell that has a � sign and subtract 10 units

from each cell with a � sign. (Because the signs alternate, the row and column totals will still

be satisfied,) This is shown in Table 15 (above).

The improved solution is shown in Table 8S.16 (above). The total costs are:

TABLE 8S.16 Revised solution

Warehouse

Factory

11774

88312

516108

A B C D

2

3

80

80

90

90 110

10

120

90

70

450

450

150

200

100

Demand

Supply

160

TABLE 8S.15 Evaluation path for cell 1-C (see A)

and reallocation of 10 units around the path (see B)

C

1

2

3

D

17

88

516110

10

100

60(1)(2)

(2)(1)A.

C D

17

88

516

110

10

10

0110

210 110

210

70

90

0

100

60(1)(2)

(2)(1)

1

2

3

B.

Total Cost

1-C . . . . . . . . . . . . . . . . . . . . 10($7) � $ 70

1-D . . . . . . . . . . . . . . . . . . . . 90($1) � 90

2-B . . . . . . . . . . . . . . . . . . . . 90($3) � 270

2-C . . . . . . . . . . . . . . . . . . . . 110($8) � 880

3-A . . . . . . . . . . . . . . . . . . . . 80($8) � 640

3-D . . . . . . . . . . . . . . . . . . . . 70($5) � 350

$2,300




TABLE 8S.17 Evaluation of cell 1-A

Warehouse

Factory

1

1774

8 1

8

8312

516108

A B C D

2

3

80

80

90

90 110

10

120

90

70

450

450

150

2004

5

9 9

0

100

Demand

Supply

Cell 1-A

160

(1)

(2)

(2)1

1

2

2

(1)

TABLE 8S.18 Evaluation of cell 1-B

Warehouse

Factory

11774

8 7

3

8312

516108

A B C D

2

3

80

80

90

90 110

10

120

90

70

450

450

150

2007

8

115

15

210

100

Demand

Supply

Cell 1-B

160

(1)

(2)

(2)1 2

(1)

0

TABLE 8S.19 Evaluation of cell 2-A

Warehouse

Factory

1

1774

88312

516108

A B C D

2

3

80

80

90

90 110

10

120

90

70

450

450

150

200

100

Demand

Supply

160

0 15

8

1

12

785

124

17

217

Cell 2-A

1 2

(1)

(2) (1)

(2)

(1) (2)

TABLE 8S.20 Evaluation of cell 2-D

Warehouse

Factory

1

1774

88312

516108

A B C D

2

3

80

80

90

90110

10

120

90

70

450

450

150

200

100

Demand

Supply

160

0

17

15

8

1

8

7

115

16

29

Cell 2-D

1 2

(1)(2)

(1) (2)




8S-5 SPECIAL PROBLEMS Not all transportation problems are as straightforward as the one just presented. Various irreg-

ularities may occur, which require certain adjustments before a solution can be obtained. Two

of the most common irregularities are unequal supply and demand, and degeneracy, a term

previously noted as referring to an insufficient number of completed cells to allow evaluation

of every empty cell.

Unequal Supply and Demand A case in which supply and demand are not equal is illustrated in Table 8S.23 . Total supply

(capacity) of the two sources is 200 units, but total demand of the destinations is only 170.

Thus, the deficiency in demand is 30 units. This condition can be remedied by adding a

dummy destination with a demand equal to the difference between supply and demand. Since

there is no such destination, no units will actually be shipped, and unit-shipping costs for each

dummy cell are $0. Table 8S.24 illustrates this.

dummy Imaginary number

added. Equal to the difference

between supply and demand

when these are unequal.

TABLE 8S.22 Evaluation of cell 3-C

Warehouse

Factory

1

1774

88312

516108

A B C D

2

3

80

80

90

90110

10

120

90

70

450

450

150

200

100

Demand

Supply

160

0

17

14

16

15

7

5

16

1

117

15

212

Cell 3-C

1 2

(2)(1)

(2) (1)

TABLE 8S.21 Evaluation of cell 3-B

Warehouse

Factory

1

1774

88312

516108

A B C D

2

3

80

80

90

90 110

10

120

90

70

450

450

150

200

100

Demand

Supply

160

0

17 16

15

3

7

10

8

51

119

14

215

Cell 3-B

1 2

(2)(1)

(1)(2)

(2) (1)

TABLE 8S.24 A dummy column with a

demand of 30 units is added to the matrix; dummy

cell costs are zero

To:

From:

1

9 05

2 04

A B

2

80 90 30Demand

Dummy

200

200

100

100

Supply

TABLE 8S.23 Supply exceeds

demand by 30 units

To:

From:

1

95

24

A B

2

80 90170

200

100

100

Demand

Supply




Once the dummy has been added, the transportation method can then be applied in the

usual manner. The final solution will have a total number of units assigned to the dummy that

equals the original difference between supply and demand (see Table 8S.25 ). These numbers

indicate which source(s) (e.g., factories) will hold the extra units or will have excess capacity.

In the example shown, factory 1 will ship only 70 units even though it has the capacity to ship

100 units.

The total transportation costs for this distribution plan is:

Total Cost

1-A . . . . . . . . . . . . . . . . . . 70($5) � $350

1-Dummy . . . . . . . . . . . . . 30($0) � 0

2-A . . . . . . . . . . . . . . . . . . 10($4) � 40

2-B . . . . . . . . . . . . . . . . . . 90($2) � 180

$570

A similar situation exists when demand exceeds supply. In those cases a dummy row (e.g.,

factory) with a supply equal to the difference between supply and demand must be added.

Again, units will not actually be shipped, so transportation costs are $0 for each cell in the

dummy row. Units in the dummy row in the final solution indicate the destinations that will

not receive all the units they desire and the amount of the shortage for each.

Note: When using the intuitive approach, if a dummy row or column exists, make assign-

ments to dummy cell(s) last.

Degeneracy The transportation method involves evaluation of empty cells using completed cells as step-

ping stones. Degeneracy exists when there are too few completed cells to allow all neces-

sary paths to be constructed. The condition occurs if an allocation (other than the final one)

exhausts both the row (supply) and the column (demand) quantities. It can occur in an initial

solution or in subsequent solutions, so it is necessary to test for degeneracy after each iteration

using R � C � 1.

An example of a matrix with too few completed cells is shown in Table 8S.26 . Note that

there are only four completed cells, although the necessary number is five:

3 � 3 � 1 � 5. Because of this, it is impossible to develop an evaluation path for cells 1-B,

1-C, 2-A, or 3-A. The situation can be remedied by placing a very small quantity, represented

by the symbol ε, into one of the empty cells and then treating it as a completed cell. The quan-

tity is so small that it is negligible; it will be ignored in the final solution.

The idea is to select a position for ε that will permit evaluation of all remaining empty

cells. For example, putting it in cell 3-A will permit evaluation of all empty cells (see

Table 8S.27 ). Some experimentation may be needed to find the best spot for ε, because not

every cell will enable construction of evaluation paths for the remaining cells. Moreover,

degeneracy The condition

of too few completed cells to

allow all necessary paths to be

constructed.

TABLE 8S.25 Units in the dummy column

are not shipped; factory 1 will ship only 70 units

To:

From:

1

9 05

2 04

A B

2

80

10

70

90

90

30

30

Demand

Dummy

200

200

100

100

Supply

TABLE 8S.26 There are too few completed

cells; some empty cells cannot be evaluated

12 53

1 48

7 67

A B

2

3

40

50 10

20

C

20

40 50 30Demand120

120

60

40

Supply




TABLE 8S.27 Cell 3-A becomes a completed

cell with the addition of a very

small quantity, represented by the

symbol ε

avoid placing ε in a minus position of a cell path that turns out to be negative because real-

location requires shifting the smallest quantity in a minus position. Since the smallest quantity

is ε, which is essentially zero, no reallocation will be possible.

The cell evaluations (not shown) are all positive, so this solution is optimal. (Note that cell

3-A is considered to be a completed cell and is therefore not evaluated.)

12 53

1 48

7 67

A B

2

3

40

50 10

20

C

20

40

e

50 30Demand120

120

60

40

Supply

8S-6 LOCATION DECISIONS The transportation model can be used to compare location alternatives in terms of their

impact on the total distribution costs for a system. The procedure involves working through

a separate problem for each location being considered and then comparing the resulting total

costs.

If other costs, such as production costs, differ among locations, these can easily be included

in the analysis, provided they can be determined on a per-unit basis. In this regard, note that

merely adding or subtracting a constant to all cost values in any row or column will not affect

the optimum solution; any additional costs should only be included if they have a varying

effect within a row or column.

8S-7 OTHER APPLICATIONS We have seen how the transportation model can be used to minimize the costs associated with

distributing goods, and we have seen how the model can be used for comparing location alter-

natives. The model is also used in a number of other ways. For example, in a slight variation

of the model, profits can be used in place of costs. In such cases, each of the cell profits can be

subtracted from the largest profit, and the remaining values (opportunity costs) can be treated

in the same manner as shipping costs.

Summary of Procedure 1. Make certain that supply and demand are equal. If they are not, determine the difference between

the two amounts. Create a dummy source or destination with a supply (or demand) equal to the

difference so that demand and supply are equal.

2. Develop an initial solution using the intuitive, lowest-cost approach.

3. Check that the number of completed cells is equal to R � C � 1. If it isn’t, the solution is degen-

erate and will require insertion of a minute quantity, ε, in one of the empty cells to make it a com-

pleted cell.

4. Evaluate each of the empty cells. If any evaluations turn out to be negative, an improved solution

is possible. Identify the cell that has the largest negative evaluation. Reallocate units around its

evaluation path.

5. Repeat steps (3) and (4) until all cells have zero or positive values. When that occurs, you have

achieved the optimal solution.




Some of the other uses of the model include production planning (see Chapter 11), prob-

lems involving assignment of personnel or jobs to certain departments or machines (see Chap-

ter 16), capacity planning, and transshipment problems. 1

The use of the transportation model for capacity planning parallels its use for location

decisions. An organization can subject proposed capacity alternatives to transportation analy-

sis to determine which one would generate the lowest total shipping cost. For example, it is

perhaps intuitively obvious that a factory or warehouse that is close to its market—or has

low transportation costs for some other reason—should probably have a larger capacity than

other locations. Of course, many problems are not so simple, and they require actual use of

the model.

1 Transshipment relates to problems with major distribution centers that in turn redistribute to smaller market des-

tinations. See, for example, W. J. Stevenson and C. Ozgur, Introduction to Management Science with Spreadsheets

(New York: McGraw-Hill, 2006).

8S-8 COMPUTER SOLUTIONS Although manual solution of transportation problems is fairly straightforward, computer solu-

tions are generally preferred, particularly for moderate or large problems. Many software

packages call for data input in the same tabular form used in this supplement. A more general

approach is to format the problem as a standard linear programming model (i.e., specify the

objective function and a set of constraints). That approach enables one to use the more gen-

eral version of an LP package to solve a transportation problem. Let’s consider this general

approach.

The decision variables for a transportation model are the quantities to be shipped. Because

each cell represents a potential transportation route, each must have a decision variable. We

can use the symbol x 1A to represent the decision variable for cell 1A, x 1B for cell 1B, and so

on. The objective function consists of the cell costs and these cell symbols:

Minimize 4x1A � 7x1B � 7x1C � 1x1D � 12x2A � 3x2B � 8x2C

� 8x2D � 8x3A � 10x3B � 16x3C � 5x3D

Because the amounts allocated in any row or column must add to the row or column total,

each row and column must have a constraint. Thus, we have

Supply (rows) x 1A � x 1B � x 1C � x 1D � 100

x 2A � x 2B � x 2C � x 2D � 200

x 3A � x 3B � x 3C � x 3D � 150

Demand (columns) x 1A � x 2A � x 3A � 80

x 1B � x 2B � x 3B � 90

x 1C � x 2C � x 3C � 120

x 1D � x 2D � x 3D � 160

We do not need a constraint for the total; the row and column constraints take care of this.

If supply and demand are not equal, add an extra (dummy) row or column with the neces-

sary supply (demand) for equality to the table before writing the constraints. Another approach to transportation problems is to use spreadsheet software. The Excel

templates can also be used to solve transportation problems. Figure 8S.3 illustrates the Excel

worksheet for the preceding problem.

LO 8S.3 Set up transportation problems in the general linear programming format.

SCREENCAM TUTORIAL

LO 8S.4 Interpret computer solutions.




FIGURE 8S.3 Excel template for the computer solution.

degeneracy, 8S-13

dummy, 8S-12

intuitive lowest-cost approach,

8S-3

modified distribution method

(MODI), 8S-8

stepping-stone method, 8S- 5

KEY TERMS

SOLVED PROBLEMS

Use the intuitive lowest-cost approach to develop an initial solution to the following problem, and

then determine if the solution is optimal. If it is not, develop the optimal solution. Problem 1

To:

From:

1

69

35

A B

2

80 90

75

75

Demand

Supply

a. First check to see if supply and demand are equal. Total demand � 170 and total supply � 150;

therefore, it will be necessary to create a dummy origin (supply) with zero transportation costs in

each of its cells (see following table).

Solution




To:

From:

1

69

35

00

A B

2 75

20

75

Dummy

80 90Demand

Supply

170

170

b. Find the cell that has the lowest unit transportation cost. When a dummy is present and you are

using the intuitive method, leave the dummy allocation until last. Thus, aside from the $0 costs in

the dummy row, the lowest cost is $3 for cell 2-B. Assign as many units as possible to this cell.

The column total is 90, and the row total is 75. Because 75 is the smaller of the two, assign 75

to cell 2-B. This exhausts the row total, so cross out the 75, and cross out the cell costs for the

second row. Revise the column total to 15.

To:

From:

1

69

0

35

0

A B

75

20

75

752

Dummy

Supply

80 9015

Demand170

170

The next smallest cell cost is $6 in cell 1-B. We want to assign as many units as possible to this

route (cell). The remaining demand for column B is 15, and the row total is 75. We assign the

smaller of these, 15, to cell 1-B, thereby exhausting the column and reducing the row supply to

60. Next we draw a line through the cell costs of column B.

To:

From:

1

9

5

6

3

00

A B

75

15

20

75

75

60

2

Dummy

Supply

80 9015

Demand170

170

Still ignoring the dummy row, the next lowest cell cost is $9 in 1-A, where a supply of 60 and a

demand of 80 exist. We assign the smaller of these, 60 units, to cell 1-A, so row 1 costs must be

crossed out. Also, the revised column A total becomes 20.




To:

From:

1

9

5

6

3

00

A B

75

1560

20

75

75

60

2

Dummy

Supply

80 9015

Demand170

17020

At this point, only the dummy row remains. Cell dummy-A has a supply and a demand of 20;

assigning this amount to the cell exhausts both the supply and the demand, finishing the initial

allocation.

To:

From:

1

0

5

9

0

3

6

A B

75

15 7560

60

2

Dummy

Supply

20

Demand170

17020 15

75

9080

20

c. The next step is to evaluate this solution to determine if it is the optimal solution. First we must

check for degeneracy: R � C � 1 � 3 (rows) � 2 (columns) � 1 � 4, the required number of

filled or completed cells. Because there are four completed cells, the solution is not degenerate.

The evaluation can use either the stepping-stone method or MODI. Suppose we use MODI. We

always begin by placing a zero at the left edge of row 1. This is the row 1 index number. We can

use it to find the column index number for any cell in row 1 that is a completed cell. Because

both cell 1-A and cell 1-B are completed, we can obtain both of their column index numbers.

For completed cells, the sum of the row and the column index numbers must equal the cell cost.

For cell 1-A, 0 � Column index � 9, so the column A index must be 9. Similarly, for cell 1-B,

0 � Column index � 6, so the column index must be 6. Next, these index numbers are used to

find values of row index numbers for completed cells in their respective columns. Thus, for cell

2-B, 6 � Row index � 3, so the row index must be � 3. For dummy-A, 9 � Row index � 0, so

the row index must be � 9.

To:

From:

1

0

5

9

9

0

3

6

6A B

75 75

15 7560

2

Dummy

Supply

20

80 90

20

Demand170

170

29

23

0

Next, we use these index numbers to evaluate the empty cells. We determine a cell evaluation by

subtracting from the cell cost the sum of its row and column index numbers.

Thus, for cell 2-A, we have 5 � ( � 3 � 9) � � 1, and for cell dummy-B, we have

0 � ( � 9 � 6) � 3.




To:

From:

1

0

5

9

9

0

3

6

6A B

75 75

15 7560

2

Dummy

Supply

20

80 90

20

Demand170

170

29

23

0

21

13

The appearance of a negative cell evaluation tells us this solution can be improved on. When

negatives appear, we select the cell with the largest negative as the local point of our effort to

improve the solution. Because there is only one such cell, it is the one we select.

d. The � 1 in cell 2-A tells us that for every unit we can move into this cell, we can reduce the total

cost of the solution by $1. Obviously, we would like to put as many units as possible into this cell,

but we are limited by the row and column totals (row 2 and column A). We are also constrained

by the need for all row and column totals to be maintained. To put units into cell 2-A, we must

take them from some other cell, which will be in another row or column. Fortunately, this matter

is simple to resolve: Determine the stepping-stone path for cell 2-A, and use it to guide the real-

location. That path is illustrated in the following table.

To:

From:

1

0

5

9

0

3

6

A B

7575

1575

60

2

Dummy

Supply

20

80 90

20

Demand170

170

(2) (1)

(1) (2)

The amount that can be shifted into cell 2-A is the smallest quantity that appears in a negative position of the path. There are two such quantities, 60 and 75. Because 60 is the smaller of these,

it is shifted around the entire path (i.e., added where there is a � sign and subtracted where there

is a � sign). The result is our revised solution:

To:

From:

1

0

5

9

0

3

6

A B

15 75

7575

2

Dummy

Supply

20

60

80 90

20

Demand170

170

e. This improved solution may be optimal, or it may need further revision. To find out, we must

evaluate it.




First, we check for degeneracy: R � C � 1 � 3 � 2 � 1 � 4. Because there are four completed

cells, the solution is not degenerate.

Next, we develop MODI index numbers using the completed cells. We always begin by

assigning an index of 0 to the first row. Then, for any completed cell in the first row, the

column index number is the same as the cell cost. Because cell 1-B has a cell cost of 6,

the column B index number is 6. Using this value and the cell cost for completed cell 2-B

enables us to find the index number for row 2: Cell cost � Column index � Row index. Thus,

3 � 6 � � 3. This row 2 index number can be used with the cell cost of completed cell 2-A to

find the column A index number, 5 � ( � 3) � 8, and the index of 8 can be used with the cell

cost of dummy-A to find the dummy index number: 0 � 8 � � 8. These index numbers are

shown in the following table.

To:

From:

1

0

5

9

8

0

3

6

6A B

15 75

75 75

602

Dummy

Supply

20

80 90

20

Demand170

170

28

23

0

Evaluations for the empty cells can now be determined.

Cell 1-A: 9 (0 � 8) � 1

Cell dummy-B: 0 � ( � 8 � 6) � 2

Because there are no negative evaluations, the current solution is optimal: Ship 75 units from 1

to B, 60 units from 2 to A, and 15 units from 2 to B. The 20 units in dummy-A means that desti-

nation A will be 20 units short of satisfying its demand of 80 units.

The total cost of the optimal solution is $795:

75(6) � 450

60(5) � 300

15(3) � 45

20(0) � 0

795

A firm that specializes in nonferrous casting must decide between locating a new foundry in Chicago

or Detroit. Transportation costs from each proposed location to existing warehouses and transporta-

tion costs for existing locations for a certain type of casting are shown in the following table. Each

will be able to supply 2,500 units per month. Explain how to determine which location should be

chosen on a transportation cost basis.

From

Detroit to

Cost

per unit

From

Chicago To

Cost

per unit

A . . . . . $10 A . . . . . $12

B . . . . . 8 B . . . . . 13

C . . . . . 15 C . . . . . 5

Problem 2




To:

From:

1

7

17

12

10

14

6

2

A B C

25 10Demand

(hundreds of

units/month)

Foundry

Warehouse

40

20

30

Supply

(hundreds of

units/month)

Solution This decision requires solving two transportation problems, one using Detroit costs and the other

using Chicago costs.

a. Set up the two transportation matrices.

1

7

17

12

10

14

6

2

A B C

Warehouse

20

10 8 15Detroit

25 10Demand 40

25

30

Supply

1

7

17

12

10

14

6

2

A B C

Warehouse

20

12 13 5Chicago

25 10Demand 40

25

30

Supply

b. Solve each problem

c. The problem that yields the lowest total cost will indicate the preferred location.

1. What information do you need to use the transportation model?

2. What check must you make before proceeding to develop an initial solution? What corrective action

might need to be taken?

3. Would it ever make sense to have a situation that required both a dummy row and a dummy

column? Explain briefly.

DISCUSSION AND REVIEW QUESTIONS




4. What is linear about the objective function in a transportation problem?

5. Why do the � and � signs alternate in cell evaluation paths?

6. How do you know if a solution is optimal?

7. What does a zero value for an evaluation path indicate?

8. If a solution is not optimal:

a. How do you decide which empty cell to shift units into?

b. How do you know which cells to include in the shifting operation?

c. How do you decide how many units to shift?

9. What is meant by the term degeneracy? How is it overcome?

10. How is the transportation model useful in location decisions? How does use of the transportation

model in location decisions differ from its use as a tool to develop an optimal distribution plan for

a given set of sources and destinations?

11. How are total costs determined for a given distribution plan?

12. What interpretation is given to quantities allocated to dummy destinations? Dummy sources?

13. What is the MODI method? How does it differ from the stepping-stone method?

PROBLEMS 1. Solve this linear programming (LP) problem using the transportation method. Find the optimal

transportation plan and the minimum cost. Also, decide if there is an alternate solution. If there is

one, identify it.

Minimize 8x11 � 2x12 � 5x13 � 2x21 � x22

� 3x23 � 7x31 � 2x32 � 6x33

Subject to x11 � x12 � x13 � 90

x21 � x22 � x23 � 105

x31 � x32 � x33 � 105

x11 � x21 � x31 � 150

x12 � x22 � x32 � 75

x13 � x23 � x33 � 75

All variables � 0

2. A toy manufacturer wants to open a third warehouse that will supply three retail outlets. The new

warehouse will supply 500 units of backyard playsets per week. Two locations are being studied,

N1 and N2. Transportation costs for location N1 to stores A, B, and C are $6, $8, and $7 per unit,

respectively; for location N2, the costs are $10, $6, and $4, respectively. The existing system is

shown in the following table. Which location would result in the lower transportation costs for the

system?

1738

5 10 9

A B C

2

600400 350

400

500

Demand

(Units/week)

Capacity

(Units/week)To:

From:

Warehouse

Store

3. A large firm is contemplating construction of a new manufacturing facility. The two leading

locations are Toledo and Cincinnati. The new factory would have a supply capacity of 160 units per

week. Transportation costs from each potential location and existing locations are shown in the fol-

lowing table. Determine which location would provide the lower transportation costs.

From

Toledo to

Cost

per Unit

From

Cincinnati to

Cost

per Unit

A $18 A $ 7

B 8 B 17

C 13 C 13




5. Obtain the optimal distribution plan for the following transportation problem. Develop the initial

solution using the intuitive lowest-cost approach.

Use the stepping-stone method for cell evaluations.

1

5

3

1

4

7

2

2

A B C

60

8 7 43

To:From:

30 45Demand 75

50

40

Supply

150

150

6. For the following transportation table:

a. Find the optimal solution. Use the intuitive method for the initial solution and MODI for evaluation.

b. What is the total cost?

c. Is there an alternate optimal solution? If so, what is it?

1

3

3

1

6

3

2

2

A B C

50

7 6 43

To:From:

55 55Demand 45

65

40

Supply

1014

2017

11 11 12

10

12

220220220

140

150

210

Demand

(Units/week)

A B C

1

2

3

Supply

(Units/week)

4. A large retailer is planning to open a new store. Three locations in California are currently under

consideration: South Coast Plaza (SCP), Fashion Island (FI), and Laguna Hills (LH). Transporta-

tion costs for the locations and costs, demands, and supplies for existing locations and warehouses

(origins), are shown below. Each of the locations has a demand potential of 300 units per week.

Which location would yield the lowest transportation costs for the system?

TO

From

Warehouse SCP FI LH

1 $ 4 $7 $5

2 11 6 5

3 5 5 6

9

7

14 18

15

10

500

660

340

200

400Demand

(Units/week)

Supply

(Units/week)A B

1

2

3




7. Given the following transportation problem, do each of the following:

a. Develop an initial feasible solution using the intuitive approach.

b. Use the stepping-stone method to evaluate the solution.

c. Repeat part b using MODI.

d. Find the optimal solution.

1

23

18

24

12

27

14

2

42 34 313

To:From:

Warehouse

90 80Demand 30

33

16

A B C D

80

26

50

130

40

Supply

8. Determine the optimal distribution plan for the following problem, and compute the total cost for

that plan. Use the MODI method for cell evaluation.

1

17

14

18

24

25

18

2

30 16 223

To:From:

Warehouse

41 34Demand 35

16

28

A B C D

56

30

20

32

48

Supply

9. Refer to problem 7. Due to a temporary condition, origin 3 has experienced a capacity reduction of

20 units per period. Using the intuitive method to develop an initial solution, determine a tempo-

rary distribution plan that will minimize transportation costs.

10. Refer to problem 8. The market supplied by destination D is experiencing a period of rapid

growth. Projected demand is for 60 units per period. A new factory with a capacity of 50 units is

planned for one of two locations, Baltimore or Philadelphia. Transportation costs are as follows.

Which location would result in the lower total cost? Use the MODI method for cell evaluation.

From

Baltimore to

Cost

per unit

From

Philadelphia To

Cost

per unit

A . . . . . $18 A . . . . . $31

B . . . . . 16 B . . . . . 25

C . . . . . 22 C . . . . . 19

D . . . . . 27 D . . . . . 20

11. Use the following transportation table to do the following:

a. Find the optimal solution. Use MODI for evaluation.

b. What is the total cost?

c. Is there an alternate optimal solution? If so, what is it?




Cleveland

7

6

2

4

7

8

Chicago

4 4 5Buffalo

From:

70 90Demand 120

Rochester

To:

Detroit Toronto

100

80

100

Supply

12. A soft drink manufacturer has recently begun negotiations with brokers in areas where it intends

to distribute new products. Before making final agreements, however, the firm wants to determine

shipping routes and costs. The firm has three plants with capacities as follows:

Plant

Capacity

(cases per week)

Metro 40,000

Ridge 30,000

Colby 25,000

The estimated shipping costs per case for the various routes are:

Warehouse

Demand

(cases per week)

RS1 24,000

RS2 22,000

RS3 23,000

RS4 16,000

RS5 10,000

TO

From RS1 RS2 RS3 RS4 RS5

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

Determine the optimal shipping plan that will minimize total shipping costs under these

conditions:

a. Route Ridge-RS4 is unacceptable.

b. All routes are acceptable.

c. What is the additional cost of the Ridge-RS4 route not being acceptable?

Stevenson, William J., and Ceyhun Ozgur. Introduction to Management Science with Spreadsheets. New York: McGraw-Hill, 2006.

SELECTED BIBLIOGRAPHY AND FURTHER

READING


operations management by stevenson, chapter 8s

Documents

transportation problems

problem supply

optimal shipping plan

shipping supply points

total shipping cost

lowestcost plan

ones demand

solved problems