Solve The Transportation Problem Engineering Essay

Introduction

Transport various quantities of a single homogeneous commodity to different destinations in such a way that total transportation costs minimum.

A scooter production company produces scooters at the units situated at various places (called origins) and supplies them to the places where the depot (called destination) are situated.

Here the availability as well as requirements of the various depots are finite and constitute the limited resources. This type of problem is known as distribution or transportation problem in which the key idea is to minimize the cost or the time of transportation.

In previous lessons we have considered a number of specific linear programming problems. Transportation problems are also linear programming problems and can be solved by simplex method but because of practical significance the transportation problems are of special interest and it is tedious to solve them through simplex method.

Objectives

Solve the transportation problem

Test the optimality of the solution.

Initial Basic Feasible Solution :

There are three different methods to obtain the initial basic feasible solution viz.

North-West corner rule

Lowest cost entry method

Vogel’s approximation method

North-west corner method (NWCM)

This is the most systematic and earliest method for obtaining initial feasible solution. Steps involved in this method are stated as:

Step 1: Construct an empty m x n matrix, completed with rows & columns.

Step 2: Indicate the row totals and columns totals at the end.

Step 3: starting with (1,1) cell at the north-west corner of the matrix, allocate maximum possible quantity keeping in views the allocation can neither be more than the quantity required by the respective warehouses nor more than the quantity available at supply Centre.

Step 4: adjust the supply and demand numbers in the respective rows & columns allocations.

Step 5: If the supply for the first row is exhausted then move down to the first cell in the second row and first column and go to step 4.

Step 6: If the demand for the first column is satisfied, then move to the next cell in the next cell in the second column and first row and go to step 4.

Step 7: If for any cell, supply equals demand then the next allocation can be made in cell either in the next row or column.

Step 8: Continue the procedure until the total available quantity is fully allocated to the cells as required.

Lowest cost entry method (LCEM)

This method takes into consideration the lowest cost and therefore takes less time to solve the problem; various steps of this method can be summarized as under:

Step 1: Select the cell with the lowest transportation cost among all the rows or columns of the transportation table. If the minimum cost is not unique then select arbitrarily any cell with lowest cost.

Step 2: allocate as many units as possible to the cell determined in step 1 and eliminate that row in which either capacity or requirement is exhausted.

Step 3: Adjust the capacity and requirement for the next allocation.

Step 4: Repeat steps 1 to 3 for the reduced table until the entire capacity is exhausted to fill the requirement at different destinations.

Vogel’s approximation method (VAM)

This method is preferred over the other two methods because the initial basic feasible solution obtained with VAM is either optimal or very close to the optimal solution. Therefore, the amount of time required to calculate the optimum solution is reduced. In Vogel’s approximation method the basis of allocation of unit cost penalty i.e. that column or row which has the highest unit cost penalty. Various steps in integration process are:

Step 1: Construct the cost, requirement and availability matrix with column and row information.

Step 2: Compare the penalty for each row and column in the transportation table. The penalty for a given row and column is merely the difference between the smallest cost and the next smallest cost element in that particular row or column.

Step 3: Identify the row and column with largest penalty. In this identifies row (column), choose the cell which has the smallest cost and allocate the maximum possible quantity to this cell. Delete the row (column) in which capacity/requirement is exhausted. Whenever the largest penalty among rows and column is not unique, make an arbitrary choice.

Step 4: Repeat step 1 to 3 for the reduced table until the entire capacities are reduced to fill the requirement at different warehouses.

Step 5: From step 4 we will get initial feasible solution. Now for IFS find the total transportation cost by multiplying the cell allocation by unit cost.

Though this method takes more time as compared to other two methods, but still it gives better solution and saves more time in reaching the optimal solution.

Some Solutions

Feasible Solution (F.S.)

A set of non-negative allocations xij

Ã¢â€°Â¥ 0 which satisfies throw and column restrictions is known as feasible solution.

Basic Feasible Solution (B.F.S.)

A feasible solution to a m-origin and n-destination problem is said to be basic feasible solution if the number of positive allocations are (m+n-1).

If the number of allocations in a basic feasible solutions are less than (m+n-1), it is called degenerate basic feasible solution

(DBFS) (Otherwise non-degenerate).

Optimal Solution

A feasible solution (not necessarily basic) is said to be optimal if it minimizes the total transportation cost.

Assignment 1:

Suppose that England, France, and Spain produce all the wheat, barley, and oats in the world. The world demand for wheat requires 125 million acres of land devoted to wheat production. Similarly, 60 million acres of land are required for barley and 75 million acres of land for oats. The total amount of land available for these purposes in England, France, and Spain is 70 million acres, 110 million acres, and 80 million acres, respectively. The number of hours of labor needed in England, France and Spain to produce an acre of wheat is 18, 13, and 16, respectively. The number of hours of labor needed in England, France, and Spain to produce an acre of barley is 15, 12, and 12, respectively. The number of hours of labor needed in England, France, and Spain to produce an acre of oats is 12, 10, and 16, respectively. The labor cost per hour in producing wheat is $9.00, $7.20, and $9.90 in England, France, and Spain, respectively. The labor cost per hour in producing barley is $8.10, $9.00, and $8.40 in England, France, and Spain respectively. The labor cost per hour in producing oats is $6.90, $7.50, and $6.30 in England, France, and Spain, respectively. The problem is to allocate land use in each country so as to meet the world food requirement and minimize the total labor cost.

(a) Formulate this problem as a transportation problem by constructing the appropriate parameter table.

(b) Reconsider the problem in the preceding example. Starting with the northwest corner rule, Find basic feasible solution.

Solution 1:

Let England, France, and Spain be the three sources, where their supplies are the millions of acres of land that are available for growing these crops. Let Wheat, Barley, and Oats be the three destinations, where their demands are the millions of acres of land that are needed to fulfill the world demand for these respective crops. The unit cost (in millions of dollars) is the labor cost per million acres, so the number of hours of labor needed is multiplied by the cost per hour. The parameter table is as follows:

wheat

barley

oats

Available

England

18*9

15*8.10

12*6.90

France

13*7.20

12*9

10*7.50

110

Spain

16*9.90

12*8.40

16*6.30

demand

125

260

(b) Draw the network representation of this problem.

The network presentation of this problem is given below.

(c) Obtain an optimal solution.

We can use the Excel Solver to solve this problem and obtain the following solution.

Allocation Quantities

Destination

Wheat

=

Source

France

110

=

110

Spain

=

Totals

125

=

Total cost = $25.02 billion

Demand

125

For this problem, the initial BF solution obtained by the northwest corner rule is shown below.

Optimality Test:

Since cij – ui – vj = 0 if xij is a basic variable,

cij = ui + vj for each (i, j) such that xij is basic.

Because the number of unknowns (the ui and vj) exceeds the number of these equations by one, we can set one unknown equal to an arbitrary value, say 0. These equations can then be solved as outlined below.

x21: 93.6 = u2 + v1. Set u2 = 0, so v1 = 93.6,

x22: 108 = u2 + v2. v2 = 108.

x11: 162 = u1 + v1. Know v1 = 93.6, so u1 = 68.4.

x32: 100.8 = u3 + v2. Know v2 = 108, so u3 = -7.2.

x33: 100.8 = u3 + v3. Know u3 = -7.2, so v3 = 108.

Since cij – ui – vj represents the rate at which the objective function will change as a no basic variable xij is increased, we now can check whether increasing any no basic variable will decrease the total cost Z.

No basic variable

cij – ui – vj

x12

121.5 – 68.4 – 108 = -54.9

x13

82.8 – 68.4 – 108 = -93.6

x23

75 – 0 – 108 = -33

x31

158.4 -(-7.2) – 93.6 = 72

Because some of these (cij – uij – vj) values are negative, the initial BF solution is not optimal.

Iteration 1:

We select the non-basic variable x13 to be the entering basic variable because it has the largest negative value of (cij – ui – vj).

When x13 is increased from 0 by any particular amount, a chain reaction is set off that requires alternately decreasing and increasing current basic variables by the same amount in order to continue satisfying the supply and demand constraints. This chain reaction is depicted in the next figure, where the + sign inside a box in cell (1, 3) indicates that the entering basic variable is being increased there and the + or – sign next to other circles indicate that a basic variable is being increased or decreased there.

Each donor cell (indicated by a minus sign) decreases its allocation by exactly the same amount as the entering basic variable and each recipient cell (indicated by a plus sign) is increased. The entering basic variable will be increased as far as possible until the allocation for one of the donor cells drops all the way down to 0. Since the original allocations for the donor cells are

x11 = 70, x22 = 55, x33 = 75,

x22 will be the one that drops to 0 as x13 is increased (by 55). Therefore, x22 is the leaving basic variable.

Since each of the basic variables is being increased or decreased by 55, the values of the basic variables in the new BF solution are

x11 = 15, x13 = 55, x21 = 110, x32 = 60, x33 = 20.

Optimality Test After Iteration 1:

Since Source 1 now has two basic variables (tied for the maximum number), let us set u1 = 0 this time. The cij = ui + vj equations then would be solved as follows.

x11: 162 = u1 + v1. Set u1 = 0, so v1 = 162,

x13: 82.8 = u1 + v3. v3 = 82.8.

x21: 93.6 = u2 + v1. Know v1 = 162, so u2 = -68.4.

x33: 100.8 = u3 + v3. Know v3 = 82.8, so u3 = 18.

x32: 100.8 = u3 + v2. Know u3 = 18, so v2 = 82.8.

We next calculate (cij – ui – vj) for the non-basic variables.

Non-basic variable

cij – ui – vj

x12

121.5 – 0 – 82.8 = 38.7

x22

108 – (-68.4) – 82.8 = 93.6

x23

75 – (-68.4) – 82.8 = 60.6

x31

158.4 – 18 – 162 = -21.6

We still have one negative value of (cij – ui – vj), so the current BF solution is not optimal.

Iteration 2:

Since x31 is the one non-basic variable with a negative value of (cij – ui – vj), x31 becomes the entering basic variable.

The resulting chain reaction is depicted next.

The donor cells have allocations of x11 = 15 and x33 = 20. Because 15 < 20, the leaving basic variable is x11.

Since the basic variables x21 and x32 were not part of this chain reaction, their values do not change. However, x31 and x13 increase by 15 while x11 and x33 decrease by 15. Therefore, the values of the basic variables in the new BF solution are

x13 = 70, x21 = 110, x31 = 15, x32 = 60, x33 = 5

Optimality Test After Iteration 2:

Because Source 3 now has the largest number of basic variables, we set u3 = 0 this time. The resulting calculations are shown below.

x31: 158.4 = u3 + v1. Set u3 = 0, so v1 = 158.4,

x32: 100.8 = u3 + v2. v2 = 100.8.

x33: 100.8 = u3 + v3. v3 = 100.8.

x13: 82.8 = u1 + v3. Know v3 = 100.8, so u1 = -18.

x21: 93.6 = u2 + v1. Know v1 = 158.4, so u2 = -64.8.

Non-basic variable

cij – ui – vj

x11

162 – (-18) – 158.4 = 21.6

x12

121.5 – (-18) – 100.8 = 38.7

x22

108 – (-68.4) – 100.8 = 72

x23

75 – (-64.8) – 100.8 = 39

Since all of these values of (cij – ui – vj) are nonnegative, the current BF solution is optimal.

Thus, the optimal allocation of land to crops is

million acres in England for oats,

110

million acres in France for wheat,

million acres in Spain for wheat,

million acres in Spain for barley,

million acres in Spain for oats.

The total cost of this grand enterprise would be

Z = $25.02 billion.

A contractor, Susan Meyer, has to haul gravel to three building sites. She can purchase as much as 18 tons at a gravel pit in the north of the city and 14 tons at one in the south. She needs 10, 5, and 10 tons at sites 1, 2, and 3, respectively. The purchase price per ton at each gravel pit and the hauling cost per ton are given in the table below. Susan wishes to determine how much to haul from each pit to each site to minimize the total cost for purchasing and hauling gravel.

Hauling Cost per Ton at Site

Pit

1

2

3 Price per Ton

North

$30

$60

$50

$100

South

$60

$30

$40

$120

Now suppose that trucks (and their drivers) need to be hired to do the hauling, where each truck can only be used to haul gravel from a single pit to a single site. Each truck can haul 5 tons, and the cost per truck is five times the hauling cost per ton given above. Only full trucks would be used to supply each site.

(a) Formulate this problem as an assignment problem by constructing the appropriate cost table, including identifying the assignees and tasks.

The tasks are the loads needed at sites 1, 2, and 3. The assignees are the three trucks from the North pit and the two trucks from the South pit. Considering the purchase price for the gravel and the hauling cost per truck, the cost table is constructed as follows.

Task (Site)

North 1

650

800

750

North 2

650

800

750

Assignee

North 3

650

800

750

South 1

900

750

800

South 2

900

750

800

(b) Obtain an optimal solution.

We use the Excel Solver to obtain the following optimal solution with a minimum cost of $3500.

Task (Site)

North 1

North 2

Assignee

North 3

South 1

South 2

(c) Reformulate this assignment problem as an equivalent transportation problem with two sources and three destinations by constructing the appropriate parameter table.

The parameter table for the formulation as an equivalent transportation problem is given below.

Destination

Supply

Source

North

650

800

750

South

650

800

750

Demand

(d) Obtain an optimal solution for the problem as formulated in part (c).

We use the Excel Solver to obtain the following optimal solution with a minimum cost of $3500.

Destination

Supply

Source

North

South

Demand

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