cb312 ch4

Systems Analysis 91 Dr. Emad Elbeltagi

CHAPTER 4

T HE T RANSPORTATION AND A SSIGNMENT PROBLEMS

In this chapter,we will discuss the transportation and assignment problems which are twospecial kinds of linear programming.The transportation problem deals with transportinggoods from their sources to their destinations. The assignment problem, on the otherhand, deals with assigning people or machines to jobs.

4.1 The Transportation Problem

Example4.1

Consider the following snow removal problem: there are a number of districts in a city.After a snowfall, the snow in each area must be moved out of the district into aconvenient location. Ina city, these locations are large grates (leading to the sewersystem), a couple large pits, and a couple entry points to the river. Each of thesedestinations hasa capacity. The goal is to minimize the distance traveled to handle all ofthe snow.

This problem is an example of a transportation problem. In such a problem, there are a setof nodes called sources(4 sources where snow is collected), and a set ofnodes calleddestinations(3 destinations where snow will be transported). All arcs go from a source toa destination. There is a perunit cost on each arc(the cost of transporting one unit from asource to adestination). Each source has a supply of material, and eachdestination has ademand. We assume that the total supply equals the total demand (possiblyadding a fakesource or destination as needed). For the snow removal problem, the network mightlooklike that in Fi g. 4.1.


Transportation problemsare often used intransportation planning. For instance,in anapplication where goods are at a warehouse, one problem might be to assign customers toa warehouse so as to meet their demands. In such a case, the warehouses are the sources,the customers are the destinations,and the costs represent the perunit transportationcosts.

Figure 4.1: Snowt ransportationnetwork

Example4.2

One of the main products ofQaha Company is canned peas. The peas are prepared atthree canneries (Qaha, Benha, and Tanta) and are thenshipped by trucks to fourdistributing warehouses inCairo, Alexandria, Portsaid, and Aswan. Because shippingcosts are a majorexpense, management has begun a study to reduce them. For theupcoming season, an estimatehas beenmade of what the output will be from eachcannery, and how much each warehouse willrequire to satisfy its customers. Theshipping costs from each cannery to each warehouse havealso been determined. This issummarized in Table 4.1 and Fig. 4.2. Formulatethis transportation problem as a linearprogramming problemby definingthe variables, theobjective and the constraints.

S1

S2

S3

d1

S4

d2

d3

(1000)

(2000)

(2500)

(1500)

(3000)

(2000)

(2000)

10

23

3

4

22

9

7

71

5

Sources Destinations

Supply Demand


Table 4.1: Shipping data for example 4.2

WarehouseShipping cost per truck load

Supply1 2 3 4

Cannery1 464 513 654 867 75

2 352 416 690 791 125

3 995 682 388 685 100

Demand 80 65 70 85

Figure 4.2: Network representation of example 4.2

You shouldfi nd it an easy exercise to model this as a linearprogram. If we letxij be the

number of truckloads shipped from canneryi to warehousej , the problem is to:

minimizeZ = 464x11 + 513x12 + 654x13 + 867x14 + 352x21 + 416x22 + 690x23 +

791x24 + 995x31 + 682x32 + 388x33 + 685x34

subject tox

11 + x 12 + x 13 + x 14 = 75

S1

S2

S3

d1

d2

d3

d4

(75)

(125)

(100)

(80)

(65)

(70)

(85)

464

513654

867352

416690

791959

682388

685

DestinationSource

Supply Demand


x21 + x 22 + x 23 + x 24 = 125

x31 + x 32 + x 33 + x 34 = 100

x11 + x 21 + x 31 = 80

x12 + x 22 + x 32 = 65

x13 + x 23 + x 33 = 70

x14 + x 24 + x 34 = 85

xij "e0 for all i and j .

This is an example of thetransportation model. As has been pointed out, this problemhas aspecific structure. All the coeffi cientsof the variables in the constraint equationsare1 and every variable appears in exactly twoconstraints. All constraint equations areformed using the equal sign not the"e or the"d signs. This is the special structurethatdistinguishes this problem as a transportation problem.

4.1.1 The Transportation Problem Model

In general, the transportation model is concerned with distributing (literally orfi guratively) a commodity from a group of supply centers, calledsourcesto a group ofreceiving centers, calleddestinationsto minimize total cost.In general, sourcei (i = 1, 2,3, &.., m) has a supply ofs

i units, and destinationj (j = 1, 2, 3, & &,n) has a demand ford

j units. Thecost of distributing items from a source to a destination is proportional to thenumber of units.This data can be conveniently represented in a table like that for thesample problemof example 4.2. Each source has a fixed supply of units,s

i , where theentire supply must be distributed tot he destinations. Each destination has a fixeddemand for units,d

j , and the entire demand must be received from the sources.The costof distributing units from any particular sourcei to any particular destinationj , c

ij , isdirectly proportional to the number of units distributed,x

ij , (i = 1, 2, &..,m; j = 1, 2, &.,n) from sourcei to destinationj .

Any transportation model can be described completely by aparameter table as given inTable 4.2.In transportation problems everys

i and dj has integer valuesand also all othervariables.


Table 4.2:Transportation problemrepresentation

Cost per d istributed u nitSupplyDestination

1 2 & n1 c

11 c12 & c1n s1Source 2 c

21 c22 & c2n s2: : : : :

m cm1 cm2 & cmn sm

Demand d1 d2 & dn

A typical transportation problem is shown in Table 4.2, using a matrix with the rowsrepresenting sources and columns representing destinations. The algorithms for solvingthe problem are based on this matrix representation. The costs of shipping from sourcesto destinations are indicated by the entries in the matrix. If shipment is impossiblebetween a given source and destination, a large costof M is entered. This discourages thesolution from using such cells. Supplies and demands are shown along the margins of thematrix.

In general, Let Z represents the total distribution cost andxij (i = 1, 2, &.., m; j = 1, 2,

&., n) be the number of units to bedistributed from sourcei to destination j .Accordingly, the formulation of the transportation problem becomes:

Minimize Z =

==

n

jijij

m

ixC

11

Subject to miSi

n

jijx ........,,2,1

1==

=

njdj

m

iijx ........,,2,1

1==

=

jandiallforxij 0


For a transportation problem to posses a feasible solution, the following propertyindicates when this will occur. A necessary and sufficient condition for a transportationproblem to have any feasible solution is that:

jn

j

m

ii ds

==

=

11

This property may be verified by observing that the constraints require that both

jn

j

m

ii ds

==

=

11and

jn

j

m

ii ds

==

=

11equal

==

n

jij

m

ix

11

As shown, the number of functional constraints equals the number of sources (m) anddestinations (n). However, number of basic variables =m + n 1. The reason behind that,the functional constraints are equality constraints, and this set ofm + n equations has oneextra (redundant) equation that can be deleted without changing the feasible region (i.e.,any one of the constraints is automatically satisfied whenever the otherm + n 1constraints are satisfied. This can be verified as any supply constraint exactly equals thesum of the demand constraints minus the sum of other supply constraints.Therefore, anybasic feasible (BF) solution will have only m + n 1 basic variables (non-zerovariables) while all other variables will have zero value. Also, the sum of allocationsfor each row or each column equals its supply or demand.

We will generally assume that thetotal supply equals the total demand. If this is not truefor a particular problem,dummy sources or destinations can be added to make it true.The text refers to such a problem as abalanced transportation problem. These dummycenters may have zero distribution costs, or costs may be assigned to represent unmetsupply or demand.

For example, suppose that cannery 3, in Example 4.2,makes only 75 truckloads. Thetotal supply is now 25 units little.A dummy supply node can be added with supply 25 to


.

.

.

.

.

.

.

.

.

.

.

.

A =

Supp

lyco

nst

rain

tsD

ema

nd

con

stra

ints

balance the problem and the cost from the dummy to each warehouse can be added torepresent the cost of not meeting the warehouses demand.

In the pea shipping example, a basic solution might be to ship 20 truckloads from cannery1 to warehouse 2 and the remaining 55 to warehouse 4, 80 from cannery 2 to warehouse 1and 45 to warehouse 2 and, finally, 70 truckloads from cannery 3 to warehouse 3 and 30to warehouse 4. Even though the linear programming formulation of the pea shippingexample has seven constraintsother than non-negativity, a basic solution has only sixbasic variables! This is because the constraints are linearly dependent: the sum of the firstthree is identical to the sum of the last four. As a consequence, the feasible region definedby the constraints would remain the same if we only kept six of them. In general, a basicsolution to the transportation model will have a number of basic variables equal to thenumber of sources plus the number of destinations minus one.

Note that the resulting table of constraint coefficientshas a special structure as shownbelow in Table 4.3. Any linear programming problem that fits this special formulation isclassified as a transportationproblem; this is why the transportation problem is generallyconsidereda special type of linear programming problem.

Table 4.3: Constraint coefficients of the transportation problem

For many applications as the supply and demand quantities have integer values, all thebasic variables in every basic feasible solution also have integer values.

x11 x12 & &. x1n x21 x22 & &. x2n & &. xm1 xm2 & &. xmn1 1 & &. 1

1 1 & &. 1

1 1 & &. 1

1 1 11 1 1

& &.1 1 1


Example4.3

Assume three sources and three destinations with the cost of supply as given in Fig. 4.3.No transport between source S3 and destination D1, and Source S1 and destination D3.The matrix representation of this example is sown inTable 4.4.

Table 4.4: Matrix representation of the transportation problem

DestinationsSupply

1 2 3

Sources1 3 1 M 52 4 2 4 73 M 3 3 3

Demand 7 3 5

The network model of the transportation problem is shown in Fig. 4.3. Sources areidentified as the nodes on the left and destinations on the right. Allowable shipping linksare shown as arcs, while disallowed links are not included.

Figure 4.3: Network flow model of the transportation problem

On each supply node the positive external flow indicates supply flow entering thenetwork. On each destination node a demand is a negative fixed external flow indicatingthat this amount must leave the network.

S1

S2

S3

D1

D2

D3

(5)(7)3

1

4

2

4

3

3

(7)

(3)

(3)

(5)


Example 4.4

A contractor must transmit workers to four work sites each day. The travel time inminutes between each dispatch location and work site is shown in Fig. 4.4. In order tomaximize the number of productive work hours per day of each worker, the contractorwishes to minimize the total worker travel time. Formulate a mathematical model tominimize total travel time.Also, set the problem in the tabular format.

Figure 4.4: Network flow diagram of Example 4.4

Solution

Assume the variables arexij , number of worker to be dispatched from locationi to work

site j , wherei = 1 and 2 andj = 1, 2, 3, and 4. Since the total supply equals total demand,the total travel time equals:

minimize Z = 40 x11 + 20 x12 + 20 x13 + 50 x14 + 2 0 x21 + 50 x22 + 10 x23 + 60 x24

subject tox

11 + x 12 + x 13 + x 14 = 30 (supply)x

21 + x 22 + x 23 + x 24 = 50 (supply)x

11 + x 21 = 20 (demand)x

12 + x 22 = 20 (demand)x

13 + x 23 = 30 (demand)x

14 + x 24 = 10 (demand)x

ij "e0 for all i and j .

Dispatch location

Work site

1

2

1

2

3

4

(30)

(50)

(20)

(20)

(30)

(10)

40

20

20

50

20

50

10

60


The above formulation can be represented as shown in the following table:

DestinationSupply

1 2 3 4

Source1 40 20 20 50 302 20 50 10 60 50

Demand 20 20 30 10

Example 4.5

Solve Example 4.4 considering that there is no workers are dispatched from location1 toworksite 1.

Solution

The same previous table will be constructed again with very large time to be assignedbetween location1 and worksite1 called the big M method . This discourages thesolution from using such cell.

DestinationSupply

1 2 3 4

Source1 M 20 20 50 302 20 50 10 60 50

Demand 20 20 30 10

Example4.6

Solve Example 4.4 considering that the demand at work site 1 is 15 instead of 20workers.

Solution

In this case there is no balance as the number of workers at locations 1 and 2 equals 80workers, while the number of workers needed are 75 only. Then, a fictitious destination dummy destination will be added to balance both sides(Fig. 4.5).



The mathematical model in this case becomes:

minimize Z = 40 x11 + 20 x12 + 20 x13 + 50 x14 + 2 0 x21 + 50 x22 + 10 x23 + 60 x24

subject tox

11 + x 12 + x 13 + x 14 + x 15 = 30 (supply)x

21 + x 22 + x 23 + x 24 + x 25 = 50 (supply)x

11 + x 21 = 15 (demand)x

12 + x 22 = 20 (demand)x

13 + x 23 = 30 (demand)x

14 + x 24 = 10 (demand)x

15 + x 25 = 5 (demand)x


1

2

1

2

3

4

(30)

(50)

(15)

(20)

(30)

(10)

40

20

20

50

20

50

10

60

Dispatch location

Work site

5 (5)

00


This formulation can be represented in the table format as follow:

DestinationSupply

1 2 3 4 5

Source1 40 20 20 50 0 302 20 50 10 60 0 50

Demand 15 20 30 10 5

4.1.2 The Simplex Method for the Transportation Problem

Becausethe transportation problem is just a special type of linear programming problem,it can be solved by applying the simplex method as described in Chapter 3.However, thespecial structure of the transportation problem, as given in table 4.3, allows for using asimplified simplex method that achieving great computational savings.This simplifiedsimplex method is called transportation simplex method.

After constructing the table of constraint coefficients as shown in Table 4.3,convertingthe objective function to maximization form, and using the Big M method to introduceartificial variablesz

1, z2, &.., zm+n into the (m+n) equality constraints to get a typicalsimplex tableau s shown in Table 4.5.

Table 4.5: Original simplex tableau for the transportation problem

Basicvariable Eq. No.

Coefficient ofRight side

Z &. xij &.. zi &.. zm+j & &.

Z 0 -1 cij M M 0

1

zi i 0 1 1 & &. si

Zm+j m + j 0 1 1 dj

m + n


All entries not shownin Table 4.5 are zeros. Before starting the simplex method(the firstiteration), it is necessary to eliminate the non- zero coefficients of the artificial basicvariables in row 0.Accordingly, after any subsequent iteration, row 0 would have theform shown in Table 4.6.

Table 4.6: Row 0 of simplex tableau for the transportation problem

The valuesui and vj havethe following interpretation:

ui = multiple of original row 0 that have subtracted directly or indirectly from

original row 0 duringall iterations leading to the current simplex tableau.

vj = multiple of original rowm + j that have subtracted directly or indirectly from

original row 0 during all iterations leading to the current simplex tableau.

As mentioned before, the number of basic variables equals the number of supply sourcesand the number of demand destinations minus 1. So, any feasible solutionshould havenomore than(m + n 1) variables with non-zero values.

The procedure for constructing an initial basic feasible solution selects them + n 1basic variables one at a time. After each selection, a value that will satisfy one constraintis assigned to that variable, accordingly, eliminating the constraint row or column fromfurther consideration. Thus after an entire m +n 1 selections, an entire basic solutionhas been constructed to satisfy all constraints. Many methods have been proposed forselecting the basic variables. In this chapter, the Vogel s approximation method is used.

Basicvariable

Eq.No.

Coefficient ofRight side

Z &. xij &. zi &. zm+j &

Z 0 -1 cij ui - vj M - ui M - vj

= =

--

m

i

n

jjjji vdus

1 1


4.1.3 Solving the Transportation Problem Using Vogel s Approximation Method

General procedure for constructing ani nitial basic feasible solution(BFS):To begin: all source rows and destination columns of the transportation simplex tableauare initially under consideration for providing a basic variable.

1. From the rows and columns still under consideration, select the next basicvariable (allocation) according to some criterion.

2. Make the allocation large enough to exactly use up the remaining supply in itsrow or the remaining demand in its column (whichever is smaller).

3. Eliminate that row or column from further consideration. (If the row and columnhave the same remaining supply and demand, then arbitrarily select the rowas theone to be eliminated).

4. If only one row or one column remains under consideration, then the procedure iscompleted by selecting every remaining variable associated with that row orcolumn tobe basic with the only feasible allocation. Otherwise, return tothe firststep.

Vogel s approximation method: For each row and column remaining underconsideration, calculate its difference, the arithmetic difference between the smallestand next-to-the-smallest unit costc

ij still remainingin that row or column. In that rowor column having the largest difference, select the variable having the smallestremaining unit cost.Ties for the largest difference, or for the smallest remaining unitcost, may be broken arbitrarily.

Example 4.7

Find an initial basic feasible solution for the problem presented inExample 4.4 usingVogel s approximation method.


Solution

(1) Destination Supply Rowdiff.

1 2 3 4

Source1 40

20 20 50 30 02 20 50 10 60 50 10

Demand 20 20 30 10 x12 = 20

Col. Diff. 2030 10 10 Eliminate col. 2

First step: The highest difference is in column 2. The least value in this column is 20.Accordingly, the first basic variableto be assigned a valueis x

12. Thenx12 = 20, the valueis shown in the demand row. So,column 2 will be eliminated as its all demand issatisfied. Also, the supply value of row number 1 will be equal to 10 as a 20 of theoriginal value has been assigned to variablex

12.


1 3 4

Source 1 40 20 50 10 202

20 10 60 50 10Demand 20 30 10 x

21 = 20Col. Diff.

20 10 10 Eliminate col.1

Second step: The highest difference is in column 1and row 1; this tie may be brokenarbitrary by either choosing row1 or column 1. Choosing column 1, the least value in thiscolumn is 20. Accordingly, the second basic variableto be assigned a valueis x

21. Thenx

21 = 20, the value is shown in the demand row. So, column 1 will be eliminated as its alldemand is satisfied. Also, the supply value of row number 2 will be equal to 30 as a 20 ofthe original value has been assigned to variablex

21.


3 4

Source 1 20 50 10 302

10 60 30 50Demand 30 10 x

23 = 30Col. Diff. 10 10 Eliminate row 2


Third step: The highest difference is in row 2. The least value in this row is 10.Accordingly, the third basic variableto be assigned a valueis x

23. Then x23 = 30, thevalue is shown in the demand row. So, row2 will be eliminated as its all supply is used(i.e, equal zero). Also, all the demand of column 3 is satisfied.Accordingly, the last basicvariable, x

14, equals 10. Hence, the basic variables are:x

12 = 20, x21 = 20, x23 = 30, x13 = 0, and x14 = 10

The sum of all basic variables = 20 + 20 + 30 + 10 = 80 = si = dj

Z = 20 x 20 + 20 x 20 + 30 x 10 + 10 x 50 = 1600 minutes.

Example 4.8

Solve Example 4.5 usingVogel s approximation method.

Solution


1 2 3 4

Source1 M 20 20 50 30 02

20 50 10 60 50 10Demand 20 20 30 10 x

21 = 20Col. Diff.

M -20 30 10 10 Eliminate col.1


2 3 4

Source1 20 20 50 30 0

2 5010 60 30 40

Demand 20 30 10 x23 = 30

Col. Diff. 30 10 10 Eliminate row 2



2 3 4

Source1 20 20 50 30

Demand 20 0 10 x12 = 20, x13 = 0x

14 = 10Col. Diff.

Hence, the basic variables are:x21 = 20, x23 = 30, x12 = 20, and x14 = 10

The sum of all basic variables = 20 + 30 + 20 + 10 = 80 = si = dj

Z = 20 x 20 + 50 x 10 + 20 x 20 + 10 x 30 = 1600 minutes.

Example 4.9

Solve Example 4.6 usingVogel s approximation method.

Solution


1 2 3 4 5

Source1 40

20 20 50 0 30 20

2 20 50 10 60 0 50 10

Demand 15 20 30 10 5 x12 = 20

Col. Diff. 2030 10 10 0 Eliminate col. 2


1 3 4 5

Source1 40 20 50 0 10 202

20 10 60 0 50 10Demand 15 30 10 5 x

21 = 15Col. Diff.

20 10 10 0 Eliminate col.1



3 4 5

Source1 20 50

0 10 202 10 60 0 35 10

Demand 30 10 5 x15 = 5

Col. Diff. 10 10 0 Eliminate col.5


3 4

Source1 20 50 5 302

10 60 35 50Demand 30 10 x

23 = 30Col. Diff. 10 10 Eliminate col. 3


4

Source1 50 52 60 5

Demand 10 x14 = 5, x24 = 5

Col. Diff.

Hence, the basic variables are:x

12 = 20, x14 = 5, x15 = 5, x21 = 15, x23 = 30, and x24 = 5

The sumof all basic variables = 20 + 5 + 5 + 15 + 30 + 5 = 80 = si = dj

Z = 20 x 20 + 50 x5 + 0 x 5 + 20 x 15 + 10 x 30+ 60 x 5= 1550 minutes.


4.2 The Assignment Problem

A special case of the transportation problem is the assignment problem which occurswhen eachsupply is one and each demand isone. In this case, theintegrity implies thatevery supplier will beassigned one destination and every destination will have onesupplier. The costs give the chargefor assigning a supplier and destination to each other.

Example 4.10

A company has three new machines of different types. There are four different plants thatcould receive the machines. Each plant can only receive one machine, and each machinecan only be given to one plant. The expected profit that can be made by each plant ifassigned a machine is as follows:

Plant1 2 3 4

Machine1 15 16 12 112 15 0 13 203 5 7 10 6

This is a transportation problem with all supplies and demands equal toone, so it is anassignmentproblem. Note that a balanced problem must have the same number ofsupplies and demands, so we mustadd a dummy machine (corresponding to receiving nomachine) and assign a zero cost for assigningthe dummy machine to a plant.Theproblem is, then, to find the minimum cost matching of machines to plants.

The network flow diagram in Fig. 4.6 is very similar to the transportation model exceptthe external flows are all ones. The only relevant parameter for the assignment model isthe cost of assigning a machine to aj ob or plant. The solution for an assignment problemas shown will have a total flow of one in every column and row, and is the assignmentthat minimizes total cost.



Considering the variable,xij , of assigning machinei (i = 1, 2, and 3) to plantj (j = 1, 2, 3,

and 4), the mathematical formulation for this example can be stated as follow:

maximixeZ = 13x11 + 1 6x12 + 12x13 + 11x14 + 15x21 + 0x22 + 13x23 + 20x24 +

5x31 + 7x32 + 10x33 + 6x34

subject tox

11 + x12 + x 13 + x 14 = 1x

21 + x 22 + x 23 + x 24 = 1x

31 + x 32 + x 33 + x 34 = 1x

11 + x 21 + x 31 = 1x

12 + x 22 + x 32 = 1x

13 + x 23 + x 33 = 1x

14 + x 24 + x 34 = 1x


4.2.1 The AssignmentProblem Model

In the assignmentproblem there aren resources orassignee(e.g., employee, machine) isto be assigned uniquely to a particular activity orassignment(e.g., task, job). There is acost c

ij associated with assigneei (i = 1, 2,3, &.,n) assigned (performing) assignmentj (j

M1

M2

M3

P1

P2

P3

P4

(1)

(1)

(1)

(1)

(1)

(1)

(1)

13

1612

1115

013

20

5 710

6

Machine Plant


= 1, 2, 3, & &,n), so that the objective is to determine how all the assignments should bemade in order to minimize the total costz, considering that:

- The number of assignees = the number of assignments (tasks).- Each assignee is to be assigned to exactly one task.- Each task is to be performed by exactly one assignee.

Accordingly and asthe assignment problem is a special case from the transportationproblem, the decision variablex

ij of assigning machine (assignee) i to job (t ask) j alwaystake a value ofone or zero as follow:

1, if assigneei performsassignmentjx

ij =

0, if not

Thus, xij is a binary variable(0 or 1), which represents a yes or no decisions. In this case,

the yes or no decisions are: should assigneei performs assignmentj ? The assignmentproblem can be described completely by aparameter table as given inTable 4.7. Theclassic statement of the assignment problem uses a matrix with therows representingassignees and columns representing assignments. The algorithms for solving the problemare based on this matrix representation. The costs of performing assignmentj by assigneei is indicated by the entries in the matrix.

Table 4.7: Cost and requirements table for the transportation problemCost of assigningi to j

SupplyAssignment1 2 & n

1 c11 c12 & c1n 1

Assignee 2 c21 c22 & c2n 1

: : : : :

n cn1 cn2 & cnn 1

Demand 1 1 & 1


Minimize Z =

==

n

jijij

n

ixC

11

Subject to nin

jijx ........,,2,11

1==

=

njn

iijx ........,,2,11

1==

=

jandiallforxij 0

The first set of functional constraints specifies that each assignee performs exactly oneassignment, whereas the second set requires each assignment to be performed by exactlyone assignee. Comparing this formulation with the formulation of the transportationproblem given in section 4.1.1, we found that this is a special case of the transportationproblem where thesources now are the assignees and the destinations are theassignments, number of sources equal number of destinations, every supply =1 and everydemand =1.If it is impossible to perform a given assignment by a specific assignee, alarge cost of M is entered. This discourages the solution from using such cells.

We will generally assume that the totalnumber of assigneeequals the totalnumber ofassignments (i.e.,i = j ). If this is not true for a particular problem,dummy assigneesorassignmentscan be added to make it true.Such problem refers to as a balancedassignmentproblem. These dummy centers may have zeroassignmentcosts.

Example 4.11

A contractor has threedifferent types of excavation equipment that are needed to beassigned to three different excavation sites. The production cost for each 1 m3 of theexcavated soil is as given in the following table.

Job site1 2 3

Machine1 8 3 72 - 10 33 6 5 4


The contractor desires to make the optimum assignment to minimize total cost. Formulatethe mathematical model to optimally assign the equipment to the excavation sits.

Solution


Control variablesx11 to x33 are defined between each assigneei and assignmentj as show

in Figure 4.7. Also, as the number of assignees equals the numberof assignments, thenthe feasible solution is met.

The mathematical model in this case becomes:

minimizez = 8 x11 + 3 x12 + 7 x13 + 10 x22 + 3 x23 + 6 x31 + 5 x32 + 4 x33

subject tox

11 + x 12 + x 13 = 1 (supply)x

21 + x 22 + x 23 = 1 (supply)x

31 + x 32 + x 33 = 1 (supply)x

11 + x 21 + x 31 = 1 (demand)x

12 + x 22+ x 32 = 1 (demand)x

13 + x 23+ x 33= 1 (demand)x


1

2

3

1

2

3

(S1=1)

(d1=1)8

37

103

56

4

(S2=1)

(S3=1)

(d2=1)

(d3=1)

x11

X31x12

x13

X22

X23

X32

X33


The above formulation can be represented as shown in the following table:

AssignmentSupply

1 2 3

Assignee1 8 3 7 12 M 10 3 13 6 5 4 1

Demand 1 1 1

4.2.2 Solving the Assignment Problem Using Vogel s Approximation Method

As explained in section 4.1.1, the number of constraints equals m +n and the number ofbasic variables equalm + n 1. In the assignment problem,m = n, then the number ofconstraints = 2n and the number of basic variables = 2n 1. But, exactlyn only of thesex

ij should equalone. Therefore, there always aren - 1 basic variables withxij = 0(degenerate variables). The assignment problem is solvedthe same way as describedpreviously in solving the transportation problem.

Example 4.12

Solve example 4.11 using Vogel s approximation method to determine an initial basicfeasible solution.

(1) Assignment Supply Rowdiff.

1 2 3

Assignee

1 8 3 7 1 4

2 M 103 1 7

3 6 5 4 1 1

Demand 1 1 1 x23 = 1

Col. Diff. 2 2 1 Eliminate row 2


The highest difference is in row 2. The least value in this row is 3. Accordingly, the firstbasic variable isx

23, then x23 = 1. So, row 2 will be eliminated as itsall demand issatisfied. Also, the demand value of column number 3 will be equal to 0(i.e., x

13 = 0 andx

33 = 0).

(2) Assignment Supply Rowdiff.

1 2 3

Assignee

1 83 7 1 4

3 6 5 4 1 1

Demand 1 1 0 x12 = 1

Col. Diff. 2 2 3 Eliminate row1

By eliminating row then the needs of column two has been satisfied (i.e.,x21 = 0 andx32 =

0). Then the last basic variable isx31 = 1.

Hence, the basic variables are:x12 = 1, x23 = 1, x31 = 1, x32 = 0, andx33 = 0.

Minimum z = 8 x0 + 3 x 1 + 7 x 0 + 10 x 0 + 3 x 1 + 6 x 1 + 5 x 0 + 4 x 0 = 12

4.3 Solved Examples

4.3.1 Example 1

Formulate the following production problem as a transportation model. The demandsfora given item are 150, 250, 200 units for the next three months. The demand may besatisfi ed by

- Excessproduction in an earlier month held in stock for later consumption,- Productionin the current month,- Excessproduction in a later month backordered for preceding months.


The variable production cost per unit in any month is $6.00. A unit produced for laterconsumptionwill incur a storage cost at the rate of $1 per unit per month. On the otherhand, backordereditems incur a penalty cost of $3.00 per unit per month. The productioncapacity in each of thenext three months is 200 units. The objective is to devise aminimum cost production plan.

Solution


1 2 3

Source1

6 7 8 200 12 9 6 7 200 1

3 12 9 6 200 3

Demand 150 250 200 x11 = 150

Col. Diff.3 1 1 Eliminate col.1


2 3

Source1 7 8 50 12 6 7 200 1

3 96 200 3

Demand 250 200 x33 = 200

Col. Diff. 1 1 Eliminate row 3

7



2 3

Source1 7 8 50 12 6 7 200 1

Demand 250 0 x12 = 50, x22 = 200

Col. Diff. 1 1


z = 6 x 150 + 7 x 50 + 6 x 200 + 6 x 200 = LE 7650

4.3.2 Example 2

A car renting companyneeds to redeploy its automobiles to correct imbalances in thesystem. Currentlythe companyhas too many cars in New York (with 10 cars excess) andChicago (12cars excess). Pittsburgh would like up to 6 cars, Los Angeles up to 14 cars,and Miami up to 7 cars(note that more cars are demanded than are available). The costoftransporting a car from onecity to another is given by:

Pittsburgh Los Angeles MiamiNew York 50 250 100Chicago 25 200 125

Formulate this problem as a transportation problem. Clearly give the nodes and arcs. Foreach node, give the supplyor demand at the node. For each arc, give the cost.

Solution



PITT LA MIA

SourceNY 50 250 100 10 50CH 25 200 125 12 100

Dummy 00 0 5 0

Demand 6 14 7 x32 = 5

Col. Diff. 25200 100


PITT LA MIA

SourceNY 50 250 100 10 50CH 25

200 125 12 100Dummy 0 0 0 0 0

Demand 6 9 7 x22 = 9

Col. Diff. 25200 100 Eliminate col. 2


PITT MIA

SourceNY 50

100 10 50CH 25 125 3 100

Dummy 0 0 0 0Demand 6 7 x

13 = 7Col. Diff. 25

100 Eliminate col. 3


PITT

SourceNY 50 3 50CH 25 3 100

Dummy 0 0 0Demand 6 x

11 = 3, x21 = 3Col. Diff. 25



z = 50 x 3 + 100 x 7 + 25 x 3 + 200 x 9 + 0 x 5 = LE 2725

4.3.3 Example 3

The following table represents the cost of performing projects 1, 2 and 3 in thousandsEgyptian pounds submitted by four contracting companies A B, C, and D. It is required todetermine the best assignment of companies to projects to minimize the construction costof all projects.

CompaniesA B C D

Projects1 90 40 60 802 70 60 80 503 80 70 50 50

Solution

As the number of projects is less than the number of companies, a dummy project will beadded with zero cost.

(1) Company Demand Rowdiff.

A B C D

Project

1 90 40 60 80 1 202 70 60 80 50 1 103 80 70 50 50 1 04

0 0 0 0 1 0Supply 1 1 1 1 x

41 = 1

Col. Diff.70 40 50 50

Eliminate col. 1and row 4



B C D

Project1

40 60 80 1 202 60 80 50 1 103 70 50 50 1 0

Supply 1 1 1 x12 = 1

Col. Diff.20 10 0 Eliminate col.2


C D

Project1 60 80 0 202 80

50 1 303 50 50 1 0

Supply 1 1 x24 = 1

Col. Diff. 10 0 Eliminate row 2

Then x33 = 1. The optimum solution is:

Company A performs the dummy project 4 with cost of LE 0Company B performs project 1 with cost of LE 40000CompanyC performs project 3 with cost of LE 50000Company D performs project 2 with cost of LE50000

4.4 Exercises

1. A manufacturingfi rm produces widgets and distributes them tofive wholesalers.Sales forecasts indicate that monthly deliverieswill be 2700, 2700, 9000, 4500and 3600 widgets to wholesalers 1-5 respectively. The monthlyproductioncapacities are 4500, 9000 and 11,250 at plants 1, 2 and 3, respectively. The directcostsof producing each widget areEL2 at plant 1,LE1 at plant 2 andLE1.80 atplant 3. The transport cost of shipping a widget from a plant to a wholesaler isgiven below.


Wholesaler 1 2 3 4 5Plant 1Plant 2Plant 3

0.050.080.10

0.070.060.09

0.110.100.09

0.150.120.10

0.160.150.16

Formulate a linear Programmingmodel for this production and distributionproblem.

2. The amount of reinforcement steel available at three warehouses of a constructioncompany is 20, 12, and 12 tons respectively. While, the amounts of reinforcementsteel needed at four construction sites are 16, 7, 11, and 10 ton respectively. Thefollowing table shows the transportation cost in Egyptian Pounds (LE) betweenthe different warehouses and the construction sites. It is required to draw thenetwork flow diagram and determine how many tons of reinforcement steelshould be shipped from each warehouse to each of the construction sites in orderto minimize the total shipping cost.

Constructionsite

Warehouse

Transportation cost (LE)Construction

site 1Construction

site 2Construction

site 3Construction

site 4Warehouse 1Warehouse 2Warehouse 3

472

216

125

334

3. Resolve problem no. 1 assuming the transportation table as follow with noreinforcement steel to be shipped from warehouse 1 to site 3 and from warehouse3 to site 1.

Constructionsite

Warehouse

Transportation cost (LE)Construction

site 1Construction

site 2Construction

site 3Construction

site 4Warehouse 1Warehouse 2Warehouse 3

47-

216

-

25

334


4. Resolve problem no. 1 assuming that the reinforcing steel available at the threewarehouses is 20, 12, and 18 tons respectively. While, the amounts ofreinforcement steel needed at four construction sites are 16, 7, 11, and 10 tonrespectively.

5. A contractor owns three excavators of different capacity.He/she wants to dispatchthese excavators into three different jobs. The performance of the excavators ismeasured as the time consumed to perform each job. The data of the time tests areshown as given in the table below.

ExcavatorJob time (hour)

Job 1 Job 2 Job 3

123

5410

4106

983

a. Draw a network flow diagram.b. Formulate a mathematical model to obtain the minimum time.c. Solve this model using the Vogel s approximation model.

6. A contractor has two concrete pumps of different capacity. It is required to assignthese pumps at three tasks to obtain the optimum performance. The optimumperformance was measured in the time in hours it took to perform each task. Theresults of the time tests are given in the table below. Solve this problemt o obtaint he optimal time.

PumpTask time (hour)

Task 1 Task 2 Task 3

12

53

46

66

cb312 ch4

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