ch. ch. 44 the transportation and the transportation and assignment...

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� In this chapter we will discuss the transportation andassignment problems which are two special kinds of linearprogramming.

� The transportation problem deals with transporting goodsfrom their sources to their destinations .

� The assignment problem, on the other hand, deals withassigning people or machine to jobs.

Ch. Ch. 4 4 THE TRANSPORTATION AND THE TRANSPORTATION AND ASSIGNMENT PROBLEMSASSIGNMENT PROBLEMS

Example: 1The amount of reinforcement steel available at threewarehouses of a construction company is 20, 12, and 12 tonsrespectively. While, the amounts of reinforcement steelneeded at four construction sites are 16, 7, 11, and 10 tonrespectively. The following table shows the transportation costin Egyptian Pounds (LE) between the different warehousesand the construction sites. It is required to draw the networkflow diagram and determine how many tons of reinforcementsteel should be shipped from each warehouse to each of theconstruction sites in order to minimize the total shipping cost.

The transportation problemThe transportation problem

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Solution

- Formulate the problem:Minimize:

Z= 4X11+2 X12+ X13+3X14+7X21+ X22+2X23+3X24+2X31+6X32+5X33+4X34

Subject to:

X11+X12+X13+X14 = 20X21+X22+X23+X24 = 12 SUPPLYX31+X32+X33+X34 = 12

X11+X21+X31 = 16X12+X22+X32 = 17 DEMANDX13+X23+X33 = 11X14+X24+X34 = 10


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- This is an example of the transportation model. - This problem has a specific structure. All the

coefficients of the variables in the constraint equations are 1

- Every variable appears in exactly two constraints- All constraint equations are formed using the equal

sign not the ≥ or the ≤ signs- This is the special structure that distinguishes this

problem as a transportation problem


The Transportation Problem Model- the transportation model is concerned with

distributing a commodity from a group of supply centers, called sources to a group of receiving centers, called destinations to minimize total cost

- In general, source i (i = 1, 2, 3, ….., m) has a supply of si units, and destination j (j = 1, 2, 3, ……, n) has a demand for dj units

- The cost of distributing units from source i to destination j, cij, is directly proportional to the number of units distributed, xij


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The Transportation Problem Model- the transportation Table


Cost per distributed unit

SupplyDestination

1 2 … n

1 c11 c12 … c1n s1

Source 2 c21 c22 … c2n s2

: : : : :

m cm1 cm2 … cmn sm

Demand d1 d2 … dn

The Transportation Problem Model- Let Z represents the total distribution cost and xij (i =

1, 2, ….., m; j = 1, 2, …., n) be the number of units to be distributed from source i to destination j


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The Transportation Problem Model- A necessary condition for a transportation problem to

have any feasible solution is that:


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

The Transportation Problem Model- Number of variables = m x n- Number of constraints = m + n- Number of basic variables = m + n – 1- This as any supply constraint equals the sum of demand

constraints minus the sum of other supply constraints- Therefore, any basic feasible (BF) solution will have

only m + n – 1 basic variables (non-zero variables) while all other variables will have zero value. Also, the sum of allocations for each row or each column equals its supply or demand


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The Transportation Problem Model- One important condition is that all supplies should

equal all demands (balanced transportation problem)- If this is not true for a particular problem, dummy

sources or destinations can be added to make it true- These dummy centers have zero distribution costs - If shipment is impossible between a given source and

destination, a large cost of M is entered. This discourages the solution from using such cells


The Transportation Problem Model- Constraint coefficients of the transportation problem


x11 x12 ……. x1n x21 x22 ……. x2n ……. xm1 xm2 ……. xmn

1 1 ……. 1

1 1 ……. 1

1 1 ……. 1

1 1 1

1 1 1

…….

1 1 1

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Vogel’s Approximation Method - For each row and column remaining under

consideration, calculate its difference, the arithmetic difference between the smallest and next-to-the-smallest unit cost cij still remaining in that row or column.

- In that row or column having the largest difference, select the variable having the smallest remaining unit cost

- Ties for the largest difference, or for the smallest remaining unit cost, may be broken arbitrarily


- Using Vogel’s Approximation to solve for number of tons to be shipped from each warehouse:

- First check that sources are equal to the demand:

- Sources = 20+12+12 = 44 = demand = 16+7+11+10


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- Substitute in the objective function to get Z

Example: 2

Resolve problem no. 1 assuming the transportation table asfollow with no reinforcement steel to be shipped fromwarehouse 1 to site 3 and from warehouse 3 to site 1.


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Solution

- Formulate the problem:Minimize:

Z= 4X11+2 X12+3X14+7X21+ X22+2X23+3X24+6X32+5X33+4X34

Subject to:

X11+X12+X14 = 20X21+X22+X23+X24 = 12 SUPPLYX32+X33+X34 = 12

X11+X21 = 16X12+X22+X32 = 17 DEMANDX23+X33 = 11X14+X24+X34 = 10


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- Sources = 20+12+12 = 44 = demand = 16+7+11+10



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Example: 3

Resolve problem no. 1 assuming that the reinforcing steelavailable at the three warehouses is 20, 12, and 18 tonsrespectively. While, the amounts of reinforcement steelneeded at four construction sites are 16, 7, 11, and 10 tonrespectively.




- Sources = 20+12+18 = 50 , - While , demand = 16+7+11+10 = 44

- So we add a dummy construction site C5 to accommodate the difference of 6 tons.


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Solution

- Formulate the problem:Minimize:Z= 4X11+2 X12+ X13+3X14+7X21+ X22+2X23+3X24+2X31+6X32+5X33+4X34

Subject to:

X11+X12+X13+X14 + X15 = 20X21+X22+X23+X24 + X25 = 12 SUPPLYX31+X32+X33+X34 + X35 = 12

X11+X21+X31 = 16X12+X22+X32 = 17 DEMANDX13+X23+X33 = 11X14+X24+X34 = 10X15+X25+X35 = 6


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� A special case of the transportation problem wheneach supply is one and each demand is one

� As such, every supplier will be assigned onedestination and every destination will have onesupplier

The The Assignment ProblemAssignment Problem

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ExampleA contractor owns three excavators of different capacity. He/shewants to dispatch these excavators into three different jobs. Theperformance of the excavators is measured as the time consumed toperform each job. The data of the time tests are shown as given inthe table below.

- Draw a network flow diagram.- Formulate a mathematical model to obtain the minimum time.- Solve this model using the Vogal’s approximation model

The assignment problemThe assignment problem


1

2

3

1

2

3

(S1=1) (d1=1)8

3

7

10

3

5

6

4

(S2=1)

(S3=1)

(d2=1)

(d3=1)

x11

X31

x12

x13

X22

X23

X32

X33

X21 4

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- Formulate the problem:

Minimize:

Z= 5X11+ 4 X12+ 9X13+4X21+ 10 X22+8X23+10X31+6X32+3X33

Subject to:

X11+X12+X13 = 1X21+X22+X23= 1 SUPPLYX31+X32+X33= 1

X11+X21+X31 = 1X12+X22+X32 = 1 DEMANDX13+X23+X33 = 1


The Assignment Problem Model- In the assignment problem there are n resources or

assignee(e.g., employee, machine) is to be assigned uniquely to a particular activity or assignment(e.g., task, job)

- There is a cost cij associated with assignee i (i = 1, 2,3, …., n) assigned (performing) assignment j (j = 1, 2, 3, ……, n)

- the objective is to determine how all the assignments should be made in order to minimize the total cost


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The Assignment Problem Model- 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- The decision variable xij of assigning machine

(assignee) i to job (task) j always take a value of one or zero


The Assignment Problem Model


Cost of assigning i to j

SupplyAssignment

1 2 … n

1 c11 c12 … c1n 1

Assignee 2 c21 c22 … c2n 1

: : : : :

n cn1 cn2 … cnn 1

Demand 1 1 … 1

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The Transportation Problem Model- Let Z represents the total assignment cost and xij (i =

1, 2, ….., n; j = 1, 2, …., n) be a binary number representing assigning machine i to job j


The assignment problemThe assignment problemUsing Vogel’s Approximation

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Example:

� A cement company has four factories A, B, C, and D. Itsmajor distribution centers are located in three cities O, P, andR. The unit transportation cost from factories to thedistribution centers is as follows:

TRANSPORTATION PROBLEMTRANSPORTATION PROBLEM

Distribution CentersFactories

RPO

243A

596B

10812C

71113D

Optimality test

- The capacities of the four factories and the threedistribution centers are as follow:

� The company needs to determine the best transportationprogram which will minimize the cost of transportation.

Ch. Ch. 4 4 TRANSPORTATION PROBLEMSTRANSPORTATION PROBLEMS

Distribution capacity (ton)

Distribution center

Production capacity (ton)

Factory

10500O8500A

12000P11000B

7500R6000C

4500D

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Solution


Row Diff.SupplyDistribution Centers

FactoriesRPO

18500243A

111000596B

2600010812C

4450071113D

X AP = 850075001200010500Demand

Eliminate Row A343Col. Diff.



FactoriesRPO

111000596B

2600010812C

4450071113D

X BO = 105007500350010500Demand

Eliminate Col. O216Col. Diff.


FactoriesRP

450059B

26000108C

44500711D

X BR = 50075003500Demand

Eliminate Row B21Col. Diff.

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Solution



FactoriesRP

26000108C

44500711D

X DR = 450070003500Demand

Eliminate Row D33Col. Diff.


FactoriesRP

6000108C

X CP = 3500, X CR = 250025003500Demand

Col. Diff.

The preliminary solution

Z min= 8500 x 4 + 10500 x 6 + 500 x 5 + 3500 x 8 + 2500 x 10 + 4500 x 7 = 184000 L.E

SupplyDistribution Centers

FactoriesRPO

8500(2)

0(4)

8500(3)

0A

11000(5)

500(9)

0(6)

10500B

6000(10)

2500(8)

3500(12)

0C

4500(7)

4500(11)

0(13)

0D

75001200010500Demand


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Test the solution

LOOP METHOD for TESTING THE SOLUTION of LOOP METHOD for TESTING THE SOLUTION of TRANSPORTATION MODELTRANSPORTATION MODEL


FactoriesRPO

8500(2)

0(4)

8500(3)

0A

11000(5)

500(9)

0(6)

10500B

6000(10)

2500(8)

3500(12)

0C

4500(7)

4500(11)

0(13)

0D

75001200010500Demand

- It is be noted that the quantity in cells XAO ,XAR ,XBP ,XCO

,XDO , and XDP have a zero value.

- To test the obtained solution, the unoccupied cell will be

tested using one unoccupied cell and the other cells are

occupied. The cost of lost chance for the unoccupied cells

will be determined.

- To check the cell AR we can catch the cells AR, AP, CP, and

CR.


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Cell ARAdding of one unit to cell AR will increase the cost by + 2

Subtracting of one unit from cell AP will decrease the cost by - 4

Adding of one unit to cell CP will increase the cost by + 8

Subtracting of one unit from cell CR will decrease the cost by- 10

: :

Cost of lost chance - 4

- Testing of other unoccupied cells will give positive values or zero values to

the cost of lost chance for each unoccupied cell which mean increasing

the value of cost for positive values and the solution doesn’t change for

zero values.


The final solution

Z min = 6000 x 4 + 2500 x 2 +10500 x 6 + 500 x 5 + 6000 x 8 + 4500 x 7 = 174000 L.E


FactoriesRPO

8500(2)

2500(4)

6000(3)

0A

11000(5)

500(9)

0(6)

10500B

6000(10)

0(8)

6000(12)

0C

4500(7)

4500(11)

0(13)

0D

75001200010500Demand


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Optimality Test

� A basic feasible solution is optimal if and only if (cij

– ui - vj) ≥ 0 for every (i, j) such that xij is a non-basic

� Thus, the only work required by the optimality test isthe derivation of the values of the ui and vj for thecurrent basic solution and then the calculation ofthese (cij – ui - vj)

� Since (cij – ui - vj) is required to be zero if xij is abasic variable, the ui and vj satisfy the set ofequations


Optimality Test

� cij = ui + vj for each i and j such that xij is a basicvariable

� There are (m+n–1) basic variables, and accordingly(m+n-1) equations

� The number of unknowns (ui and vj) is (m + n), oneof these variables can be assigned a value arbitrarily

� Select the ui that has the largest number of allocationsin its row and assign it the value of zero


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Example:

� The following table shows the transportation cost of shippingone unit of a given product from the sources 1, 2 and 3 todestinations 1, 2, 3 and 4. The demand of each destinationand the supply of each source are also given in the table

TRANSPORTATION TRANSPORTATION PROBLEMSPROBLEMS

DestinationsSupply

1 2 3 4

Sources

1 1 2 - 3 20

2 7 1 2 3 12

3 - 6 5 4 12

Demand 16 7 11 10

� An initial basic feasible solution for this problem wasdetermined as shown in the next table. Check if this solutionis optimal or not and if not, determine the optimal solution.


Destinations

1 2 3 4

Sources

1 x11=16 x12=4

2 x22=1 x23=11

3 x32=2 x34=10

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1 2 3 4Suppl

yui

11 2 M 3

20 016 4

27 1 2 3

12 -11 11

3M 6 5 4

12 42 10

Demand 16 7 11 10 Z=99

vj 1 2 3 0

� Check the optimality by checking the value (cij – ui - vj) forall non-basic variables, these values should be positive orzero

� If any of these values are negative, then the solution is notoptimal and another iterative is required

� Because one of these values (c33 – u3 – v3) = -2 (negative), weconclude that, the current basic feasible solution is notoptimal. Thus, the transportation simplex method must nextgo to the iterative step to find a better basic feasible solution


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1 2 3 4 Supply ui

11 2 M M-3 3 3

20 016 4

27 7 1 2 3 4

12 -11 11

3M M-5 6 5 -2 4

12 42 10

Demand 16 7 11 10 Z=99

vj 1 2 3 0 Not optimal

� Find an entering basic variable, a leaving basic variable andthen identifying the new basic feasible solution

� The entering basic variable must have negative (cij–ui – vj)

� If there more than one negative value exists, select the onehaving the largest absolute negative value

� In this example,x33 is the entering basic variable

� Increasing the non-basic variable from zero sets off a chainreaction of changes in other basic variables in order tocontinue satisfying the supply and demand constraints

� The basic variable to be decreased to zero becomes theleaving basic variable. Withx33 being the entering basicvariable, the chain reaction is shown in the next Table


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� Find an entering basic variable, a leaving


1 2 3 4 Supply ui

11 2 M M-3 3 3

20 016 4

27 7 1 2 3 4

12 -1+ 1 - 11

3M M-5 6 5 -2 4

12 4- 2 + 10

Demand 16 7 11 10

vj 1 2 3 0

TRANSPORTATION PROBLEMSTRANSPORTATION PROBLEMS

1 2 3 4 Supply ui

11 2 M M-3 3 1

20 016 4

27 7 1 2 3 2

12 -13 9

3M M-3 6 2 5 4

12 22 10

Demand 16 7 11 10 Z=95

vj 1 2 3 2 optimal

ch. ch. 44 the transportation and the transportation and assignment...

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