Transportation Winston 2

7/30/2019 Transportation Winston 2

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Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.1

Transportation, Assignment andTransshipment Problems

fromOperations Research: Applications & Algorithms,

4th edition, by Wayne L. Winston


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Description

A transportation problem basically deals with the

problem, which aims to find the best way to fulfill

the demand of n demand points using the

capacities of m supply points. While trying to findthe best way, generally a variable cost of shipping

the product from one supply point to a demand

point or a similar constraint should be taken into

consideration.


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7.1 Formulating Transportation

ProblemsExample 1: Powerco has three electric

power plants that supply the electric needs

of four cities.

The associated supply of each plant and

demand of each city is given in the table 1.

The cost of sending 1 million kwh ofelectricity from a plant to a city depends on

the distance the electricity must travel.


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Transportation tableau

A transportation problem is specified by

the supply, the demand, and the shipping

costs. So the relevant data can be

summarized in a transportation tableau.The transportation tableau implicitly

expresses the supply and demand

constraints and the shipping cost between

each demand and supply point.


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Table 1. Shipping costs, Supply, and Demand

for Powerco Example

From To

City 1 City 2 City 3 City 4 Supply

(Million kwh)

Plant 1 $8 $6 $10 $9 35

Plant 2 $9 $12 $13 $7 50

Plant 3 $14 $9 $16 $5 40Demand

(Million kwh)

45 20 30 30

Transportation Tableau


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Solution

1. Decision Variable:

Since we have to determine how much electricity

is sent from each plant to each city;

Xij = Amount of electricity produced at plant i

and sent to city j

X14 = Amount of electricity produced at plant 1

and sent to city 4


7/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.7

2. Objective function

Since we want to minimize the total cost of shipping

from plants to cities;

Minimize Z = 8X11+6X12+10X13+9X14

+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34



3. Supply Constraints

Since each supply point has a limited productioncapacity;

X11+X12+X13+X14



4. Demand Constraints

Since each supply point has a limited productioncapacity;

X11+X21+X31 >= 45X12+X22+X32 >= 20

X13+X23+X33 >= 30

X14+X24+X34 >= 30



5. Sign Constraints

Since a negative amount of electricity can not be

shipped all Xijs must be non negative;

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)



LP Formulation of Powercos Problem

Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34

S.T.: X11+X12+X13+X14 = 30

X14+X24+X34 >= 30

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)


12/86Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.

12

General Description of a Transportation

Problem

1. A set of m supply points from which a good is

shipped. Supply point i can supply at most si

units.

2. A set of n demand points to which the good is

shipped. Demand pointj must receive at least di

units of the shipped good.

3. Each unit produced at supply point i and shippedto demand pointj incurs a variable costofcij.



13

Xij = number of units shipped from supply point i to

demand point j

),...,2,1;,...,2,1(0

),...,2,1(

),...,2,1(..

min

1

1

1 1

njmiX

njdX

misXts

Xc

ij

mi

i

jij

nj

j

iij

mi

i

nj

j

ijij



14

Balanced Transportation Problem

If Total supply equals to total demand, theproblem is said to be a balanced

transportation problem:

nj

j

j

mi

i

i ds11



15

Balancing a TP if total supply exceeds total

demand

If total supply exceeds total demand, we

can balance the problem by adding dummy

demand point. Since shipments to thedummy demand point are not real, they are

assigned a cost of zero.



16

Balancing a transportation problem if total

supply is less than total demand

If a transportation problem has a total

supply that is strictly less than total

demand the problem has no feasiblesolution. There is no doubt that in such a

case one or more of the demand will be left

unmet. Generally in such situations a

penalty cost is often associated with unmetdemand and as one can guess this time the

total penalty cost is desired to be minimum



17

7.2 Finding Basic Feasible

Solution for TPUnlike other Linear Programming

problems, a balancedTP with m supply

points and n demand points is easier to

solve, although it has m + n equality

constraints. The reason for that is, if a set

of decision variables (xijs) satisfy all but

one constraint, the values for xijs willsatisfy that remaining constraint

automatically.


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Methods to find the bfs for a balanced TP

There are three basic methods:

1. Northwest Corner Method

2. Minimum Cost Method

3. Vogels Method


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1. Northwest Corner Method

To find the bfs by the NWC method:

Begin in the upper left (northwest) corner of the

transportation tableau and set x11 as large as

possible (here the limitations for setting x11 to a

larger number, will be the demand of demandpoint 1 and the supply of supply point 1. Your

x11 value can not be greater than minimum of

this 2 values).


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According to the explanations in the previous slide

we can set x11=3 (meaning demand of demand

point 1 is satisfied by supply point 1).5

6

2

3 5 2 3

3 2

6

2

X 5 2 3


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After applying the same procedure, we saw that we

can go south this time (meaning demand point 2

needs more supply by supply point 2).

3 2 X

3 2 1

2

X X X 3

3 2 X

3 2 1 X

2

X X X 2


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Finally, we will have the following bfs, which is:

x11

=3, x12

=2, x22

=3, x23

=2, x24

=1, x34

=2

3 2 X

3 2 1 X

2 X

X X X X


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2. Minimum Cost Method

The Northwest Corner Method dos not utilize shippingcosts. It can yield an initial bfs easily but the total

shipping cost may be very high. The minimum cost

method uses shipping costs in order come up with a

bfs that has a lower cost. To begin the minimum costmethod, first we find the decision variable with the

smallest shipping cost (Xij). Then assignXijits largest

possible value, which is the minimum ofsi and dj


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After that, as in the Northwest Corner Method we

should cross out row i and column j and reduce the

supply or demand of the noncrossed-out row orcolumn by the value of Xij. Then we will choose the

cell with the minimum cost of shipping from the

cells that do not lie in a crossed-out row or column

and we will repeat the procedure.


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An example for Minimum Cost Method

Step 1: Select the cell with minimum cost.

2 3 5 6

2 1 3 5

3 8 4 6

5

10

15

12 8 4 6


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Step 3: Find the new cell with minimum shipping

cost and cross-out row 2

2 3 5 6

2 1 3 5

2 8

3 8 4 6

5

X

15

10 X 4 6


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cost and cross-out row 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

X

X

15

5 X 4 6


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cost and cross-out column 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5

X

X

10

X X 4 6


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cost and cross-out column 3

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4

X

X

6

X X X 6


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Step 7: Finally assign 6 to last cell. The bfs is found

as: X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4 6

X

X

X

X X X X


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3. Vogels Method

Begin with computing each row and column a penalty.The penalty will be equal to the difference between

the two smallest shipping costs in the row or column.

Identify the row or column with the largest penalty.

Find the first basic variable which has the smallestshipping cost in that row or column. Then assign the

highest possible value to that variable, and cross-out

the row or column as in the previous methods.

Compute new penalties and use the same procedure.


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An example for Vogels Method

Step 1: Compute the penalties.

Supply Row Penalty

6 7 8

15 80 78

Demand

Column Penalty 15-6=9 80-7=73 78-8=70

7-6=1

78-15=63

15 5 5

10

15


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Step 3: Identify the largest penalty and assign the

highest possible value to the variable.

Supply Row Penalty

6 7 8

5 5

15 80 78

Demand

Column Penalty 15-6=9 _ _

_

_

15 X X

0

15


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Step 4: Identify the largest penalty and assign the

highest possible value to the variable.

Supply Row Penalty

6 7 8

0 5 5

15 80 78

Demand

Column Penalty _ _ _

_

_

15 X X

X

15


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Step 5: Finally the bfs is found as X11=0, X12=5,

X13=5, and X21=15

Supply Row Penalty

6 7 8

0 5 5

15 80 78

15

Demand

Column Penalty _ _ _

_

_

X X X

X

X


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7.3 The Transportation Simplex

Method

In this section we will explain how the simplex

algorithm is used to solve a transportation problem.


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How to Pivot a Transportation Problem

Based on the transportation tableau, the followingsteps should be performed.

Step 1. Determine (by a criterion to be developed

shortly, for example northwest corner method) the

variable that should enter the basis.

Step 2. Find the loop (it can be shown that there is

only one loop) involving the entering variable and

some of the basic variables.Step 3. Counting the cells in the loop, label them as

even cells or odd cells.


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Illustration of pivoting procedure on the Powerco

example. We want to find the bfs that would result if

X14 were entered into the basis.

Nothwest Corner bfs and loop for Powerco

35 35

10 20 20 50

10 30 40

45 20 30 30

E O E O E O

(1, 4) (3, 4) (3, 3) (2, 3) (2, 1) (1, 1)

0

12

34

5


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New bfs after X14 is pivoted into basis. Since There

is no loop involving the cells (1,1), (1,4), (2,1),

(2,2), (3,3) and (3, 4) the new solution is a bfs.

35-20 0+20 35

10+20 20

20-20

(nonbasic) 50

10+20 30-20 40

45 20 30 30

After the pivot the new bfs is X11=15, X14=20,

X21=30, X22=20, X33=30 and X34=10.


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Copyright 2004 Brooks/Cole, a division of Thomson Learning, Inc.

44

Two important points!

In the pivoting procedure:

1. Since each row has as many +20s as20s, the

new solution will satisfy each supply and demand

constraint.

2. By choosing the smallest odd variable (X23) to

leave the basis, we ensured that all variables willremain nonnegative.


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Since the example is a minimization problem, the

current bfs will be optimal if all the ijs are

nonpositive; otherwise, we enter into the basis withthe most positive ij.

After determining cBVB-1 we can easily determine

ij. Since the first constraint has been dropped,

cBVB-1 will have m+n-1 elements.cBVB

-1= [u2 u3um v1 v2vn]

Where u2 u3um are elements ofcBVB-1

corresponding to the m-1 supply constraints, and v1v2vn are elements ofcBVB-1 corresponding to the n

demand constraints.


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To determine cBVB-1 we use the fact that in any

tableau, each basic variableXij must have ij=0.

Thus for each of the m+n-1 variables in BV,

cBVB-1aijcij=0

For the Northwest corner bfs of Powerco problem,BV={X11, X21, X22, X23, X33, X34}. Applying the

equation above we obtain:

11= [u2 u3 v1 v2 v3 v4] -8 = v1-8=0

0

0

0

1

0

0

1


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21= [u2 u3 v1 v2 v3 v4] -9 = u2+v1-9=0

22= [u2 u3 v1 v2 v3 v4] -12 = u2+v2-12=0

23= [u2 u3 v1 v2 v3 v4] -13 = u2+v3-13=0

0

0

0

1

0

1

0

0

1

0

0

1

0

1

0

00

1


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u1=0

u1+v1=8

u2+v1=9

u2+v2=12

u2+v3=13

u3+v3=16

u3+v4=5

By solving the system above we obtain:v1=8, u2=1, v2=11,v3=12, u3=4, v4=1


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For each nonbasic variable, we now compute

ij = ui+vjcij

We obtain:

12 = 0+116 = 5 13 = 0+1210 = 2

14 = 0+19 = -8 24 = 1+17 = -531 = 4+814 = -2 32 = 4+119 = 6

Since 32 is the greatest positive ij, we would nextenter X32 into the basis. Each unit that is entered

into the basis will decrease Powercos cost by $6.


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We have determined that X32 should enter the basis. As

shown in the table below the loop involving X32 and

some of the basic variables is (3,2), (3,3), (2,3), (2,2).

The odd cells in the loop are (2,2) and (3,3). Since the

smallest value of these two is 10 the pivot is 10.

35 35

10 20 20 50

10 30 40

45 20 30 30


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The resulting bfs will be:

X11=35, X32=10, X21=10, X22=10, X23=30 and X34=30

The uis and vjs for the new bfs were obtained by

solving

u1=0

u2+v2=12

u3+v4=5

u1+v1=8

u2+v3=13

u2+v1=9

u3+v2=9


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7.4 Sensitivity AnalysisIn this section we discuss the following three aspects

of sensitivity analysis for the transportation problem:

1. Changing the objective function coefficient of anonbasic variable.

2. Changing the objective function coefficient of a

basic variable.

3. Increasing a single supply by and a single demand

by .


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1. Changing the objective function coefficient

of a nonbasic variable.

Changing the objective function coefficient of a

nonbasic variableXij will leave the right hand side of

the optimal tableau unchanged. Thus the current

basis will still be feasible. Since we are not changingcBVB

-1, the uis and vjs remain unchanged. In row 0

only the coefficient ofXij will change. Thus as long

as the coefficient ofXij in the optimal row 0 is

nonpositive, the current basis remains optimal.

L t t t th f ll i ti b t


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Lets try to answer the following question about

Powerco as an example:

For what range of values of the cost of shipping 1million kwh of electricity from plant 1 to city 1 will the

current basis remain optimal?

Suppose we change c11from 8 to 8+ .

Now 11 = u1+v1-c11=0+6-(8+ )=2- .

Thus the current basis remains optimal for2- =-2, and c11>=8-2=6.


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2. Changing the objective function coefficient of a

basic variable.

Since we are changing cBVB-1, the coefficient of each

nonbasic variable in row 0 may change, and to determine

whether the current basis remain optimal, we must find

the new uis and vjs and use these values to price out all

nonbasic variables. The current basis remains optimal as

long as all nonbasic variables price out nonpositive.


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Thus, the current basis remains optimal for2


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7.5. Assignment Problems

Example: Machineco has four jobs to be completed.

Each machine must be assigned to complete one job.

The time required to setup each machine for

completingeach job is shown in the table below.Machinco wants to minimize the total setup time needed

to complete the four jobs.


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The Model

According to the setup table Machincos problem can be

formulated as follows (for i,j=1,2,3,4):

10

1

1

1

1

1

1

1

1..

10629387

5612278514min

44342414

43332313

42322212

41312111

44434241

34333231

24232221

14131211

4443424134333231

2423222114131211

ijij orXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXXts

XXXXXXXX

XXXXXXXXZ


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For the model on the previous page note that:

Xij=1 if machine i is assigned to meet the demands of

jobj

Xij=0 if machine i is assigned to meet the demands ofjobj

In general an assignment problem is balancedtransportation problem in which all supplies and

demands are equal to 1.

Although the transportation simplex appears to be very


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Although the transportation simplex appears to be very

efficient, there is a certain class of transportation

problems, called assignment problems, for which the

transportation simplex is often very inefficient. For thatreason there is an other method called The Hungarian

Method. The steps of The Hungarian Method are as

listed below:

Step1. Find a bfs. Find the minimum element in each row

of the mxm cost matrix. Construct a new matrix by

subtracting from each cost the minimum cost in its row.

For this new matrix, find the minimum cost in eachcolumn. Construct a new matrix (reduced cost matrix) by

subtracting from each cost the minimum cost in its

column.


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Step2. Draw the minimum number of lines (horizontal

and/or vertical) that are needed to cover all zeros in the

reduced cost matrix. If m lines are required , an optimalsolution is available among the covered zeros in the

matrix. If fewer than m lines are required, proceed to step

3.

Step3. Find the smallest nonzero element (call its value

k) in the reduced cost matrix that is uncovered by the

lines drawn in step 2. Now subtract k from eachuncovered element of the reduced cost matrix and add k

to each element that is covered by two lines. Return to

step2.


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7.6 Transshipment Problems

A transportation problem allows only shipments that godirectly from supply points to demand points. In many

situations, shipments are allowed between supply points

or between demand points. Sometimes there may also

be points (called transshipment points) through whichgoods can be transshipped on their journey from a

supply point to a demand point. Fortunately, the optimal

solution to a transshipment problem can be found by

solving a transportation problem.


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The following steps describe how the optimal solution to

a transshipment problem can be found by solving a

transportation problem.

Step1. If necessary, add a dummy demand point (with a

supply of 0 and a demand equal to the problems excess

supply) to balance the problem. Shipments to the dummyand from a point to itself will be zero. Let s= total

available supply.

Step2. Construct a transportation tableau as follows: A

row in the tableau will be needed for each supply pointand transshipment point, and a column will be needed for

each demand point and transshipment point.


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Each supply point will have a supply equal to its

original supply, and each demand point will have a

demand to its original demand. Let s= total available

supply. Then each transshipment point will have a supply

equal to (points original supply)+s and a demand equal

to (points original demand)+s. This ensures that anytransshipment point that is a net supplier will have a net

outflow equal to points original supply and a net

demander will have a net inflow equal to points original

demand. Although we dont know how much will beshipped through each transshipment point, we can be

sure that the total amount will not exceed s.

Transportation Problem


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Powerco Power Plant: Formulation

Define Variables:Let Xij = number of (million kwh) produced at plant i and

sent to city j.

Objective Function:Min z = 8*X11 + 6*X12 + 10*X13 + 9*X14

+ 9*X21 + 12*X22 + 13*X23 + 7*X24

+ 14*X31 + 9*X32 + 16*X33 + 5*X34

Supply Constraints:X11 + X12 + X13 + X14 < = 35 (Plant 1)

X21 + X22 + X23 + X24 < = 50 (Plant 2)

X31 + X32 + X33 + X34 < = 40 (Plant 3)

Transportation


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pProblem

Powerco Power Plant: Formulation (Contd.)

Demand Constraints:X11 + X21 +X31 > = 45 (City 1)

X12 + X22 +X32 > = 20 (City 2)

X13 + X23 +X33 > = 30 (City 3)

X14 + X24 +X34 > = 30 (City 4)

Nonnegativity Constraints:Xij > = 0 (i=1,..,3; j=1,..,4)

Balanced?Total Demand = 45 + 20 + 30 + 30 = 125

Total Supply = 35 + 50 + 40 = 125

Yes, Balanced Transportation Problem! (If problem is

unbalanced, add dummy supply or demand as required.)



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Powerco Power Plant: LINGO Formulation

model:

! A 3 Plant, 4 Customer

Transportation Problem;

SETS:

PLANT / P1, P2, P3/ : CAPACITY;CUSTOMER / C1, C2, C3, C4/ : DEMAND;

ROUTES( PLANT, CUSTOMER) : COST, QUANTITY;

ENDSETS

! The objective;

[OBJ] MIN = @SUM( ROUTES: COST * QUANTITY);



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Transportation ProblemPowerco Power Plant: LINGO Formulation (Contd.)

! The demand constraints;

@FOR( CUSTOMER( J): [DEM]

@SUM( PLANT( I): QUANTITY( I, J))>= DEMAND( J));

! The supply constraints;

@FOR( PLANT( I): [SUP]

@SUM( CUSTOMER(J): QUANTITY(I,J))


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Transportation ProblemPowerco Power Plant: LINGO Solution

Optimal solution found at step: 7Objective value: 1020.000

QUANTITY( P1, C1) 0.0000000

QUANTITY( P1, C2) 10.00000

QUANTITY( P1, C3) 25.00000QUANTITY( P1, C4) 0.0000000

QUANTITY( P2, C1) 45.00000

QUANTITY( P2, C2) 0.0000000

QUANTITY( P2, C3) 5.000000

QUANTITY( P2, C4) 0.0000000QUANTITY( P3, C1) 0.0000000

QUANTITY( P3, C2) 10.00000

QUANTITY( P3, C3) 0.0000000

QUANTITY( P3, C4) 30.00000

All Quantities

are Integers!



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Transportation ProblemPowerco Power Plant: LINGO Solution (Contd.)

Row Slack or Surplus Dual Price

OBJ 1020.000 1.000000

DEM( C1) 0.0000000 -9.000000

DEM( C2) 0.0000000 -9.000000

DEM( C3) 0.0000000 -13.00000DEM( C4) 0.0000000 -5.000000

SUP( P1) 0.0000000 3.000000

SUP( P2) 0.0000000 0.0000000

SUP( P3) 0.0000000 0.0000000

All constraints are binding!(Typical of Balanced Transportation Problems; results in

simple algorithms)


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Powerco Power Plant: Sensitivity Analysis (Contd.)V i bl V l R d d C t


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Variable Value Reduced Cost

QUANTITY( P1, C1) 0.00 2.00 (c11)

QUANTITY( P1, C2) 10.00 0.00 (c12)

QUANTITY( P1, C3) 25.00 0.00 (c13)

QUANTITY( P1, C4) 0.00 7.00 (c14)

QUANTITY( P2, C1) 45.00 0.00 (c21)

QUANTITY( P2, C2) 0.00 3.00 (c22)

QUANTITY( P2, C3) 5.00 0.00 (c23)QUANTITY( P2, C4) 0.00 2.00 (c24)

QUANTITY( P3, C1) 0.00 5.00 (c31)

QUANTITY( P3, C2) 10.00 0.00 (c32)

QUANTITY( P3, C3) 0.00 3.00 (c33)QUANTITY( P3, C4) 30.00 0.00 (c34)

z = c11* Q11 + c12* Q12 + c13* Q13 + c14* Q14

+ c21* Q21 + c22* Q22 + c23* Q23 + c24* Q24

+ c31* Q31 + c32* Q32 + c33* Q33 + c34* Q34

Nonbasic

Basic

Basic

Nonbasic

Basic

Nonbasic

BasicNonbasic

Nonbasic

Basic

NonbasicBasic

A i t P bl


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Assignment ProblemsPersonnel Assignment:

Time (hours)

Job 1 Job 2 Job 3 Job 4

Person 1 14 5 8 7

Person 2 2 12 6 5Person 3 7 8 3 9

Person 4 2 4 6 10

Assignment Problem Formulation


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Assignment Problem Formulation

Define Variables:

Let Xij = 1 if ith person is assigned to jth job

Xij = 0 if ith person is not assigned to jth job

Objective Function:

Min z = 14*X11 + 5*X12 + + 10*X44Personnel Constraints:

X11 + X12 + X13 + X14 = 1

X21 + X22 + X23 + X24 = 1

X31 + X32 + X33 + X34 = 1X41 + X42 + X43 + X44 = 1

Demand Constraints:

X11 + X21 + X31 + X41 = 1

X12 + X22 + X32 + X42 = 1

X13 + X23 + X33 + X43 = 1X14 + X24 + X34 + X44 = 1

Binary Constraints:

Xij = 0 or Xij = 1


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Assignment Problem Algorithm


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Assignment Problem Algorithm

9 0 3 00 10 4 1

4 5 0 4

0 2 4 6

Subtract Column Minimum from Each Column:

10 0 3 0

0 9 3 05 5 0 4

0 1 3 5

Subtract Minimum uncrossed value from uncrossed valuesand add to twice-crossed values:

Draw lines to cross out zeros and read solution from zeros

0

00

0

Solution:

Hungarian Method


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gStep 1: Find the minimum element in each row of the m x m cost

matrix. Construct a new matrix by subtracting from each cost the

minimum cost in its row. For this new matrix, find the minimum cost ineach column. Construct a new matrix (called the reduced cost matrix)

by subtracting from each cost the minimum cost in its column.

Step 2: Draw the minimum number of lines (horizontal and/or vertical)

that are needed to cover all the zeros in the reduced cost matrix. If mlines are required, an optimal solution is available among the covered

zeros in the matrix. If fewer than m lines are needed, proceed to Step

3.

Step 3: Find the smallest nonzero element (call its value k) in the

reduced cost matrix that is uncovered by the lines drawn in Step 2.Now subtract k from each uncovered element of the reduced cost

matrix and add k to each element that is covered by two lines. Return

to Step 2

Transportation Winston 2

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Transcript of Transportation Winston 2

Transportation winston

Dear Mr. Winston Review. What genre is “Dear Mr. Winston”? “Dear Mr. Winston” is a Realistic Fiction.

Mississauga Road, Olde Base Line Road, Winston Churchill ... · Winston Churchill Boulevard and Old Base Line Road for a comprehensive transportation network review Public Open House

Winston Watt

Winston Smith- In Chapter 2 Winston is engaged in a conflict with himself. As we know from previous chapters Winston has always prided himself upon.

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Chapter 7 Transportation, Assignment & Transshipment Problems to accompany Operations Research: Applications and Algorithms 4th edition by Wayne L. Winston.

Winston Churchill will · Title: Winston Churchill will Keywords: winston churchill, will Created Date: 20150129123400Z

Fluorescence, Phosphorescence, and Chemiluminescence 2... · ∥Department of Chemistry, Winston-Salem State University, Winston-Salem, North Carolina 27110, United States ⊥Department

Transportation and the United States Economy: Implications ...€¦ · Transportation and the United States Economy: Implications for Governance Clifford Winston Brookings Institution

John Winston

Winston Churchill 2

SIT-STAND...Anti-Fatigue Mat Mounts Laptop Holders Winston Workstation Dual Winston Workstation Triple Winston Workstation Quad Winston-E Single Winston-E Dual Winston-E Triple 4 |

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Winston Smith

DON FLOW AND HIS VISION FOR WINSTON-SALEM LAMPLIGHTER... · 2019. 1. 2. · 3:00 PM – Program Start Winston-Salem Symphony Stevens Center Transportation for ticket holders for those

Workshop on Intercultural Skills Enhancement · Winston 1A, Winston 1B, Winston 1C, Winston 3A, Winston 3B, Winston 3C, and Winston 2. 2 | WORKSHOP ON INTERCULTURAL SKILLS ENHANCEMENT

Winston churchill

Review 2 Algebra 1 - Winston-Salem/Forsyth County Schools

Chapter 2 of Winston Anderson's Book