transportation and assignment problemsise.tamu.edu/inen420/inen420_2005spring/slides/chapter...

1

L. Ntaimo (c) 2005 INEN420 TAMU

Transportation and Assignment Problems

Based onChapter 7

Introduction to Mathematical Programming: Operations Research, Volume 14th edition, by Wayne L. Winston and Munirpallam Venkataramanan

Lewis Ntaimo

2


7.1Transportation Problems

1. Example formulation2. General formulation3. Balancing a transportation problem4. Finding a basic feasible solution5. The transportation simplex algorithm

3


7.1 Powerco Problem

Powerco has 3 electric power plants that supply the needs of 4 cities. Each power plant can supply the following numbers of kilowatt-hours (kwh) of electricity: plant 1 – 35 million; plant 2 – 50 million; plant 3 – 40 million (see Table 1). The peak power demands in these cities, which occur at the same time (2pm), are as follows (in kwh): city 1 – 45 million; city 2 – 20 million; city 3 – 30 million; city 4 - 30 million. The costs of sending 1 million kwh of electricity from plant to city depend on the distance the electricity must travel. Formulate an LP to minimize the cost of meeting each city’s peak power demand.

Table 1: Shipping Costs, Supply, and Demand for Powerco

To Supply

From City 1 City 2 City 3 City 4 (million kwh)

Plant 1 $8 $6 $10 $9 $35

Plant 2 $9 $12 $13 $7 $50

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

Demand 45 20 30 30

(million kwh)

4


7.1 Graphical Representation

Demand Points

Plant 1

Plant 3

City 1

City 2

City 3

City 4

Supply Pointsd1 = 45

Plant 2

xij

s1 = 35 d2 = 20

s2 = 50d3 = 30

s3 = 40

d4 = 30

Decision Variables: xij # of (million) kwh produced at plant i and sent to city j

Constraints: Supply (Capacity) constraints

Demand constraints

5


7.1 Powerco Problem Formulation

)4,3,2,1;3,2,1( 0 30 30 20 45

40 50 35 s.t.

516914 713129 9106 8Min

342414

332313

322212

132111

34333231

24232221

14131211

34333231

24232221

14131211

==≥≥++≥++≥++≥++

≤+++≤+++≤+++

+++++++++++

jixxxxxxxxxxxxx

xxxxxxxxxxxx

xxxxxxxxxxxx

ij

Supply Constraints

Demand Constraints

Minimize total shipping costs

0. toequal ablesother vari ,30,10,5,45,10,10,1020

:solution Optimal

343223211312

allxxxxxxz =======

6


7.1 Graphical Representation

Demand Points

Plant 1

Plant 3

City 1

City 2

City 3

City 4


Plant 2

x21= 45

x12= 10

x34= 30

x13= 25

x23= 5

x32= 10

s1 = 35 d2 = 20

s2 = 50d3 = 30

s3 = 40

d4 = 30

7


7.1 General Formulation of a Transportation Problem

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

s)constraint (demand ),...,2,1(

s)constraint(supply ),...,2,1( t.s.

Min

1

1

1 1

njmix

njdx

misx

xc

ij

j

n

iij

i

n

jij

m

i

n

jijij

==≥

=≥

=≤

∑

∑

∑∑

=

=

= =

8


7.2 A Balanced Transportation Problem

∑∑==

=n

jj

m

ii ds

11

In a “balanced transportation” problem, the total supply is equal to the total demand:

• All constraints must be binding

• It becomes relatively easy to find a basic feasible solution

• Simplex pivots do not involve multiplication, they reduce to additions and subtractions

•Therefore, it is desirable to formulate a transportation problem as a balanced transportation problem

9


7.2 A Balanced Transportation Problem

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

s)constraint (demand ),...,2,1(

s)constraint(supply ),...,2,1( t.s.

Min

1

1

1 1

njmix

njdx

misx

xc

ij

j

n

iij

i

n

jij

m

i

n

jijij

==≥

==

==

∑

∑

∑∑

=

=

= =

10


7.2 Balancing a Transportation Problem if Total Supply Exceeds Total Demand

Create a “dummy demand point” that has demand equal to the amount of excess supply

Shipments to the dummy demand point:

(1) Are assigned a cost of zero because they are not real shipments

(2) Indicate unused supply capacity

11


Demand Points

7.2 Balancing a Transportation Problem if Total Supply Exceeds Total Demand

Plant 1

Plant 3

City 1

City 2

City 3

City 4


Plant 2

xij

Dummy5

s1 = 35 d2 = 20

1154

1

=∑=j

jd

1253

1=∑

=iis s2 = 50

d3 = 30

s3 = 40

d4 = 30

d5 = 10c15 = c25 =c35 = 0

12


7.2 Balancing a Transportation Problem if Total Supply is Less than Total Demand

In this case the problem has no feasible solution: demand cannot be satisfied

However, it is sometimes desirable to allow the possibility of leaving some demand unmet:

(1) A penalty (cost) is often associated with the unmet demand

(2) To balance the problem, add a “dummy (or shortage) supply point”

13


7.3 Transportation Tableau

A transportation problem is specified by supply, the demand, and the shipping costs, so the relevant data can be summarized in a “transportation tableau”:

j 1 2 n

c11 c1n

c21

c12

c22 c2n

cm1 cmncm2

.

.

.

.

.

.

.

.

.

.

.

.

. . .

. . .

. . .

. . .

i1 s1

2 s2

Cell: (row i, col j)

smm

d2 dnd1

14


7.3 Powerco Transportation Tableau

• If xij is a bv, its value is placed in the lower left-hand corner of the ijth cell

8 6 9 1010 3525

9 12 13 75045 5

14 9 16 5 4010 30

45 20 30 30

15


7.4 Finding BFS’s for Transportation Problems

Consider a transportation with m supply points and n demand points

- Such a problem contains m + n equality constraints

- Recall: In the Big M method and Two-phase simplex method it is difficult to find a bfs if all of the LP’s constraints are equalities

- Fortunately, the structure of the transportation problems makes it easy to find a bfs

Important Observation

If a set of values for the xij’s satisfies all but one of the constraints of a balanced transportation problem, then the values for the xij’s will automatically satisfy the other constraints.

16


7.4 Powerco Problem Formulation

)4,3,2,1;3,2,1( 0 30 30 20 45

40 50 35 s.t.

516914 713129 9106 8Min

342414

332313

322212

312111

34333231

24232221

14131211

34333231

24232221

14131211

==≥=++=++=++=++

=+++=+++=+++

+++++++++++

jixxxxxxxxxxxxx

xxxxxxxxxxxx

xxxxxxxxxxxx

ij

Supply Constraints

Demand Constraints


Omit this constraint

0. toequal ablesother vari ,30,10,5,45,10,10,1020

:solution Optimal

343223211312

allxxxxxxz =======

17


7.5 Powerco ExampleRecall: s1 = 35, s2 = 50, s3 = 40,

d1 = 45, d2 = 20, d3 = 30, d4 = 30

Let a set of xij’s satisfy all constraints except the first supply constraint.

Then this set of xij’s must supply

d1 + d2 + d3 + d4 = 125 million kwh to cities 1 to 4

and supply

s2 + s3 = 125 – s1 = 90 million kwh from plants 2 and 3.

Thus plant 1 must supply

125 – (125 – s1) = 35 million kwh,

So the xij’s must satisfy the first supply constraint!

Therefore, we arbitrarily assume that the first constraint is omitted from consideration.

18


7.5 Loop

Definition:

An ordered sequence of at least 4 different cells is called a loop if

1. Any 2 consecutive cells lie in either the same row or same column

2. No 3 consecutive cells lie in the same row or column

3. The last cell in the sequence has a row or column in common with the first cell in the sequence

Path: (1,1)-(1,2)-(2,3)-(2,1)Loop or Path?

Loop: (2,1)-(2,4)-(4,4)-(4,1) Loop: (1,1)-(1,2)-(2,2)-(2,3)-

(4,3)-(4,5)-(3,5)-(3,1)Path: (1,1)-(1,2)-(1,3)-(2,3)-(2,1)

19


7.5 Theorem

In a balanced transportation problem with m supply points and ndemand points, the cells corresponding to a set of (m + n – 1) variables contain no loop iff the (m + n – 1) variables yield a basic solution

This follows from the fact that a set of (m + n – 1) cells contains no loop iff the

(m + n – 1) columns corresponding to these cells are linearly independent.

Example: Loop: (1,1)-(1,2)-(2,2)-(2,1)

4

5

3 2 4

Because (1,1)-(1,2)-(2,2)-(2,1) is a loop, the Theorem tells us that

{x11, x12, x22, x21} cannot yield a bfs for this transportation problem.

20


7.6 The Northwest Corner Method for Finding a BFS for a Balanced Transportation Problem

Begin in the upper left (or northwest) corner of the transportation tableau

Set x11 as large as possible. Clearly x11 = min{s1, d1}.

If x11= s1, cross out row 1 of the transportation tableau; no more bv’s will come from row 1. Also set d1 = d1 - s1.

If x11= d1, cross out the column 1 of the transportation tableau; no more bv’s will come from column 1. Also set s1 = s1 - d1.

If x11= s1 = d1, cross out either row 1 or column 1 (but NOT both).

If you cross out row 1, set d1 = 0.

If you cross out column 1, set s1 = 0.

Continue applying this procedure to the most northwest corner cell in the tableau that does not lie in a crossed-out row or column. Eventually, will come to a point where there is only one cell that can be assigned a value.

Assign this cell a value equal to its row or column demand, and cross out both the cell’s row and column.

A BFS has now been obtained.

21


7.6 Powerco Example: Finding a BFS8 6 9 10

35 - 3535

9 12 13 750

14 9 16 540

45 - 35 20 30 30

8 6 9 10x3550

9 12 13 7

4014 9 16 5

10 20 30 30

22



X35

9 12 13 710 50 - 10

14 9 16 540

10 - 10 20 30 30

8 6 9 10x35

9 12 13 74010

14 9 16 540

X 20 30 30

23



X35

9 12 13 710 40 - 2020

14 9 16 540

X 320 - 20 30 0

8 6 9 10x35

9 12 13 72010 20

14 9 16 540

X 3X 30 0

24



X35

9 12 13 710 20 20

20 - 2014 9 16 5

40

X 3X 30 - 20 0

8 6 9 10x35

9 12 13 7X10 20 20

14 9 16 540

X 3X 10 0

25


Powerco Example: Finding a BFS8 6 9 10

X35

9 12 13 7X10 20 20

14 9 16 510 40 -10

X 3X 10 -10 0

8 6 9 10x35

9 12 13 7X10 20 20

14 9 16 53010

X 3X X 0

26



X35

9 12 13 7X10 20 20

14 9 16 530 - 30

1030

X 3X X 0 - 30

8 6 9 1035 X

9 12 13 7X10 20 20

14 9 16 5X10 30

X X X X

27


7.6 Powerco Example: Basic Feasible Solution

8 6 9 1035 35

9 12 13 710 5020 20

14 9 16 510 30

40

45 20 30 30

BFS: x11 = 35

x22 = 10

x22 = 20

x23 = 20

x33 = 10

x34 = 30

NOTE: The BFS does NOT form a LOOP

28


7.8 The Transportation Simplex Method

Pricing Out Nonbasic Variables Recall:

.variablesbasic ofset dropped)been has constraintsupply first that the(assuming

LP, original in the for column

,for t coefficien objective where,

,

by given is cost) (reduced 0 Row s tableau'in the variable theoft coefficien The1

=

=

=

−= −

BV

xa

xc

caBcc

x

ijij

ijij

ijijBVij

ij

E.NONPOSITIV

are variablesnonbasic theallfor s' theif optimal be willbfscurrent theproblem, ONMINIMIZATI a solving are weSince

ijc

. POSITIVEmost with the variablenonbasic thebasis theinto ENTER weOtherwise,

ijc

29


7.8 The Transportation Simplex Method Cont..

s.constraint demand the toingcorrespond of elements theare ,...,,

s;constraintsupply 1 the toingcorrespond of elements theare ,... , where

,] ... ... [

:elements )1( have willdropped,been has constraintfirst theBecause

. computeeasily can we, gdetermininAfter

121

1 ,32

21 321

1

nBcvvv

m-Bcuuu

vvvuuuBc

n-m

cBc

BVn

BVm

nmBV

ij-

BV

−

−

− =

+

30


7.8 The Transportation Simplex Method Cont..

solve! easy to very are (1) equations theproblem,ation transportaFor

(1) 0

),(in variablesbasic 1 theofeach for Thus .0

havemust variablebasiceach u,any tableain fact that theuse we, determine To

1

1

=−

+=−

ijijBV

ij

ij-

BV

caBc

BVn-m c

xBc

. variablesbasic allfor

,0

:equations of system following thesolvemust we,for solve toThus,

. variablesbasic allfor

toreduces (1) that see will we,0set weIf

1

1

1

ijji

-BV

ijjiijijji

cvuu

nmBc

cvuccvu u

=+=

+

−+=⇒=+=

31


7.8 Powerco Example: Solution of (1)We start with the BFS obtained by applying the NW Corner method:

8 6 9 1035 35

9 12 13 710 5020 20

14 9 16 510 30

40

45 20 30 30

slides) 2next (See

:following obtain the we0 Applying

}, , , , ,{ bfs, For this

1

343323222111

=

=

ijij-

BV -caBc

xxxxxx BV

32


7.8 Powerco Example: Solution of (1) Model

088

000100

] [ 143213211 ==−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

= -vvvvvuuc

099

000101

] [ 1243213221 =+=−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

= -vuvvvvuuc

01212

001001

] [ 2243213222 =+=−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

= -vuvvvvuuc

01313

010001

] [ 3243213223 =+=−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

= -vuvvvvuuc

01616

010010

] [ 3343213233 =+=−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

= -vuvvvvuuc

055

100010

] [ 4343213234 =+=−

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

= -vuvvvvuuc

0 Applying

}, , , , ,{1

343323222111

=

=

ijij-

BV -caBc

xxxxxx BV

33


7.8 Powerco Example: Solution of (1)We start with the BFS obtained by applying the NW Corner method:

8 6 9 1035 35

9 12 13 710 5020 20

14 9 16 510 30

40

45 303020

5 16 13 12 9 8

,0 :solvingby find We

43

33

32

22

12

11

1

1

=+=+=+=+=+=+

=

vuvuvuvuvuvu

uBc -

BV

34


7.8 Powerco Example: Solution of (1)

69114 21484

5711 8910 210120

56110 :obtain weand

compute now we variable,nonbasiceach For

32

31

24

14

13

12

=−+=−=−+=

−=−+=−=−+==−+==−+=

−+=

cccccc

cvu c ijjiij

5 16 13 12 9 8


43

33

32

22

12

11

1

1

=+=+=+=+=+=+

=

vuvuvuvuvuvu

uBc -

BV

1 4 12 11 1 8

:obtain We

4

3

3

2

2

1

======

vuvvuv Entering Nonbasic Variable

$6.by cost sPowerco' decrease willbasis theinto entered is that ofunit Each :Note

basis. theinto enter next wouldwe

, positivemost theis Because

32

32

32

x

x

cc ij

35


7.9 How to Pivot in a Transportation Problem

Step 1: Determine the variable that should enter the basis

Step 2: Find the loop involving the entering variable and some of thebasic variables. (It can be shown that there is only one loop)

Step 3: Counting only cells in the loop, label those found in Step 2 thatare an even number (0, 2, 4, …) cells away from the entering variable as even cells. Also label those that are an odd numberof cells away from the entering variable as odd cells.

Step 4: Find the odd cell whose variable assumes the smallest value.Call this value θ. The variable corresponding to this cell will leavethe basis.

To perform the pivot, decrease the value of each odd cell by θand increase the value of each even cell by θ. The values of variablesnot in the loop remain unchanged. The pivot is now complete!

36


7.9 How to Pivot in a Transportation Problem Cont…

Degenerate Solution:

In Step 4 if θ = 0, then the entering variable will equal 0, and an odd variable that has a current value of 0 will leave the basis.

In this case, a degenerate bfs existed before and will result after the pivot.

If more that one odd cell in the loop equals θ, you may arbitrarily chooseone of these to leave the basis, again, a degenerate bfs will result.

37


7.10 Summary of the Transportation Simplex Method

Step 1: If the problem is unbalanced, balance it.

Step 2: Use the northwest corner method to find a bfs.

Step 3: Use the fact that u1 = 0 and ui + vj = cij for all basic variables to find[u1 u2 … um v1 v2 … vn] for the current bfs.

Step 4: If ui + vj - cij ≤ 0 for all nonbasic variables, then the current bfs is optimal. If this is not the case, then we enter the variable with the most positiveui + vj - cij into the basis using the pivoting procedure. This yields a new bfs.

Step 5: Using the new bfs, return to steps 3 and 4.

Step 4’: If ui + vj - cij ≥ 0 for all nonbasic variables, then the current bfs is optimal. Otherwise, enter the variable with the most negative ui + vj - cij into the basis using the pivoting procedure. This yields a new bfs.

For a maximization problem proceed as stated, but replace step 4 by Step 4’:

38


7.11 Example: Powerco Problem

69114 21484

5711 8910 210120

56110 :obtain and

compute we variable,nonbasiceach For

32

31

24

14

13

12

=−+=−=−+=

−=−+=−=−+==−+==−+=

−+=

cccccc

cvu c ijjiij



32

32

x

cc ij

We start with the initial bfs obtained by using theNW corner method:

Table 18 6 10 9

35 35

9 12 13 710 20 5020

10

5 16 13 12 9 8

,0 :solvingby foundalready have We

43

33

32

22

12

11

1

1

=+=+=+=+=+=+

=

vuvuvuvuvuvu

uBc -

BV

1 4 12 11 1 8 :Solution

4

3

3

2

2

1

======

vuvvuv

45 303020

14 9 16 530 40

2) Table (see 10by and increase and 10,by and decrease

pivot will the20, and 10 Because(2,2). and (3,3) are loop in this cells odd The

(2,2).-(2,3)-(3,3)-(3,2) :1 Tablein shown is sbv' theof some and involving The

23

322233

2233

32

xxxx

xx

xloop

==

39



616122 81482

1771 2970 210120

56110 :obtain and


33

31

24

14

13

12

−=−+−=−=−+−=

=−+=−=−+==−+==−+=

−+=

cccccc

cvu c ijjiij



12

21

x

cc ij

Table 28 6 10 9

35

10

35

5 9 13 12 9 8


43

23

32

22

12

11

1

1

=+=+=+=+=+=+

=

vuvuvuvuvuvu

uBc -

BV

7 2

12 11 1 8 :Solution

4

3

3

2

2

1

=−=

====

vuvvuv

45 303020

9 12 13 710 30 50

14 9 16 510 30 40

3). Table (see 10by and increase and 10,by and decrease pivot will the

cell, oddan in entry smallest theis 10 Because(1,1). and (2,2) are loop in this cells odd The


2112

1122

22

12

xxxx

x

xloop

=

40



416120 61480

1751 51261

4950 210120

:obtain and


33

31

24

22

14

13

−=−+=−=−+=−=−+=−=−+=−=−+==−+=

−+=

cccccc

cvu c ijjiij



13

13

x

cc ij

Table 38 6 10 9

5 6 6 13 9 8


43

23

21

32

12

11

1

1

=+=+=+=+=+=+

=

vuvuvuvuvuvu

uBc -

BV

5 0 12 6 1 8 :Solution

4

3

3

2

2

1

======

vuvvuv

45 303020

9

25

20

10 35

12 13 730 50

14 9 16 510 30 40

4). Table (see 25by and increase and 25,by and decrease pivot will the

cell, oddan in entry smallest theis 25 Because(1,1). and (2,3) are loop in this cells odd The


2113

1123

11

13

xxxx

x

xloop

=

41



316103 51463

2723 31263

4950 2860

:obtain and


33

31

24

22

14

11

−=−+=−=−+=−=−+=−=−+=−=−+=−=−+=

−+=

cccccc

cvu c ijjiij

Table 48 6 10 9

10 25 35

9 12 13 745 5 50

14 9 16 510 30 40

.020,1$ )30(5)10(9)5(13)45(9)25(10)10(6

and,30,10,5,45,25,10

:is problem Powerco theosolution t optimal theThus obtained.been

hassolution optimalan 0, all Because

343223211312

=+++++=

======

≤

z

xxxxxx

cij45 20 30 30

9 10 13 6 5 9


23

31

32

21

43

12

1

1

=+=+=+=+=+=+

=

vuvuvuvuvuvu

uBc -

BV

2 3 10 6 3 6 :Solution

4

3

3

2

2

1

======

vuvvuv

42


7.12 Assignment Problems

1. Problem Description2. Example Formulation3. The Hungarian Method4. Example using the Hungarian Method5. Transshipment Problems

43


7.12 On the integrality Property of the Transportation Problem

• We already know the fact that solutions to the transportation problem are integral. This is a useful property of the transportation problem.

• In general, solutions to integer programs with the integer restrictions relaxed are fractional (linear programming (LP) relaxation).

e.g. x1 + x2 = 1, x1 – x2 = 0, has the unique solution (x1 = 0.5, x2 = 0.5)

• But solutions to the transportation problem are integral.

• In general, if there is at most one 1 and at most one –1 in any column of the constraint matrix (unimodular constraint matrix), then every basic feasible solution is integer (so long as RHS is integral.)

44


7.12 A special case of the Transportation Problem: Supplies and Demands are All Equal to 1.

Supply Demand Supply Demand

11 411

1

1

1

1

2

3

4

5

6

1

1

1

0

1

1

0

2

1

1

2

22

6

1

1

1

3

5

1

1

1

A solution to this transportation problem may be viewed as an “assignment”

45


7.12 The Assignment Problem

•3 people: nodes 1, 2, 3

1

2

3

P

4

5

6

T/J

1

1

1

1

1

11

0

1

1

0

21

2

2

•3 tasks (jobs): nodes 4, 5, 6

•Each person must be assigned to a task

•Each task has a person assigned

•Cost of assigning a person to a task

•Objective: meet constraints while minimizing total cost

Formulate this as an integer program or IP.

(Note that all variables are required to be integer).

46


7.13 Assignment Problem

• Suppose that n people (let us label them 1 to n) are assigned to n tasks (also let us label them labeled 1 to n). Then,

– let xij = 1 if person i is assigned to task j

– let xij = 0 if person i is not assigned to task j

– let cij be the cost of assigning person i to task j

• Formulate the assignment problem

47


7.13 The Assignment Problem

In general the LP formulation is given as

Minimize

s.t.

1 1

1

1

1 1

1 1

0

, , ,

, , ,

or 1,

n n

ij iji j

n

ijj

n

iji

ij

c x

x i n

x j n

x ij

= =

=

=

= ∀ =

= ∀ =

= ∀

∑∑

∑

∑

…

…

Each supply is 1

Each demand is 1

48


7.14 Example Problem

Machineco has 4 machines and 4 jobs to be completed. Each machine must be assigned to complete one job. The time required to set up each machine for completing each job is shown in Table 1. Machineco wants to minimize the total setup time needed to complete the 4 jobs. Formulate this problem as an LP.

Table 1Time (Hours)

Machine Job 1 Job 2 Job 3 Job 41 14 5 8 72 2 12 6 53 7 8 3 94 2 4 6 10

49


7.14 Example Problem Formulation

Machineco must determine which machine should be assigned to each job. Let the decision variables be defined as:

jix

jix

ij

ij

job of demands meet the toassignednot is machine if 0

job of demands meet the toassigned is machine if 1

=

=

14131211

14131211

14131211

14131211

10642 9387

56122 78514Min

xxxxxxxxxxxxxxxxz

+++++++++++++++=

1 1 1

)contraints (Machine 1 s.t.

44434241

34333231

24232221

14131211

=+++=+++=+++=+++

xxxxxxxxxxxxxxxx

Ensures that each machine is assigned to one job

).4,3,2,1;4,3,2,1(,0or 1 1 1 1

)contraints (Job 1

44342414

43332313

42322212

41312111

=====+++=+++=+++=+++

jixxxxxxxxxxxxxxxxxx

ijij

Ensures that each job is assigned to one machine (each job is completed)

50


7.15 Relaxing Integrality Requirements

Since all the supplies and demands for the assignment problem are integers, our discussion of the transportation simplex method implies that all variables in an optimal solution of a transportation problem must be integers

Since each RHS of each constraint is equal to 1, each xij must be a nonnegative integer that is no larger than 1 , so each xij must be equal to 0 or 1

Therefore, we can remove the integrality requirements on the decision variables

51


7.15 Solution Approach

A high degree of degeneracy in an assignment problem may cause the transportation simplex to be an inefficient way of solving assignment problems.

For this reason, and the fact that the algorithm is even much simpler than the transportation simplex, the Hungarian method is usually used to solve assignment (min) problems

Note: If your assignment is a max problem, convert it to a min problem before applying the Hungarian Method.

52


7.15 The Hungarian Method

Step 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 in each column. Construct a new matrix (called the reduced cost matrix) by subtracting from each cost the minimum cost in its column.

Step 2. Cover all the zeros in the reduced cost matrix using the minimum number of lines needed. If m lines are required, then an optimal solution is available among the covered zeros in the matrix. If fewer than m lines are needed, then proceed to Step 3.

Step 3. Find the smallest nonzero element (k) in the reduced cost matrix that is uncovered by the lines drawn in Step 2. Now subtract k from each uncovered element and add k to each element that is covered by two lines. Return to Step 2.

53


7.15 Remarks on the Hungarian Method

To solve an assignment problem in which the goal is to maximize the objective function, multiply the profits matrix by -1 and solve the problem as a minimization problem

If the number of rows and columns in the cost matrix are unequal, then the assignment problem is unbalanced. The Hungarian method may yield an incorrect solution if the problem is unbalanced. Thus, any assignment problem should be balanced (by the addition of one or more dummy points) before it is solved by the Hungarian method.

54


7.16 Machineco Example

14 5 8 7

2 12 6 5

7 8 3 9

2 4 6 10

Job 1 Job 2 Job 3 Job 4

Machine 1

Machine 2

Machine 3

Machine 4

Table 2. Machineco Cost Matrix

55


7.16 Machineco Example

14 5 8 7

2 12 6 5

7 8 3 9

2 4 6 10

5

Row Minimum

2

3

2

Table 3. Machineco Cost Matrix

9 0 3 0

0 10 4 1

4 5 0 4

0 2 4 6

Table 5. Cost Matrix After Column Mins are Subtracted

3 lines with smallest uncovered element equals 1

9 0 3 2

0 10 4 3

4 5 0 6

0 2 4 8

Table 4. Cost Matrix After Row Mins are Subtracted

10 0 3 0

0 9 3 0

5 5 0 4

0 1 3 5

Table 6. Four lines required: optimal solution found

0 00 2.1,1,1,1 :Assignment Optimal 41332412 ==== xxxx

Column Minimum

56


7.17 Transshipment Problems

A transportation problem allows only shipments that go directly 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 which goods 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.

57


7.17 Widgetco Example: See problem description on page 400

Memphis New York

Denver Boston

Los Angeles

Chicago

150

200

130

130

Demand

Total = 260Supply

Total = 350

58


7.18 Converting a Transshipment Problem into a Balanced Transportation problem

(assuming supply >= demand)

Step1. If necessary, add a dummy demand point (with a supply of 0 and a demand equal to the problem’s excess supply) to balance the problem. Shipments to the dummy and 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 point and transshipment point, and a column will be needed for each demand point and transshipment point.

59


7.18 Converting a Transshipment Problem into a Balanced Transportation problem

Step2. Cont..

Each supply point will have a supply equal to it’s 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 (point’s original supply) + s and a demand equal to (point’s original demand) + s. This ensures that any transshipment point that is a net supplier will have a net outflow equal to point’s original supply and a net demander will have a net inflow equal to point’s original demand.

Although we don’t know how much will be shipped through each transshipment point, we can be sure that the total amount will not exceed s.

60


7.19 Widgeco Example: As a Balanced Transportation Problem

130 20

130 70

220 130

350

N.Y. Chicago L.A. Boston Dummy

8 13 25 28 0

Memphis 150

15 12 26 25 0Denver 200

0 6 16 17 0N.Y 350

6 0 14 16 0Chicago 350

350 350 130 130 90

61


References

Winston, Wayne L. and M. Venkataramanan, Introduction to Mathematical Programming, 4th Edition, Duxbury Press, Belmont, CA, 2003.

62


Powerco Problem Formulation back

)4,3,2,1;3,2,1( 0 30 30 20 45

40 50 35 s.t.

516914 713129 9106 8Min

342414

332313

322212

132111

34333231

24232221

14131211

34333231

24232221

14131211

==≥≥++≥++≥++≥++

≤+++≤+++≤+++

+++++++++++

jixxxxxxxxxxxxx

xxxxxxxxxxxx

xxxxxxxxxxxx

ij

Supply Constraints

Demand Constraints


u1

u2u3

v1

v2

v3

v4

transportation and assignment problemsise.tamu.edu/inen420/inen420_2005spring/slides/chapter...

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