Chapter 7 Transportation Problems ( 運輸問題 ) Assignment Problems ( 指派問題 ) Transshipment Problems ( 轉運問題 )

Chapter 7

Transportation Problems ( 運輸問題 )

Assignment Problems ( 指派問題 )

Transshipment Problems ( 轉運問題 )

2

簡介 Transportation Problems ( 運輸問題 )

如何以最低之運輸成本，將貨物由來源地送至目的地 Assignment Problems ( 指派問題 )

如何以適當之方式，做一對一之指派為運輸問題之特例

Transshipment Problems ( 轉運問題 )在運輸問題中，允許貨物由來源地經由其他來源地或目的地轉運，最後送至目的地

上述之線性規劃模式仍然可以單體法求解，但有特別之求解方式，遠較單體法來得有效率。

3

7.1 Transportation Problems

A transportation problem( 運輸問題 ) deals with the problem, which aims to find the best way to fulfill the demand of n demand points( 需求點 ) using the capacities of m supply points ( 供應點 ).

While trying to find the 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.

p.360

4

Example 1: Powerco Formulation

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 following table.

p.360

Table 1

From

To

City 1 City 2 City 3 City 4Supply

(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

5

Example 1: Solution

Decision VariablesPowerco must determine how much power is sent from each plant to each city so xij = amount of electricity produced at plant i and sent to city j

ConstraintsA supply constraint( 供應限制 )ensures that the

total quality produced does not exceed plant capacity. Each plant is a supply point.

A demand constraint ( 需求限制 ) ensures that a location receives its demand. Each city is a demand point.

Since a negative amount of electricity can not be shipped all xij’s must be non negative

6

From

To

City 1 City 2 City 3 City 4 Supply (Million kwh)

Plant 1 8x11 6x12 10x13 9x14 ≤ 35

Plant 2 9x21 12x22 13x23 7x24 ≤ 50

Plant 3 14x31 9x32 16x33 5x34 ≤ 40

Demand (Million kwh)

≥ 45 ≥ 20 ≥ 30 ≥ 30

Min z = 8x11+6x12+10x13+9x14+9x21+12x22+13x23

+7x24 +14x31+9x32+16x33+5x34

s.t. x11+x12+x13+x14 ≤ 35 (Supply Constraints)

x21+x22+x23+x24 ≤ 50

x31+x32+x33+x34 ≤ 40

x11+x21+x31 ≥ 45 (Demand Constraints)

x12+x22+x32 ≥ 20

x13+x23+x33 ≥ 30

x14+x24+x34 ≥ 30

xij ≥ 0 (i= 1,2,3; j= 1,2,3,4)

7

General description of a Transportation Problem

A set of m supply points from which a good is shipped. Supply point i can supply at most si units.

A set of n demand points to which the good is shipped. Demand point j must receive at least di units of the shipped good.

Each unit produced at supply point i and shipped to demand point j incurs a variable cost of cij.

Table 2 (p.364).

p.362

8

Table 2 Supply

C11 C12…

C1n S1

C21 C22…

C2n S2

: : :

Cm1 Cm2…

Cmn Sm

Demand D1 D2 … Dn

p.364A Transportation Tableau

9

Let xij = number of units shipped

from supply point i to demand point j

)n,...,2,1j;m,...,2,1i( 0x

s)constraint (demand )n,...,2,1j( dx

s)constraint (supply )m,...,2,1i( sx .t.s

xc min

ij

mi

1i

jij

nj

1j

iij

mi

1i

nj

1j

ijij

≥

≥

≤

∑

∑

∑∑

10

Balanced Transportation Problem

If

then total supply equals to total demand,the problem is said to be a balanced transportation problem ( 平衡運輸問題 ).

nj

1j

j

mi

1i

i ds

p.363

11


Table 3

City 1 City 2 City 3 City 4 Supply

Plant 18 6 10 9

3510 25

Plant 29 12 13 7

5045 5

Plant 314 9 16 5

4010 30

Demand 45 20 30 30 125

p.364

12

If total supply exceeds total demand

If

total supply exceeds total demand, we can balance the problem by adding dummy demand point ( 虛擬需求點 ). Since shipments to the dummy demand point are not real, they are assigned a cost of zero.

Figure 2 (p.364).

∑∑n=j

1=jj

m=i

1=ii d> s

13

If total supply exceeds total demand

=125

City 1 City 2 City 3 City 4 Supply

Plant 18 6 10 9

3510 25

Plant 29 12 13 7

5045 5

Plant 314 9 16 5

4010 30

120 = Demand 40 20 30 30

p.364

14

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 feasible solution. 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 unmet demand and as one can guess the total penalty cost is desired to be minimum.

Example 2 (p.365).

15

Example 2 : Handing Shortages

Two reservoirs are available to supply the water needs of three cities. Each reservoir can supply up to 50 million gallons of water per day. Each city would like to receive 40 million gallons per day. For each million gallons per day of unmet demand, there is a penalty. At city 1, the penalty is $20; at city 2, the penalty is $22; and at city 3, the penalty is $23. The cost of transporting 1 million gallons of water from each reservoir to each city is shown in Table 4. Formulate a balanced transportation problem that can be used to minimize the sum of shortage and transport costs.

p.365

16

Example 2 :Table 4

From

To

City 1 City 2 City 3

Reservoir 1 $7 $8 $10

Reservoir 2 $9 $7 $8

Daily supply = 50+50 = 100 million gallons per day

Daily demand = 40+40+40 =120 million gallons per day

Table 5

From

To

City 1 City 2 City 3 Supply

Reservoir 1 $8 $6 $10 50

Reservoir 2 $9 $12 $13 50

Dummy $20 $22 $23 20

Demand 40 40 40

17

Exercise 1 ：p.371

A company supplies goods to three customers, who each require 30 units. The company has two warehouses. Warehouses 1 has 40 units available, and warehouses 2 has 30 units available. The cost of shipping 1 unit from warehouse to customer are shown in Table 7. There is penalty for each unmet customer unit of demand: With customer 1, a penalty cost of $90 incurred; with customer 2, $80; and with customer 3, $110. Formulate a balanced transportation problem to minimize the sum of shortage and shipping costs.

Table 7

Cust. 1 Cust. 2 Cust. 3

Warehouse 1 $15 $35 $25

Warehouse 2 $10 $50 $40

18

Solution ：

Cust. 1 Cust. 2 Cust. 3 Supply

Warehouse 1 $15 $35 $25 40

Warehouse 2 $10 $50 $40 30

Shortage $90 $80 $110 20

Demand 30 30 30

19

Exercise 2 ：

Referring to exercise 1, suppose that extra units could be purchases and shipped to either warehouse for a total cost of $100 per unit and that all customer demand must be met. Formulate a balanced transportation problem to minimize the sum of purchasing and shipping costs.

20

Solution ：

C1 C2 C3 Dummy Supply

W1 $15 $35 $25 0 40

W2 $10 $50 $40 0 30

W1 Extra $115 $135 $125 0 20

W2 Extra $110 $150 $140 0 20

Demand 30 30 30 30 120

21

運輸問題求解

1.將問題建成運輸問題模式2.平衡運輸問題3.求解過程分為二階段；

第一階段為求起始解，第二階段為求最佳解4.起始解之解法

Northwest Corner Method ( 西北角法 )Minimum Cost Method ( 最小成本法 )Vogel’s Method ( 佛格法 )

5.最佳解之解法The Transportation Simplex Method( 運輸單體法 )Stepping Stone Method( 階石法 )

22

7.2 Finding Basic Feasible Solution for Transportation Problems

Unlike other Linear Programming problems, a balanced transportation problem 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 (xij’s) satisfy all but one constraint, the values for xij’s will satisfy that remaining constraint automatically.

p.373

23

An ordered sequence of at least four different cells

is called a loop if Any two consecutive cells lie in either the same

row or same columnNo three consecutive cells lie in the same row or

columnThe last cell in the sequence has a row or

column with the first cell in the sequence

24

The Northwest Corner Method dos not utilize

shipping costs. 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. Often the minimum cost method will yield a costly bfs.

Vogel’s Method for finding a bfs usually avoids extremely high shipping costs.

25

Northwest Corner Method ( 西北角法 )

Begin in the upper left (northwest) corner of the transportation tableau and set x11 as large as possible. x11 can clearly be no larger than the smaller of s1 and d1.

Continue applying this procedure to the most northwest cell in the tableau that does not lie in a crossed-out row or column.

Assign the last cell a value equal to its row or column demand, and cross out both cells row and column.

Table 15 ~ Table 20.

p.376

26

Northwest Corner Method

Table 15

D1 D2 D3 D4 Supply

S1 5

S2 1

S3 3

Demand 2 4 2 1 9

p.376

27

Northwest Corner Method

Table 20

D1 D2 D3 D4 Supply

S1 2 3 5

S2 1 1

S3 0 2 1 3

Demand 2 4 2 1 9

p.377

28

The Minimum Cost Method ( 最小成本法 ) Find the decision variable with the smallest

shipping cost (xij). Then assign xij its largest possible value, which is the minimum of si and dj

Next, as in the Northwest Corner Method cross out row i and column j and reduce the supply or demand of the noncrossed-out row or column by the value of xij.

Choose the cell with the minimum cost of shipping from the cells that do not lie in a crossed-out row.

Table 21 ~ Table 26.

p.378

29

The Minimum Cost Method

Table 21

D1 D2 D3 D4 Supply

S1

2 3 5 65

S2

2 1 3 510

S3

3 8 4 615

Demand 12 8 4 6 30

p.378

30

The Minimum Cost Method

Table 26

D1 D2 D3 D4 Supply

S1

2 3 5 65

5

S2

2 1 3 510

2 8

S3

3 8 4 615

5 4 6

Demand 12 8 4 6 30

p.380

31

The Vogel’s Method ( 佛格法 )

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 smallest

shipping cost in that row or column. 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. Table 28 ~ Table 32.

p.380

32

The Vogel’s Method ( 佛格法 )

①計算每列剩餘格中最小與次小之差 (penalty)

②計算每行剩餘格中最小與次小之差 (penalty)

③找出 Max{,}

④找出產生之列或行剩餘格中之最小成本

33

The Vogel’s Method

Table 28

D1 D2 D3 Supply

S1

6 7 810

S2

15 80 7815

Demand 15 5 5 25

p.381

34

The Vogel’s Method

Table 32

D1 D2 D3 Supply

S1

6 7 810

0 5 5

S2

15 80 7815

15

Demand 15 5 5 25

p.382

35

Exercise 1 ：

To find A bfs by the following methods .(1)Northwest Corner Method (2)Minimum Cost Method(3)Vogel’s Method

C1 C2 C3 Supply

W1 $15 $35 $25 40

W2 $10 $50 $40 30

Shortage $90 $80 $110 20

Demand 30 30 30

p.382

36

Solution ： Northwest Corner Method

C1 C2 C3 Supply

W1 30 10 40

W2 20 10 30

Shortage 0 20

Demand 30 30 30

37

Solution ： Minimum Cost Method

C1 C2 C3 Supply

W1 0 10 30 40

W2 30 30

Shortage 20 20

Demand 30 30 30

38

Solution ： Vogel’s Method

C1 C2 C3 Supply

W1 40

W2 30

Shortage 20

Demand 30 30 30

39

Exercise 2 ：


W1 15 35 25 0 40

W2 10 50 40 0 30

W1 Extra 115 135 125 0 20

W2 Extra 110 150 140 0 20

Demand 30 30 30 20 120

To find A bfs by the following methods .

(1)Northwest Corner Method (2)Minimum Cost Method(3)Vogel’s Method

p.382

40

Solution ： Northwest Corner Method


W1 30 10 40

W2 20 10 30

W1 Extra 20 0 20

W2 Extra 20 20

Demand 30 30 30 20

41

Solution ： Minimum Cost Method


W1

15 35 25 040

W2

10 50 40 030

W1 Extra115 135 125 0

20

W2 Extra110 150 140 0

20

Demand 30 30 30 20

42

Solution ： Vogel’s Method


W1

15 35 25 040

W2

10 50 40 030

W1 Extra115 135 125 0

20

W2 Extra110 150 140 0

20

Demand 30 30 30 20

43

7.3 The Transportation Simplex Method

Method of Multipliers ( 乘數法 ) Modified Distribution Method (MODI, 修正分配法 )

Step 1 Determine the enter the basis.

Step 2 Find the 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.

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

p.382

44

To perform the pivot, decrease the value of each odd cell by θ and increase the value of each even cell by θ. The variables that are not in the loop remain unchanged. The pivot is now complete. If θ=0, 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 than one odd cell in the loop equals θ, you may arbitrarily choose one of these odd cells to leave the basis; again a degenerate bfs will result

45

Two important points to keep in mind in the pivoting procedureSince each row has as many +20s as –20s,

the new solution will satisfy each supply and demand constraint.

By choosing the smallest odd variable (x23) to leave the basis, we ensured that all variables will remain nonnegative.

46

單體法求解步驟 ( 複習一下 !!)

1. 建立起始可行解2. 最佳測試 , 決定 entering var.

3. 決定 leaving var.

4. 建立下一個可行解

47

Summary of the Transportation Simplex Method

Step1. Balance problem.

Step2. Find a bfs. (initial solution)

Step3. Let u1=0, cij=ui+vj for all BV. Find ui ,vj.

Step4. Let =ui+vj-cij

　　　 If 　 for all NBV, then the current bfs is optimal

else the var with the most positive is a new bfs.

Step5. using the new bfs return to step 3 and 4.

Step4’.

ijc

p.387

0c ≤ij

ijc

0≥c ij

For a min problem

For a max problem

48

vj v1 v2 v3 v4

uiSi Dj D1 D2 D3 D4 supply

u1 S1

8 6 10 935

35

u2 S2

9 12 13 750

10 20 20

u3 S3

14 9 16 540

10 30

demand 45 20 30 30

u1=0u2=1u3=4v1=8v2=11v3=12v4=1

u1=0

u1+v1=8

u2+v1=9

u2+v2=12

u2+v3=13

u3+v3=16

u3+v4=5

ui+vj =cij

6=c

2-=c

5-=c

-8=c

2=c

5=c

32

31

24

14

13

12

Each unit of x32 that is entered into the basis will decrease cost by $6

Z=8(35)+9(10)+12(20)+13(20)+16(10)+5(30)=1180

p.387

= ui+vj-cijijc

49

p.388

Table 35 vj = 8 11 12 1

ui = Si Dj D1 D2 D3 D4 supply

0 S1

8 6 10 935

35

1 S2

9 12 13 750

10 20 20

4 S3

14 9 16 540

10 30

demand 45 20 30 30

x32 is the entering variable.

P.S. 閉迴路中，除起點與終點為 NBV ，其餘轉角點均為BV

Min{20,10}=10 x33 is the leaving variable.

50

Table 36 vj = 8 11 12 7


0 S1

8 6 10 935

35

1 S2

9 12 13 750

10 10 30

-2 S3

14 9 16 540

10 30

demand 45 20 30 30



以 ui+vj =cij 計算 = ui+vj-cij 選擇正最大為進入變數ijc

Z=8(35)+9(10)+12(10)+13(30)+910)+5(30)=1120

51

Table 37 vj = 8 6 12 2


0 S1

8 6 10 935

25 10

1 S2

9 12 13 750

20 30

3 S3

14 9 16 540

10 30

demand 45 20 30 30



Z=8(35)+6(10)+9(20)+13(30)+9(10)+5(30)=1050

52

Table 38 vj = 6 6 10 2


0 S1

8 6 10 935

10 25

3 S2

9 12 13 750

45 5

3 S3

14 9 16 540

10 30

demand

45 20 30 30

z=6(10)+10(25)+9(45)+13(5)+9(10)+5(30)=1020

53

Exercise ： use the transportation simplex method to solve the problem.

vj =

ui =

D1 D2 D3 D4 supply

S1

3 11 3 1017

4 13

S2

1 9 2 824

13 11

S3

7 4 10 519

16 3

demand 13 16 15 16

z=3(4)+10(13)+1(11)+2(11)+4(16)+5(3)=256

54

Exercise ： use the transportation simplex method to solve p.382 #1 #2

55

Stepping Stone Method ( 階石法 )

1. 對於每一個 NBV xij 對應一個邊際成本2. 邊際成本：以該 NBV 為 entering variable, 其運輸

量每增加 1 單位總運輸成本的增加量。

ijcijc

56

D1 D2 D3 D4 supply

S1

3 11 3 1017

x11 4 13

S2

1 9 2 824

13 11

S3

7 4 10 519

16 3

demand 13 16 15 16

D1 D2 D3 D4 supply

S1

3 11 3 1017

x12 4 13

S2

1 9 2 824

13 11

S3

7 4 10 519

16 3

demand 13 16 15 16

57

D1 D2 D3 D4 supply

S1

3 11 3 1017

4 13

S2

1 9 2 824

13 11 x24

S3

7 4 10 519

16 3

demand 13 16 15 16

D1 D2 D3 D4 supply

S1

3 11 3 1017

4 13

S2

1 9 2 824

13 11

S3

7 4 10 519

16 x33 3

demand 13 16 15 16

58

D1 D2 D3 D4 supply

S1

3 11 3 1017

4 13

S2

1 9 2 824

13 11

S3

7 4 10 519

x31 16 3

demand 13 16 15 16

D1 D2 D3 D4 supply

S1

3 11 3 1017

4 13

S2

1 9 2 824

13 x22 11

S3

7 4 10 519

16 3

demand 13 16 15 16

59

0>1=C-C+C-C+C-C=C

0>10=C-C+C-C+C-C=C

0>12=C-C+C-C=C

0<1-=C-C+C-C=C

0>2=C-C+C-C=C

0>1=C-C+C-C=C

32341413232222

34141323213131

1314343333

2313142424

2334141212

2123131111

◎ x24 ： entering variable

D1 D2 D3 D4 supply

S1

3 11 3 1017

4 13

S2

1 9 2 824

13 11 x24

S3

7 4 10 519

16 3

demand 13 16 15 16

Min{13,11}=11 x23 ： leaving variable

60

D1 D2 D3 D4 supply

S1

3 11 3 1017

4+11 13-11

S2

1 9 x23 2 X24 824

13 11-11 0+11

S3

7 4 10 519

16 3

demand 13 16 15 16

x24 ： EV

x23 ： LV

D1 D2 D3 D4 supply

S1

3 11 3 1017

15 2

S2

1 9 2 824

13 11

S3

7 4 10 519

16 3

demand 13 16 15 16

Z=245<256

61

Next x11 ：EVx14 ：LV why ？

D1 D2 D3 D4 supply

S1

3 11 3 1017

15 2

S2

1 9 2 824

13 11

S3

7 4 10 519

16 3

demand 13 16 15 16

D1 D2 D3 D4 supply

S1

3 11 3 1017

2 15

S2

1 9 2 824

11 13

S3

7 4 10 519

16 3

demand 13 16 15 16

Z=245

degeneracy

62

7.5. Assignment Problems

Assignment problems( 指派問題 ) are a certain class of transportation problems for which transportation simplex is often very inefficient.

In general an assignment problem is balanced transportation problem in which all supplies and demands are equal to 1.

The assignment problem’s matrix of costs is its cost matrix( 成本矩陣 ).

All the supplies and demands for this problem are integers which implies that the optimal solution must be integers.

Using the minimum cost method a highly degenerate bfs is obtained.

p.393

63

Example 4: Machine Assignment Problem

Machineco has four jobs to be completed. Each machine must be assigned to complete one job. The time required to setup each machine for

completing each job is shown.

Machineco wants to minimize the total setup time needed to complete the four jobs.

Table 43

Time (Hours)

Job1 Job2 Job3 Job4

Machine 1 14 5 8 7

Machine 2 2 12 6 5

Machine 3 7 8 3 9

Machine 4 2 4 6 10

p.393

64

Solution :

Machineco must determine which machine should be assigned to each job. i, j =1,2,3,4 xij=1 (if machine i is assigned to meet the demands of job j)

xij=0 (if machine i is not assigned to meet the demands of job j)

1X or 0x

1xxxx

1xxxx

1xxxx

)constraint (Job1xxxx

1xxxx

1xxxx

1xxxx

)constraintMachine(1xxxx .t.s

x10x6xx2x9x3x8x7

x5x6x12x2x7x8x5x14Zmin

ijij

44342414

43332313

42322212

41312111

44434241

34333231

24232221

14131211

4443424134333231

2423222114131211

65

指派問題之求解方法

Simplex method ( 單體法 ) Transportation Simplex Method ( 運輸單體法 ) Hungarian Method ( 匈牙利法 )

66

Initial Solution — Minimum Coat Method

Table 44 J1 J2 J3 J4 supply

M1

14 5 8 71

M2

2 12 6 51

M3

7 8 3 91

M4

2 4 6 101

demand 1 1 1 1

67



M1

14 5 8 71

1

M2

2 12 6 51

1

M3

7 8 3 91

1

M4

2 4 6 101

1

demand 1 1 1 1

68



M1

14 5 8 71

1 0 0

M2

2 12 6 51

1

M3

7 8 3 91

1

M4

2 4 6 101

1 0

demand 1 1 1 1

The current BFS is highly degenerate.

69

Optimal Solution —Transportation Simplex Method


M1

14 5 8 71

1 0

M2

2 12 6 51

1

M3

7 8 3 91

1

M4

2 4 6 101

1 0 0

demand 1 1 1 1

70

The Hungarian Method ( 匈牙利法 )

Step1 Find a bfs. 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 (reduced cost matrix) by subtracting from each cost the minimum cost in its column.

p.395

71

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 optimal solution 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 each uncovered element of the reduced cost matrix and add k to each element that is covered by two lines. Return to step2.

72

匈牙利法之步驟

1.建立成本表：方陣2.簡化列：列－ row min

3.簡化行：行－ col min

4.最佳測試性：最少劃 0 線 , 若線條數 = 列數 , 則進行指派5.進一步簡化成本表

(a) 未劃線元素 - 最小未劃線元素(b) 劃二線元素 + 最小未劃線元素(c) 返回 4.

73

row min

5

2

3

2

Table 46

14 5 8 7

2 12 6 5

7 8 3 9

2 4 6 10

Table 47

9 0 3 2

0 10 4 3

4 5 0 6

0 2 4 8

Table 48

9 0 3 0

0 10 4 1

4 5 0 4

0 2 4 6

以最少之線條劃過所有的 0

Table 49

10 0 3 0

0 9 3 0

5 5 0 4

0 1 3 5

未劃線元素 - 最小未劃線元素劃二線元素 + 最小未劃線元素

p.396

最少劃 0線

若線條數 = 列數則進行指派

0 0 0 2 col min

74

Exercise ： p.398 #1 Optimal Solution -The Hungarian Method

Table 50J1 J2 J3 J4 J5 Row Min

P1 22 18 30 18 0 0P2 18 M 27 22 0 0P3 26 20 28 28 0 0P4 16 22 M 14 0 0P5 21 M 25 28 0 0

Col Min 16 18 25 14 0

J1 J2 J3 J4 J5P1 6 0 5 4 0P2 2 M 2 8 0P3 10 2 3 14 0P4 0 4 M 0 0P5 5 M 0 14 0

75

J1 J2 J3 J4 J5P1 6 0 5 4 0P2 2 M 2 8 0P3 10 2 3 14 0P4 0 4 M 0 0P5 5 M 0 14 0

J1 J2 J3 J4 J5P1 6 0 7 4 2P2 0 M 2 6 0P3 8 0 3 12 0P4 0 4 M 0 2P5 3 M 0 12 0

#(col)=5#(line)=4

The smallest uncovered element is 2.

#(col)=5#(line)=5

An optimal solution is available.

Subtract 2 from uncovered costs.Add 2 to all twice covered cost.

76

J1 J2 J3 J4 J5

P1 6 0 7 4 2

P2 0 M 2 2 0

P3 8 0 3 12 0

P4 0 4 M 0 2

P5 3 M 0 12 0

J1 J2 J3 J4 J5

P1 6 0 7 4 2

P2 0 M 2 2 0

P3 8 0 3 12 0

P4 0 4 M 0 2

P5 3 M 0 12 0

J1 J2 J3 J4 J5

P1 6 0 7 4 2

P2 0 M 2 2 0

P3 8 0 3 12 0

P4 0 4 M 0 2

P5 3 M 0 12 0

J1 J2 J3 J4 J5

P1 6 0 7 4 2

P2 0 M 2 2 0

P3 8 0 3 12 0

P4 0 4 M 0 2

P5 3 M 0 12 0

Person 3 is not assigned any job.Total time =18+18+14+25=75

Assignment

77

Free Breast Fly Back

Hall 3 3 0 2

Spitz 0 6 1 1

Montgomery 0 3 4 6

Jastremski 3 1 2 0

Column Min 0 1 0 0

Table 51Free Breast Fly Back Row Min

Hall 54 54 51 53 51Spitz 51 57 52 52 51Montgomery 50 53 54 56 50Jastremski 56 54 55 53 53

Exercise ： p. 398 #2 Optimal Solution -The Hungarian Method

78

Free Breast Fly BackHall 3 2 0 2Spitz 0 5 1 1Montgomery 0 2 4 6Jastremski 3 0 2 0

Free Breast Fly BackHall 4 2 0 2Spitz 0 4 0 0Montgomery 0 1 3 5Jastremski 4 0 2 0

#(col)=4#(line)=3

#(col)=4#(line)=4

79

Exercise ： p. 399 #3a Optimal Solution -The Hungarian Method

max z=7x11 + 5x12 + 8x13 + 2x14 +... + 5x41 + 5x42 + 6x43 + 7x44

s.t x11 + x12 + x13 + x141 (TC) x21 + x22 + x23 + x241 (FP) x31 + x32 + x33 + x341 (HF) x41 + x42 + x43 + x441 (ML) x11 + x21 + x31 + x411 (JA) x12 + x22 + x32 + x421 (CC) x13 + x23 + x33 + x431 (GP) x14 + x24 + x34 + x441 (JR) xij0

JA CC GP JR

TC -7 -5 -8 -2

FP -7 -8 -9 -4

HF -3 -5 -7 -9

JR -5 -5 -6 -7

Transform Max problem to Min problem, Multiplying benefits by (‑1).

Table 52

JA CC GP JR

TC 7 5 8 2

FP 7 8 9 4

HF 3 5 7 9

JR 5 5 6 7

80

JA CC GP JR

TC 1 3 0 6

FP 2 1 0 5

HF 6 4 2 0

JR 2 2 1 0

JA CC GP JR Row Min

TC -7 -5 -8 -2 -8

FP -7 -8 -9 -4 -9

HF -3 -5 -7 -9 -9

JR -5 -5 -6 -7 -7

JA CC GP JR

TC 0 2 0 6

FP 1 0 0 5

HF 5 3 2 0

JR 1 1 1 0

Col 1 1 0 0Min

JA CC GP JR

TC 0 2 0 7

FP 1 0 0 6

HF 4 2 1 0

JR 0 0 0 0

JA CC GP JR

TC 0 2 0 7

FP 1 0 0 6

HF 4 2 1 0

JR 0 0 0 0

JA CC GP JR

TC 0 2 0 7

FP 1 0 0 6

HF 4 2 1 0

JR 0 0 0 0

JA CC GP JR

TC 0 2 0 7

FP 1 0 0 6

HF 4 2 1 0

JR 0 0 0 0

81

7.6 Transshipment Problems

A transportation problem( 轉運問題 ) allows only shipments that go directly from supply points to demand points.

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.

p.400

82

Step 1 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.

83

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 are sure that the total amount will not exceeds.

84

Example 5 ：

Widgetco manufactures widgets at to factories, one in Memphis and one in Denver. The Memphis factory can produce as many as 150 widgets per day, and the Denver factory can produce as many as 200 widgets per day. Widgets are shipped by air to customers in Los Angeles and Boston. The customers in each city require 130 widgets per day. Because of the deregulation of airfares, Widgetco believes that it may be chapter to first fly some widgets to New York or Chicago and then fly them to their final destinations. The costs of flying a widget are shown in Table 58. Widgetco wants to minimize the total cost of shipping the required widgets to its customers.

p.400

85

Example 5 ：p.400

TABLE 58Shipping Cost for Transshipments

To

From Memphis Denver N.Y. Chicago L.A. Boston

Memphis 0 — 8 13 25 28

Denver — 0 15 12 26 25

N.Y. — — 0 6 16 17

Chicago — — 6 0 14 16

L.A. — — — — 0 —

Boston — — — — — 0

86

Figure 9A graphical representation of possible shipments

Memphis N.Y.

Denver Chicago Boston

L.A.

87


•source ： Memphis(150), Denver(200)•Destination ： L.A.(130), Boston(130)•Transshipment ： N.Y., Chicago (Source & Destination)

Table 59 N.Y. Chicago L.A. Boston Dummy supply

Memphis8 13 25 28 0

150130 20

Denver15 12 26 25 0

200130 70

N.Y.0 6 16 17 0

350220 130

Chicago6 0 14 16 0

350350

demand 350 350 130 130 90

p.400

88

Figure 10Optimal Solution to Widgetco

Memphis N.Y.

Denver Chicago Boston

L.A.130 130

130

89

Exercise ： p.403 #1a

Table 60

LA Detroit Atlanta Houston Tampa supplyLA 0 140 100 90 225 1100

Detroit 145 0 111 110 119 2900

Atlanta 105 115 0 113 78Houston 89 109 121 0 -Tampa 210 117 82 - 0demand 2400 1500

LA Detroit Atlanta Houston Tampa Dummy supplyLA 0 140 100 90 225 0 5100Detroit 145 0 111 110 119 0 6900Atlanta 105 115 0 113 78 0 4000Houston 89 109 121 0 M 0 4000Tampa 210 117 82 M 0 0 4000

4000 4000 4000 6400 5500 100

90

Exercise ： p.403 #1b

LA Detroit Atlanta Houston Tampa DummyLA 0 140 100 90 225 0 5100Detroit 145 0 111 110 119 0 6900Atlanta 105 115 0 113 78 0 4000Houston 89 109 121 0 M 0 4000Tampa 210 117 82 M 0 0 4000

4000 4000 4000 6400 5500 100

LA Detroit Atlanta Houston Tampa DummyLA 0 M 100 90 225 0 5100Detroit M 0 111 110 119 0 6900Atlanta 105 115 0 113 78 0 4000Houston 89 109 121 0 M 0 4000Tampa 210 117 82 M 0 0 4000

4000 4000 4000 6400 5500 100

91

Exercise ： p.403 #2Table 61

Well1 Well2 Mobile Galv. N.Y. L.A.Well 1 0 - 10 13 25 28 150Well 2 - 0 15 12 26 25 200Mobile - - 0 6 16 17Galv. - - 6 0 14 16N.Y. - - - - 0 15L.A. - - - - 15 0

140 160

Mobile Galv. N.Y. L.A. Dummy supplyWell 1 10 13 25 28 0 150Well 2 15 12 26 25 0 200Mobile 0 6 16 17 0 350Galv. 6 0 14 16 0 350N.Y. M M 0 15 0 350L.A. M M 15 0 0 350demand 350 350 490 510 50 1050

92

Exercise ： p.403 #3

Table 61Mobile Galv. NY LA Dummy Supplies

Well 1 10 13 25 28 0 150Well 2 15 12 26 25 0 200Mobile 0 6 16 17 0Galv. 6 0 14 16 0NY M M 0 15 0LA M M 15 0 0Demands 140 160 50

Mobile Galv. NY LA Dummy SuppliesWell 1 10+12 13+10 M M 0 150Well 2 15+12 12+10 M M 0 200Mobile 0 6 16 17 0 350Galv. 6 0 14 16 0 350NY M M 0 15 0 350LA M M 15 0 0 350Demands 350 350 490 510 50

93

Exercise ： p.403 #4

Mobile Galv. NY LA Dummy SuppliesWell 1 10+12 13+10 M M 0 150Well 2 15+12 12+10 M M 0 200Mobile 0 6 16 17 0Galv. 6 0 14 16 0NY M M 0 15 0LA M M 15 0 0Demands 140 160 50

Mobile Galv. NY LA Dummy SuppliesWell 1 10+12 13+10 M M 0 150Well 2 15+12 12+10 M M 0 200Mobile 0 6 16 17 0 180Galv. 6 0 14 16 0 150NY M M 0 15 0 350LA M M 15 0 0 350Demands 180 150 140 160 50

94

Transportation Problems-- 總供給小於總需求

D1 D2 D3 Supply

S1 3 7 8 18

S2 9 4 6 7

S3 10 11 5 15

Demand 21 16 13

D1 D2 D3 Supply

S1 3 7 8 18

S2 9 4 6 7

S3 10 11 5 15

S4 0 0 0 10

Demand 21 16 13

總供給 = 40, 總需求 =50

95

Transportation Problems-- 總供給大於總需求

D1 D2 D3 Supply

S1 3 7 8 21

S2 9 4 6 16

S3 10 11 5 13

Demand 18 7 15

D1 D2 D3 D4 Supply

S1 3 7 8 0 18

S2 9 4 6 0 7

S3 10 11 5 0 15

Demand 21 16 13 10

總供給 = 50, 總需求 =40

96

Example: 總供給大於總需求

若來源貨物無法送出時，分別導致每單位 3 、 1 、 2 的存貨成本

D1 D2 D3 D4 Supply

S1 3 7 8 0 18

S2 9 4 6 0 7

S3 10 11 5 0 15

Demand 21 16 13 10 50

D1 D2 D3 D4 Supply

S1 3 7 8 3 18

S2 9 4 6 1 7

S3 10 11 5 2 15

Demand 21 16 13 10 50

97

Example: 總供給大於總需求

若 S1 之貨物必須全部送出，以便挪出空間作為他用

D1 D2 D3 D4 Supply

S1 3 7 8 0 18

S2 9 4 6 0 7

S3 10 11 5 0 15

Demand 21 16 13 10 50

D1 D2 D3 D4 Supply

S1 3 7 8 M 18

S2 9 4 6 0 7

S3 10 11 5 0 15

Demand 21 16 13 10 50

98

建立轉運供需表之步驟

1.建立平衡之轉運供需表， ΣSi=ΣDj=T

2.增設來源地於列；增設目的地於行。3.填入各路徑之單位成本。通常本站到本站之單位成本為 0 。

任兩點來回之單位成本可能相同，也可能不相。4.可供轉運之起點供給量增加 T ；可供轉運之終點供給量增加 T ；

所有新增之點 ( 無論供給量或需求量 ) 其量均為 T 。

99

來源地 (S1,S2) 與目的地 (D1,D2) 之供給與需求為： s1=75, s2=60, d1=55, d2=85 。其運輸成本表為如下，則轉運供需表為何？

Example ：來源地與目的地不為轉運點

S1 S2 D1 D2

S1 0 7 16 25

S2 7 0 12 16

D1 15 13 0 8

D2 23 17 9 0

D1 D2 supply

S1 16 25 75

S2 12 16 60

S3 0 0 5

demand 55 85 140

100

Example ：來源地與目的地均為轉運點

S1 S2 D1 D2

S1 0 7 16 25

S2 7 0 12 16

D1 15 13 0 8

D2 23 17 9 0

S1 S2 D1 D2 supply

S1 0 7 16 25 75+140

S2 7 0 12 16 60+140

D1 15 13 0 8 0+140

D2 23 17 9 0 0+140

S3 0 0 0 0 5

demand 0+140 0+140 55+140 85+140 700

Chapter 7 Transportation Problems ( 運輸問題 ) Assignment Problems ( 指派問題 ) Transshipment Problems ( 轉運問題 )

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