Kuliah_10_Transportaion Routing & Scheduling

8/10/2019 Kuliah_10_Transportaion Routing & Scheduling

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Senator Nur Bahagia@

ROUTING AND SCHEDULING


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Separate and Single Origin andDestination Point


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Shortest Route Method

A BE I

C

H

G

F

DJ

Origin

Destination

90

90

8484

60

48

348

150

48126

132

126156

132

66

132

138


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Step Solved Its Closest Total Time nth Nearest Its Min. Its LastNodes Conc.un Sl Involved Node Time Contion

1 A B 90 B 90 AB*A D 138A C 348

2 A C 138 C 138 ACA D 348B C 90+66=156

B E 90+84=174

3 A D 348B E 90+84=174 E 174 BE*C D 138+156=294C F 138+90=228

4 A D 348C D 138+156=294C F 138+ 90=228 F 228 CFE F 174+132=306E I 174+84 =258


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Step Solved Its Closest Total Time nth Nearest Its Min. Its LastNodes Conc.un Sl Involved Node Time Contion

5 A D 348

C D 138156=294E I 174+84=258 I 258 EI*F G 228132=360F H 228+60=288

6 A D 348

C D 138+156=294F G 228+132=360

F H 228+ 60=288 H 288 FH

I H 258+132=390

I J 258+126=384

7 A D 348C D 138+156=294F G 228+132=360H G 288+ 48=336 G 336 HGI J 258+126=384


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8/36Senator Nur Bahagia@

Shortest Route Solution

A BE I

C

H

G

F

DJ

Origin

Destination

90

90

8484

60

48

348

150

48

126

132

126156

132

66

132

138



Multiple Origins andDestination Points



P Q

1 2 3 8

Multiple Origins and Destination Points



Data

1 2 3 4 5 6 7 8

A B C D E F G H Supply

P 12 24 21 20 21.5 19 17 20 100

Q 24 15 28 20 18.5 19.5 24 28 45

Dmd 22 14 18 17 15 13 15 20



Component of Model

Performance Criteria : Min. Cost Variable Decision :

Number of product to be supplied from

plant i (Si) Number of product to be transported from

plant i to retailer j( Xij)

Constraints: Supply

Demand



Model Formulation

Min Z=12X11+ 24X12+ 21X13+ 20X14+ 21.5X15+ 19X16 +

17X17+ 20X18 + 24X21+ 15X22+ 28X23 + 20X24 +18.5X25+ 19.5 X26+ 24 X27+ 28X28

Subject to:

1). X11+ X12+ X13+ X14+ X15+ X16+X17+ X18



Model Formulation

3). X11+ X21= 22

4). X12+ X22 = 14

5). X13+ X23= 186). X14+ X24= 17

7). X15+ X25= 15

8). X16+ X26= 13

9). X17+ X27= 15

10). X18 + X28 = 20

All Variable Non Negative



Optimal Solution

1 2 3 4 5 6 7 8

A B C D E F G H Supply

P 22 - 18 17 - 13 15 4 89

Q - 14 - - 15 - - 16 45

Dmnd 22 14 18 17 15 13 15 20 134

Minimal Cost = $ 2583.50


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Additional Cost Data

Cross Dock 4 5 6 Cap.

P 11

Q 10

Cross Dock - 6 5 5 30

Material Handling Cost at Cross Dock: $2/unit


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Component of Model

Performance Criteria : Min. Cost Variable Decision :

Number of product to be supplied from

plant i (Si) Number of product to be transported from

plant i to Cross Dock ( Xic)

Number of product to be transported from

plant i to retailer j( Xij)

Number of product to be transported from

Cross Dock c to retailer j( Xcj)


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Component of Model

Constraints:

Supply

Demand

Cross Dock


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Model Formulation

Min Z=12X11+ 24X12+ 21X13+ 20X14+ 21.5X15+ 19X16 +17X17+ 20X18 + 24X21+ 15X22+ 28X23 + 20X24 +

18.5X25+ 19.5 X26+ 24 X27+ 28X28+ 11X1c+10X2c

+2Xc+ 6 Xc4+ 5 Xc5+ 5Xc6

Subject to:1). X11+ X12+ X13+ X14+ X15+ X16+X17+ X18


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Subject to:

6). X11+ X21= 22

7). X12+ X22 = 14

8). X13+ X23= 18

9). X14+ X24+ Xc4= 1710). X15+ X25+ Xc5= 15

11).X16+ X26+ Xc6= 13

12). X17+ X27= 15

13). X18 + X28 = 20

All variables Nonnegative


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Optimal Solution

1 2 3 4 5 6 7 8

A B C D E F G H C.D

X 22 - 18 2 - 13 15 - 19

Y - 14 - - - - - 20 11

C.D - - - 15 15 - - -

Minimal Cost = $ 2542


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P Q

1 4 5 8

R

R

6

Optimal Solution

3 7 2


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Coincident Origin and

Destination Points


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Travelling Salesman

W

C

B

A

D

48

31

34

2326

47

17

34

34

67

W D C B A W


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Routing and Scheduling

Resources:

Capacity(gal): 4000 5000 6000

Available : 12 3 4

Destination

12 Customer

Distance


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Clark and Wright

Saving Matrix Method

Identify the distance matrix

Identify the saving matrix

Assign customer to vehicles or route Sequence customers within routes


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18 12

2

25

10 22

2

32

10 24

1200

10 21

1900

2

381800

1600

2

221700

Load,

q

2

14

2

9

10 51

10 49

10 41

10 37

10 35

10 31

72 1664 2050 2744 3046 2934 3920 46

2

36

1400

2

52P

1284 1070 20

2

42

17002

50P

11

1100

1200

1700

2

212

23

1500

1400

20 16

64 10

72 14

20 17

18 5

64 1850 2544 2842 3134 3720 44

28 7

50 1744 2036 2932 3120 36

P6

84 876 12

50 11

38 22

58 10

50 13 54 16

P5

68 6

P7

P4

68 1272 6

16 23

16 43

24 35

20 41

44 16

20 26

20 30

16 36

20 37

26 27 30 25 44 10 50 7

P3

38 9

P0

P2

22 21

16 30 20 28

P1

34 10

26 19

P8

P9

P10

92 10aCustomers are identified and ranked for convenience according to the savings Sy,z

Matrix setup


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18

2

25

10

2

32

10

1200

10

1900

2

381800

1600

2

221700

Load,

q

2

14

2

9

10

10

10

10

10

10

72645044463420

2

36

1400

2

52P

128470

2

42

17002

50P

11

1100

1200

1700

2

212

23

1500

1400

20

64

72

20

18

645044423420

28

5044363220

P6

8476

50

38

58

50 54

P5

68

P7

P4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P3

38

P0

P2

22

16 20

P1

34

26

P8

P9

P10

92

Matrix Saving:

Syz= doy+ doz-dyz


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18

1

10

1

10

5600

10

15100

5600

21700

Load,

q

2

2

10

10

10

10

10

10

72645044463420

5100

P12

1

8470

1

P11

1200

46001

14600

20

64

72

20

18

645044423420

28

5044363220

P6

8476

50

38

58

50 54

P5

68

P7

P4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P3

38

P0

P2

22

16 20

P1

34

26

P8

P9

P10

1

92

Initial Solution Matrix

z

y

1


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18

2

25

10

2

32

10

1200

10

1900

2

381800

1600

2

221700

Load,

q

2

14

2

9

10

10

10

10

10

10

72645044463420

2

36

1400

2

52P

128470

242

17002

50P

11

1100

1200

1700

2

212

23

1500

1400

20

64

72

20

18

645044423420

28

5044363220

P6

8476

50

38

58

50 54

P5

68

P7

P4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P3

38

P0

P2

22

16 20

P1

34

26

P8

P9

P10

92

Inilitial Solution

1

1

1


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18

2

25

10

2

32

10

1200

10

1900

2

381800

1600

2

221700

Load,

q

2

14

2

9

10

10

10

10

10

10

72645044463420

2

36

1400

2

52P

128470

242

17002

50P

11

1100

1200

1700

2

212

23

1500

1400

20

64

72

20

18

645044423420

28

5044363220

P6

8476

50

38

58

50 54

P5

68

P7

P4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P3

38

P0

P2

22

16 20

P1

34

26

P8

P9

P10

92

Intermediate Solution

1

1

1

1

1

1


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18

1

10

1

10

5600

10

15100

5600

21700

Load,

q

2

2

10

10

10

10

10

10

72645044463420

5100

P12

1

8470

1

P11

1200

46001

14600

20

64

72

20

18

1

645044423420

1

28

5044363220

P6

8476

1

50

38

58

50 54

P5

1

68

P7

P4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P3

38

P0

P2

22

16 20

P1

1

34

26

P8

P9

P10

1

92

Revised Solution Matrix

z

y


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18

1

10

1

10

5600

10

1

5100

5600

21700

Load,

q

2

2

10

10

10

10

10

10

72645044463420

5100

P12

1

8470

1

P11

1200

17001

1

2900

2900

20

64

72

20

18

1

645044423420

28a

5044363220

P6

8476

1

50

38

58

50 54

P5

1

68

P7

P4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P3

38

P0

P2

22

16 20

P1

1

34

26

P8

P9

P10

1

92

Intermediate Step

Previous eliminations

Conditions a, step 3



aCell 2, 3 is next choice to combine

tours

z

y


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1

15600

1

5100

5600

21700

Load,

q

2

1

5100

P12

1

1

P11

5800

1

15800

1

1

P6

1

P5

1

P7

P4

P3

P0

P2

P1

1

P8

P9

P10

1

Final Routing Plan

z

y

Documents

Kuliah_10_Transportaion Routing & Scheduling