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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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
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Multiple Origins andDestination Points
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P Q
1 2 3 8
Multiple Origins and Destination Points
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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
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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 retailer j( Xij)
Constraints: Supply
Demand
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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
Subject to:
1). X11+ X12+ X13+ X14+ X15+ X16+X17+ X18
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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
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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
Conditions a, step 3
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
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