kuliah 12 transportaion management

7/27/2019 Kuliah 12 Transportaion Management

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Transportation Management



Transport Functionality

Product Movement Product Storage



Transport Principles

Economy of Scale Economy of Distance

Cost per unit of weight Cost per unit of weight

decrease as the size of decrease as

a shipment increase distance increase



Transport Regulation

Economic Regulation Safety & Social



Transportation Mode

1. Rail

2. Highway

3. Water

4. Pipeline

5. Air



Operating Transportation Mode

PerformanceTransportati

on Mode Speed Availability Dependability Capability

Rail 3 2 3 2

Truck 2 1 2 3

Water 4 4 4 1

Pipe 5 5 1 5

Air 1 3 5 4

*) 1=Highest



Transportation Mode Cost &Services

Transportati

on Mode Cost*

Delivery

Time**

Consistency

***

Loss and

Damage***

Rail 3 3 4 5

Truck 2 2 3 4

Water 5 5 5 2

Pipe 4 4 2 1

Air 1 1 1 3

*) 1=Highest **) 1= Fastest ***) 1=Least



Intermodal Transportation

Combines two or more modes to take advantage of the inherent economies of each and thus an

integrated services at lower cost

Piggyback/TOFC/COFC

Containerships

Coordinated Air Truck



History of Intermodal Transportation

• There are two main types of dry cargo: bulk

cargo and break bulk cargo. Bulk cargoes,

like grain or coal, are transportedunpackaged in the hull of the ship, generally

in large volume. Break-bulk cargoes, on the

other hand, are transported in packages, andare generally manufactured goods.

http://en.wikipedia.org/wiki/Bulk_cargo


http://en.wikipedia.org/wiki/Break_bulk_cargo

http://en.wikipedia.org/wiki/Break_bulk_cargo





Piggy Back / Trailer on Flat Car (TOFC)



Containerships



Container on Flatcars



Choice of Transport Mode

Trade of

Inventory Cost Transportation Cost



Choice of Transport Mode

Plant WarehouseInventory: 100.000 100.000 unit

Price at Plant : $30/unitCarrying Cost : 30% of price/unit/year Transport Rate/R Transit Time/T Number of Mode ($/unit) (days) shipment/yearRail 0.10 21 10

Piggyback 0.15 14 20Truck 0.20 5 20Air 1.40 2 40

It is estimated that for every day that transit time can bereduced from the current 21 days, average inventory levels can

be reduced by 1%, which mode to be selected if demand (D) was700.000/year ?



Criteria

Total cost = Inventory Cost + Transportation Cost» Plant

» In-Transit

» Warehouse

• Transportation Cost (C.Trans) = R x D• In-Transit Inventory Cost (C.In-Trans)= ICDT/365

• Plant Inventory Cost (C.Plant) = ICQP/2

•

Warehouse Inventory Cost (C.Ware

) = IC’QW

/2• QP = 100.000/N IC = 0.3 x 30

• QW = 100.000/N IC’ = 0.3 x (30 +R)



SolutionMode

Cost Rail Piggyback Truck Air

Transpor-

tation

(0.1)(700.000)

= 70.000

(0.15)(700.000)

= 105.000

(0.20)(700.000)

= 140.000

(1.40)(700.000)

= 980.000

Plant

Inventory

(0.3x30)(100.000)

= 900.000

(0.3x30)(50.000)

(0.93) = 418.500

(0.3x30)(50.000)

(.84) = 378.000

(0.3x30)(25.000)

(0.81) = 182.250

In-transit

Inventory

(0.3x30)(700.000)

(21)/365

= 363.465

(0.3x30)(700.000)

(14)/365

= 241.644

(0.3x30)(700.000)

(5)/365

= 86.301

(0.3x30)(700.000)

(2)/365

= 34.521

Warehouse

Inventory

(0.3x30.1)(100.000)

(1.0)

= 903.000

(0.3x30.15)(50.000)

(0.93)

= 420.593

(0.3x30.2)(50.000)

(0.84)= 380.520

(0.3x31.4)(25.000)

(0.81)= 182.250

Total 2.235.465 1.185.737 984.821 1.387.526



Exercise: Choice of Transport Mode

Plant WarehouseInventory: 100.000 50.000 unit

Price at Plant : $20/unitCarrying Cost : 30% of price/unit/year Transport Rate/R Transit Time/T Number of Mode ($/unit) (days) shipment/yearRail 0.10 25 10

Piggyback 0.15 15 20Truck 0.25 8 20Air 2.00 2 40

It is estimated that for every day that transit time can bereduced from the current 25 days, average inventory levels can

be reduced by 1%, which mode to be selected if demand (D) was500.000/year ?



Transport Participants

Public

Internet.com

Government

Shipper Carrier Consignee



Structure of Transport Industry

Retailer

Private

Fleets

Manager

Retailer

Retailer

Retailer

Air

Express

Ocean

Motor

Carriers

Rail

.Operation

.Order

processing.Tracking

.Billing

Warehousing

Others

Pooling

Yards

Inter-change

Point Processor

EndConsumer

Retailer

Shippers CarriersMiddlemen CustomersInterchanges



Transportation Service

1. Traditional Carriers2. Package Service

3. Ground Package Service

4. Air Package Service



Non-operating Intermediaries

1. Freight Forwarders• Profit business that consolidate small shipments from various

customers into bulk shipment and then utilize a common

carriers for transport

2. Shipper Association/Cooperative – Voluntary nonprofit entities where members

collaborate to gain economies of transportation

3. Brokers

• Coordinate transportation arrangement for

shippers, consignees, and carriers.



International Trade Specialist

1. International FW2. Non vessel-Operating Common Carrier

(NVOCCs)

3. Custom House Brokers

4. Export Management Company

5. Export Trading Company



International FW

1. Advice on Acceptance of LC2. Booking Space on Carriers

3. Preparing an Export Declaration

4. Preparation an air Waybill or Bill of Lading

5. Obtaining Consular Document

6. Arranging for Insurance

7. Preparing and Sending Shipping Notice and

Documents8. Serving as General Consultant on Export

Matters



Documentation

1. Bill of Lading2. Freight Bill

3. Freight Claims

4. Loss, Damage and Delay Claims

5. Overcharges



International Transport Document

Exporting• Bill of lading

• Dock Receipt

• Delivery Instruction

• Export Declaration

• Letter of Credit

• Consular Invoice

• Commercial Invoice

• Certificate of Origin• Insurance Certificate

• Transmittal Letter

Importing

• Arrival Notice

• Custom Entry

•

Carriers Certificateand Release Order

• Delivery Order

• Freight Release

•Special CustomInvoice



Transport Management

1. Transport Planning2. Vehicle Routing and Scheduling

3. Delivery Execution and Shipment

Tracking

4. Performance Management



Economic Driver Cost

1. Distance2. Volume

3. Density

4. Stow-ability

5. Handling

6. Liability

7. Market



Rate

1. Volume Related Rates2. Distance Related Rates

• Uniform rates

•

Proportional rates• Tapering rates

• Blanket rates

3. Demand Related rates



Line Haul rates

1. By Product2. Class Rates

3. Contract rates

4. Freight all kinds

5. By Shipment Size6. Miscellaneous Rates

7. Cube rates

8. Import-Export Rates

9. Deffered Rates10. Released Value Rates

11. Ocean Freight Rates



Special Service Charges

1. Special Line Haul Services• Diversion and Re-consignment

• Transit Privileges

• Protection

2. Terminal Services

• Pickup and Delivery

• Switching

• Demurrage and Detention



Operation Management

1. Equipment Scheduling2. Load planning

3. Routing

4. Carrier Administration

5. Carrier Selection



Freight Consolidation

1. Reactive consolidation2. Market Area

3. Cost Cutting Strategy

4. Schedule Delivery

5. Pooled Delivery

6. Proactive Consolidation

7. Preorder Planning

8. Multi-firm Consolidation



Pricing Strategy

1. Cost of Service2. Value of Service

3. Combination Pricing

4. Net-Rates Price



Vehicle Routing and Scheduling

• Separate and Single Origin and

Destination Point• Multiple Origin and Destination Points

• Coincident Origin and Destination Points



Separate and Single Origin andDestination Point



Shortest Route Solution

A B E I

C

H

G

F

D J

Origin

Destination

90

90

8484

60

48

348

150

48

126

132

126156

132

66

132

138



Step Solved Its Closest Total Time n’th Nearest Its Min. Its Last Nodes Conc.un Sl Involved Node Time Con’tion

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




5 A D 348C D 138156=294E I 174+84=258 I 258 EI*F G 228132=360F H 228+60=288

6 A D 348C D 138+156=294

F 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




8 G J 336+150=486

H J 288+126=414I J 258+126=384 J 384 IJ*

Shortest Route : A B E I J( 384)



Shortest Route Solution

A B E I

C

H

G

F

D J

Origin

Destination

90

90

8484

60

48

348

150

48

126

132

126156

132

66

132

138



Multiple Origin andDestination Points



P Q

1 2 3 8

Multiple Origin 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 <=100

2). X21 + X22 + X23 + X24 + X25 + X26 +X27 + X28 <= 45



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

O i l S l i



P Q

1 4 5 82 3 6 7

Optimal Solution



P Q

1 4 5 8

R

R

6

Transshipment Model



Additional Cost Data

Cross Dock 4 5 6

P 11

Q 10

Cross Dock - 6 5 5



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)



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 <=100

2). X21 + X22 + X23 + X24 + X25 + X26 +X27 + X28 <=45

3). X1c + X2c - Xc = 0

4). Xc - Xc4 + Xc5 + Xc6 = 0

5). Xc <= 30



Subject to:

6). X11 + X21 = 22

7). X12 + X22 = 14

8). X13 + X23 = 18

9). X14 + X24 + Xc4 = 17

10). X15 + X25 + Xc5 = 1511). X16 + X26 + Xc6 = 13

12). X17 + X27 = 15

13). X18 + X28 = 20

All variables Nonnegative



Optimal Solution

1 2 3 4 5 6 7 8A 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



Routing and Scheduling

Resources:

Capacity(gal): 4000 5000 6000

Available : 12 3 4

Destination

• 12 Customer

• Distance



Clark and Wright

Saving Matrix Method

• Identify the distance matrix

• Identify the saving matrix

•

Assign customer to vehicles or route• Sequence customers within routes

Load,

P



12

2

25

22

2

32

24

1200

21

1900

2

38 1800

1600

2

22 1700

q

2

14

2

9

51

49

41

37

35

31

16 20 2730 29 39 46

2

36

1400

2

52 P

1210 20

242

17002

50 P

11

1100

1200

1700

221

2

23

1500

1400

16

10

14

17

5

18 25 28 31 37 44

7

17 20 29 31 36

P 6

8 12

11

22

10

13 16

P 5

6

P 7

P 4

12 6

23

43

35

41

16

26

30

36

37

27 25 10 7

P 3

9

P 0

P 2

21

30 28

P 1

10

19

P 8

P 9

P 10

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

Matrix setup

Load,

P



18 12

2

25

10 22

2

32

10 24

1200

10 21

1900

2

38 1800

1600

2

22 1700

q

2

14

2

9

10 51

10 49

10 41

10 37

10 35

10 31

72 16 64 20 50 2744 30 46 29 34 39 20 46

2

36

1400

2

52 P

1284 10 70 20

242

17002

50 P

11

1100

1200

1700

221

2

23

1500

1400

20 16

64 10

72 14

20 17

18 5

64 18 50 25 44 28 42 31 34 37 20 44

28 7

50 17 44 20 36 29 32 31 20 36

P 6

84 8 76 12

50 11

38 22

58 10

50 13 54 16

P 5

68 6

P 7

P 4

68 12 72 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

P 3

38 9

P 0

P 2

22 21

16 30 20 28

P 1

34 10

26 19

P 8

P 9

P 10

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

Matrix Saving:

Syz = doy + doz -dyz

Load,

P



18

2

25

10

2

32

10

1200

10

1900

2

381800

1600

2

221700

q

2

14

2

9

10

10

10

10

10

10

72645044463420

2

36

1400

2

52 P

128470

242

17002

50 P

11

1100

1200

1700

221

2

23

1500

1400

20

64

72

20

18

645044423420

28

5044363220

P 6

8476

50

38

58

50 54

P 5

68

P 7

P 4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P 3

38

P 0

P 2

22

16 20

P 1

34

26

P 8

P 9

P 10

92

Matrix Saving:

Syz = doy + doz -dyz



18

1

10

1

10

5600

10

1

5100

5600

21700

Load,

q

2

2

10

10

10

10

10

10

72645044463420

5100

P 12

1

8470

1

P 11

1200

46001

14600

20

64

72

20

18

645044423420

28

5044363220

P 6

8476

50

38

58

50 54

P 5

68

P 7

P 4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P 3

38

P 0

P 2

22

16 20

P 1

34

26

P 8

P 9

P 10

1

92

Initial Solution Matrix

z

y

1

Load,

P0



18

2

25

10

2

32

10

1200

10

1900

2

381800

1600

2

221700

q

2

14

2

9

10

10

10

10

10

10

72645044463420

2

36

1400

2

52 P

128470

242

17002

50 P

11

1100

1200

1700

221

2

23

1500

1400

20

64

72

20

18

645044423420

28

5044363220

P 6

8476

50

38

58

50 54

P 5

68

P 7

P 4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P 3

38

P 0

P 2

22

16 20

P 1

34

26

P 8

P 9

P 10

92

Inilitial Solution

1

1

1

Load,

P0



18

2

25

10

2

32

10

1200

10

1900

2

381800

1600

2

221700

q

2

14

2

9

10

10

10

10

10

10

72645044463420

2

36

1400

2

52 P

128470

242

17002

50 P

11

1100

1200

1700

221

2

23

1500

1400

20

64

72

20

18

645044423420

28

5044363220

P 6

8476

50

38

58

50 54

P 5

68

P 7

P 4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P 3

38

P 0

P 2

22

16 20

P 1

34

26

P 8

P 9

P 10

92

Intermediate Solution

1

1

1

1

1

1

L d



18

1

10

1

10

5600

10

15100

5600

21700

Load,

q

2

2

10

10

10

10

10

10

72645044463420

5100

P 12

1

8470

1

P 11

1200

46001

14600

20

64

72

20

18

1

645044423420

1

28

5044363220

P 6

8476

1

50

38

58

50 54

P 5

1

68

P 7

P 4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P 3

38

P 0

P 2

22

16 20

P 1

1

34

26

P 8

P 9

P 10

1

92

Revised Solution Matrix

z

y

L d



18

1

10

1

10

5600

10

15100

5600

21700

Load,

q

2

2

10

10

10

10

10

10

72645044463420

5100

P 12

1

8470

1

P 11

1

1

5800

20

64

72

20

18

1

645044423420

28a

5044363220

P 6

8476

1

50

38

58

50 54

P 5

1

68

P 7

P 4

6872

16

16

24

20

44

20

20

16

20

26 30 44 50

P 3

38

P 0

P 2

22

16 20

P 1

1

34

26

P 8

P 9

P 10

1

92

Intermediate Step

Previous eliminations

Conditions a, step 3



aCell 2, 3 is next choice to combine tours

z

y

L d



1

15600

15100

5600

21700

Load,

q

2

1

5100

P 12

1

1

P 11

5800

1

15800

1

1

P 6

1

P 5

1

P 7

P 4

P 3

P 0

P 2

P 1

1

P 8

P 9

P 10

1

Final Routing Plan

z

y

Load



1

15600

15100

5600

21700

Load,

q

2

1

5100

P 12

1

1

P 11

5800

1

15800

1

1

P 6

1

P 5

1

P 7

P 4

P 3

P 0

P 2

P 1

1

P 8

P 9

P 10

1

Final Routing Plan

z

y