Logistics Project

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Project Logistics- PART II

Prof. Milind Jagtap

Prof. Milind Jagtap

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Problem FormulationConstruction projects that involves material delivery from a single material yard to the various project sites which are

i) Geographically apartii) Order sizes are differentiii) Delivery timing(due dates) are differentiv) Consumption rates at site are differentv) Sites are in different stages of construction

How to ensure material requirements of different project sites by routing and scheduling minimal number of vehicles in a logistics network

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Principle of Savings Matrix Method

Longer the two project sites from the main material yard, but closer to each other, will potentially generate more logistics cost savings.

),(

),(),(),(),(

21S

BAWBWABAS

:Example

DistDistDist

Routing & Scheduling in Transportation

Outline Vehicle Scheduling Problem

SAVINGS MATRIX METHODIdentify Distance MatrixIdentify Savings MatrixRank SavingsAssign project sites to Vehicles Sequence project sites within RoutesConstruction and Improvement ProceduresExample

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Assume that There are orders from 5 different project sites.There are 2 carriers available to a transporter each

capable of carrying 200 tons of load.

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Vehicle Scheduling Problem

Assume that the site locations and order sizes are as shown above.

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X Coordinate Y Coordinate Order Size

Main Yard 0 0

1 0 12 48

2 6 5 60

3 7 15 43

4 9 12 92

5 15 3 80


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W

1

2

3

4

5

Location of W arehouse and Customers


Location of Material yard and project sites

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• Following are the steps of the Savings Matrix Method:

1. Identify distance matrix

2. Identify savings Matrix

3. Rank savings

4. Assign project sites to vehicles

5. Sequence project sites within routes

Savings Matrix Method

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Identify Distance Matrix

• First, the Euclidean distances are computed. The formula and a sample computation is shown below. The other distances are computed similarly and shown on the next slide.

22),(Dist BABA yyxxBA

Dist(1,2) = Square root [(0-6)**2+ (12-5)**2)]

= Sqrt(36+49)= Sqrt(85)=9.2

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Identify Distance Matrix

MY S1 S 2 S 3 S4 S 5MY 0 12.0 7.8 16.6 15.0 15.3S1 0 9.2 7.6 9.0 17.5S 2 0 10.0 7.6 9.2S 3 0 3.6 14.4S 4 0 10.8S 5 0

Distance Matrix

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),(

),(),(),(),(

21S

BAWBWABAS

:Example

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Identify Savings Matrix

• The savings are computed for all pairs of sites using the data from the distance matrix. The formula and a sample computation is shown below. The other savings are computed similarly.

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Identify Savings Matrix

S1 S 2 S3 S 4 S5S1 0 10.6 20.9 18.0 9.8S 2 0 14.3 15.2 13.9S 3 0 27.9 17.4S 4 0 19.5S 5 0

Savings Matrix

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Rank Savings

• The next step is to rank the savings.The idea is to merge those two sites to the same vehicle, whose merging gives the highest savings.

• The savings are ranked from high to low. • From the savings matrix shown on the previous

slide, the highest savings of 27.9 is obtained by merging sites 3 and 4 to the same vehicle.

• Next highest savings of 20.9 is obtained by merging sites 1 and 3 to the same vehicle.

• Similarly the other savings are ranked and shown on the next slide.

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Rank Savings

Rank (3,4) (1,3) (4,5) (1,4) (3,5)(2,4) (2,3) (2,5) (1,2) (1,5)

S 1 S 2 S 3 S 4 S 5S 1 0 10.6 20.9 18.0 9.8S 2 0 14.3 15.2 13.9S 3 0 27.9 17.4S 4 0 19.5S5 0

Savings Matrix

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Rank (3,4) (1,3) (4,5) (1,4) (3,5)(2,4) (2,3) (2,5) (1,2) (1,5)

Assign project sites to Vehicles

Next, merge the sites. The pair giving the highest savings is merged first if the capacity is available.

W

1

2

3

4

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Location of material yard and project sites

SiteOrderSize

1 482 603 434 925 80

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Rank (3,4) (1,3) (4,5) (1,4) (3,5)(2,4) (2,3) (2,5) (1,2) (1,5)


To merge the lowest ranked pair (3,4), the capacity required = 43+92= 135 < 200 = capacity available. So, merge 3 and 4.

W

1

2

3

4

5


SiteOrderSize

1 482 603 434 925 80

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Rank (3,4) (1,3) (4,5) (1,4) (3,5)(2,4) (2,3) (2,5) (1,2) (1,5)


To merge the next pair (1,3), capacity required = 43+92+40= 175 < 200 = capacity available. So, merge 1 and 3 (and 4).

W

1

2

3

4

5


SiteOrderSize

1 482 603 434 925 80

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Rank (3,4) (1,3) (4,5) (1,4) (3,5)(2,4) (2,3) (2,5) (1,2) (1,5)


Merging (4,5), (3,5), (2,4) and (2,3) requires more capacity than available. The pair (1,4) is already merged. So, the pairs are crossed out.

W

1

2

3

4

5


SiteOrderSize

1 482 603 434 925 80

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Rank (3,4) (1,3) (4,5) (1,4) (3,5)(2,4) (2,3) (2,5) (1,2) (1,5)


The next pair (2,5) are merged and assigned to a new vehicle as the capacity available = 200 > 60 + 80 = 140 = capacity required.

W

1

2

3

4

5


Site OrderSize

1 482 603 434 925 80

The next step is sequencing site assigned to the same vehicle. A question is in what sequence will the first vehicle visit site 1, 3 and 4 and return to the main yard? Similarly, another question is in what sequence will the other vehicle visit site 2 and 5.

This problem is popularly called the traveling salesman problem.

We shall use the nearest neighbor rule which states that always visit the site that is nearest.

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Sequence project sites

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W

1

2

3

4

5

Location


First, consider the problem of sequencing project sites 1, 3 and 4 who are assigned to the same vehicle. The relevant distances are copied from the distance matrix and shown below.

MY S1 S3 S 4MY 0 12.0 16.6 15.0S 1 0 7.6 9.0S 3 0 3.6S 4 0

Distance Matrix

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W

1

2

3

4

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Location


Among 1, 3 and 4 is 1 is the nearest to Main yard. So, the vehicle will first travel from main yard to site 1. The row (=from) corresponding to main yard and the column (=to) corresponding to site 1 are crossed out.

MY S 1 S 3 S4MY 0 12.0 16.6 15.0S 1 0 7.6 9.0S 3 0 3.6S 4 0

Distance Matrix

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W

1

2

3

4

5

Location


Between 3 and 4 is 3 is the nearest to 1. So, the vehicle will travel from 1 to 3. The row (=from) corresponding to site 1 and the column (=to) corresponding to site 3 are crossed out.

MY S 1 S 3 S4MY 0 12.0 16.6 15.0S1 0 7.6 9.0S3 0 3.6S4 0

Distance Matrix

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W

1

2

3

4

5

Location

Sequence Customers

The only possible tour is then W-1-3-4-W. Next, consider the problem of sequencing site 2 and 5 who are assigned to the same vehicle. The relevant distances are copied from the distance matrix and shown below.

MY S2 S5MY 0 7.8 15.3S2 0 9.2S5 0

Distance Matrix

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W

1

2

3

4

5

Location

Sequence project site

Between 2 and 5 is 2 is the nearest to the main yard. So, the vehicle will travel from the main yard to site 2. The only tour is then W-2-5-W. Note: due to symmetry both W-2-5-W and W-5-2-W have the same distance traveled.

W S2 S5MY 0 7.8 15.3S2 0 9.2S5 0

Distance Matrix

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Construction and Improvement Procedures

• The nearest neighbor rule just discussed is a tour construction procedure which can construct a tour when there is no tour.

• The nearest neighbor rule is only a heuristic and does not guarantee optimality. The tour obtained by the heuristic may provide improvement opportunities.

• If a tour intersects its own path, the tour can be improved. An improvement procedure will be discussed now.

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• For example, consider the locations and the tour shown on the right.

• From the Main yard, 1 is the nearest. From 1, 2 is the nearest, etc. So, the nearest neighbor rule produces the tour W-1-2-3-4-W.

• However, the tour intersects itself. The arc (1,2) intersects arc (4,W).

W

12

3

4

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• The improvement procedure has three steps.

• Step 1: Remove the intersecting arcs. The result is two disjointed paths

W

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• Step 2: Arbitrarily choose one of the two disjointed paths and reverse the path. In this picture, 2-3-4 is reversed to get 4-3-2. One could as well reversed 1-W to W-1.

• Step 3: There is only one way to get a tour from the two resulting paths. Construct the tour. (continued..)

W

12

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• Step 3 continues: For example, here the tour is constructed by adding arcs (1,4) and (2,W)

• Note: The resulting tour may include a new intersection. In such a case, apply the procedure again!. W

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Logistics Project

Presentations & Public Speaking

Transcript of Logistics Project