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  • SYSTEMATIC LAYOUT

    PLANNING (SLP)

    Additional Slides

  • ACTIVITY RELATIONSHIP

    ANALYSIS

    2

  • Graph Based Process

    In summary,

    The graph based approach provides a structured approach for developing the REL diagram

    Graph based approach is widely used in activity-based lock layouts

    It emphasizes the importance of constructing a planar graph of the REL chart if the block diagram is to be constructed to satisfy the relationships

    It is used to determine by just inspection which activities/departments are adjacent with each other in order to construct the block layout. (There are multiple solutions in constructing the block layout.)

    3

  • RELATIONSHIP DIAGRAMMING

    4

  • Steps in RDP (Stage 1)

    Step 1

    The numerical values are assigned to the closeness rating as: A= 10 000, E= 1000, I= 100, O= 10, U= 0, X= 10 000

    Step 2

    TCR (Total Closeness Rating) for each department is computed. TCR refers to the sum of the absolute values for the relationships with a particular department.

    Step 3

    The department with the greatest TCR is selected as the first placed department in the sequence of placement.

    Step 4

    Department having X relationship with the first placed department is labelled as the last placed department.

    Step 5

    Second department in the sequence of placement is determined to satisfy the highest closeness rating with the first placed department. With respect to the closeness priorities A>E>I>O>U

    5

  • Steps in RDP (Stage 1)

    Step 6

    Department having X relationship with the second placed department is labelled as the next-to-the-last or last placed department.

    Step 7

    Next department in the sequence of placement is the one with the highest closeness relationship (A>E>I>O>U) with the already placed departments.

    Step 8 Procedure continues until all departments have been placed.

    *Note: If ties exist during this process, TCR values are utilized to break the ties

    arbitrarily.

    6

  • Steps in RDP (Stage 2)

    *Note: If ties exist during this process, first location with the largest WPV is selected.

    Step 9

    Calculate Weighted Placement Value (WPV) of locations to which the next department in the order will be assigned. WPV refers to the sum of the numerical values for all pairs of adjacent department(s).

    When a location is fully adjacent, its weight equals to 1.0, and when it is partially adjacent its weight equals to 0.5.

    Step 10

    Evaluate all possible locations in counter clock-wise order, starting at the western edge of the partial layout.

    Step 11

    Assign the next department to the location with the largest WPV.

    7

  • Example

    Given the relationship chart below, determine the order of

    placement of the departments and the relative location of each

    department.

  • Example

    Below is the table of Total Closeness Rating (TCR) values for each department. The first department placed in the layout is the one with the greatest TCR value. If there is a tie, then choose the one with more A (E, etc.).

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 -

  • Example

    If a department has an X relationship with the first one, it is placed last in

    the layout. If a tie exists, choose the one with the smallest TCR value.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - - 8

  • Example

    The second department is the one with an A relationship with the first one

    (or E, I, etc.). If a tie exists, choose the one with the greatest TCR value.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - 5 - - 8

  • Example

    The next department is the one with an A (or E, I, etc.) relationship with the

    already placed departments. If a tie exists, choose the one with the

    greatest TCR value.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - 5 - 9 - - 8

  • Example

    The next department is the one with an A (or E, I, etc.) relationship with the

    already placed departments. If a tie exists, choose the one with the

    greatest TCR value.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - 5 - 9 - 3 - - 8

  • Example

    The next department is the one with an A (or E, I, etc.) relationship with the

    already placed departments. If a tie exists, choose the one with the

    greatest TCR value.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - 5 - 9 - 3 - 1 - - 8

  • Example

    The next department is the one with an A (or E, I, etc.) relationship with the

    already placed departments. If a tie exists, choose the one with the

    greatest TCR value.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - 5 - 9 - 3 - 1 - 4 - - 8

  • Example

    The next department is the one with an A (or E, I, etc.) relationship with the

    already placed departments. If a tie exists, choose the one with the

    greatest TCR value.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - 5 - 9 - 3 - 1 - 4 - 2 - - 8

  • Example

    The sequence of the placement of the departments has been identified.

    Next, is to determine the relative location of each department.

    Dept. Department Summary

    TCR 1 2 3 4 5 6 7 8 9 A E I O U X

    1 - A A E O U U A O 3 1 0 2 2 0 31,020

    2 A - E A U O U E U 2 2 0 1 3 0 22,010

    3 A E - E A U U E A 3 3 0 0 2 0 33,000

    4 E A E - E O A E U 2 4 0 1 1 0 24,010

    5 O U A E - A A O A 4 1 0 2 1 0 41,020

    6 U O U O A - A O O 2 0 0 4 2 0 20,040

    7 U U U A A A - X A 4 0 0 0 3 1 50,000

    8 A E E E O O X - X 1 3 0 2 0 2 33,020

    9 O U A U A O A X - 3 0 0 2 2 1 40,020

    Placing Order: 7 - 5 - 9 - 3 - 1 - 4 - 2 - 6 - 8

  • WPV1 = 1(10,000) = 10,000

    WPV2 = 0.5(10,000) = 5,000

    WPV3 = 1(10,000) = 10,000

    WPV4 = 0.5(10,000) = 5,000

    WPV5 = 1(10,000) = 10,000

    WPV6 = 0.5(10,000) = 5,000

    WPV7 = 1(10,000) = 10,000

    WPV8 = 0.5(10,000) = 5,000

    Example

    7

    3

    1 5

    8 6

    4 2

    Department 5?

    *Relationship of dept. 7 - 5 = A (10,000)

    10

    5 7 7 6

    4 3 2

    1

    9 8 7

    5

    WPV1 = 1(10,000) = 10,000

    WPV2 = 0.5(10,000) = 5,000

    WPV3 = 1(10,000) + 0.5(10,000) = 15,000

    WPV4 = 1(10,000) + 0.5(10,000) = 15,000

    WPV5 = 0.5(10,000) = 5,000

    WPV6 = 1(10,000) = 10,000

    WPV7 = 0.5(10,000) = 5,000

    WPV8 = 1(10,000) + 0.5(10,000) = 15,000

    WPV9 = 1(10,000) + 0.5(10,000) = 15,000

    WPV10 = 0.5(10,000) = 5,000

    *Relationship of dept. 7 - 9 = A (10,000)

    Relationship of dept. 5 - 9 = A (10,000)

    Department 9?

    Compute the Weighted Placement Value of each location. If the location is fully

    adjacent, its weight equals to 1.0, and if it is partially adjacent its weight equals to 0.5.

  • 1

    WPV1 = 15,000

    WPV2 = 15,000

    WPV3 = 5,000

    WPV4 = 10,000

    WPV5 = 5,000

    WPV6 = 15,000

    WPV7 = 0

    WPV8 = 0

    WPV9 = 0

    WPV10 = 5,000

    WPV11 = 10,000

    WPV12 = 5,000

    Example

    Department 3?

    Compute the Weighted Placement Value of each location. If the location is fully

    adjacent, its weight equals to 1.0, and if it is partially adjacent its weight equals to 0.5.

    12

    5

    Department 1?

    9

    7

    9

    8

    7 6

    5 4 3

    2

    11 10 14

    3

    13 12 11 10

    9

    8 7

    6 5 4

    3 2

    1

    9

    5 7

    WPV1 = 10,000

    WPV2 = 5,000

    WPV3 = 10,015

    WPV4 = 5

    WPV5 = 10

    WPV6 = 5

    WPV7 = 15

    WPV8 = 0

    WPV9 = 0

    WPV10 = 0

    WPV11 = 5

    WPV12 = 5,010

    WPV13 = 10,005

    WPV14 = 5,000

  • Example

    Department 4?

    Compute the Weighted Placement Value of each location. If the location is fully

    adjacent, its weight equals to 1.0, and if it is partially adjacent its weight equals to 0.5.

    Department 2?

    WPV1 = 6,000

    WPV2 = 10,500

    WPV3 = 5,000

    WPV4 = 10,000

    WPV5 = 10,000

    WPV6 = 10,000

    WPV7 = 5,000

    WPV8 = 10,000

    WPV9 = 5,000

    WPV10 = 0

    WPV11 = 0

    WPV12 = 5,000

    WPV13 = 10,000

    WPV14 = 5,000

    14

    3

    13 12 11 10

    9

    8 7

    6 5 4 3

    2

    1

    9

    5 7

    1

    WPV1 = 1,500

    WPV2 = 1,500

    WPV3 = 500

    WPV4 = 1,000

    WPV5 = 500

    WPV6 = 0

    WPV7 = 10,500

    WPV8 = 5,000

    WPV9 = 10,000

    WPV10 = 5,000

    WPV11 = 10,500

    WPV12 = 6,000

    WPV13 = 1,500

    WPV14 = 5

    14

    3

    13 12 11 10

    9

    8

    7 6 5 4 3

    2

    1

    9

    5 7

    1 4

  • Example

    Department 6?

    Compute the Weighted Placement Value of each location. If the location is fully

    adjacent, its weight equals to 1.0, and if it is partially adjacent its weight equals to 0.5.

    Department 8?

    WPV1 = 10

    WPV2 = 5

    WPV3 = 10

    WPV4 = 10

    WPV5 = 15

    WPV6 = 10

    WPV7 = 5

    WPV8 = 5,010

    WPV9 = 10,005

    WPV10 = 5,000

    WPV11 = 10,500

    WPV12 = 10,500

    WPV13 = 5,000

    WPV14 = 0

    WPV15 = 10

    WPV16 = 5

    14

    3

    13 12 11 10

    9

    8

    7 6 5 4 3 2

    1 9

    5 7

    1 2 4

    16 15

    16

    3

    2 1

    5 7

    6

    9 4 1

    2 3 5 4 6 7

    8

    10

    9

    13 12 11

    15 14

    17 18

    WPV1 = 1,000

    WPV2 = 500

    WPV3 = 6,000

    WPV4 = 5,500

    WPV5 = -4,500

    WPV6 = -4,000

    WPV7 = 500

    WPV8 = -4,000

    WPV9 = -9,495

    WPV10 = -4,990

    WPV11 = 5

    WPV12 = 10

    WPV13 = 5

    WPV14 = -9,480

    WPV15 = 1,005

    WPV16 = 500

    WPV17 = 7,000

    WPV18 = 500

  • Example

    Final Layout

    3

    2 1

    5 7

    6

    9 4

    8

  • ALDEP (Automated Layout Design Program)

    the size of the facility and the size of the departments are expressed in terms of blocks.

    Score is determined using the numerical values assigned to the closeness rating).

    A = 43 = 64 I = 41 = 4 U = 0

    E = 42 = 16 O = 40 = 1 X = 45 = 1.024

    The area per grid is subjective but it is recommended that it is divisible to all the departments areas.

    The sweep width defines the width in number of blocks/grids. For example, let sweep width = 3 then the

    offices is allocated 3x4 =12 grids plus the two remaining

    blocks

    23

  • SPACE REQUIREMENT AND

    AVAILABILITY

    24

  • 1. Production-Center Method

    The production center consists of a single machine plus all the

    associated equipment and space required for its operation. Work space

    (front, rear, left side, right side), additional maintenance space, and storage

    space are added to the space requirements for the machine

    2. Converting Method

    The present space requirements are converted to those required for

    the proposed layout

    3. Roughed-out layout method

    Templates or models are placed on the layout to obtain an estimate of

    the general configuration and space requirement

    Methods of Determining Space Requirements

  • 4. Space-standards method

    Use of industry standards, or past successful applications

    5. Ratio trend and projection method

    Establish a ratio of floor area to some other factor that can be

    measured and predicted for the proposed layout.

    Ex.: Square-feet per machine

    Square-feet per operator

    Square-feet per direct labor hour

    Square-feet per unit produced

    Methods of Determining Space Requirements

  • In determining the space requirements of the facility, the number of equipment

    and number of employees must first be determined. Some parameters needed

    to compute for the resources needed:

    Production Rate

    Batch Production Quantity

    Methods of Determining Space Requirements

  • Determination of Employee Requirements

    The number of machine operators required depends on the number of machines tended by one or more operators. The determination of the number of machines to be assigned to one operator can take two approaches:

    Deterministic -Multiple activity chart

    Probabilistic -Queuing models, Monte Carlo Simulation

    A deterministic approach is to employ the multiple activity chart. This chart shows the multiple activity relationships graphically against a time scale. The chart is useful in analyzing multiple activity relationships, specially, when non-identical machines are supervised by a single operator.

  • Multiple Activity Chart

    Charts on which activities of workers, product and machines are recorded on

    a common time scale to show their relationships.

    There are various charts with different names but serve almost the same

    purpose:

    1) Work-Machine Process Chart (Man-Machine Chart)

    Seeking most effective relationship between operator(s) and

    machine(s), i.e. minimum total %idle time

    2) Gang Process Chart (Multi-Man Chart)

    Multiple activity chart applied to a group of workers, seeking most

    effective relationship between several workers

  • Worker-Machine Relationship

    Worker-machine relationships can be of three types:

    1) synchronous servicing

    2) random (asynchronous) servicing

    3) combination of both - real-life

  • Types of Worker-Machine Relationships

  • Types of Worker-Machine Relationships