Docstoc

IENG 471 Lecture 05

Document Sample
IENG 471 Lecture 05 Powered By Docstoc
					IENG 471 - Lecture 05

        Group Technology –
        Production Flow Analysis




8/17/2011          IENG 471 Facilities Planning   1
Agenda

   Assignment
   Direct Clustering Algorithm
   King’s Algorithm
   Questions & Issues




8/17/2011        IENG 471 Facilities Planning   2
Assignments
   Current Assignment: HW 4
         Reading:
             CH 3
         HW:        (HW 4) See Assignment Link
             CH 2: 13, 16, 17, 24, 38, 40, 41.


   Next Assignment: HW 5 (Last - Exam I)
         Reading:
             CH 3
         HW:
             CH 3: # 14, 15, 24, 25, 26, 37 use King’s Alg instead of
              DCA

8/17/2011                     IENG 471 Facilities Planning               3
Layout Impact on Facility’s Machines

    Job Shop Layout
          Group by individual machines
          Add all equipment fractions, then round up*
    Mass Production Layout
          Group by individual products
          Round up* all equipment fractions, then add
    Cellular Layout
          Group by similar part-process families
          Add family equipment fractions, then round up*,
             then add
         * Multiply number of each machine type by footprint size after rounding
            to find the actual space required

 8/17/2011                       IENG 471 Facilities Planning                      4
  Layout Patterns


                                       High
                                                 Product
                      Product Volume             Layout

                                       Medium


                                                            Family / GT          Bulky, Difficult to Move Equip.
                                                            Cell Layout          (Precision fixtures)

                                                                                         Process
                                       Low




                                                 Fixed Location
                                                 Layout                                  Layout

Bulky, Difficult to Move Prod.                  Low                Medium                 High
(Planes, ships, etc.)
                                                               Product Variety
                                                Can Have Combinations (HYBRIDS)!

    8/17/2011                                     IENG 471 Facilities Planning                             5
Examples

 For the following situations, what type of
    General Layout would you suggest?
       The assembly of bodies for GM midsize SUVs

       Fabrication & Assembly of custom made sheet metal parts

       Fabrication of computer cases for a line of desktop PCs, plus
            custom made sheet metal parts

       Assembly of three distinct families of electronic cards for inkjet
            printers

       Production of high quality, custom office furniture



8/17/2011                     IENG 471 Facilities Planning                   6
Group Technology (GT)

              Philosophy:
                 Use the similarity of current products to simplify
                  the design and manufacturing of new products
              Some Applications:
                 Identify and reuse similar process plans
                 Identify and reuse similar CNC programs
                 Identify the equipment that may be best used in a
                  particular machine cell
                 Identify and eliminate redundant inventory
              Requirement:
                 A taxonomy of part characteristics


 8/17/2011                    IENG 471 Facilities Planning             7
Benefits of GT
    Facilitates formation of part          Allows rationalization and
       families and machine cells               improvement in tool design
      Quick retrieval of designs,             Aids production planning
       drawings, & process plans                and scheduling procedures
      Reduces design
       duplication                             Improves cost estimation
      Provides reliable workpiece              and facilitates cost
       statistics                               accounting procedures
      Facilitates accurate                    Provides for better machine
       estimation of machine tool               tool utilization and better
       requirements and logical                 use of tools, fixtures, &
       machine loadings                         people
      Permits rationalization of              Facilitates NC part
       tooling setups, reduces                  programming.
       setup time, and reduces
       production throughput time

                                                           (Ham)
 8/17/2011                 IENG 471 Facilities Planning                       8
How to Identify Groups
    Similar Design Attributes
       Size of parts
       Geometric shape of parts
       Materials
       Technique: Parts Classification & Coding

    Similar Manufacturing Attributes
       Common processing steps (routings)
       Common tools and fixtures
       Technique: Production Flow Analysis

    Similarity groupings are called Part Families

 8/17/2011             IENG 471 Facilities Planning   9
PFA Introduction
              PFA is Production Flow Analysis
                 A subset of Group Technology (GT)

              Goals:
                 Reduce material transport efforts
                 Reduce set up efforts
                 Reduce work in process inventory

              Steps:
                 Identify OP-Codes for each Component
                     Routing/Process Planning information for each part
                 Incidence Matrix
                 Blocking (Triangularization) Algorithm
                 Cluster Identification
 8/17/2011                     IENG 471 Facilities Planning                10
OP-Codes
   An OP-Code forms an index to an Operation Plan


   An Operation Plan is a generalized sequence of
      steps, perhaps common to multiple parts

   An OP-Code Sequence is a method of condensing
      the Operation Plan into a compact structure
      suitable for data processing




8/17/2011              IENG 471 Facilities Planning   11
OP-Code Example
                                            OP Code           Operation Plan
                                            01 Saw01          Cut to size
                                            02 Lathe02        Face end
                                                              Center drill
                                                              Drill
                                                              Ream
                                                              Bore
                                                              Turn straight
                                                              Turn groove
                                                              Chamfer
            Rotational Part                                   Cut off
                                                              Face
            (sectional view)                                  Chamfer
                                            03 Grind01        Grind
                                            04 Insp06         Inspect dimension
                                                              Inspect finish
8/17/2011                      IENG 471 Facilities Planning                       12
OP-Code Example
  OP Code      Operation Plan
  01 Saw01     Cut to size
  02 Lathe02   Face end
               Center drill
               Drill                                   OP Code
               Ream
               Bore                                     Sequence
               Turn straight                            01   Saw01
               Turn groove                              02   Lathe02
               Chamfer                                  03   Grind01
               Cut off                                  04   Insp06
               Face
               Chamfer
  03 Grind01   Grind
  04 Insp06    Inspect dimension
               Inspect finish

8/17/2011               IENG 471 Facilities Planning                   13
Incidence Matrix
    Rows represent OP-Codes                  (index = i)
    Columns represent Components             (index = j)
    Cell Entries are:                        (Mij)
       1 (or mark) - if the component requires the operation
       0 (or blank) - if the component does NOT require the operation
                                               A-112          A-115
                     Saw01                         1
                    Lathe01
                    Lathe02                        1
                     Drill01                                   1
                      Mill02                                   1
                      Mill05
                     Grind05                       1
 8/17/2011                     IENG 471 Facilities Planning              14
King’s Algorithm – Step 1
              Calculate the total (binary) weight of each column j:
                                 Wj =  2i Mij                      After Chang, Wysk, &
                                              i                              Wang (1998) p.500

                    A1    A2   A3   A4   A5        A6   A7   A8    A9   A0    2i                i
              S01   1          1    1     1                   1    1                   2        1
              L01                   1                                                  4        2
              L02   1          1          1                   1    1                   8        3
              D01         1                                             1             16        4
              M02         1                                                           32        5
              M05                                  1    1               1             64        6
              G05   1                                                               128         7
              G06                                                       1           256         8

             Wj
                    138




                                                                        336
                          48

                               10



                                         10

                                                   64

                                                        64

                                                             10

                                                                   10
                                    6




             Rank   5     3    2    1     2        4    4     2    2    6
 8/17/2011                          IENG 471 Facilities Planning                           15
King’s Algorithm – Step 2
    Sort the columns into rank order, then go to Step 3:
                    A4   A3   A5   A8   A9     A2    A6     A7    A1    A0

              S01   1    1    1    1     1                        1
              L01   1
              L02        1    1    1     1                        1
              D01                               1                       1
             M02                                1
             M05                                      1      1          1
              G05                                                 1
              G06                                                       1

             Wj
                                                                  138

                                                                        336
                         10

                              10

                                   10

                                        10

                                               48

                                                      64

                                                            64
                    6




             Rank   1    2    2    2     2      3     4      4    5     6

 8/17/2011                         IENG 471 Facilities Planning               16
King’s Algorithm – Step 3
    Calculate the total (binary) weight of each row i:
                       Wi =  2j Mij
                                       j

                        A4   A3   A5   A8    A9     A2    A6     A7    A1    A0     Wi          Rank
                 S01    1    1    1     1     1                        1                  574     4
                  L01   1                                                                   2     1
                  L02        1    1     1     1                        1                  602     5
                 D01                                 1                       1           1088     7
                 M02                                 1                                     64     2
                 M05                                       1      1          1           1408     8
                 G05                                                   1                  512     3
                 G06                                                         1           1024     6




                                                                             1024
             2j
                                                           128

                                                                 256

                                                                       512
                                        16

                                             32

                                                    64
                        2

                             4

                                  8




             j          1    2    3     4     5      6     7      8    9     10

 8/17/2011                              IENG 471 Facilities Planning                             17
King’s Algorithm – Step 4
              If all rows are in rank order STOP; otherwise, sort the rows into
               rank order, and then go to Step 1:

                    A4   A3   A5   A8   A9     A2    A6     A7    A1   A0   Wi          Rank
              L01   1                                                               2     1
             M02                                1                                  64     2
              G05                                                 1               512     3
              S01   1    1    1    1     1                        1               574     4
              L02        1    1    1     1                        1               602     5
              G06                                                      1         1024     6
              D01                               1                      1         1088     7
             M05                                      1      1         1         1408     8




 8/17/2011                         IENG 471 Facilities Planning                          18
King’s Algorithm – Step 1 (2nd time)
    Calculate the total (binary) weight of each column j:
                       Wj =  2i Mij
                                   i

                    A4   A3   A5   A8    A9     A2    A6     A7    A1   A0    2i          i
              L01   1                                                                2    1
             M02                                 1                                   4    2
              G05                                                  1                 8    3
              S01   1    1    1     1     1                        1                16    4
              L02        1    1     1     1                        1                32    5
              G06                                                       1           64    6
              D01                                1                      1          128    7
             M05                                       1      1         1          256    8

             Wj
                                                132

                                                       256

                                                             256



                                                                        448
                    18

                         48

                              48

                                    48

                                         48




                                                                   56

             Rank   1    2    2     2     2      4     5      5    3    6
 8/17/2011                          IENG 471 Facilities Planning                         19
King’s Algorithm – Step 2 (2nd time)
              Sort the columns into rank order, then go to Step 3:

                     A4   A3   A5   A8    A9     A1    A2     A6    A7    A0
               L01   1
              M02                                       1
              G05                                 1
              S01    1    1    1     1     1      1
               L02        1    1     1     1      1
              G06                                                         1
              D01                                       1                 1
              M05                                              1    1     1

             Wj
                                                        132

                                                              256

                                                                    256

                                                                          448
                     18

                          48

                               48

                                    48

                                           48

                                                 56




             Rank    1    2    2     2     2      3     4      5    5     6
 8/17/2011                          IENG 471 Facilities Planning                20
King’s Algorithm – Step 3 (2nd time)
    Calculate the total (binary) weight of each row i:
                       Wi =  2j Mij
                                       j

                        A4   A3   A5   A8    A9     A1    A2     A6    A7    A0     Wi          Rank
                  L01   1                                                                   2      1
                 M02                                       1                              128      5
                 G05                                 1                                     64      2
                 S01    1    1    1     1     1      1                                    126      4
                  L02        1    1     1     1      1                                    124      3
                 G06                                                         1           1024      6
                 D01                                       1                 1           1152      7
                 M05                                              1    1     1           1792      8




                                                                             1024
             2j
                                                           128

                                                                 256

                                                                       512
                                        16

                                             32

                                                    64
                        2

                             4

                                  8




             j          1    2    3     4     5      6     7      8    9     10
 8/17/2011                              IENG 471 Facilities Planning                             21
King’s Algorithm – Step 4 (2nd time)
              If all rows are in rank order STOP; otherwise, sort the rows into
               rank order, and then go to Step 1:


                    A4   A3   A5   A8   A9     A1    A2     A6    A7   A0   Wi          Rank
              L01   1                                                               2      1
              G05                               1                                  64      2
              L02        1    1    1     1      1                                 124      3
              S01   1    1    1    1     1      1                                 126      4
             M02                                      1                           128      5
              G06                                                      1         1024      6
              D01                                     1                1         1152      7
             M05                                             1    1    1         1792      8




 8/17/2011                         IENG 471 Facilities Planning                          22
King’s Algorithm – Step 1                                                 (3 rd

time)
    Calculate the total (binary) weight of each column j:
                       Wj =  2i Mij
                                   i

                    A4   A3   A5   A8    A9     A1    A2     A6    A7    A0       2i          i
              L01   1                                                                    2    1
              G05                                1                                       4    2
              L02        1    1     1     1      1                                       8    3
              S01   1    1    1     1     1      1                                      16    4
             M02                                       1                                32    5
              G06                                                        1              64    6
              D01                                      1                 1             128    7
             M05                                              1    1     1             256    8

             Wj
                                                       160

                                                             256

                                                                   256

                                                                         510
                    18

                         24

                              24

                                    24

                                         24

                                                26




             Rank   1    2    2     2     2      3     4      5    5     6
 8/17/2011                          IENG 471 Facilities Planning                             23
King’s Algorithm – Step 2 (3rd time)
              Sort the columns into rank order, then go to Step 3:
                                   NO CHANGE IN SORTED ORDER!
                    A4   A3   A5    A8    A9     A1    A2     A6    A7    A0
              L01   1
              G05                                 1
              L02        1    1      1     1      1
              S01   1    1    1      1     1      1
             M02                                        1
              G06                                                         1
              D01                                       1                 1
             M05                                               1    1     1

             Wj
                                                        160

                                                              256

                                                                    256

                                                                          510
                    18

                         24

                              24

                                    24

                                          24

                                                 26




             Rank   1    2    2      2     2      3     4      5    5     6
 8/17/2011                           IENG 471 Facilities Planning               24
King’s Algorithm – Step 3 (3rd time)
    Calculate the total (binary) weight of each row i:
                       Wi =  2j Mij
                                       j

                        A4   A3   A5   A8    A9     A1    A2     A6    A7    A0     Wi          Rank
                  L01   1                                                                   2      1
                 G05                                 1                                     64      2
                  L02        1    1     1     1      1                                    124      3
                 S01    1    1    1     1     1      1                                    126      4
                 M02                                       1                              128      5
                 G06                                                         1           1024      6
                 D01                                       1                 1           1152      7
                 M05                                              1    1     1           1792      8




                                                                             1024
             2j
                                                           128

                                                                 256

                                                                       512
                                        16

                                             32

                                                    64
                        2

                             4

                                  8




             j          1    2    3     4     5      6     7      8    9     10

 8/17/2011                              IENG 471 Facilities Planning                              25
King’s Algorithm – Step 4 (3rd time)
    If all rows are in rank order STOP; otherwise, sort the rows into
     rank order, and then go to Step 1:
     SINCE THE ROWS ARE IN RANK ORDER, WE STOP! (yea!)


                   A4   A3   A5   A8   A9     A1    A2     A6    A7   A0   Wi          Rank
             L01   1                                                               2      1
             G05                               1                                  64      2
             L02        1    1    1     1      1                                 124      3
             S01   1    1    1    1     1      1                                 126      4
             M02                                     1                           128      5
             G06                                                      1         1024      6
             D01                                     1                1         1152      7
             M05                                            1    1    1         1792      8


 8/17/2011                        IENG 471 Facilities Planning                          26
Partitioning
    Ideally, the cells form mutually exclusive blocks (as below). These
     blocks define the Families:
       Family A consists of Components A1, A3, A4, A5, A8, and A9; which
           can be machined in a cell performing Operations G05, L01, L02, and
           S01
          Family B consists of Components A0, A2, A6, and A7; which can be
           machined in a cell performing Operations D01, G06, M02, and M05
                               A4   A3    A5     A8     A9     A1   A2   A6   A7   A0
                         L01   1
                         G05                                   1
                         L02         1      1     1      1     1
                         S01   1     1      1     1      1     1
                         M02                                        1
                         G06                                                       1
                         D01                                        1              1
                         M05                                             1    1    1
 8/17/2011                      IENG 471 Facilities Planning                            27
Partitioning
         Often, the cells do NOT form mutually exclusive blocks (as below).
          In this case, the capability for Operation G06 must be common to
          both machining cells:
              Family A consists of Components A1, A3, A4, A5, A8, and A9; which can be
               machined in a cell performing Operations G05, G06, L01, L02, and S01
              Family B consists of Components A0, A2, A6, and A7; which can be
               machined in a cell performing Operations D01, G06, M02, and M05

                                 A4   A3    A5     A8     A9     A1   A2   A6   A7   A0
                           L01   1
                          G05                                    1
                           L02         1      1     1      1     1
                           S01   1     1      1     1      1     1
                          M02                                         1
                          G06                                    1                   1
                           D01                                        1              1
                          M05                                              1    1    1
 8/17/2011                        IENG 471 Facilities Planning                            28
Strategies for Overlapping PFA
Blocks:
          Provide for transporting some components between cells requiring
           the overlapping operation(s)
              Pick the component(s) with the smallest volume(s) to transport to
               reduce handling costs
              Locate the cells with operation overlap as near to each other as
               possible to reduce handling costs
          Avoid scheduling concurrent production runs of the components that
           require overlapping operation(s)
              Assumes that the equipment providing the overlapping capability
               can be easily moved between cells
              This solution may improve capacity if the overlapping operation
               is a bottleneck
          Put equipment capable of the overlapping operation(s) into each cell
           requiring it
              Assumes the additional equipment capability is cost justifiable
              This solution will improve capacity if the overlapping operation is
               a bottleneck

 8/17/2011                       IENG 471 Facilities Planning                        29
Extended GT (PFA)
             Requirements:
                Need for grouping similar items together, and identifying
                  separable items
                 Matrix of related entities:
                    Tooling
                    Equipment
                    Parts
                    Integrated Circuits
                    Modular Components

             Solution Method: Triangularization
                   Direct Clustering Algorithm
                   King’s Method
                   Kusiak’s Triangularization Method
                   Ullman’s Design Structure Matrix

8/17/2011                      IENG 471 Facilities Planning              30
Relationship (Incidence) Matrix
    Rows and Columns:
       Parts requiring operations on different machines
       Tools (in a CNC magazine) needed to produce
        part families
       Departments requiring technicians (shared head
        count)
       Departments requiring adjacent location
       ICs requiring modularization
    An entry in the incidence matrix means that there
     is a strong relationship between the row and
     column items


 8/17/2011             IENG 471 Facilities Planning        31
Space Considerations
      Equipment footprint
      Utilities space
      Maintenance space
      Operating space
          (Operator space is CH 4)
    Material space
            Tooling space
            Scrap space
            Work-In-Process (WIP)                           Depends on cell
                                                             production rate and on
            Receiving Space
                                                             production control policy
            Shipping Space                                  for material flow
    Plus Material Transportation!


 8/17/2011                    IENG 471 Facilities Planning                               32
Production Control of Flow
    Push Production:
          Traditional
          Build to schedule
              Cell Efficiency Effects – Optimal
              Overall Efficiency Effects – Sub-Optimal
    Pull Production:
          JIT
          Kanban
              Production Card (POK)
              Withdrawal Card (WLK)
              Minimizes WIP (and a host of other problems!)


 8/17/2011                    IENG 471 Facilities Planning     33

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:12
posted:8/18/2011
language:English
pages:33