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Push vs Pull Process Control

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					Push vs. Pull Process Control


              IE 3265 POM
              Slide Set 9
              R. Lindeke, Sp 2005
Basic Definitions
    MRP (Materials Requirements Planning). MRP is the
     basic process of translating a production schedule for
     an end product (MPS or Master Production
     Schedule) to a set of time based requirements for all
     of the subassemblies and parts needed to make that
     set of finished goods.

    JIT Just-in-Time. Derived from the original Japanese
     Kanban system developed at Toyota. JIT seeks to
     deliver the right amount of product at the right time.
     The goal is to reduce WIP (work-in-process)
     inventories to an absolute minimum.
Why Push and Pull?
    MRP is the classic push system. The MRP system
     computes production schedules for all levels based
     on forecasts of sales of end items. Once produced,
     subassemblies are pushed to next level whether
     needed or not.

    JIT is the classic pull system. The basic mechanism
     is that production at one level only happens when
     initiated by a request at the higher level. That is, units
     are pulled through the system by request.
Comparison

   These methods offer two completely different approaches to basic
    production planning in a manufacturing environment. Each has
    advantages over the other, but neither seems to be sufficient on
    its own. Both have advantages and disadvantages, suggesting
    that both methods could be useful in the same organization.

   Main Advantage of MRP over JIT: MRP takes forecasts for end
    product demand into account. In an environment in which
    substantial variation of sales are anticipated (and can be
    forecasted accurately), MRP has a substantial advantage.

   Main Advantage of JIT over MRP: JIT reduces inventories to a
    minimum. In addition to saving direct inventory carrying costs,
    there are substantial side benefits, such as improvement in quality
    and plant efficiency.
Comparisons (cont.)

           Advantages                        Disadvantages
                            JIT  PULL
                                    Every job is a „High Stress‟ Rush
Limited and known Final Inventory
                                    order
Worker only consume their time &
                                    Balanced systems MUST be in
Raw Materials on what is actually
                                    place
needed
                                  Setup times will greatly impact
Quality MUST be High – each piece throughput
has a definite place to go – else Any problem will lead to unhappy
immediate feedback is given       customers (either internal or
                                  external)
   Comparisons (cont.)

           Advantages                         Disadvantages

                            MRP  PUSH
Allows Managers to manage – that
                                    Can lead to large inventories
is, plan and control things
Requires intricate knowledge of     Can generate large quantities of
Production Times & Product Flow     scrap before errors are discovered
Can lead to economies of scale in   Requires diligence to maintain
purchasing and production           effective product flow
Allows for the planning and
completion of complex assemblies    Requires maintenance of large and
as sub-components are delivered     complex databases
only by scheduled need
Focusing on JIT
    JIT (Just In Time) is an outgrowth of the Kanban system developed by
     Toyota.
    Kanban refers to the posting board (and the inventory control cards
     posted there) where the evolution of the manufacturing process would
     be recorded.
    The Kanban system is a manual information system that relies on
     various types of inventory control cards.
    It‟s development is closely tied to the development of SMED: Single
     Minute Exchange of Dies, that allowed model changeovers to take
     place in minutes rather than hours.
    The Fundamental Idea of JIT – and Lean Manufacturing Systems in
     General (an Americanization of the Toyota P. S.) – is to empower the
     workers to make decisions and eliminate waste wherever it is found
The Tenets of JIT/Lean

   Empower the workers:
          Workers are our intelligent resources – allow them to
           exhibit this strength
          Workers ultimately control quality lets them do their job
           correctly (Poka-Yoke)
          Don‟t pit workers against each other – eliminate “piece-
           work” disconnected from quality and allow workers to
           cooperate in teams to design jobs and expectations
The Tenets of JIT/Lean

   Eliminate Waste
    –   Waste is anything that takes away from the operations
        GOAL (to make a profit and stay in business!)
            Reduce inventory to only what is absolutely needed
            Improve Quality – scrap and rework are costly and disrupt flow
            Only make what is ordered
            Make setups and changes quickly and efficiently
            Employ only the workers needed
            Eliminate Clutter – it wastes time
Features of JIT Systems

Small Work-in-Process Inventories.

  Advantages:
      1. Decreases Inventory Costs
      2. Improves Efficiency
      3. Reveals quality problems (see Figure 7-10)

  Disadvantages:
      1. May result in increased worker idle time
      2. May result in decreased throughput rate
River/Inventory Analogy
Illustrating the Advantages of Just-in-Time




        Revealing fundamental „problems‟ is the noted
            competitive advantages of JIT/Lean
Features of JIT Systems (continued)

Kanban Information Flow System
     Advantages
1. Efficient tracking of lots
2. Inexpensive implementation of JIT
3. Achieves desired level of WIP – based on Number of
     Kanbans in the system

     Disadvantages
1. Slow to react to changes in demand
2. Ignores predicted demand patterns (beyond 2 months
     or so)
Focus on The Kanban

   Typically it is a 2-card system
   The P (production) Card and W (withdrawal) Card
   Limits on product inventory (number of P & W cards)
    are set by management policy
   The count is gradually lowered until problems
    surface
   The actual target level (card count) is based on short
    term forecasting of demands
Focus on The Kanban
Focus on The Kanban – the worker as
manager

   P cards cycle from their accumulation post at Center
    1 to product (when a defined trigger point is reached)
    and then to output queue
   When trigger level is reached, Ct 1 worker pulls
    product from Ct 1 Wait point queue and replaces the
    Ct 1 W-cards with Ct 1 P-Cards which then are
    loaded to the Ct 1 processors – the worker puts Ct 1
    W-Cards to his/her acc. Post for W-cards
   Finished Product is pushed into the Ct 1 output
    queue
Focus on The Kanban – the worker as
manager

   A second worker (Ct 2‟s worker) watches for accumulation of Ct
    2 W-Cards
   When it reaches their trigger level, he/she pulls product into Ct
    2 Holding area after replacing Ct 1 P-Cards with their W-Cards
    – and returns Ct 1 P-Cards to their Acc. Post for Ct. 1 workers
    benefit
   They also watch for accumulation of Ct. 2 P-Cards on their acc.
    Post and when trigger count is reached they pull product from
    holding area and replace Ct 2 W-Cards w/ Ct 2 P-Cards then
    push it into the processors
   And around and around they go!
Focus on The Kanban

   So how many cards?
    – speaking of which, a         DL  w
                             y
    card is associated               a
    with a container (lot)   y   is # of Kanbans
    of product so the
                             D is 'average' demand
    number of P & W
    cards at a station       L is lead time (proc+wait+travel+others)
    determines the           w is buffer stock/ set by policy
    inventory level of a
    product!                 typically 10% of DL

                             a is container cap. < 10% of daily demand
Focus on The Kanban

   Lets look at an example:
          950 units/month (20 productive days) → 48/day
          Container size: a = 48/10 = 4.8 → 5
          “L” data:
            –   A. setup is 45 minutes (.75 hour)
            –   B. Setup is 3 minutes (.05 hr)
            –   Wait time: .3 hr/container
            –   Transport time: .45 hr/container
            –   Prod Time: 0.09 hr/each = .45 hr/container
Focus on The Kanban
La  .45  .75  .3  .45  1.95hr             Requires 3*2 = 6*5 = 30
                                               pieces in inventory – also,
D  48/8  6 / hr
                                               with 45mins set up 10 times
DLa  6  1.95  11.7; wa  .1  11.7  1.17   a day means that we
                                               consume 450 min or 7.5
    11.7  1.17
ya              2.57  3                     hours/day just setting up!
         5
Lb  .45  .05  .3  .45  1.25hr
                                                 Here only 2*2 = 4*5 =
DLb  6  1.25  7.5                             20 pieces and also
                                                 only .05*10 = 50 min
wb  .1  7.5  .75                              for setup (.833 hr) per
     7.5  .75                                   day
yb             1.65  2
         5
So, setup reduction impacts Factory
Capabilities & Inventory

   Lets look at the effect of
                                                       K  hQ
    studies comparing cost of
    setup vs. inventory cost –
                                          G  Q, K          I aK 
    like EOQ                                           Q    2
   Then lets see what we can             Like in EOQ
    invest to reduce inventory
    levels                                but last term is a 'Penalty'
            We will spend money on
             reducing setup cost (time)   factor for investing in setup
             and see if reduced
             inventory will offset our
             investment                   reduction rather than other
   This is the driving force for
    SMED                                  projects
Focus on the Penalty Factor

   We can effectively model this “a(K)” function as a
    „logarithmic‟ investment function
   By logarithmic we imply that there is a an increasing
    cost to continue to reduce setup cost
   We state, then, that there is a sum of money that can
    be invested to yield a fixed percentage of cost
    reduction
   That is (for example) for every investment of $200
    the organization can get a 2% reduction in Setup
    cost
Focus on the Penalty Factor

   Lets say that the investment is $
    buys a fixed percent reduction in    a  q j K0   j 
    K0
   If we get actually get 10% setup     j   is 'number' of investments
    cost reduction for $, then an
    investment of $ will mean:          q is the decimal equivalent of the
     –   Setup cost drops to: 0.9K0
                                         amount of reduction the $
   A second $ investment will lead
    to a further 10% reduction or:       investment will buy:
     –   .9K-.1*.9K = .81K0
   This continues: K3 = .729K0         q=(1-%setup cost reduction)
   Generalizing:
Focus on the Penalty Factor

   With that “shape” we
    can remodel the a(K)                                     
    logarithmically:                             b
            a(K) = b[ln(K0) – ln(K)]
                                                       ln  1 
            where:                                        q
                                                           

   Reverting back to     G  K   2 K  h  I  b ln( K 0 )  ln  K  
                                                                         
    G(Q,K) function – and now, minimize w.r.t. K
    substituting Q*:
                                        meaning: G'(K) = 0
Focus on the Penalty Factor

   Finding the K* after the minimization:
                             2 2
                     2I b
                  K 
                       h
                 I is MARR for the company

   To determine what we should do, determine G(K)
    using K0 and K*
Lets try one:

   K0: $1000
   : $95 for each 7.5% reduction in setup cost
   Annual quantity: 48000
   Holding cost: $4.50
                                  95          95
   MARR is 13%           b                     1218.55
                                                .07796
                             ln  1
                                 1  .075  
                                              

                                  2  .13  1218.55 
                                          2            2

                          K                            $0.232
                                        48000  4.5
Continuing:

   Investment to get to K*   a  K   1218.55 ln 1000   ln .232  
                                                                        
                               $10195.91
Testing for decision
                              G  K 0   2 K 0 h  $20784.61
   No investment (K = K0):                     2 K 0
                              note : EOQ                     4619 pieces
                                                         h

                              GK            2 K  h  Ia  K  
                                                                   
   At Min K*:
                               316.58  1325.47  $1642

                              EOQ  K      
                                                  2 K            70
                                                                h
SMED
Some terms:

   SMED = single minute exchange of dies
    which means quick tooling change and low
    setup time (cost)
   Inside Processes  setup functions that
    must be done „inside‟ the machine or done
    when the machine is stopped
          Minimally these would include unbolting departing
           fixtures/dies and positioning and bolting new fixture/dies
           to the machine
More Terms:

   Outside Setup  activities related to tooling
    changes that can be done „outside‟ of the
    machine structure
    –   These would include:
            Bringing Tooling to Machine
            Bringing Raw Materials to Machine
            Getting Prints/QC tools to machine
            Etc.
Focus on SMED

   When moving from “No Plan” or Step 1 to
    Step 2 (separating Inside from Outside
    activities) investments would be relatively low
    to accomplish a large amount of time (cost)
    saving
    –   Essentially a new set of change plans and a small
        amount of training to the Material Handlers so that
        they are alerted ahead of time and bring the
        tooling out to the machine before it is needed
Moving to Step 3 and Step 4

   Require investments in Tooling
   Require Design Changes
   Require Family tooling and adaptors
   Require common bolstering attachments
   In general requiring larger and larger
    investments in hardware to achieve smaller
    and smaller time (cost) savings in setup
Therefore, we can say SMED is:


              In reality the essence of
              a Logarithmic setup
              reduction plan!
Lets Look into Line Balancing

   This is a process to optimize the assignment of
    individual tasks in a process based on a planed
    throughput of a manufacturing system
     –   It begins with the calculation of a system “Takt” or
         Cycle time to build the required number of units
         required over time
     –   From takt time and a structured sequential analysis
         of the time and steps required to manufacture
         (assembly) a product, compute the number of
         stations required on the line
     –   Once station count is determined, assign feasible
         tasks to stations one-at-a-time filling up to takt time
         for each station using rational decision/assignment
         rules
Line Balancing

   Feasible tasks are ones that have all
    predecessors completed (or no
    predecessors) and take less time
    that the remaining time at a station
   Feasibility is also subject to physical
    constraints:
           Zone Restriction – the task are physically
            separated taking to much movement time
            to accomplish both within cycle (like
            attaching tires to front/back axles on a
            bus!)
           Incompatible tasks – the Grinding/Gluing
            constraint
Some of the Calculations:

   Takt (Cycle) Time:
                Prod. time/day        Total Time - Allowances (T)
         C                       
              Target output/day            Req'r Output (Q)

         CT        (min/unit)
                Q
   Minimum # Workstations req‟r:


           Nt     (Total Job Task Time)
                                       C
Efficiency Calculations:
                       N
                i   ti
    Effline           i 1
                 C N
    here t i is the time actually consummed at the stations
                 J

                t
                j 1
                       j

    Eff StK 
                 C
    t j is the time of any task actually assigned to St K
Lets Try One:

                               Times:
                         A 25s; B 33s;C 33s;
 A         B         C   D 21s; E 40s; F40s;
                            G 44s; H 19s
      Production
     Requirement         G              H
      is 400/shift

 D           E       F
Calculation of Takt Time & Optimal
Station count


   480  .12 * 480 422.4
C                       1.056min  63.4s
        400         400


N  t   25  33  33  21  40  40  44  19  4.04
        i

    C                       63.4
 N 5stations
To Perform Assignment we need
Assignment Rules:

   Primary Rule:
    –   Assign task by order of those having largest
        number of followers
   Secondary Rule:
    –   Assign by longest task time
Primary Assignment Rule

        Task      # Followers

        A,D           4

        E,B           3

        F,C           2

         G            1
Line Balancing Assignments

                                       Feas.       Task w/
                           Remaining                           Task w/ L.
Station   Task   T. Time               Remaining   Most
                           Time                                Time
                                       Task        followers
           A       25        38.4        B, D         D            B
  1
           D       21        17.4          -

  2        E       40        23.4          -

  3        B       33        30.4          -

  4        F       40        23.4          -

  5        C       33        30.4          -

           G       44        19.4         H
  6
           H       19         .4           -
The Line Balance


  A     B      C

       WS 3    WS 5



WS 1                  G          H
                          WS 6


   D     E     F

       WS 2   WS 4
Checking Efficiencies:

                  T        255
       Eff L                  67%
                 C N 6  63.4
                   46
       Eff S 1         72.6%
                 63.4
                   33
       Eff S 3         52.1%
                 63.4
                   63
       Eff S 6         99%
                  63.4
Dealing with Efficiencies

   We investigate other Rules – application to improve
    layout
           1st by followers then by longest time then most followers
           Alternating!
   Consider line duplication (if not too expensive!)
    which lowers demand on a line and increases Takt
    time
   The problem of a long individual task
           In Koeln, long time stations were duplicated – then the system
            automatically alternated assignment between these stations

				
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