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
Limited and known Final Inventory
Every job is a ‘High Stress’ Rush
order
Worker only consume their time &
Raw Materials on what is actually
needed
Balanced systems MUST be in
place
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
is, plan and control things
Can lead to large inventories
Requires intricate knowledge of
Production Times & Product Flow
Can generate large quantities of
scrap before errors are discovered
Can lead to economies of scale in
purchasing and production
Requires diligence to maintain
effective product flow
Allows for the planning and
completion of complex assemblies
as sub-components are delivered
only by scheduled need
Requires maintenance of large and
complex databases
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 “piecework” 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
card is associated
with a container (lot)
of product so the
number of P & W
cards at a station
determines the
inventory level of a
product!
y
y
DL  w
a
is # of Kanbans
D is 'average' demand
L is lead time (proc+wait+travel+others)
w is buffer stock/ set by policy
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
D  48/ 8  6 / hr
DLa  6  1.95  11.7; wa  .1  11.7  1.17
11.7  1.17
 2.57  3
5
Lb  .45  .05  .3  .45  1.25hr
ya 
DLb  6  1.25  7.5
wb  .1  7.5  .75
7.5  .75
yb 
 1.65  2
5
Requires 3*2 = 6*5 = 30
pieces in inventory – also,
with 45mins set up 10 times
a day means that we
consume 450 min or 7.5
hours/day just setting up!
Here only 2*2 = 4*5 =
20 pieces and also
only .05*10 = 50 min
for setup (.833 hr) per
day
So, setup reduction impacts Factory
Capabilities & Inventory


Lets look at the effect of
studies comparing cost of
setup vs. inventory cost –
like EOQ
Then lets see what we can
invest to reduce inventory
levels


We will spend money on
reducing setup cost (time)
and see if reduced
inventory will offset our
investment
This is the driving force for
SMED
K  hQ
G  Q, K  

 I aK 
Q
2
Like in EOQ
but last term is a 'Penalty'
factor for investing in setup
reduction rather than other
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
K0
If we get actually get 10% setup
cost reduction for $, then an
investment of $ will mean:
–

A second $ investment will lead
to a further 10% reduction or:
–


Setup cost drops to: 0.9K0
a  q j K0   j 
j
is 'number' of investments
q is the decimal equivalent of the
amount of reduction the $
investment will buy:
.9K-.1*.9K = .81K0
This continues: K3 = .729K0
Generalizing:
q=(1-%setup cost reduction)
Focus on the Penalty Factor

With that “shape” we
can remodel the a(K)
logarithmically:



a(K) = b[ln(K0) – ln(K)]
where:
b

ln  1 
 q 
G  K   2 K  h  I  b ln( K 0 )  ln  K  
Reverting back to
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
b

 1218.55
MARR is 13%


ln  1

 1  .075  
2  .13  1218.55 
 $0.232
48000  4.5
2
K 
.07796
2
Continuing:

Investment to get to K*
a  K   1218.55 ln 1000   ln .232  
 $10195.91
Testing for decision


No investment (K = K0):
At Min K*:
G  K 0   2 K 0 h  $20784.61
note : EOQ 
2 K 0
GK 
2 K  h  Ia  K  
h
 4619 pieces
 316.58  1325.47  $1642
EOQ  K


2 K 
h
 70
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
C

Total Time - Allowances (T)
Target output/day
CT

Q
Req'r Output (Q)
(min/unit)
Minimum # Workstations req’r:
Nt
(Total Job Task Time)


C
Efficiency Calculations:
N
Eff line 
i   ti
i 1
C N
here t i is the time actually consummed at the stations
J
Eff StK 
t
j 1
j
C
t j is the time of any task actually assigned to St K
Lets Try One:
Times:
A
B
C
Production
Requirement
is 400/shift
D
E
A 25s; B 33s;C 33s;
D 21s; E 40s; F40s;
G 44s; H 19s
G
F
H
Calculation of Takt Time & Optimal
Station count
480  .12 * 480 422.4
C

 1.056min  63.4s
400
400
t  25  33  33  21  40  40  44  19 

N

 4.04
i
C
 N 5stations
63.4
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.
Remaining
Task
Task w/
Most
followers
Task w/ L.
Time
D
B
Task
T. Time
Remaining
Time
A
25
38.4
B, D
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
H
19
.4
-
Station
1
6
The Line Balance
A
B
WS 3
C
WS 5
G
WS 1
H
WS 6
D
E
WS 2
F
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