Process Enhancement Limits Lss Analysis1

Research
Publication Date: November 2010

The Limits of process enhancements: Lean six sigma
approach to real processes
Index

The Limits of process enhancements: Lean six sigma approach to real processes 1
Introduction: production demand variability scenarios for generic processes.........................................1
Global Process Productivity ...................................................................................................................2
Process Cycle Efficiency ........................................................................................................................3
Output Production for real processes.....................................................................................................4
Example of applicability: Industrial production plant...............................................................................6
Appendix: A Process’s area of opportunity for a defect..........................................................................6

Key Findings
• Real process robustness versus production demands

• Process assets productivity related to production demands

• Quantitative statistical model of cumulative errors in real processes

• Resources allocation in industrial production demand scenarios

Introduction: production demand variability scenarios for generic processes
Lean six sigma is a methodology that allows to develop solutions to reducing waste upon the
analysis of the variability of a process. Using this methodology and statistical analysis, we can try
to offer solutions to production stress scenarios in real processes.
Hence, when a process is being built, it makes sense to ask about how process behaviour will be
against production demands variations depending on the process resources or available assets.
This question should be identical to the problem of describing how to obtain the required
production demand in a given lead time, the latter being the time corresponding to the critical path
for the global process.
In other words, in some generic production variation scenario, the lead time shall change. An
alternative way to characterize that change is in terms of the available resources of it. To se this,
firstly notice that assets increments and lead time variations are simply related ρ ≡ 1 /(1 − τ )) ,
where ρ is the asset -or resources- factor and, τ , is the lead time factor that defines the
process change - for duplicating the process assets will naturally correspond to half the lead time
REF: Project Enhancements limits

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have no liability for errors, omissions or inadequacies in the information contained herein or for interpretations
thereof. The opinions expressed herein are subject to change without notice. ©2010 Xerox Corporation. All right
reserved. José Luis Rosales, GDO  Print Output Consultant (Spain)  November 2010

Process enhancements Analysis

for that process change-. Secondly, see that the production increment of an optimal –ideal-
process,℘ , should increase linearly with the available resources, therefore, the a priori solution
to the dependency between assets and production should be simply ℘ideal = ρ .

Unfortunately, processes are not ideal ones. The reasons for that are the inefficacies of the
design when applied to the real world, the existence of many out of control variables, as well as
internal process input and output wastes. Thus, the projected output experiences a lot of
variability and it never reaches its projected optimal rate. In the end, the relation ℘( ρ ) shall be
non linear in a real process (also known from the work of Putnam on project sizing design.)
We can also ask about process robustness in the sense that there could be a maximum limit for
process outputs enhancements. Of course, one can simply say that this problem will always be a
cost/benefits question for improving process resources and process design when production
demands require it to do. But, as a regular rule, the on-going process can be considered to work
rather optimally -at standard production rates- and one should logically ask how the given
process can adapt itself to the new production demands upon adding resources that cope with
new production circumstances. It is in this sense that we require a definition of process
robustness, i.e., how large the production demand that the process can afford for will be, before it
breaks down. In order for the analysis to take place, and to find the analytical behavior of the
function ℘( ρ ) , we will set up the more affordable question to obtaining the relation between the
value of the process assets and their productivity change, ℑ( ρ ) .

For this computation, the probabilities theory, analyzing Poisson-like errors in any process chain
included in the global process, will be needed. Of course, it will depend on the lead time change
factor - for the issue of a project- or, alternatively, in production processes, on the asset change
factor ,in some industrial design. Let us consider identical small probabilities for uncontrollable
errors that take place into individual sub-processes but that would be eventually cumulated
through the sequence of the global production process.

Global Process Productivity
In the process the Area of Opportunity for defects (see appendix) spread out through the time.
That is, in order to obtain the process outputs, such as services, implant, etc, a time goal is
required i.e., the lead time .
Process results are achieved in an actual averaged time <Tactual>, and the defect per unit
(considered as the Process lead time goal) is given, if Tgoal= (1- τ ) <Tactual>, by
< T actual > − T goal < T actual > − (1 − τ ) < T actual > τ
DPU = = =
T goal (1 − τ ) < T actual > (1 − τ )

Here, the quantity τ represents the proportionality of the expected lead time reduction of the
achieved time versus the goal. For instance, some reduction proportion of τ = 20% in the actual

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©2010 Xerox Corporation. All right reserved. José Luis Rosales, Xerox GDO Consultant (Spain) November 2010


time for achieving the process results, gives DPU =0.2/(1-0.2)=25%. This quantity can be
measured for every process or, more generally speaking, for every product output in our industry.
It is a feature of the process line and it measures its productivity.
A process consists on many things, for instance, different process products should be implanted
in different phases, sales, solution’s architecture and design, service level agreement elaboration
and process metric and governance are areas of opportunities for defects in the whole process
life cycle.
In some abstract manner, let’s simplify the model in a way that an opportunity for defect be
associated to every process phase, i.e., given that the process defect per unit is DPU= τ /(1- τ ),
if the process has N sub-processes (or parts), we’ll get the defect free probability of each life
cycle sub-process as
Probability (OK)(sub-process)= 1-DPU/N= 1- τ /(1- τ )N,
and the probability of a process to be defect free for a time to revenue reduction given by τ

P(OK =lim→∞{ −τ /( −τ)N}N =exp{ /( −τ)}……… 1
) N 1 1 τ
− 1
In order to understand this result, recall that
ρ ≡ 1 /(1 − τ )) ,
is the relative increment of the process resources, so we get 1 in the equivalent form (depending
on the resources).
PCE
Y ≡ P(OK ) = e −( ρ −1) = ………….….…… 2
PCEGOAL

It should be the probability of success of the given new lead time scenario. This can also be
interpreted as the relative efficiency of the actual process versus the process goal.

Process Cycle Efficiency

The time to revenue reduction objective can be defined as the time reduction proportion
necessary to achieve the results and products of a process at the Goal Lead Time.
Equation 2 gives the First Time Yield of a process as a function the time to revenue reduction
objective. We also obtained a relation for the FTY in terms of the increment of resources needed
to achieve the time to revenue goal.
We notice that

3


PCE GOAL ⎡ e + ( ρ −1) ⎤
Y −1 ≡ = e + ( ρ −1) = ρ ⎢ ⎥ ……..….. 3
PCE ⎣ ρ ⎦
Moreover, increasing the assets in a process results in a Process Cycle Efficiency increment
since, by definition, PCEGOAL is

= ρ [PCE ]
CVAGOAL CVAGOAL CVAGOAL
PCE GOAL = = …
PCTGOAL (1 − τ ) PCT CVA

………………………………………………………….… 4
Where PCE is the actual Process Cycle Efficiency .
Comparing 3 and 4 obtains
CVAGOAL 1
= exp[( ρ − 1)] ……………………….… 5
CVA ρ
This means that, in order to achieve the time to revenue process reduction, the productivity
increment per process asset should be given by
1
Δℑ= exp ρ −1)] −1 ……………………………………. 6
[(
ρ
To achieve the time to revenue reduction goal in a real process, one should reduce the errors rate
of each process resource in this amount.

Output Production for real processes
We have seen that in a real process, any additional production demand that does not makes the
global process to break down can be solved upon adding more resources and this gives that the
productivity per resource decrease in a factor given by 6 ; thus, a real process suffers a cost of
productivity and the production rate should be corrected by a factor
℘ = ρ − ρ Δ ℑ = 2 ρ − exp [( ρ − 1) ] ……………… 7
Or else, taken into account increments, one gets, instead :
Δ℘ = 2Δρ − exp[Δρ ] + 1 ……………………………… 8

This is a very important result. In Figure 1 the production incremental rate versus the resources
necessary increments is drawn.

4


0,5

0,4

0,3

0,2

0,1
Output increment

0
-0,6 -0,4 -0,2 0 0,2 0,4 0,6 0,8 1 1,2 1,4
-0,1

-0,2

-0,3

-0,4

-0,5

-0,6
asset increment factor

Figure 1

Notice that, effectively, in order for the process to respond to the required demand, the right
procedure is to add more resources; this is only a first approximation. Notwithstanding with this, a
limit for this approach arises: After the limit is reached, one can not increase production upon
adding more resources to the process. The utter value for the production improvements of a real
process is
Δ℘Max = 2Log(2) − 1 ≈ 39%

The value of the asset change factor for that maximum is
Δρ Max = Log (2) ≈ 69%

Figure 2, represents necessary resources increments for a given production demand change
0,8

0,6
Needed Resources Increments

0,4

0,2

0
-0,6 -0,4 -0,2 0 0,2 0,4 0,6

-0,2

-0,4

-0,6
Production increment factor

Figure 2

5


Example of applicability: Industrial production plant.
The aim of this exercise is to compute how to allocate the resources of some industrial plant in
order to cope with production demands scenarios. Let’s put an example consisting on two
production independent chain process, say the one producing good A and B respectively. They
have 50 % of the resources each. We wanted to see if the plant can respond to a production
demand peak of 130% of good A. The solution assumes the possibility to manage and move
resources from production chain B to A in order to cope with the demand.
We see that for P’(A)=130 % one has to increment the number of resources in a factor given by
R’(A)= ρ (130%)*50=1.40*50%=70%

And this implies R’(B)=30%.
Consequently, for process chain of good B, the production decays as P’(B)= 1+ Δ℘ (1-
30/50)=53% - as predicted by 8-.
Any kind of production demand might be analyzed using the above procedure.

Appendix: A Process’s area of opportunity for a defect
The Poisson Distribution provides a sense of the area of opportunity for the events or defects it tracks.
When Defect Per Unit =1, for instance, some units (about 36,8%) are expected to contain zero defects.
In order to proof this statement, let’s set up an example unit as containing 10 areas of opportunity, each
representing a region where a single defect either will or not will be found.
For the overall Defect Per Unit (DPU=1) each of these 10 Opportunities For Defect (OFD) regions must
actually contain a defect only one time every 10, i.e., probability (defect)=0.1; in other words, the chances
that any single OFD is defect free (OK) must be 9 in 10, or probability(OK)=0.9.
For the unit to be defect free, all 10 OFD must be defect free. The concept of joint probability provides that
the chances of all 10 independent OFD being defect free is the product of the probability for each one
being defect free.
P(defect free unit)= 0.910 = 0.349
In general, if the unit consist on N elements and DPU=z, then,
Probability( OK) (each element)=1-DPU/N=1-z/N
And, probability (defect free unit)= (1-z/N)N
Now suppose N>> 1, then,

P (ok ) = lim N →∞ (1 − z / N ) N = exp(− z )
Which is the First Time Yield.

6

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