Planning Production of a Set of
Semiconductor Components with
Uncertain Wafer and Component Yield
Frank W. Ciarallo
Assistant Professor
Biomedical, Industrial and Human Factors
Engineering
Overview

Industry and Process Motivation

Problem Description/Formulation

Solution Approaches

Results and Summary
Semiconductor Manufacturing: The
Industry


Fabrication facility (a “fab”) can cost $1 billion to $4
billion
Major roles in industry
–
–

Component Design
Component Manufacturing
Business Strategies
–
–
–
“Integrated” (Design and Manufacture, i.e. Intel)
“Fabless” (Design, i.e. Cisco)
“Foundry” (Manufacture, i.e. Flextronics)
Semiconductor Manufacturing: The
Process
Wafer Fabrication
Problem Motivation: Semiconductor
Component Manufacturing
Wafer

Components are produced
from single crystal silicon
Wafers consisting of a gridlike array of Sites.

Wafers move through a
fabrication facility in groups
called Jobs.
Site
Job
Characteristics of Production Process

Uncertain yield at the component, wafer and
job level.
–

Release 100 components into production, only 60
usable components are produced because of
yield loss
Sources of yield loss
–
–
–
Defects in wafers or masks
Contamination from the air in the fab
Errors in alignment, chemical concentrations,
process steps, etc.
Is this just a Quality Control problem?

Relatively new processes are used in
production
–
–

Competitive advantage and profit come from
pushing the technology envelope
“Mature” technology goes into commodity (low
profit) products, or is not longer marketable
Uncertain yields are an inherent feature of
this technology driven industry
Characteristics of Production Control

Long production lead times (cycle times)
–
Time from release of wafers to completion is “many” weeks
–
Side effect of cost of fabrication equipment and production
environment:
High Capital Cost
Unproven Technology
High Utilization
Rates
Uncertainty in
Production
Congestion!
Set Production Environment


Production of Multiple Components in “Sets”
ASIC or R&D Environment with Smaller
Production Volumes
–
–
From a single group of wafers a set(s) of
components is desired.
Because of lead times, more than one group of
wafers is undesirable.
Model of Set Completion






To compute set completion probability
A "set" of components consists of ai of component i for
i =1,...,m.
Each component type, i, has Bernoulli probability of
being good: pi.
N wafers with m sites per wafer.
Each wafer has a probability of being good, .
Demand for a set of components must be met with
probability . (Service level constraint).
Definition of an “Allocation”


An allocation defines a possible solution
Each site on a wafer is assigned to one of
the component types.
–
–
Identical allocations are the same on each wafer.
Non-Identical allocations vary between wafers.
Problem Formulation

Let X(Z,N) be the number of sets produced given
–
–
–
–

Allocation matrix (vector) Z (defines a solution)
N wafers
Number of sets demanded is D
Must meet this demand with probability 
Solve:
Maximize Pr  X  Z, N   D
subject to Wafer Size Constraints
iteratively to find the smallest N that satisfies the service level
constraint.
Related Research

Singh, Abraham, Akella (1988)
–
–

Connors and Yao (1993)
–
–
–

Component, wafer, job loss
Non-Identical allocations with practical constraints
Ignores correlation between component types
Avram and Wein (1992)
–
–
–

Continuously valued allocations
No wafer yield
Uses long run average number of sets produced as an objective
Linear Programming deterministic model based on mean yields
Stochastic model supplies approximate objective for LP
Seshadri and Shanthikumar (1997)
–
Extended Avram and Wein’s work considering additional release policies
Discrete Wafer Model

Probability of completing the requirement
N
N!
N k
k
g x  
 1   
F  kSx 
k 1 k ! N  k  !
I
Across
Components
m

  f j x j     x j  p ij 1  p j x i
F y   f j y j
j 1
xj
j
ia j  i

Mathematical Difficulties
–
–
Not continuous.
May be maximized with a non-identical allocation.

Wafer
Level
Non-identical allocations difficult to evaluate.
Component
Level
Small example





Number of Sites
2 wafers
given to Type 1
2 sites per wafer
p1=0.1
 = 0.9
2 component types
Set requirement: 1 of each
Number of Sites
component
given to Type 2
p2 = 0.9
0
0
0
0
Discrete Wafer, Continuous Allocation
Model

Probability of completing the requirement
g
I
N
N!
N k
 k 1   
F  kSx 
k 1 k ! N  k  !
x  
m
F y   
f jC
j 1

Improvements
–
Continuous
y j 
pj
 

f jC x j  0
1

0
a j 1
a j 1
1    x a
j
1    x a
j
j
j
d
d
Small example: continued
Type 1




p1=0.1
2 wafers
2 sites per wafer
Type 2
 = 0.9
p2 = 0.9
0.25
Set requirement:
1 of each component
0.2
x1 = fraction of wafer
assigned to component type 0.15
1
0.1
2
1
0
1
gI

0.05
0.2
0.4
x1
0.6
0.8
1
Example Problem Insight




Small Problems are tricky!
Continuous allocation model is easy to work
with
Continuous allocation, Discrete wafer model
has discontinuities
Best solutions could require non-identical
wafer allocations
Solution I: Marginal Allocation




Use a "Greedy" procedure to build up an
identical allocation.
In the no wafer loss problem yields close to
optimal solution (product-form objective
function).
It’s quick; can be used to initialize more
complicated procedures.
Yields an integer valued allocation.
Solution II: Based on a Fixed Number of
Wafers

Exploit the efficiency with which we can solve the
no-wafer-loss problem.
– Choose some fixed number of wafers, S
–
–
solve problem optimally for that number of wafers
apply that fractional allocation to all wafers identically.
Min gI[S x]
S
m
 




f
S
1

x

'
k
f S x 
  k 1 
subject to i i 
, i =1, ..., m - 1
m

f i S xi 


f m S 1   xk 
  k 1 
'
m
Insight into Solution II
Set
Completion
Probability
1.0
0.8
Pr{ X > 3}
0.6
0.4
0.2
0.0
500
Probability
for Number
of Sites
Available
1000
1500
2000
Number of Sites
Binomial Probability
0.3
0.2
0.1
0.0
Number
of Sites
Available
Solution III: Continuous Wafer
Approximation

An Approximation to the (realistic) Discrete Wafer Model
 N  1
Nt
g x   t  0
 t 1   FtSxdt
 t  1N  t  1
IC





N
Integrand equal to terms in summation at integer values of k.
Approximation is bad for small numbers of wafers.
As number of wafers increases, the Discrete Wafer expression
converges to the Continuous Wafer expression.
IC
g is continuous and differentiable.
Optimize using a non-linear programming code.
Evaluating continuously valued
allocations

Solutions II and III generate continuously valued allocations
1. Nearest Neighbor Method (Branch and Bound)
–
–
Finds the best allocation available by rounding up or down to the
nearest integer.
Computationally very expensive
2. Overall Fractional Allocation Target Method
–
–
–
Generates Non-Identical Allocations that closely match the target
fractional allocation across all wafers (rather than on each wafer)
A large percentage of each wafer is identical. (90%)
Remaining percentage varies across wafers. (10%)
Defining Test Problems

Practically interesting problems have the following features:
–
–
–
–
–
A non-trivial number of component types
A non-trivial number of wafers
One (or a few) components which are required in large numbers,
and have a reasonably high yield rate
One (or a few) components which are required in small numbers,
and have a very low yield rate
Other components with a varying range of usages and yield rates
Experiment A


Each problem has exactly 20 wafers available
Compare set completion probabilities given 20 wafers
Solution I
Solution II
Solution III
Avg Set Comp. Prob.
0.8378
0.8683
0.8710
Avg % difference from best
-4.9%
-0.35%
-0.04%
-20.06%
-6.34%
-0.44%
6%
28%
70%
Experiment A
Worst case % difference
% of problems best
–
–
–
Solution I: Marginal Allocation
Solution II: Fixed Number of Wafers
Solution III: Continuous Wafers/Nonlinear Search
Experiment B


Same problems as A
Find minimum # wafers to achieve 95% service level
Experiment B
Heuristic I
Heuristic II
Heuristic III
22.0
22.0
21.2
Avg % diff. from best
3.87%
3.27%
0.21%
Worst case % diff.
10.53%
14.81%
5.26%
% of problems best
26%
56%
96%
Avg # Wafers Required
–
–
–
Solution I: Marginal Allocation
Solution II: Fixed Number of Wafers
Solution III: Continuous Wafers/Nonlinear Search
Comparison to Connors/Yao Work
Experiment-A
HJ-I
Average probability of set completion (2 jobs)
0.82251206
Average % difference from the best
-5.182%
Worst-case % difference from the best
-34.089%
% of problems best
10%
2 problems had the same set completion probability.
HJ-II
0.8659371
-0.0129%
-0.338%
94%
Experiment-B
HJ-I
Average number of jobs needed
3.6
Average % difference from the best
9.23333333%
Worst-case % difference from the best
50%
% of problems best
72%
36 problems used same minimum number of wafers.
HJ-II
3.3
0%
0%
100%
•Includes yield loss at the job level
HJ-I Connors and Yao rule.
HJ-II Heuristic based on the exact objective function.
HJ-II clearly outperforms HJ-I when the wafer/job yields are low.
Summary

Solutions and Insight come from study of the problem, detailed
modeling, and careful experimentation
–
–

Insight: Generate identical allocations from approximate models,
use practical procedure to get “mostly” identical allocation for
implementation
Insight: Multi-component wafers can greatly reduce the number of
wafers required to complete a set requirement
Future Work:
–
–
Prove the continuous wafer approximation has a unique minimum
Explore the economics of using multi-component wafers versus
single component wafers