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

Components are produced from single crystal silicon Wafers consisting of a grid-like array of Sites.

Wafers move through a fabrication facility in groups called Jobs.

Job

SiteWafer

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:

iteratively to find the smallest N that satisfies the service level constraint.

Pr ,

subject to Wafer Size Constraints

Maximize X N DZ

Related Research

Singh, Abraham, Akella (1988)– Continuously valued allocations– No wafer yield

Connors and Yao (1993)– Component, wafer, job loss– Non-Identical allocations with practical constraints– Ignores correlation between component types

Avram and Wein (1992)– 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

Mathematical Difficulties– Not continuous.– May be maximized with a non-identical allocation.

Non-identical allocations difficult to evaluate.

1

!1

! !

NN kI k

k

Ng F kS

k N k

x x

1

m

j jj

F f y

y

j

j

j

x

ai

ixj

ij

jjj pp

i

xxf 1

Component Level

Across Components

Wafer Level

Small example

2 wafers 2 sites per wafer = 0.9 2 component types Set requirement: 1 of each

component

0

0

0

0

Number of Sites given to Type 1

p1=0.1

Number of Sites given to Type 2

p2 = 0.9

Discrete Wafer, Continuous Allocation Model

Probability of completing the requirement

Improvements– Continuous

1

!1

! !

NN kI k

k

Ng F kS

k N k

x x

m

jj

Cj yfF

1

y

1

01

1

0

1

1

j

j j j

j j j

pa x a

Cj j

a x a

d

f x

d

Small example: continued

2 wafers 2 sites per wafer = 0.9 Set requirement:

1 of each component x1 = fraction of wafer

assigned to component type 1

0.2 0.4 0.6 0.8 1x1

0.05

0.1

0.15

0.2

0.25

Ig

2

0

1

1

Type 1

p1=0.1

Type 2

p2 = 0.9

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, – solve problem optimally for that number of wafers– apply that fractional allocation to all wafers identically.

MinS

gI[S x]

subject to f i

' S xi f i S xi

f m' S 1 xk

k1

m

f m S 1 xkk1

m

, i =1, .. ., m - 1

S

Insight into Solution II

2000150010005000.0

0.2

0.4

0.6

0.8

1.0

Number of Sites

Pr{

X >

3}

2000150010005000.0

0.1

0.2

0.3

k

Bin

om

ial P

rob

abil

ity

Set Completion Probability

Probability for Number

of Sites Available Number

of Sites Available

Solution III: Continuous Wafer Approximation

An Approximation to the (realistic) Discrete Wafer Model

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. is continuous and differentiable. Optimize using a non-linear programming code.

gIC x N 1 t 1 N t 1 t0

N

t 1 N tF tSx dt

g IC

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: Marginal Allocation– Solution II: Fixed Number of Wafers– Solution III: Continuous Wafers/Nonlinear Search

Experiment A 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%

Worst case % difference -20.06% -6.34% -0.44%

% of problems best 6% 28% 70%

Experiment B

Same problems as A Find minimum # wafers to achieve 95% service level

– Solution I: Marginal Allocation– Solution II: Fixed Number of Wafers– Solution III: Continuous Wafers/Nonlinear Search

Experiment B Heuristic I Heuristic II Heuristic III

Avg # Wafers Required 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%

Experiment-A HJ-I HJ-IIAverage probability of set completion (2 jobs) 0.82251206 0.8659371Average % difference from the best -5.182% -0.0129%Worst-case % difference from the best -34.089% -0.338%% of problems best 10% 94%2 problems had the same set completion probability.

Experiment-B HJ-I HJ-IIAverage number of jobs needed 3.6 3.3Average % difference from the best 9.23333333% 0%Worst-case % difference from the best 50% 0%% of problems best 72% 100%36 problems used same minimum number of wafers.

•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.

Comparison to Connors/Yao Work

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