1. 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

2. Overview Industry and Process Motivation Problem Description/Formulation Solution Approaches Results and Summary

3. 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)

4. Semiconductor Manufacturing: The Process

5. Wafer Fabrication

6. 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 Site Wafer

7. 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.

8. 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

9. 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!

10. 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.

11. Model of Set Completion To compute set completion probability A "set" of components consists of a i of component i for i =1,...,m. Each component type, i, has Bernoulli probability of being good: p i. 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).

12. 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.

13. 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.

14. 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

15. 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. Component Level Across Components Wafer Level

16. 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

17. Discrete Wafer, Continuous Allocation Model Probability of completing the requirement Improvements – Continuous

18. Small example: continued 2 wafers 2 sites per wafer  = 0.9 Set requirement: 1 of each component x 1 = fraction of wafer assigned to component type 1 2 0 1 1 Type 1 p1=0.1 Type 2 p2 = 0.9

19. 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

20. 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.

21. 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.

22. Insight into Solution II Set Completion Probability Probability for Number of Sites Available Number of Sites Available

23. 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.

24. 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%)

25. 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

26. 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

27. 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

28. 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

29. 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