Multidimensional Assignment Problems for Semiconductor...

Outline Relevance Our problem: MVA Results

Multidimensional Assignment Problems for SemiconductorPlants

Trivikram Dokka, Yves Crama, Frits Spieksma

ORSTAT, KULeuven

April 1, 2014

Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP


About merging vectors

Our problem - a prologue

Let u = (12 91 7), and v = (47 32 12).





Let u = (12 91 7), and v = (47 32 12).

How do we merge u and v?





Let u = (12 91 7), and v = (47 32 12).

How do we merge u and v?

Well, we say thatu ∨ v = (max(u1, v1),max(u2, v2),max(u3, v3)) = (47 91 12)

Oh, and the cost of a vector is represented by a function c(u) : Zp+ → R+.



Our Problem

Instance:

m sets: V1,V2, . . . ,Vm

Each Vi consists of n vectors each of size p, 1 ≤ i ≤ m

Each entry of a vector is a non-negative integer

Objective:

partition the given m sets into n m-tuples, such that each m-tuplecontains one vector from each set Vi

minimize the total cost of this partition

We will abbreviate the name of this problem as MVA.



An instance of our problem MVA

Let m = 3, and let the three sets be denoted by V1, V2, and V3. The length ofeach vector, p, equals 3, and n = 4, and let us specify c as the sum of theentries of a vector, ie, c(u) =

∑pi=1 ui .

V1 V2 V3

(12 91 7) (47 31 12) (83 3 37)(54 29 64) (5 44 73) (37 2 80)(92 32 26) (40 15 71) (38 13 68)(2 97 43) (32 32 32) (12 91 7)



An instance of our problem MVA

Let m = 3, and let the three sets be denoted by V1, V2, and V3. The length ofeach vector, p, equals 3, and n = 4, and let us specify c as the sum of theentries of a vector, ie, c(u) =

∑pi=1 ui .

V1 V2 V3

(12 91 7) (47 31 12) (83 3 37)(54 29 64) (5 44 73) (37 2 80)(92 32 26) (40 15 71) (38 13 68)(2 97 43) (32 32 32) (12 91 7)

A particular m-tuple could consist of the second vector of V1 ((54 29 64)),the first vector of V2 ((47 31 12)), and the fourth vector of V3 ((12 91 7)),coming out at: (54 91 64).



1 Relevance

2 Our problem: MVAOn the cost functionHeuristics for MVAAn instance

3 ResultsAnalysis of Heuristics

Monotone and Submodular Case

HardnessPolynomial Special caseQuestions


Outline Relevance Our problem: MVA Results Wafer-to-wafer integration

A wafer

Emerging Technology Through Silicon Vias(TSV) based Three-Dimensional Stacked Integrated Circuits (3D-SIC)

Benefits • smaller footprint • higher interconnect density • higher performance • lower power consumption

compared to planar IC’s Si Wafer



Stacking wafers

Stacking

From lot 1 From lot 2 From lot 3

Stack



Yield optimization: bad dies and good dies

Defect map

(0,..,0,1,1,0,…0,1,0,…,0,1,0,…,0,1,0,…,0,1,0,1)



Yield optimization: superimposing dies

Stacking

From lot 1 From lot 2 From lot 3

Defect map of resulting stack: (0,..,0,1,1,0,…0,1,0,…,0,1,0,…,0,1,0,…,0,1,0,1)

Yield = no. of zeros in defect map vector



Yield optimization: an example

Stack 1

Stack 2 Total number of bad dies in stack 1 + stack 2 = 23



Yield optimization: an example

Stack 1

Stack 2 Total number of bad dies in stack 1 + stack 2 = 17



Previous work

S. Reda, L. Smith, and G. Smith. Maximizing the functional yield ofwafer-to-wafer integration. IEEE Transactions on VLSI Systems,17:13571362, 2009.

M. Taouil and S. Hamdioui. Layer redundancy based yield improvementfor 3D wafer-to-wafer stacked memories. IEEE European TestSymposium, pages 4550, 2011.

M. Taouil, S. Hamdioui, J. Verbree, and E. Marinissen. On maximizingthe compound yield for 3D wafer-to-wafer stacked ICs. In IEEE, editor,IEEE International Test Conference, pages 183192, 2010.

J. Verbree, E. Marinissen, P. Roussel, and D. Velenis. On thecost-effectiveness of matching repositories of pre-tested wafers forwafer-to-wafer 3D chip stacking. IEEE European Test Symposium,pages 3641, 2010.

Eshan Singh. Wafer ordering heuristic for iterative wafer matching inw2w 3d-sics with diverse die yields. In 3D-Test First IEEE InternationalWorkshop on Testing Three-Dimensional Stacked Integrated Circuits,2010. poster.



Yield optimization is a special case of MVA

Observe that in the yield optimization application, all vectors are0, 1-vectors, and that the cost-function c is additive, ie, c(u) =

∑pi=1 ui .

Instances from practice may have m = 10, n = 75, and p = 1000.

We refer to this special case of MVA as the Wafer-to-Wafer Integrationproblem (WWI).


Outline Relevance Our problem: MVA Results On the cost function Heuristics for MVA An instance

Cost Functions

Monotonicity

If u, v ∈ Z p+ and u ≤ v , then 0 ≤ c(u) ≤ c(v).

Subadditivity

If u, v ∈ Z p+, then c(u ∨ v) ≤ c(u) + c(v).

Submodularity

If u, v ∈ Z p+, then c(u ∨ v) + c(u ∧ v) ≤ c(u) + c(v).

Modularity

If u, v ∈ Z p+, then c(u ∨ v) + c(u ∧ v) = c(u) + c(v).



Heuristics

Sequential Heuristics

Sequential Heuristic (Hseq): Solve a bipartite assignment problem betweenHi−1 and Vi . Let Hi be the resulting assignment for V1 × . . .× Vi ;i = 2, . . . ,m. Return Hm.

Heavy Heuristic (Hheavy ): Rearrange the sets such that c(V1) is theheaviest. Apply Hseq .

Hub Heuristics

Single-hub Heuristic (Hshub): Choose a hub h ∈ 1, . . . ,m. Solve anassignment problem between Vh and Vi (call the resulting solutions Mhi ).Construct a feasible solution by combining the solutions Mhi .

Multi-hub Heuristic (Hmhub): Apply Hshub for each possible choice of hub andoutput the best solution among all.



Example

00

01

00

10

10

01

V1 V2 V3



Example: the optimum

00

01

00

10

10

01

V1 V2 V3


Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions

Results

Results Overview

When c is monotone and subadditive: every heuristic is anm-approximation algorithm.

When c is monotone and submodular, both the sequential heuristic, aswell as the multi-hub heuristic have a worst-case ratio of 1

2 m.

When c is additive, the Heaviest-first has a better performance:ρheavy (m) ≤ 1

2 (m + 1)− 14 ln(m − 1).

WWI-3 is APX-hard.

WWI with fixed p is solvable in polynomial time.



Overview of results

Monotone • ratio:

unbounded

Monotone and Submodular • ratio: O(m/2)

Monotone and Modular (Additive) • ratio: O(m/2 – ln(m)/4)

Submodular • ratio: unbounded



Monotone and Submodular Case



Analysis of Hseq

Notation:

c(Hr ) = value of partial solution restricted to V1 × . . .Vr ,

c(Am−2,m) = value of the partial solution corresponding to an optimalassignment between Hm−2 and Vm,

c(Vi ) = total weight of the set Vi , i = 1, . . . ,m.



Analysis of Heuristic Hseq

Recall: Am−2,m = solution of optimal assignment between Vm and Hm−2

Case 1: c(Vm−1) ≤ 12 cOPT

m

c(Hm) ≤ c(Am−2,m) + c(Vm−1)

c(Am−2,m) ≤ 12

(m − 1) cOPT (W ) ≤ 12

(m − 1) cOPTm

where W = V1 × . . .× Vm−2 × Vm

c(Hm) ≤ (m − 1

2+

12

) cOPTm =

m2

cOPTm .



Analysis of Heuristic Hseq

Mm−1,m = solution of optimal assignment between Vm−1 and Vm

Case 2: c(Vm−1) ≥ 12 cOPT

m

c(Hm) ≤ c(Hm−1) + c(Mm−1,m)− c(Vm−1)

≤ m − 12· cOPT

m−1 + cOPTm − 1

2· cOPT

m

≤ (m − 1

2+

12

) cOPTm

≤ m2

cOPTm



Analysis of Heuristic Hseq: Example

Theorem

When the cost-function c is monotone and submodular, the sequentialheuristic has a performance ratio of ρseq(m) = 1

2 m. This bound is tight evenwhen the input of MVA-m is restricted to binary vectors.

Tight example

c(u) = f (∑p

i=1 ui ), where f : R→ R is defined by f (x) = x when x ≤ 2,and f (x) = 2 when x ≥ 2.

f is monotone nondecreasing and concave, and c is monotone andsubmodular.

p = n = m − 1, Vi = ei , 0, . . . , 0 for i = 1, . . . ,m, where ei is the i th

unit vector.

c(Hm) = m and cOPTm = 2.



Hardness

Theorem

WWI-3 is APX-hard even when all vectors in V1 ∪ V2 ∪ V3 are 0, 1 vectorswith exactly two nonzero entries per vector.

Sketch

L-reduction from 3-bounded MAX-3DM to WWI-3.



Binary inputs and fixed p case

Theorem

Binary MVA can be solved in polynomial time for each fixed p.

Binary MVA - MIP

A mixed integer formulation of MVA with variables:for each t = 1, . . . , 2p,

xt = number of m-tuples of type t in the assignment, .

for each i = 1, . . . ,m; j = 1, . . . , n; t = 1, . . . , 2p,

z ijt = 1 if vij is assigned to an m-tuple of type t .



Binary inputs and fixed p case

Binary MVA - MIP

min2p∑

t=1

c(bt )xt (1)∑j: bt≥vij

z ijt = xt for each t , i (2)

∑t : bt≥vij

z ijt = 1 for each j , i (3)

xt integer for each t (4)

z ijt ≥ 0 for each j , t , i . (5)

Claim: Binary MVA - MIP can be solved in polynomial time for everyfixed p.



Future work and extensions

Questions

1 What is the exact approximation ratio of the multi-hub heuristic in case ofadditive costs? We know that it lies between m/4 and m/2.

2 What is the exact approximation ratio of the heaviest-first sequentialheuristic in case of additive costs? We know that it lies between Ω(

√m)

and O(m − ln m).3 Does there exist a polynomial-time algorithm with constant (i.e.,

independent of m) approximation ratio for MVA-m?4 Can we design practical exact algorithm based on Binary MVA - MIP for

reasonable n,p and m?



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