multidimensional assignment problems for semiconductor...
TRANSCRIPT
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
Outline Relevance Our problem: MVA Results
About merging vectors
Our problem - a prologue
Let u = (12 91 7), and v = (47 32 12).
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results
About merging vectors
Our problem - a prologue
Let u = (12 91 7), and v = (47 32 12).
How do we merge u and v?
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results
About merging vectors
Our problem - a prologue
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+.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results
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)
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results
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).
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results
1 Relevance
2 Our problem: MVAOn the cost functionHeuristics for MVAAn instance
3 ResultsAnalysis of Heuristics
Monotone and Submodular Case
HardnessPolynomial Special caseQuestions
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
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
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Wafer-to-wafer integration
Stacking wafers
Stacking
From lot 1 From lot 2 From lot 3
Stack
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Wafer-to-wafer integration
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)
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Wafer-to-wafer integration
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
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Wafer-to-wafer integration
Yield optimization: an example
Stack 1
Stack 2 Total number of bad dies in stack 1 + stack 2 = 23
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Wafer-to-wafer integration
Yield optimization: an example
Stack 1
Stack 2 Total number of bad dies in stack 1 + stack 2 = 17
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Wafer-to-wafer integration
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Wafer-to-wafer integration
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).
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
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).
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results On the cost function Heuristics for MVA An instance
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results On the cost function Heuristics for MVA An instance
Example
00
01
00
10
10
01
V1 V2 V3
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results On the cost function Heuristics for MVA An instance
Example: the optimum
00
01
00
10
10
01
V1 V2 V3
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
Monotone and Submodular Case
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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 .
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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 .
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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.
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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?
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
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Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
MAPSP2015 takes place in La Roche, 2015, June 8 - June 12
see www.mapsp2015.com
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
MAPSP2015 takes place in La Roche, 2015, June 8 - June 12
see www.mapsp2015.com
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP
Outline Relevance Our problem: MVA Results Analysis of Heuristics Hardness Polynomial Special case Questions
THANKS!
Trivikram Dokka, Yves Crama, Frits Spieksma MAPSP