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, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 19 Biomaterials processing by microfluidic patterning...
https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/511de50fb6460b31fd47ee2f9e4958a0_07lecture05.pdf
/ 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 21 Biomaterials processing by stencils > Stencils • Use PDMS stamps as dry resists ...
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.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 23 Material properties and coupled domains > The basic functionality of many MEMS devi...
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, have highly predictable and repeatable constitutive properties > Most microelectronic materials, however, exhibit some degree of process dependence in their material properties, especially deposited or thermally formed thin films > Some properties, like thin-film residual stress, can be wildly dependent on deposi...
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Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 28 Examples of constitutive properties...
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YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 30 Tensor properties > Properties that involve directions, either the relative directions of applied vector loads and vector responses, or the orientation of loads and/or responses relative to internal (crystalline) axes, require tensors for their specification > Exa...
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. > One course goal is acquiring domain knowledge, which gives you some insight into which material properties are important in a given situation. > Today: a sneak preview of what you might worry about > A useful resource: • A previous year’s assignment involved looking up material properties for many MEMS materials....
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Breakdown Strength > The maximum electric field that an insulating material in the gap between two flat electrodes can withstand without suffering dielectric breakdown > Depends on the size of the interelectrode gap > Importance: high voltage actuators, maximum performance > Can vary with film composition, defect den...
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wafer, in the absence of external loading. > Two flavors: • Intrinsic stress: related to structure • Thermal stress: accumulated from a change in temperature > Residual stress is a VERY VARIABLE PROPERTY, and must be measured. > Can play games, such as adjusting deposition conditions to ensure that intrinsic and t...
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Pyrex) to Si • Thermal stress in a film that is deposited at high T > CTE is tabulated, and one of the less variable material properties. Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Mass...
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ank tensor, as is the stress σ > The piezoresistive effect is described by a fourth-rank tensor ] J ⋅σ⋅Π+ρ=E [ e Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Te...
https://ocw.mit.edu/courses/6-777j-design-and-fabrication-of-microelectromechanical-devices-spring-2007/511de50fb6460b31fd47ee2f9e4958a0_07lecture05.pdf
3 Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 42 Piezoresistivity in Sili...
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idene fluoride) Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 44 Piezoelect...
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in the x or y direction = d31*Electric field in the z direction • Strain in the z direction = d33*Electric field in the z direction Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachuset...
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icients depend on exact PZT material, on underlying material, on frequency, and on electric field • - d31 is in the ballpark of 100 pC/N to several hundred pC/N • d33 is in the ballpark of several hundred pC/N to 1000 pC/N • Again, expansion along the field corresponds to contraction in the transverse direction Cite ...
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w.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. CL: 6.777J/2.372J Spring 2007, Lecture 5 - 50 The bottom line > You need to use test structures to characterize materials with variable properties > Measuring electrical properties requires electrical test devices > Measuring mechanica...
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77J/2.372J Spring 2007, Lecture 5 - 52 Membrane Load-Deflection Example > The problem: residual stress and stiffness of membranes affect their deflection under load > Example: pressure sensors d t 2a > Approach: Pressure (p) • Apply different pressures and measure resulting deflections • Fit to an energy-based model...
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Test Structures." Journal of Microelectromechanical Systems 6, no. 2 (1997): 107-118. > Weakness: surface micromachined beams often have some support compliance Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (ht...
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5)!(cid:17)(cid:9)(cid:18)(cid:10)!(cid:26)(cid:29)#$(cid:26)(cid:29)(cid:13)(cid:16)$(cid:25)(cid:5)(cid:9)!(cid:20) (cid:30).(cid:29)(cid:29).(cid:29)(cid:30)(cid:10) +++/(cid:18)(cid:14)(cid:17)(cid:20)(cid:8)(cid:4)(cid:5)(cid:17)(cid:3)(cid:4)(cid:5)(cid:7)(cid:8)(cid:18)(cid:9)(cid:23)(cid:5)/(cid:12)(cid:7)(cid:...
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(cid:1)Currently, in the UNFCCC negotiation process, the concrete environmental consequences of the various positions are not clear to all of us. There is a dangerous void of understanding of the short and long term impacts of the espoused …unwillingness to act on behalf of the Parties.(cid:2)(cid:3)(cid:4) –  Chri...
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31)(cid:6)/000(cid:20) +++/(cid:18)(cid:14)(cid:17)(cid:20)(cid:8)(cid:4)(cid:5)(cid:17)(cid:3)(cid:4)(cid:5)(cid:7)(cid:8)(cid:18)(cid:9)(cid:23)(cid:5)/(cid:12)(cid:7)(cid:6)(cid:10) 11 C-ROADS Scientific Review Panel •  Dr. Robert Watson - Department for Environment, Food and Rural Affairs (DEFRA) and former chai...
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31)(cid:10)&%(cid:17)(cid:14)(cid:13)(cid:10)(cid:17)(cid:3)(cid:4)%(cid:17)(cid:9)(cid:12)(cid:3)(cid:10) 16 Energy and Climate System Overview Economic Social Environmental effects Changes to: Capital stocks (vehicles(cid:2)) Efficiency (mpg(cid:2)) Utilization (driving(cid:2)) Transportation Changes to: ...
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Bayesian Networks Representation and Reasoning Marco F. Ramoni Children’s Hospital Informatics Program Harvard Medical School HST 951 (2003) Harvard-MIT Division of Health Sciences and Technology HST.951J: Medical Decision Support Introduction � Bayesian network are a knowledge representation formalism for reasoning...
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(B,A) is not in N. Undirected: if (A,B) is in N, then (B,A) is in N. Note: The link — should be « . A B C Characters: D E Adjacent set: the nodes one step away from A: Adj(A)={B|(A,B)˛ L}. Path: The set of n nodes Xi from A to B via links: Loop: A closed path: X1 = Xn. Acyclic graph: A graph with no cycles....
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of states p i of its parents P. A p(A) A p(A) 0.3 Y 0.3 Y 0.7 O 0.7 O E p(E) E p(E) L 0.8 0.8 L 0.2 H 0.2 H A E I I A E I A E L L Y L L Y H L Y H L Y L H Y H Y L H H Y H H Y L L O L L O H L O H O L L O H O H L O H H O H H p(I|A,E) p(I|A,E) 0.9 0.9 0.1 0.1 0.5 0.5 0.5 ...
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two variables are independent, I do not need to model their interaction but I can reason about them separately. � In this form of independence, however, a variable will nothing about another variable, by design. independence, called marginal tell me � There is another, more useful, form of independence, whic...
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� C) = p(A|C) · p(B|C) · p(C). C A B HST 951 Markov Equivalence � Different network structures may encode the same conditional independence statements: A and B are conditionally independent given C. can be encoded by 3 different network structures. � In all these network structures, the information flow running...
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0.3 0 0.7 1 B p(B) 0 1 0.6 0.4 E p(E) 0 1 0.1 0.9 A C p(C|A) 0 0 1 1 0.25 0.75 0.50 0.50 0 1 0 1 D F p(F|D) 0 0 1 1 0.80 0.20 0.30 0.70 0 1 0 1 A B D p(D|A,B) 0 0 0 0 1 1 1 1 0.40 0.60 0.45 0.55 0.60 0.40 0.30 0.70 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 ...
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0 4000 3500 3000 2500 2000 1500 1000 500 0 2 4 6 8 10 12 Decomposition Decomposition: D breaks the BBN into two BBNs: p(d)= S p(a)p(b)p(c|a)p(d|a,b)p(e)p(f|d)p(g|d,e)=. = (S p(a)p(b)p(c|a)p(d|a,b)) (S p(e)p(f|d)p(g|d,e)). Saving: We move from 64 to 23 + 23=16, and most of all the terms move from 7 to ...
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in each node. HST 951 Algorithm Input: A BBN with a set of variables X and a set of evidential statements e = {A=a,B=b,…}. Output: Conditional probability distribution p(X|e ) for each non evidential variable X. Initialization Step: Each evidential variable X, if x ˛ e p(x)=1, else p(x)=0. if x ˛ e l(x)=1, els...
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- Bayesians are not just dreamers after all. HST 951 Multiply Connected BBN When the BBN is a multiply connected graph. The associated undirected graph contains a loop. Each node does not break the network into 2 parts. Information may flow through more than one paths. Pearl’s Algorithm is no longer applicable....
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. For each node X, � a =p(x|e ) � � Compute p(x|e)= a /S cp(x|e ,c1,…,cn)p(e |c1,…,cn)p(c1,…,cn), xp(x). HST 951 S Complexity � The computational complexity is exponential in the size of the loop cutset, as we must generate and propagate a BBN for each combination of states of the loop cutset. � The identifica...
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0 1 1 1 1 0.4 0.6 0.5 0.5 0.7 0.3 0.2 0.8 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 Example � Loop cutset: {A}. � p(B=0)=p(B=0|A=0)p(A=1) + p(B=0|A=1)p(A=1). A 1.000 0.000 0 1 A 0.000 1.000 0 1 B 0.400 0.600 0 1 C 0.200 0.800 0 1 + B 0.100 0.900 0 1 C 0.500 0.500 0 1 ...
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The basic strategy (Lauritzen & Spiegelhalter 1988) is: 1. Convert a BBN in a undirected graph coding the same conditional independence assumptions. 2. Ensure the resulting graph is decomposable. 3. This operation clusters nodes in locally independent subgraphs (cliques). 4. These cliques are joint to each other ...
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C A B C Moralize 1.Marry parents 2.Drop arrows D E D E HST 951 Reading Independence � The translation method via moralization reads the conditional independence statements in BBN. � DAGs cannot encode any arbitrary set of conditional independence assumptions. I(D,A|(B,C)) I(C,B|(A,D)) A B C B D A ...
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Acceleration Structures for Ray Casting MIT EECS 6.837 Computer Graphics Wojciech Matusik, MIT EECS © ACM. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/. Hašan et al. 2007 1 Recap: Ray Tracing trace ray Intersec...
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. . Can we reduce this? 9 Today • Motivation – You need LOTS of rays to generate nice pictures – Intersecting every ray with every primitive becomes the bottleneck • Bounding volumes • Bounding Volume Hierarchies, Kd-trees For every pixel Construct a ray from the eye For every object in the scene Fi...
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dx = 0 (ray is parallel) AND Rox < X1 or Rox > X2 → no intersection y=Y2 y=Y1 Rd Ro x=X1 x=X2 (The same for Y and Z, of course) 18 Find Intersections Per Dimension • Basic idea – Determine an interval along the ray for each dimension – The intersect these 1D intervals (remember CSG!) – Done! y=Y2 y=...
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26 Find Intersections Per Dimension • Calculate intersection distance t1 and t2 y=Y2 y=Y1 t2 t1 Rd Ro x=X1 x=X2 27 Find Intersections Per Dimension • Calculate intersection distance t1 and t2 – t1 = (X1 - Rox) / Rdx – t2 = (X2 - Rox) / Rdx – [t1, t2] is the X interval y=Y2 y=Y1 t2 t1 Rd Ro x=X1...
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2 :-) 33 Is there an Intersection? • If tstart > tend → box is missed tstart tend y=Y2 y=Y1 x=X1 x=X2 34 Is the Box Behind the Eyepoint? • If tend < tmin → box is behind y=Y2 y=Y1 tend tstart x=X1 x=X2 35 Return the Correct Intersection • If tstart > tmin → closest intersection at tstart • Els...
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3,z3) = M (xmax,ymax,zmin) (x2,y2,z2) = M (xmin,ymax,zmin) (x1,y1,z1) = M (xmax,ymin,zmin) (x0,y0,z0) = M (xmin,ymin,zmin) (x'min, y'min, z'min) = (min(x0,x1,x2,x3,x4,x5,x6,x7), min(y0,y1,y2,y3,y4,x5,x6,x7), min(z0,z1,z2,z3,z4,x5,x6,x7)) 43 Bounding Box of a Transform Bounding box of transformed object...
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sphere 45 Bounding Volume Hierarchies • If ray hits bounding volume, must we test all primitives inside it? – Lots of work, think of a 1M-triangle mesh • You guessed it already, we’ll split the primitives in groups and build recursive bounding volumes – Like collision detection, remember? bounding sphere h...
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traverse – binary tree (=simple structure) • Disadvantages – may be difficult to choose a good split for a node – poor split may result in minimal spatial pruning 60 BVH Discussion • Advantages – easy to construct – easy to traverse – binary tree (=simple structure) • Disadvantages – may be difficult to ...
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& Stoll, IRT 2006 – Zhou et al., SIGGRAPH Asia 2008 Zhou et al. © ACM. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/. 68 Kd-tree Traversal - High Level • If leaf, intersect with list of primitives • If intersects b...
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,t_end) else: # case three: traverse both sides in turn t_hit = self.frontSideNode.traverse(orig, dir, t_start, t) if t_hit <= t: return t_hit; # early ray termination return self.backSideNode.traverse(orig, dir, t, t_end) 74 Important! travers(orig, dir, t_start, t_end): #adapted from Ingo Wald’s thesis #...
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(orig, dir,t_start,t_end) else: # case three: traverse both sides in turn t_hit = self.frontSideNode.traverse(orig, dir, t_start, t) if t_hit <= t: return t_hit; # early ray termination return self.backSideNode.traverse(orig, dir, t, t_end) 76 Early termination is powerful • If there is an intersection in th...
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is parallel to one axis 80 Important Details Questions? • For leaves, do NOT report intersection if t is not in [tnear, tfar]. – Important for primitives that overlap multiple nodes! • Need to take direction of ray into account – Reverse back and front if the direction has negative coordinate along the spli...
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87 Efficient Implementation • Not so easy, need ability to sort primitives along the three axes very efficiently and split them into two groups • Plus primitives have an extent (bbox) • Extra tricks include smarter tests to check if a triangle is inside a box bbox of triangle Node 88 Hard-core efficiency c...
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r 1:  P 1 8.3 1 4 re   u Lect inciples  of  Applied  Mathematics   Rodolfo  Rosales Spring  2014 an  1  now  th e r Mo Con  is Flu  q x ect v equal  sources  &  sink -­‐D.  a   w a  l ion t serva or.  Us s u  Ga e s s).  2-­‐D  or  3-­‐ s  in e  th orem  t...
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c mi  [ es v wa y  e o r  f r e  riv nto c   i t e n g roma ng  i lect ter  township.   E XAMPLE:  Heat  flow  in  2-­‐D  or  3-­‐D.   Then, ρ      =  r  c v  T  =  conserved  stuff  (heat)  per  unit  mass   Wh ere:   cv  = c e t  of  material   a ...
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thermal  diffusion.   nu   T),  and  kappa  =  heat  conductivity.   t  in  water,  sugar  in  coffee,  ink  in  water,  ETC.)   al  (S ion XAM E PLE:   ion t a  equ sion Diffu eat   uat q e  h s  a me a S placian  C   La * =  nu t   C Where C  =  con nu...
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ns:  does  stirring  help?   wh y why ? ur c c n  o o ti c s  c e nv o oe  d ating  something  with  a  flame  in  the  absence  of   e  h n e  wh n e p ul p ha d o  w at wh avity? gr At  room  temperature,  in  cm^2/sec   l  d herma T Diffu sion ity v i us f if er...
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6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 18. Consensus Consensus (sometimes called ‘reliable broadcast’ or ‘atomic broadcast’) is a fundamental building block for distributed systems. Informally, we say that several processes achieve consensus if they all agree on some ...
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input by defining some total order on the set of possible inputs.1 We have already seen one application of this replicated state machine idea, in the code for transactions; there the replication takes the form of redoing a sequence of actions that is remembered in a log. Suppose, for example, that we want to build a...
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of this handout explains a variety of ways to make a replicated state machine run efficiently: leases, transactions, and batching. Spec for consensus Here is the spec for consensus; we have seen it already in handout 8 on history and prophecy variables. The idea is that the outcome of consensus should be one and o...
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.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 MODULE Consensus [V] EXPORT Allow, Outcome = % data value to agree on MODULE TermConsensus [V] EXPORT Allow, Outcome = VAR outcome : (V + Null) := nil APROC Allow(v) = << outcome = nil => outcome := v [] SKIP >> APROC Outcome() ...
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before LateConsensus does (in the Agree action). We saw these specs in handout 8 on generalized abstraction functions, where prophecy variables are explained. In the code we have in mind, there are some processes, each with its own outcome variable initialized to nil. The outcome variables are supposed to reach cons...
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that can communicate messages from one process to another. In particular, the model must define what faults are possible. There are lots of ways to do this, and we have space to describe only the models that are most popular and closest to reality. There are two broad classes of models: • Synchronous, in which a n...
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faults: a faulty process makes arbitrary transitions; these are named after the Byzantine Empire, famous for treachery. The motivation for this model is usually not fear of treachery, but ignorance of the ways in which a process might fail. Clearly Byzantine failure is an upper bound on how bad things can be. Is co...
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+1 rounds of communication, and 2f bits of data communicated. For processors with Byzantine faults and digital signatures (so that a process can present unforgeable evidence that another process sent it a message), consensus requires f+1 processes. Even if the network is fully connected, it takes f+1 rounds to reac...
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one vote for 11 and one for 12, so you can’t tell that 12 had a majority. The Paxos algorithm: The idea In the rest of this handout, we describe Lamport’s Paxos algorithm for coding asynchronous consensus; Liskov and Oki independently invented this algorithm as part of a replicated data storage system.6 Its heart i...
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with a sloppy timeout-based algorithm for choosing a single leader. If the sloppy algorithm leaves us with no leader or more than one leader for a time, the consensus algorithm may not terminate during that time. But if the sloppy algorithm ever produces a single leader for long enough the algorithm will terminate, ...
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is easier to explain. It takes a total of 21/2 round trips for a successful round. If there’s only one leader that doesn’t fail, Paxos reaches consensus in one round. If the leader fails repeatedly, or several leaders fight it out, it may take arbitrarily many rounds to reach consensus. The rounds are numbered (not...
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a single value by having at most one leader process per round, and making the leader’s identity part of the round number. So N = [i, l], and leader l chooses (i, l) for n, where i is an I that l has not used before, for instance, a local clock. The leader keeps vn in a volatile variable; rather than resuming an old ...
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IF VAR a, v | sn a IS V}.majority a = v => RET v [*] RET nil FI For this to be an abstraction function, we need an invariant: (I1) Every successful round has the same value. Handout 18. Consensus 7 Handout 18. Consensus 8 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 I...
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to this query from a majority of agents give the leader enough information to make round n safe, as follows: It looks back from n, skipping over rounds with no V state, since these must be dead a = v (remember that the reported state is a V or no). When it comes to a round n' with sn' for some agent a, it chooses v...
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in a second round trip the leader commands everyone for round n. Each agent that is still neutral in round n (because it hasn’t answered the query of a round later than n) accepts by changing its state to vn in round n; in any case it reports its state to the leader. If the leader collects vn reports from a majority...
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’t all have to be up at the same time. Therefore we want a single leader, who runs one round at a time. If there are several leaders, the one running the biggest round will eventually Handout 18. Consensus 9 Handout 18. Consensus 10 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems...
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agents retransmit only in response to the leader’s retransmission. A process acting as a leader uses messages to communicate with the same process acting as an agent, so we describe the two roles of each process completely independently. In fact, the leader need not be an agent at all. The next section gives the al...
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outcome(a) >> = << VAR l | allowed(l) := allowed(l) \/ {v} >> THREAD LeaderActions(l) = VAR n phase reports v: (V+Null) := nil | := N{i := 1, l := l}, := idle, := S{}, % leader state (volatile except n) % last round started % leader’s phase % info about agents so far % used iff phase = commanding DO <<...
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Handout 18. Consensus 11 Handout 18. Consensus 12 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 [] % This round is dead if a majority has no state. Try another round. Dead(reports, n) => phase := idle >> OD THREAD AgentActions(a) = % State is in sa and outcomea , which ...
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v => RET v [*] RET nil FI % The value of round n according to s': if anyone has v then v else nil. ===================Useful functions for the invariants=================== % We write xl for LeaderActions(l).x to make the formulas more readable. FUNC IsTotal(le: (L, L) -> Bool) -> Bool = % Is le a total order? RE...
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% from neutral, and either to no (query) or to agree with the leader (command). a IS V | sn (ALL n | {a | s!a /\ sn a}.size <= 1) % (4) All the S's in the channel or in any reports agree with s. ( ALL s1 :IN {m | m IN UnreliableCh.q /\ m.x IS S | m.x} \/ {l | reportsl} | (ALL a, n | s1!a /\ s1(a)!n /\ s1n a # neutral...
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the value of round n, in messages, in s, or in a leader. Optimizations It’s possible to reduce the size of the leader and agent state and the messages transmitted to a few bytes, and in many cases to reduce the latency and the number of messages sent by combining rounds of the algorithm. RET {m, a, s1 | m IN Unre...
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here on or just after accepting: last next . . . vlast neutral from here on In a leader, there are two cases for reports. • • If phase = querying, reports consists of the sa’s, for rounds less than n, transmitted by a set of agents a. Hence it can be encoded as a set of ‘last state’ tuples (a, lasta, v). From ...
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run a separate instance of the algorithm for each action, and combine the query/report messages for all the actions. Note that these action numbers, which we are calling K’s, are not the same as the Paxos round numbers, the N’s; each action has its own instance of Paxos and therefore its own set of round numbers. In...
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successful outcome was never broadcast. If leaders follow the rule of not starting consensus on k+1 until a majority knows the outcome for k, then this can happen at most once. It may be convenient for a new leader to start by getting consensus on a SKIP action in order to get this complication out of the way before...
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state component. So there is a tradeoff between the cost of renewing a lease and the time you have to wait for it to expire after a (possible) failure. There are several variations: • • If you issue the lease to some known set of processes, you can revoke it provided they all acknowledge the revocation. If you k...
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to be changed. There’s no reliable way to do this without running consensus. In spite of this, leases are not completely irrelevant to updates. With a lease you can use a simple read-write memory as an agent for consensus, rather than a fancier one that can do compare-and- swap, since the lease allows you do the nece...
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(vlast, last, n) provided next <= n. A command for round n changes the state to (vn, n, n) provided next = n. So we need a representation that allows us to atomically test the current value of next and change the state in one of these ways. This is possible if an N fits in a single memory cell that CAS can read and ...
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way of doing this is to use a single master with passive agents that just implement simple memory; usually these are disk drives that record redundant copies of the log. The previous section on leases explains how to run Paxos with such passive agents. When a master fails, the new one has to sort out the consensus o...
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Massachusetts Institute of Technology Department of Materials Science and Engineering 77 Massachusetts Avenue, Cambridge MA 02139-4307 3.205 Thermodynamics and Kinetics of Materials—Fall 2006 October 26, 2006 Lecture 1: Fields and gradients; fluxes; continuity equation; entropy production; driving forces and fluxes 1. ...
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course of a spontaneous process. • The local entropy production can be expressed as a sum of terms, each of which is a product of a flux and a conjugate “force” (see KOM Eq. 2.15). • Familiar empirical laws are linear relationships between fluxes and their conjugate forces: Fourier’s law of heat conduction, Fick’s la...
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18.700 JORDAN NORMAL FORM NOTES These are some supplementary notes on how to find the Jordan normal form of a small matrix. First we recall some of the facts from lecture, next we give the general algorithm for finding the Jordan normal form of a linear operator, and then we will see how this works for small matrices...
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. . (2) (3) Since dim(V ) = n, it cannot happen that each dim(E(X−λi)k ) < dim(E(X−λi)k+1 ), for each k = 1, . . . , n. Therefore there is some least integer ei ≤ n such that E(X−λi )ei = E(X−λi)ei+1 . As was proved in class, for each k ≥ ei we have E(X −λi)k = E(X−λi)ei , and we defined the generalized eigenspace...
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. . . , Egen defined by λ1 λr Egen = {v ∈ V |∃e, (T − λiIV )e(v) = 0}, λi Date: Fall 2001. 1 (4) 2 18.700 JORDAN NORMAL FORM NOTES give a direct sum decomposition of V . Moreover, we have dim(Egen) equals the algebraic multiplicity of λi, mi. λi (B) The semisimple part S of T and the nilpotent part N of T ...
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T = T T �) iff T � commutes with both S and N . Moreover T � commutes with S iff for each i = 1, . . . , r, we have T � → (Egen) ⊂ Egen . λi λi (6) If (S�, N �) is any pair of a diagonalizable operator S� and a nilpotent operator N � such that T = S� + N � and S�N � = N �S�, then S� = S and N � = N . We call the un...
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⎜ ⎝ [S] B,B = 0m1×m2 λ1Im1 0m2×m1 λ2Im2 . . . . . . 0mr ×m1 0mr ×m1 ⎛ ⎜ ⎜ ⎜ ⎝ C (1) 0m2×m1 . . . 0m1×m2 C (2) . . . 0mr ×m1 0mr ×m2 [N ] B,B = . . . 0m1×mr . . . 0m2×mr . . . . . λr Imr . . . . . . . 0m1×mr . . . 0m2×mr . . . . . C (r) . . . . ⎞ ⎟ ⎟ ⎠ , ⎞ ⎟ ⎟ ⎟ ⎠ . (8) (9) Notice that D(i...
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0 . . . 1 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . (10) In other words, (11) Jae1 = e2, Jae2 = e3, . . . , Jaea−1 = ea, Jaea = 0. Notice that the powers of Ja are very easy to compute. In fact J a = 0a,a, and for d = 1, . . . , a − 1, we have a d e1 = ed+1, Ja Ja Notice that we have ker(Ja d e2 = ed+2, . . . , Ja d) = span(ea+1−d, e...
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⎟ ⎠ , (13) We say that a basis B(i) puts T (i) in Jordan normal form if C (i) is in Jordan normal form. , . . . , B(r) puts T in Jordan normal form if each B(i) puts T (i) � B(1) � We say that a basis B = in Jordan normal form. WARNING: Usually such a basis is not unique. For example, if T is diagonalizable, then...
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1 and ea1+1. We have already seen that a distinguishing feature of e1 is that it is an element of ker(J a1 ) which is not in ker(J a1−1). If a2 = a1, then this is also a distinguishing feature of ea1+1. But if a2 < a1, this doesn’t work. In this case it turns out that the distinguishing feature is that ea1+1 is in k...
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is as in the previous section. For each eigenvalue λi, choose any basis C for V and let A = [T ]C,C . Define B = A − λiIn. Let < ku ≤ n be the distinct integers such that there exists a nontrivial primitive 1 ≤ k1 for Gkj . Then the subspace Gkj . For each j = 1, . . . , u, choose a basis v[j]1, . . . , v[j]pj des...
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characteristic polynomial cA(X) is a quadratic polynomial. The first dichotomy is whether cA(X) has two distinct roots or one repeated root. Two distinct roots Suppose that cA(X) = (X − λ1)(X − λ2) with λ1 = λ2. Then for each i = 1, 2 we form the matrix Bi = A − λiI2. By performing Gauss­Jordan elimination we may fin...
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have � � λ1 0 0 λ2 , A = P � λ1 0 0 λ2 � P −1 . [A]B,B = Also S = A and N = 02×2. Now we consider an example. Consider the matrix � A = � . 38 −70 21 −39 The characteristic polynomial is X 2 − trace(A)X + det(A), which is X 2 + X − 12. This factors as (X + 4)(X − 3), so we are in the case discu...
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root: cA(X) = (X −λ1)2 . Again we form the matrix B1 = A − λ1I2. There are two cases depending on the dimension of Eλ1 = ker(B1). The first case is that dim(Eλ1 ) = 2. In this case A is diagonalizable. In fact, with respect to some basis B we have [A]B,B = � � . λ1 0 0 λ1 But, if you think about it, this means...
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, A = P � � λ 0 1 λ � � λ 0 1 λ P −1 . This is the one case where we have nontrivial nilpotent part: S = λ1I2 = , N = A − λ1I2 = B1 = P � � λ 0 0 λ � 0 0 1 0 � P −1 . Let’s see how this works in an example. Consider the matrix from the practice problems: � A = � . −5 −4 1 −1 The trace of A is...
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1 0 −3 , A = P � − 3 1 0 −3 � P −1 . The semisimple part is just S = −3I2, and the nilpotent part is: N = B1 = P � � 0 0 1 0 P −1 . (32) (33) (26) (27) (28) (29) (30) 18.700 JORDAN NORMAL FORM NOTES 7 3.2. Three­by­three matrices. This is basically as in the last subsection, except now there a...
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this: ⎛ 2 D = ⎝ −1 1 1 0 −1 ⎞ ⎠ . 1 − 2 3 (35) (36) Here we easily compute trace(D) = 6 and det(D) = 8. Finally to compute the coefficient t, we set c = 2 and we get det(2I2 − A) = det ⎝ −1 0 ⎛ 0 −1 1 ⎞ 1 −2 ⎠ = 0. 1 −1 (37) Plugging this in, we get (2)3 − 6(2)2 + t(2) − 8 = 0 (38) or t = 12, i.e. cA(X...
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