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the buffer of a two-machine line is not exactly the same as the behavior of the flow in a buffer of a long line. 2. prob [ni −1(t − 1) = 0 and ni (t − 1) = Ni ] ≈ 0 Question: When will this work well, and when will it work badly? 2.852 Manufacturing Systems Analysis 52/91 Copyright c(cid:13)2010 Stanley B. Gershwin....
https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf
(cid:13)2010 Stanley B. Gershwin. Examples Long lines 20 15 10 5 l e v e L r e f f u B e g a r e v A 0 0 50 Machines; r=0.1; p=0.01; mu=1.0; N=20.0 Distribution of material in a line with identical machines and buffers. Explain the shape. 10 20 30 40 50 Buffer Number 2.852 Manufacturing Systems Ana...
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1; p=0.01; mu=1.0; N=20.0 EXCEPT N(25)=2000.0 Same as Slide 55 except that Buffer 25 is now huge. Explain the shape. 10 20 30 40 50 Buffer Number 2.852 Manufacturing Systems Analysis 57/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Examples Long lines 20 15 10 5 l e v e L r e f f u B e g a r...
https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/155e83dd472fcab7095db53005c093bb_MIT2_852S10_long_lines.pdf
Gershwin. Examples Long lines 50 Machines; upstream r=0.1; p=0.01; mu=1.0; N=20.0; N(25)=2000.0 downstream r=0.09; p=0.01; mu=1.0, N=50.0 20 15 10 5 l e v e L r e f f u B e g a r e v A 0 0 Upstream same as Slide 58; downstream faster. Explain the shape. 10 20 30 40 50 Buffer Number 2.852 Manu...
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20 15 10 5 l e v e L r e f f u B e g a r e v A 0 0 Same as upstream half of Slide 61 except for Machine 26. Explain the shape. How was Machine 26 chosen? 10 20 30 40 50 Buffer Number 2.852 Manufacturing Systems Analysis 62/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Examples Long lines...
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40 50 Buffer Number 2.852 Manufacturing Systems Analysis 64/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Examples Long lines — Bottlenecks 50 Machines; r=0.1; p=0.01; mu=1.0; N=20.0 EXCEPT r(10)=0.02667 20 15 10 5 l e v e L r e f f u B e g a r e v A 0 0 Repair time bottleneck. Explain the shap...
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ru(i ) = ru(i − 1)X (i ) + ri (1 − X (i )) ru = ru X + r (1 − X ) so ru(i ) = rd (i ) = r . FRIT says pd (i −1) rd (i −1) + pu (i ) ru (i ) = E (i ) + 1 1 ei − 2 2pu r = 1 E + 1 e − 2 2.852 Manufacturing Systems Analysis 67/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Examples Infinitely long line...
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size on all buffer levels 25 20 15 10 5 l e v e L r e f f u B e g a r e v A n1 n2 n3 n4 n5 n6 n7 Continuous material model. ◮ Eight-machine, seven-buffer line. ◮ For each machine, r = .075, p = .009, µ = 1.2. ◮ For each buffer (except Buffer 6), N = 30. 0 0 5 10 15 20 N 6 25 30 35 40 45 50...
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5 B 5 M6 B6 M 7 B 7 M8 2.852 Manufacturing Systems Analysis 70/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Examples Buffer allocation Which has a higher production rate? ◮ 9-Machine line with two buffering options: ◮ 8 buffers equally sized; and M1 B1 M 2 B 2 M 3 B 3 M4 B4 M 5 B 5 M...
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0 5000 5000 6000 6000 6000 6000 7000 7000 7000 7000 8000 8000 8000 8000 9000 9000 9000 9000 10000 10000 10000 10000 Total Buffer Space ◮ Continuous model; all machines have r = .019, p = .001, µ = 1. ◮ What are the asymptotes? ◮ Is 8 buffers always faster? 2.852 Manufacturing Systems Analysis 72/91 Copyrigh...
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Blockage and starvation may be caused by differences in machine speeds, not only failures. ◮ Decomposition of these classes of systems is similar to that of discrete-material, deterministic-processing time lines except ◮ The two-machine lines have machines with 3 parameters (ru (i), pu(i), µu(i); rd (i), pd (i), µd...
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, t + δt), for small δt; pd (i)δt = probability that Md (i) goes from up to down in (t, t + δt) if it is not starved, for small δt; µd (i)δt = probability that a piece flows out of Bi in (t, t + δt) when Md (i) is up and not starved, for small δt. 2.852 Manufacturing Systems Analysis 75/91 Copyright c(cid:13)20...
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δt = prob αu(i; t + δt) = 0 (cid:20) or, αu (i; t) = 1 and ni (t) < Ni (cid:12) (cid:21) (cid:12) (cid:12) (cid:12) , pu(i)δt = prob Mu(i) down at t + δt (cid:20) Mu(i) up and ni < Ni at t (cid:12) (cid:12) (cid:12) (cid:12) . (cid:21) 2.852 Manufacturing Systems Analysis 77/91 Copyright c(cid:13...
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) = pi + ru (i − 1)p(i − 1; 001) Eu(i) . pd (i) = pi +1 + rd (i + 1)p(i + 1; N10) Ed (i) . in which p(i − 1; 001) is the steady state probability that line L(i − 1) is in state (0, 0, 1) and p(i + 1; N10) is the steady state probability that line L(i + 1) is in state (Ni +1, 1, 0). 2.852 Manufacturing System...
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Copyright c(cid:13)2010 Stanley B. Gershwin. Long Lines — Exponential Processing Time Model Conservation of Flow P(i ) = P(1), i = 2, . . . , k − 1. 2.852 Manufacturing Systems Analysis 81/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Long Lines — Exponential Processing Time Model Flow Rate/Idle Time T...
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· · , k − 2. 2.852 Manufacturing Systems Analysis 83/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Long Lines — Exponential Processing Time Model Boundary Conditions Md (1) is the same as M1 and Md (k − 1) is the same as Mk . Therefore ru(1) = r1 pu(1) = p1 µu(1) = µ1 rd (k − 1) = rk pd (k − 1) = pk µd (...
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Conceptually very similar to exponential processing time model. One difference: ◮ prob (xi −1 = 0 and xi = Ni ) = 0 exactly . 2.852 Manufacturing Systems Analysis 86/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Long Lines — Continuous Material Model New approximation ◮ New approximation: The observer sees bo...
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ershwin. Long Lines — Continuous Material New approximation M i−2 B i−2 M i−1 B i−1 M i B i M i+1 B i+1 M i+2 B i+2 M i+3 M (i) u M (i) d Assume that ... < µi −2 < µi −1 < µi < µi +1 < .... Assume all the machines are up and Bi is not full. Then the observer in Bi actually sees mater...
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conditions. They are the same as in the exponential processing time case except for the Interruption of Flow equations. 2.852 Manufacturing Systems Analysis 89/91 Copyright c(cid:13)2010 Stanley B. Gershwin. Long Lines — Continuous Material Interruption of Flow Considerable manipulation leads to pu (i) = pi ...
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come ◮ Assembly/Disassembly Systems ◮ Buffer Optimization ◮ Effect of Buffers on Quality ◮ Loops ◮ Real-Time Control ◮ ???? 2.852 Manufacturing Systems Analysis 91/91 Copyright c(cid:13)2010 Stanley B. Gershwin. For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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MIT OpenCourseWare http://ocw.mit.edu 18.014 Calculus with Theory Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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Chapter 1 Introduction This course will be organized around algorithmic issues that arise in machine learn­ ing. The usual paradigm for algorithm design is to give an algorithm that succeeds on all possible inputs, but the difficulty is that almost all of the optimization problems that arise in modern machine learnin...
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. Note: These are unpolished, incomplete course notes. Developed for educational use at MIT and for publication through MIT OpenCourseware. 3 4 CHAPTER 1. INTRODUCTION This course will focus on (a) nonnegative matrix factorization (b) topic modeling (c) tensor decompositions (d) sparse recovery (e) dictionary...
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3.37 (Class 3) Review: The inherent strength of all bonds (even van der Waals) is extremely high Primary (1-3eV) (cid:198) 1,000,000 – 3,000,000 psi van der Waals (0.1-0.2eV) (cid:198) 100,000 – 200,000 psi graph of energy vs. distance graph of force vs. distance interatomic distance can get bulk compressibilit...
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Can think of as either as energy or force Cleaving a material in a vacuum, then if stick it back together (no contamination) will bond with much of its original strength • Time to form a monolayer • Langmuir on order of 10^-8 atm*sec • In an ultrahigh vacuum, this time is much longer, then the only issue is surfa...
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with a punch since need to push aside material on sides (drawn as slip line flow field, volume is about 3 times) • See handouts for formulas o Macroscopic (apparent area) o Microscopic (true area of contact) o Only get about 30% contact area • Can’t solve surface roughness by just squeezing things together (even ...
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MATH 18.152 COURSE NOTES - CLASS MEETING # 1 18.152 Introduction to PDEs, Fall 2011 Professor: Jared Speck Class Meeting # 1: Introduction to PDEs 1. What is a PDE? We will be studying functions u = u(x1, x2, · · · , xn) and their partial derivatives. Here x1, x2, · · · , xn are standard Cartesian coordinates on Rn. We...
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· , n} for some function F. Here N is called the order of the PDE. N is the maximum number of derivatives appearing in the equation. Example 1.0.1. u = u(t, x) (1.0.3) is a third-order nonlinear PDE. Example 1.0.2. u = u(t, x) (1.0.4) is a second-order linear PDE. −∂2u + (1 + cos u)∂3 t xu = 0 −∂2 t u + 2∂2 xu + u = t ...
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. In this course, we will discuss some important physical systems and the PDEs that are commonly used to model them. Now let’s assume that we have a PDE that we believe is a good model for our system of interest. Then most of the time, the primary goals of PDE are to answer questions such as the following: (1) Does the...
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, first-order, nonlinear, homogeneous MATH 18.152 COURSE NOTES - CLASS MEETING # 1 E = (cid:0)E1(x, y, z), E2(x, y, z), E3(x, y, z)(cid:1), B = (cid:0)B1(x, y, z), B2(x, y, z), B3(x, y, z) 3 (cid:1) are vectors in R3 (3.0.7a) (3.0.7b) ∂tE − ∇ × B = 0, ∂tB + ∇ × E = 0, ∇ · E = 0, ∇ · B = 0 “Maxwell’s equations” in a vac...
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NOT define a linear operator: y (u + v) = ∂xu + u ∂y u + ∂xv + v ∂y v = Lu + Lv 2 2 2 2 2 2 Definition 4.0.3. A PDE is linear if it can be written as (4.0.9) Lu = f (x1, · · · , xn) for some linear operator L and some function f of the coordinates. Definition 4.0.4. If f = 0, then we say that the PDE is homogeneous. Other...
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let uI be a “fixed” solution to the inhomogeneous linear PDE (4.0.14) Lu = f (x1, · · · , xn). Then the set SI of all solutions to (4.0.14) is the translation of SH by uI : SI = {uI + uH | uH ∈ SH}. Proof. Assume that LuI = f, and let w be any other solution to (4.0.14), i.e., Lw = f. Then , and so w ∈ SI L(w − uI) = f ...
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) and v is the vector (a, b) ∈ R2. Thus, the derivative of says that ∇u · def MATH 18.152 COURSE NOTES - CLASS MEETING # 1 5 u in the direction (a, b) is 0, which implies that u is constant along lines pointing in the direction of (a, b). The slope of such a line is b . Therefore, every such line can be described as t...
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as V. This vector can be viewed as a line segment with slope x.y Therefore, if the curve C is parameterized by x → (x, y(x)) (where we are viewing y as a function of x along C) then C has slope dy , and y is therefore a solution to the following ODE: dx (6.2.2) dy dx x = . y We can use the following steps to integrate ...
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integer k ≥ 0, we define the C k norm of f on Ω by (7.1.1) (cid:107)f (cid:107) def Ck(Ω) = k (cid:88) a=0 sup |f (a)(x) , | ∈Ω x where f (a)(x) is the ath order derivative of f (x). We often omit the symbol Ω when Ω = R. Example 7.1.1. (7.1.2) (cid:107) sin(x)(cid:107)C7(R) = 8. The same notation is used in the case th...
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,2 refers to the t coordinate, while the “2” refers to the x coordinate. The next definition provides a very important example of another class of norms that are prevalent in PDE theory. Definition 7.1.2 (Lp norms). Let 1 ≤ p < ∞ be a number, and let f be a function defined on a domain Ω ⊂ Rn. We define the Lp norm of f by...
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lot of PDE results are derived using integration by parts (some- times very fancy versions of it), which provides us with integral identities. This will become more apparent as the course progresses. Let’s recall a very important version of integration by parts from vector calculus: the divergence theorem. We first need...
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3 can locally be described as the graph of a function φ(x, y) (e.g., ∂Ω = {(x, y, z) |z = φ(x, y)}), then (7.2.4) (cid:112) dσ = 1 + |∇φ(x, y)|2dxdy, ∇ φ = (∂xφ, ∂yφ) is the gradient of where of ∇φ. φ, and |∇φ| = (cid:112)(∂xφ)2 + (∂yφ)2 is the Euclidean length def Remark 7.2.1. The divergence theorem holds in all dime...
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1 2 3 4 5 6 7 8 MIT OpenCourseWare http://ocw.mit.edu 6.890 Algorithmic Lower Bounds: Fun with Hardness Proofs Fall 2014 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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Elasticity (and other useful things to know) Carol Livermore Massachusetts Institute of Technology * With thanks to Steve Senturia, from whose lecture notes some of these materials are adapted. Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spri...
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beiz, Gabriel M. Hoboken, NJ: John Wiley, 2003. I RF MEMS: Theory, Design, and Technology. SBN: 9780471201694. Image by MIT OpenCourseWare. AFM cantilevers Courtesy of Veeco Instruments, Inc. Used with permission. Switches Zavracky et al., Int. J. RF Microwave CAE, 9:338, 1999, via Rebeiz RF MEMS Cite as: Carol Live...
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Spring 2007, Lecture 6 - 4 Why things have stiffness I Unloaded beam is undeformed: Axially loaded beam is stretched: Stretching costs energy, which is stored as elastic energy. Exactly how much energy is determined by material and geometry. Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and ...
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ite 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]. C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 7 Elasticity > Elast...
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, σy, or σz > Compressive: σ < 0 > Tensile: σ > 0 > Shear stress τxy, τxz, or τyz y y z z x σx x τxy σx τxy 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 Technolo...
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1, y1 x4, y4 Adapted from Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer Academic Publishers, 2001. ISBN: 9780792372462. Image by MIT OpenCourseWare. 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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u x ε x = ( xu x ) x −Δ+ x Δ )( xu x = u ∂ x x ∂ x+ux(x) x+Δx + ux(x+Δx) 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]. ...
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proportional to stress • Most things start out linear > Elastic vs. plastic • Elastic: deformation is recovered when the load is removed • Plastic: some deformation remains when unloaded > Isotropic vs. anisotropic • Life is simpler when properties are the same in all directions; however, anisotropic silicon is a par...
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of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 18 Linear Elasticity in Isotropic Materials > Poisson ratio, ν • Some things get narrower in the transver...
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Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 20 Isotropic Linear Elasticity > For a general case of loading, the constitutive r...
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more: 6.777J/2.372J Spring 2007, Lecture 6 - 22 Plane stress > Special case: when all stresses are confined to a single plane Often seen in thin films on substrates (will discuss origin of these stresses shortly) > Zero normal stress in z direction (σz = 0) > No constraint on normal strain in z, εz ε x = ε y = ε z =...
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.372J Spring 2007, Lecture 6 - 24 Stresses on Inclined Sections > Can resolve axial forces into normal and shear forces on a tilted plane q F F F F F F FN FV FV FN F F Adapted from Figure 9.3 in: Senturia, Stephen Kluwer Academic Publishers, 2001, p. 205. ISBN: 9780792372462. Image by MIT OpenCourseWare. D. Microsys...
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Ware (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 26 Special case: biaxial stress > A special case of plane stress • Stresses σx and σy along principal axes are equal • Strains εx and εy along principal axes are equa...
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J/2.372J Spring 2007, Lecture 6 - 28 Types of strain > What we have just talked about is elastic strain • Strains caused by loading; returns to undeformed configuration when load is removed • Described by the isotropic equations of linear elasticity > There are other kinds of strain as well • Thermal strain, which is ...
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ermally Induced Residual Stress > If a thin film is adhered to a substrate, mismatch of thermal expansion coefficient between film and substrate can lead to stresses in the film (and, to a lesser degree, stresses in the substrate) > The stresses also set up bending moments • You care about this if you don’t want you...
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Month YYYY]. C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 32 Intrinsic residual stress > Any thin film residual stress that cannot be explained by thermal expansion mismatch is called an intrinsic stress > Sources of intrinsic stress • Deposition far from equilibrium • Secondary grain growth can modify stress...
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> Using this to find the stiffness of structures 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]. C. Livermore: 6.777J/2.3...
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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]. C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 37 Including Shear...
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6.777J/2.372J Spring 2007, Lecture 6 - 39 Stiffness and Compliance > The matrix of stiffness coefficients, analogous to Young’s modulus, are denoted by Cij > The matrix of compliance coefficients, which is the inverse of Cij, is denoted by Sij > Yes, the notation is cruel > Some texts use different symbols, but t...
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�� ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 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]. C. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 41 M...
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) s t i n u y r a r t i b r a ( s s e r t S Loading curve Unloading curve Strain if unloaded to zero stress Stress if unloaded to zero strain 0 1 2 4 Strain (arbitrary units) 3 5 6 Adapted from Figure 8.8 in: Senturia, Ste Kluwer Academic Publishers, 2001, p. 198. ISBN: 9780792372462. phen D. Microsystem Design. Bosto...
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. Livermore: 6.777J/2.372J Spring 2007, Lecture 6 - 45 Any thoughts on this device? Figures 2, 3, and 4 on pp. 236-237 in: Kinoshita, H., K. Hoshino, K., K. Matsumoto, and I. Shimoyama. "Thin Compound eye Camera with a Zooming Function by Reflective Optics." In MEMS 2005 Miami: 18th IEEE International Conference o...
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position) Geometry : σ = F A = F WH and ε = L Δ L stress : Uniaxial σ εE = F WH F = = E L Δ L EWH L Δ L LkF =⇒Δ= k Plug in for L=100 μm, W=5 μm, H=1 μm, E=160 GPa: EWH L k=8000 N/m Cite as: Carol Livermore, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT ...
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6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 23. Networks — Links and Switches1 This handout presents the basic ideas for transmitting digital data over links, and for connecting links with switches so that data can pass from lots of sources to lots of destinations. You may ...
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are: Parallel busses, as in the first four lines of the table. Switched networks: the telephone system and switched LANs. Multiple disks, each holding part of a data block, that can transfer in parallel. Cellular telephony, using spatial separation to reuse the same frequencies. In the latter two cases the paralle...
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memory packet 16 32/64 DMA block 1 1 1 16 1 1 packet packet ? 32 ? packet, 64-1500 B 1 1 1 µs/km 1 µs/km 1 µs/km 1 µs/km 1 µs/km 6 MHz byte or cell packet, 64-1500 B packet, 64-1500 B packet, < 1500 B byte or 48 B cell byte byte byte Flow control Many links do not have a fixed bandwidth t...
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8 (Aug. 1996), pp 63-73. Handout 23. Networks — Links and Switches 1 Handout 23. Networks — Links and Switches 2 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 increasing factor is 2, it is ‘binary exponential backoff’. It is used in the Ethernet6 and in TCP7, and is ana...
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it requires all the senders to cooperate. A sender who doesn’t play by the rules can get an unfair share of the link resource, and in many cases two such senders can cause the total throughput of the entire link to become very small. Backpressure In backpressure the link tells the sender explicitly how much it can ...
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pp 40-48. 9 M. Schroeder et al., Autonet: A high-speed self-configuring local area network using point-to-point links. IEEE Journal on Selected Areas in Communication 9, 8 (Oct. 1991), pp 1318-1335. The failure modes of the two backpressure schemes are different. A lost ‘stop’ may cause lost data. A lost credit may ...
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free to discard excess traffic. The rate is measured by taking a moving average across some time window.10 Framing The idea of framing (sometimes called ‘acquiring sync’) is to take a stream of X’s and turn it into a stream of Y’s. An X might be a bit and a Y a byte, or an X might be a byte and a Y a packet. This ...
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recognize the start of a Y no matter what the previous sequence of X’s has been. Certain values of X can be reserved to mark the beginning or the end of a Y. In FDDI12, for example, 4 bits of data are coded in 5 bits on the wire (this is called a 4/5 code). This is done because the wire doesn’t work if there are to...
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detected probabilistically. In telephony T-1 signaling there is a ‘frame’ of 193 bits, one sync bit and 192 data bits. The data bits can be arbitrary, but they are xored with a ‘scrambling’ sequence to make them pseudo-random. The encoding specifies a definite pattern (say “010101”) for the sync bits of successive f...
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channels. There are three main reasons for multiplexers: • Traffic may flow between one node and many on a single wire, for example when the one node is a busy server or the head end of a cable TV system. • One wide wire may be cheaper than many narrow ones, because there is only one thing to install and maintain,...
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of cellular telephony. Figure 2 illustrates the simplest form of CDM, in which n senders share a digital channel. Bits on the channel have length 1, each sender’s bits have length n (5 in the figure), and a sender has an n-bit ‘code’ (10010 in the figure) which it xor’s with its current data bit. The receiver xor’s ...
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of the sub-channel appears explicitly in the packet. If the packets are fixed size, they are often called ‘cells’, as in the Asynchronous Transfer Mode (ATM) networking standard. Fixed-size packets are used in other contexts, however, for instance to carry load and store messages on a programmed I/O bus. Variable si...
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as in a daisy-chained bus or a token ring like FDDI. Flow control means buffering, as we saw earlier, and there are several ways to arrange buffering around a multiplexer, shown on the left side of figure 1. Having the buffers near the arbitration point is good because it reduces the round-trip time r and hence the...
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input channel must select the proper output channel to receive each sub-channel. Again both centralized and distributed versions are possible, as the right side of figure 1 shows. A distributed implementation broadcasts the input channel to all the output channels, and an address decoder picks off the sub-channel a...
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there is a highly reliable place to put the central arbiter: at the head end (or, in a fiber-to-the-neighborhood system, in the fiber-to- coax converter. And by measuring the round-trip delays between the head end and each node, the head end can broadcast “node n can make its request now” messages with timing which en...
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node starts an election by broadcasting its ID. When a node receives the ID of another node, it forwards it unless its own ID is larger, in which case it sends its own ID. When a node receives its own ID, it becomes the leader; this works because every other node has seen the leader’s ID and determined that it is l...
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contains much of the hardware needed for the switches discussed in the next lecture, but doesn’t provide any of the performance gains that switches do. Fig. 4: A ring with a hub tolerates multiple failures In spite of these problems, two token rings are in wide use (though much less wide than Ethernet, and rapidly...
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can’t hear each other: the head end sends a collision signal whenever it hears more than one sender. Contention slots packet packet idle packet packet Contention interval time The critical parameter for a ‘CSMA/CD’ (Carrier Sense Multiple Access/Collision Detection) network like the Ethernet is the round-t...
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826—Principles of Computer Systems 2002 introduce their own crazy ideas. It’s much faster and safer to leave in the unused features. In any case, the logic for CSMA/CD must be in the chips so that they can run at the slower speeds as well, in order to ensure that the network will still work no matter how it’s wired...
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it smaller by resetting its backoff time to 0 after each successful transmission. Of course each node must chose its actual backoff time randomly in the interval [0 .. maximum backoff]. As long as all the nodes obey the rules, they share the medium fairly, with one exception: if there are very few nodes, say two, an...
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(contrary to the expectations of the Ethernet’s designers). An Ethernet hub just repeats an incoming signal to all the nodes. Hub wiring has three big advantages: It’s easier to run Ethernet wiring in parallel with telephone wiring, which runs to a hub. The hub is a good place to put sensors that can measure traffic...
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is the same, and there is much more aggregate bandwidth. Furthermore, a switch can have a number of slow ports and a few fast ones, which is exactly what you want to connect a local group of clients to a higher bandwidth ‘backbone’ network that has more global scope. In the rest this handout we describe the differe...
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IP router Email router SMTP Copper pair Central office Bandwidth 60 122 4 GB/s GB/s GB/s Latency Links .8 1 5-100 ns 6 µs 2K µs 32 275 10–400 µs 22 MB/s MB/s 100–1200 µs 32 1-6400 MB/s 50–5000 µs 16 10-1000 KB/s 80 MB/s 125 1-100 s many µs 50K Handout 23. Networks — Links and Switches 15 ...
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holes but without interleaving, each point where a link feeds a slower one adds the difference in the time a unit spends entering them; where a link feeds a faster one there is no added time because the faster link gobbles up the unit as fast as the slower one can deliver it. 3 3 2 concatenate 4 2 Fig. 7. Compos...
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si – Σ tfi = Σ (tsi – tfi ) m ax ti The moral is to use either wormholes or small units, and to watch out for alternating fast and slow links if you don’t have interleaving. However, a unit shouldn’t be too small on a variable TDM link because it must always carry the overhead of its address. Thus ATM cells, with 4...
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The new mechanism we need to make this work is routing, which converts an address into a ‘path’, a sequence of decisions about what output link to use at each switch. Routing is done with a map from addresses to output links at each switch. In addition the address may change along the path; this is coded with a seco...
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extend a broadcast LAN transparently, however, great care has to be taken in changing routes to preserve the FIFO property, even though it has very little value to most clients. This use of switching is called ‘bridging’. Addresses There are three kinds of addresses. In order of increasing cost to code the maps, an...
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traffic for other nodes to the left or right, and this will drain out in time. Handout 23. Networks — Links and Switches 19 Handout 23. Networks — Links and Switches 20 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 2 1 4 1 3 4 3 2 Fig. 9. Deadlock. The version on...
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mesh networks such as AN222 have done so, as do modern multiprocessor interconnects. Topology In the remainder of the handout, we study mechanisms for routing in more detail.23 It’s convenient to divide the problem into two parts: computing the topology of the network, and making routing decisions based on some to...
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each N P = [n, r: SEQ L] WITH {"<=":=Prefix} % Path starting at n Here t(n)(l) is the node reached from node n on link l. For the network of figure 8, t(3)(a) = 1 t(3)(d) = 4 t(1)(a) = 3 t(1)(b) = 5i etc. Note that a T is defined on every node, though there may not be any links from a node. The End function compu...
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p.n = n1 /\ End(t, p) = n2 /\ IsAcyclic(t, p)} Handout 23. Networks — Links and Switches 21 Handout 23. Networks — Links and Switches 22 6.826—Principles of Computer Systems 2002 6.826—Principles of Computer Systems 2002 Like anything else in computing, a network can be recursive. This means that a connected ...
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)) How does a network find out what its topology is? Aside from supplying it manually, there are two approaches. In both, each node learns which nodes are ‘neighbors’, that is, are connected to its links, by sending ‘hello’ messages down the links. 1. Run a global computation in which one node is chosen to learn th...
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. Otherwise a router that can reach n2 returns the router’s LAN address. In the Internet this is done by the address resolution protocol (ARP). Of course n1 caches this result and times out the cache periodically. The Autonet paper describes a variation on this, in which end-nodes use an ARP protocol to map Ethern...
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