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Foundations of Program Analysis Node find (Node prev, Node cur, int key) { while (cur.key < key) { prev = cur; cur = cur.next; } return cur; } 6.820 Armando Solar-Lezama Course Staff • Armando Solar-Lezama – Instructor L01-2 What this course is about The top N good ideas in programming langu...
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pieces to analyze separately) L01-7 Skills • Haskell • Coq • Ocaml • Spin L01-8 Grading o 6 homework assignments - Each is 15-20% of your grade - start on them early! L01-9 6 Homework Assignments o Pset 1 (out now, due in about 2 weeks!) - Practice functional programming - Build some Lambda Calc...
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6 + 9  15 The final answer did not depend upon the order in which reductions were performed September 9, 2015 L01-13 Confluence Informally - The order in which reductions are performed in a Functional program does not affect the final outcome This is true for all functional programs regardless whe...
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w in x + y + z let y = 2 * 2 x = 3 + 4 z = let x’ = 5 * 5 w = x’ + y * x’ in w in x + y + z September 9, 2015 L01-18 Lexical Scoping and -renaming plus x y = x + y plus' a b = a + b plus and plus' are the same because plus' can be obtained by systematic renaming of bounded iden...
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(x) from a to b using trapezoidal rule dx a x b Integral(a,b) = (f(a + dx/2) + f(a + 3dx/2) + ...) dx September 9, 2015 L01-22 Local Function Definitions Free variables of sum? integrate dx a b f = let sum x tot = if x > b then tot else sum (x+dx) (tot+(f x)) in (sum (a+dx/2) 0) * dx i...
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L01-26 l-Abstraction Lambda notation makes it explicit that a value can be a function. Thus, (plus 1) can be written as \y -> (1 + y) (In Haskell \x is a syntactic approximation of lx) plus x y = x + y can be written as plus = \x -> \y -> (x + y) or as plus = \x y -> (x + y) September 9, 2015 ...
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? 2. twice (append “Zha") “Gabor"  “ZhaZhaGabor” twice :: (Str  Str)  Str  Str ? September 9, 2015 L01-31 Deducing Types twice f x = f (f x) What is the most "general type" for twice? 1. Assign types to every subexpression x :: t0 f :: t1 f x :: t2 f (f x) :: t3  t...
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2015 L01-33 Now for some fun twice f x = f (f x) a = twice1 (twice2 succ) 4 b = twice3 twice4 succ 4 1. Is a=b ? yes succ (succ (succ (succ 4) 2. Are the types of all the twice instances the same? no twice1 :: (I -> I) -> I -> I twice2 :: (I -> I) -> I -> I twice3 :: ((I -> I) -> I -> I) -> (...
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Optimality Conditions for Constrained Optimization Problems Robert M. Freund February, 2004 2004 Massachusetts Institute of Technology. 1 1 Introduction Recall that a constrained optimization problem is a problem of the form (P) minx f (x) s.t. g(x) ≤ 0 h(x) = 0 x ∈ X, where X is an open set and g(x)) = (...
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2 2 Necessary Optimality Conditions 2.1 Geometric Necessary Conditions A set C ⊆ (cid:4)n is a cone if for every x ∈ C, αx ∈ C for any α > 0. A set C is a convex cone if C is a cone and C is a convex set. Suppose ¯x ∈ S. We have the following definitions: • F0 := {d : ∇f (¯x)td < 0} is the cone of “improving” dir...
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λ > 0 sufficiently small gi(¯ x) = 0 for all i ∈ I (for i (cid:10) x + λd) = x + λd) < 0), and h(¯ x + λd ∈ S for all λ > 0 sufficiently small. (A¯ On the other hand, for all sufficiently small λ > 0, f (¯ x). This contradicts the assumption that ¯x is a local minimum of (P). x − b) + λAd = 0. Therefore ¯ ∈ I, since λ is...
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d which is an improving direction but ∇f (¯x)td = 0 and/or ∇g(¯x)td = 0. x) = 0 for i ∈ I and h(¯ x + λd) ≤ gi(¯ The proof of Theorem 2 is rather awkward and involved, and relies on the Implicit Function Theorem. We present this proof at the end of this note, in Section 6. 2.2 Separation of Convex Sets We will sh...
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H + and T ⊆ H − . If, in addition, S ∪ T (cid:10)⊂ H, then H is said to properly separate S and T . n t • H is said to strictly separate S and T if ptx > α for all x ∈ S and ptx < α for all x ∈ T . • H is said to strongly separate S and T if for some (cid:1) > 0, p x ≥ α + (cid:1) t for all x ∈ S and p x ≤ α − (ci...
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y(cid:15)}. Then S ¯ is a compact set. Let f (x) = (cid:15)x − y(cid:15). Then f (x) attains its minimum over the set S ¯ at some point ¯ x ∈ S. Note that ¯ = y. x (cid:10) ¯ To show uniqueness, suppose that there is some x(cid:2) ∈ S for which (cid:15)y − 2 (¯x + x(cid:2)) ∈ S. But by the triangle . By convexity ...
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0 for all x ∈ S. To establish sufficiency, note that for any x)t(x − ¯ (y − ¯ x ∈ S, (cid:15)x−y(cid:15) 2 = (cid:15)(x−¯ x)−(y−¯ 2 x)(cid:15) = (cid:15) x−¯x(cid:15) 2+(cid:15)y−¯ x(cid:15)2−2(x−¯ x)t(y−¯ x) ≥ (cid:15)¯ 2 x−y(cid:15) . Conversely, assume that x is the minimizing point. For any x ∈ S, λx + ¯ 5 (1 −...
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by choosing λ > 0 and sufficiently small. ¯ ¯ Proof of Theorem 3: Let ¯x ∈ S be the point minimizing the distance from the point y to the set S. Note that ¯ (cid:10) x), and (cid:1) = 1 (cid:15)y − ¯ 2 1 x = y. Let p = y −x, α = 2 (y −x)t(y + ¯ ¯ x) ≤ 0, and so x)t(y − ¯ x(cid:15)2 . Then for any x ∈ S, (x − ¯ ¯ p ...
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and S2 can be strongly separated by a hyper- plane. Proof: Let T = {x ∈ (cid:4) : x = y − z, where y ∈ S1, z ∈ S2}. Then it is easy to show that T is a convex set. We also claim that T is a closed set. n 6 ∞ i=1 ⊂ T , and suppose ¯ To see this, let {xi} x = limi→∞ xi. Then xi = yi − zi for {yi}∞ ⊂ S1 and {zi}∞ ...
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2. Then x = y − z ∈ T , and so pt(y − z) > ¯α > 0 for any y ∈ S1 and z ∈ S2. Let α1 = inf{p y : y ∈ S1} and α2 = sup{ptz : z ∈ S2} (note that α ≤ α1 − α2); define α = 2 (α1 + α2) and (cid:1) = α > 0. Then for all 1 2 ¯ 1 t 0 < ¯ y ∈ S1 and z ∈ S2 we have p t y ≥ α1 = 1 2 (α1 + α2) + 1 (α1 − α2) ≥ α + α = α + (c...
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Also, S is a closed set. (For an exact proof of this, see Appendix B.3 of Nonlinear Programming by Dimitri Bertsekas, Athena Scientific, 1999.) Therefore there exist p and α such that ctp > α and pt(A y) = (Ap)ty ≤ α for all y ≥ 0. t 7 If (Ap)i > 0 for some i, one could set yi sufficiently large so that (Ap)ty > α,...
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0, . . . , 0, 1) · θ (cid:8) (cid:9) x θ > 0. From Farkas’ Lemma, there exists a vector (u; v; w1; w2) ≥ 0 such that ⎛ ⎤ ¯ A e ⎛ ⎞ ⎡ ⎞ 0 ⎟ ⎜ . ⎜ . ⎟ ⎟ ⎢ ⎢ B 0 ⎥ ⎜ v ⎟ ⎟ = ⎜ . ⎟ . ⎢ ⎜ ⎟ ⎣ H 0 ⎦ ⎝ w ⎠ ⎝ 0 ⎠ w t ⎥ ⎜ ⎥ · ⎜ −H 0 1 2 u 1 This can be rewritten as ¯ At u + Bt v + H t(w 1 − w 2) = 0, t e...
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the first equation can be rewritten as u0∇f (¯ x) + ∇g(¯ x)t u + ∇h(¯ x)t v = 0 .) Proof: If the vectors ∇hi(¯ such that ∇h(¯x)tv = 0. Setting (u0, u) = 0 establishes the result. x) are linearly dependent, then there exists v (cid:10) = 0 Suppose now that the vectors ∇hi(¯x) are linearly independent. Then we can ap...
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. , up) ≥ 0. Define up+1, . . . , um = 0. Then (u0, u) ≥ 0, (u0, u) (cid:10) x) = 0, or ui = 0. Finally, = 0, and for any i, either gi(¯ u0∇f (¯ x) + ∇g(¯ x)t u + ∇h(¯ x)t v = 0. ¯ Theorem 11 (Karush-Kuhn-Tucker (KKT) Necessary Conditions) Let x be a feasible solution of (P) and let I = {i : gi(¯ x) = 0}. Further, ...
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dependent. This contradicts the assumptions of the theorem. Example 1 Consider the problem: min 6(x1 − 10)2 +4(x2 − 12.5)2 s.t. 2 x1 2 x1 (x1 − 6)2 +(x2 − 5)2 ≤ 50 2 +3x2 2 +x2 ≤ 200 ≤ 37 In this problem, we have: 10 We also have: f (x) = 6(x1 − 10)2 + 4(x2 − 12.5)2 g1(x) = x1 + (x2 − 5)2 − 50 2 2 ...
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0 g3(¯x) = 0 ≤ 0 To check for optimality, we compute all gradients at ¯x: ⎛ ⎝ −36 ⎞ ⎠ ∇f (x) = −52 ⎛ ⎞ 14 ⎝ ⎠ 2 ⎛ ⎞ 14 ⎝ ⎠ 36 ⎛ ⎞ 2 ⎝ ⎠ 12 ∇g1(x) = ∇g2(x) = ∇g3(x) = We next check to see if the gradients “line up”, by trying to solve for u1 ≥ 0, u2 = 0, u3 ≥ 0 in the following system: ⎛ ⎝ −36 −52 ⎞ ⎛...
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. x x ≤ 1 . T The KKT conditions are: −2Qx + 2ux = 0 Tx x ≤ 1 u ≥ 0 u(1 − xT x) = 0 . One solution to the KKT system is x = 0, u = 0, with objective function value xT Qx = 0. Are there any better solutions to the KKT system? If x (cid:10)= 0 is a solution of the KKT system together with some value u , then x ...
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In this problem, we have: f (x) = (x1 − 12)2 + (x2 + 6)2 2 2 g1(x) = x + 3x1 + x2 1 − 4.5x2 − 6.5 g2(x) = (x1 − 9)2 + x2 2 − 64 h1(x) = 8x1 + 4x2 − 20 Let us determine whether or not the point x = (¯ x2) = (2, 1) is a x1, ¯ ¯ candidate to be an optimal solution to this problem. We first check for feasibility:...
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⎠ u2 + ⎝ ⎠ v1 = 4 ⎛ ⎞ 0 ⎝ ⎠ 0 Notice that (¯ v) = (¯ u2, v¯1) = (4, 0, −1) solves this system and that x is a candidate to be an optimal solution of u2 = 0. Therefore ¯ u1, ¯ u, ¯ u ≥ 0 and ¯ ¯ this problem. 3 Generalizations of Convexity Suppose X is a convex set in (cid:4) . The function f (x) : X → (cid:4) ...
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is quasiconvex on X if and only if Sα is a convex set for every α ∈ (cid:4). Proof: Suppose that f (x) is quasiconvex. For any given value of α, suppose that x, y ∈ Sα. Let z = λx+(1−λ)y for some λ ∈ [0, 1]. Then f (z) ≤ max{f (x), f (y)} ≤ α, so z ∈ Sα, which shows that Sα is a convex set. Conversely, suppose Sα ...
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y) ≥ f (x) + ∇f (x)t(y − x). Hence, if ∇f (x)t(y − x) ≥ 0, then f (y) ≥ f (x), and so f (x) is pseudoconvex. To show the second claim, suppose f (x) is pseudoconvex. Let x, y and λ ∈ [0, 1] be given, and let z = λx + (1 − λ)y. If λ = 0 or λ = 1, then f (z) ≤ max{f (x), f (y)} trivially; therefore, assume 0 < λ < 1....
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solution of (P), and suppose ¯x together with multipliers (u, v) satisfies ∇f (¯ x) + ∇g(¯ x)t u + ∇h(¯ x)t v = 0, u ≥ 0, 17 uigi(¯x) = 0, i = 1, . . . , m. If f (x) is a pseudoconvex function, gi(x), i = 1, . . . , m are quasiconvex func- tions, and hi(x), i = 1, . . . , l are linear functions, then ¯x is a glob...
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0 = gi(¯ x) for any λ ∈ (0, 1), and since the value of gi(·) does not increase by moving in the direction x − ¯ x) ≤ 0 for all i ∈ I. x)t(x − ¯ Similarly, ∇hi(¯ x)) = 0, and so ∇hi(¯ x)t(x − ¯ x) = 0 for all x, we must have ∇gi(¯ x + λ(x − ¯ i = 1, . . . , l. Thus, from the KKT conditions, ∇f (¯ x)t(x − ¯ (cid:17)...
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6) is the global minimum. Example 5 Continuing Example 3, note that f (x), g1(x), g2(x) are all con- vex functions and that h1(x) is a linear function. Therefore the problem is a convex optimization problem, and the KKT conditions are necessary and sufficient. Therefore ¯x = (2, 1) is the global minimum. 5 Constraint...
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), i = 1, . . . , l are linearly independent, and (P) has a Slater point. Then the KKT conditions are necessary to characterize an optimal solution. Proof: Let ¯x be a local minimum. The Fritz-John conditions are necessary for this problem, whereby there must exist (u0, u, v) (cid:10)= 0 such that (u0, u) ≥ 0 and ...
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KKT conditions are necessary to characterize an optimal solution. Proof: Our problem is 20 (P) minx f (x) s.t. Ax ≤ b M x = g . ¯ Suppose ¯x is a local optimum. Without loss of generality, we can partition the constraints Ax ≤ b into groups AI x ≤ bI and AI¯x ≤ b ¯ such that x < bI¯. Then at x, the set {d : A...
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) m (cid:16) l i=1 i=1 Using the Lagrangian, we can, for example, re-write the gradient conditions of the KKT necessary conditions as follows: ∇xL(¯x, u, v) = 0, (1) since ∇xL(x, u, v) = ∇f (x) + ∇g(x)tu + ∇h(x)tv. Also, note that ∇2 L(x, u, v) = ∇2f (x)+ i=1 vi∇2hi(x). Here we use the standard notation: ∇2q(...
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S together with multipliers (u, v) satisfies the KKT conditions. Let I + = {i ∈ I : ui > 0} and I 0 = {i ∈ I : ui = 0}. Additionally, suppose that every d (cid:10)= 0 that satisfies ∇gi(¯x)td = 0, ∇gi(¯x)td ≤ 0, ∇hi(¯x)td = 0, i ∈ I + , i ∈ I 0 , i = 1 . . . , l also satisfies dt∇2 L(¯x, u, v)d > 0 . xx Then ¯x is ...
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0. This idea of “invertability” of a system of equations is generalized (although only locally) by the following version of the Implicit Function Theorem, where we will preserve the notation used above: Theorem 22 (Implicit Function Theorem) Let h(x) : (cid:4)n → (cid:4)l and x = (¯ ¯ y1, . . . , y¯l, z¯1, . . . ,...
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= 0 . k=1 ∂yk Proof of Theorem 2: Let A = ∇h(¯x) ∈ (cid:4)l×n . Then A has full row rank, and its columns (along with corresponding elements of ¯x) can be re-arranged so that A = [B | N ] and ¯ x = (¯ z), where B is non-singular. Let z lie in a small neighborhood of ¯z. Then, from the Implicit Function Theorem, t...
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(θ)) n−l ∂hi(s(z(θ)), z(θ)) ∂zk(θ) ∂θ (cid:16) ∂θ + · · ∂yk ∂zk . k=1 k=1 Let rk = ∂sk(z(θ)) , and recall that ∂zk(θ) = pk. The above equation system ∂θ can then be re-written as 0 = Br + N p, or r = −B−1N p = q. Therefore, ∂xk(θ) = dk for k = 1, . . . , n. ∂θ ∂θ For i ∈ I, 24 gi(x(θ)) = gi(¯x) + θ ...
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� > 0 sufficiently small, which contradicts the local optimality of x.¯ Therefore no such d can exist, and the theorem is proved. 7 Constrained Optimization Exercises 1. Suppose that f (x) and gi(x), i = 1, . . . , m are convex real-valued functions over (cid:4)n, and that X ⊂ (cid:4)n is a closed and bounded con- vex...
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i- mality conditions for this problem? Use these conditions to show that ∗ z = |(cid:15)c(cid:15) −α |. What is the optimal solution x ∗? 4. Let S1 and S2 be convex sets in (cid:4) . Recall the definition of strong separation of convex sets in the notes, and show that there exists a hyperplane that strongly separate...
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1 j=1 whenever λ1, . . . , λk satisfy λ1, . . . , λk ≥ 0 and (cid:19) k j=1 λj = 1. 8. Let f1(·), . . . , fk (·) : (cid:4)n → (cid:4) be convex functions, and consider the function f (·) defined by: f (x) := max{f1(x), . . . , fk (x)} . Prove that f (·) is a convex function. State and prove a similar result for...
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convex function on S. Prove that ¯x is an optimal solution of this problem if and only if ∇f (¯ x) ≥ 0 for every x ∈ S. x)t(x − ¯ 12. Consider the following problem: maximizex 2 3x1 − x2 + x3 s.t. x1 + x2 + x3 ≤ 0 −x1 + 2x2 + x2 = 0 3 x ∈ (cid:4)3 . • Write down the KKT optimality conditions. • Argue why this...
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optimal objective value of this problem is z and is attained at some feasible point x . Let M be a proper subset of {1, . . . , m} and suppose that ˆx solves the problem to minimize f (x) subject to gi(x) ≤ 0 for i ∈ M . Let V := {i | gi(ˆx) > 0}. If ∗ z > f (ˆx), show that gi(x ) = 0 for some i ∈ V . (This shows t...
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j=1 aj xj = b xj ≥ 0, j = 1, . . . , n x ∈ (cid:4) . n x Write down the KKT optimality conditions, and solve for the point ¯ that solves this problem. 17. Let c ∈ (cid:4)n , b ∈ (cid:4)m , A ∈ (cid:4)m×n, and H ∈ (cid:4)n×n. Consider the following two problems: P1 : minimizex c x + 1 xT Hx t 2 s.t. Ax ≤ b ...
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P ∇f (¯ direction, that is, ¯ λ > 0 and sufficiently small. • Suppose that d ¯ = 0 and that u := −AT [Aβ AT ]−1Aβ ∇f (¯x) ≥ 0. β β Show that ¯x is a KKT point. • Show that d ¯ is a positive multiple of the optimal solution of the following problem: minimized ∇f (¯x)T d s.t. Aβ d dT d d ∈ (cid:4)n . 0 0 ≤ 1 •...
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TWO HYPOTHETICAL EXAMPLES 1. A furniture manufacturer produces two types of desks: Standard and Executive. These desks are sold to an office furniture wholesaler, and for all practical purposes, there is an unlimited market for any mix of these desks, at least within the manufacturer’s production capacity. Each des...
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1/12 ton of C. Each ton of Y yields 1/2 ton of A, 1/10 ton of B, and 1/12 ton of C. Compound X costs $250 per ton, compound Y $400 per ton. The cost of processing is $250 per ton of X and $200 per ton of Y. Amounts produced in excess of daily requirements have no value as the products undergo chemical changes if no...
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15.066J 25 Summer 2003 E Z = 800 Z = 560 (0, 8) (6, 6) Z = 400 (9, 4) (0, 0) (12, 0) S 15.066J 26 Summer 2003 Y (0, 8) X/4 + Y/2 >= 4 (16, 0) X 15.066J 27 Summer 2003 (0, 20) Y Y (0, 12) 15.066J X/4 + Y/10 >= 2 (8, 0) X X/12 + Y/12 >= 1 (12, 0) 28 X Summer 2003 Y (0, 20) Z = 9 K ...
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ally, an inactive constraint is one that does not pass through the optimal solution. 7. There may be more than one optimal solution to a linear program, i.e., there may be multiple optima. 8. A linear program is unbounded if there exist feasible plans with arbitrarily large (if a max problem) or arbitrarily sm...
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Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Lecture 4: Analysis Based On Continuity 1 This lecture presents several techniques of qualitative systems analysis based on what is frequently called to...
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17, 2003 x˙ (t) = a(x(t)), (4.2) ¯ x ⊂ Rn . Then with x(0) = x exist and are unique on the time interval t ⊂ [0, 1] for all ¯ discrete time system (4.1) with f (¯) = x(1, ¯) describes the evolution of continuous time x system (4.2) at discrete time samples. In particular, if a is continuous then so is f . x 2 L...
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n is called relatively open in X if for every y ⊂ Y there exists r > 0 such that all x ⊂ X satisfying |x − y| < r belong to Y . A boundary of a subset Y � X � Rn in X is he set of all x ⊂ X such that for every r > 0 there exist y ⊂ Y and z ⊂ X/Y such that |y − x| < r and z − x| < r. For example, the half-open interv...
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to converge to one of two equilibria. However, it is not possible for both equilibria to be locally attractive. Otherwise, according to The­ orem 4.1, Rn would be represented as a union of two disjoint open sets, which contradicts the notion of connectedness of Rn . ¯ 4.1.2 Proof of Theorem 4.1 x0). Let x1 = x1(t) ...
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Now take an xk ⊂ A arbitrary x ⊂ d(A). By the definition of the boundary, there exists a sequence ¯ x). If converging to ¯. Hence, by the continuity of f , the sequence f (¯ f (¯) ≤⊂ A, this implies f (¯ x) ⊂ d(A). Let us show that the opposite is impossible. Indeed, if f (¯) ⊂ A then, since A is proven open, there ...
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Rn is a locally Lipschitz function. If x : [0, →) ∞� Rn is a solution of (4.2) then the set lim(x) of its limit points is a closed subset of Rn , and every solution of (4.2) with initial conditions in lim(x) lies completely in lim(x). Proof First, if tk,q � → and x(tk,q , x(0)) � ¯q as k � → for every q, and ¯ xq � ...
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0 and a non-constant solution xp : (−→, +→) ∞� R2 such that xp(t + T ) = xp(t) for all t, and the set of limit points of x is the trajectory (the range) of xp; 4 (c) the limit set is a union of trajectories of maximal solutions x : (t1, t2) ∞� R2 of (4.2), each of which has a limit (possibly infinite) as t � t1 or...
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H : Sn × [0, 1] ∞� Sn is continuous then ind(H(·, 0)) = ind(H(·, 1)) (such maps H is called a homotopy between H(·, 0) and H(·, 1)); (b) if the map Fˆ : Rn+1 ∞� Rn+1 defined by is continuously differentiable in a neigborhood of S n then Fˆ(z) = |z|F (z/|z|) ind(F ) = � x�Sn det(Jx(Fˆ))dm(x), where Jx(Fˆ) is the ...
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point theorem, which states that for every continuous function G : Bn ∞� Bn, where Bn = {x ⊂ Rn+1 : |x| ∀ 1}, equation F (x) = x has at least one solution. The statement is obvious (though still very useful) when n = 1. Let us prove it for n > 1, starting with assume the contrary. Then the map G : Bn ∞� Bn which m...
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Then (4.3) has a T -periodic solution x = x(t) = x(t + T ) for all t ⊂ R. x ∞� x(T, 0, ¯ x) is a continuous function G : B n ∞� Bn . The solution Indeed, the map ¯ x = G(¯ of ¯ x) defines the initial conditions for the periodic trajectory.
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MIT OpenCourseWare http://ocw.mit.edu 18.306 Advanced Partial Differential Equations with Applications Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Lecture 01 2009 09 09 WED TOPICS: Mechanics of the course. Example pde. Initial and boundary value pr...
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the information obtained on being told the outcome of a pr obabilistic experiment : S si I(S = si ) = log2 ⎞ ⎛ 1 ⎟ ⎜ ⎝ pS (si )⎠ wher pS (si ) e is the pr obability of the event . S = si The unit of measurement (when the log is base-2) is the bit (binary information unit --- not the same as binary digit!). 1 bit of...
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10)(cid:8)(cid:5) (cid:25)(cid:8)(cid:31)(cid:6)(cid:3)(cid:28)(cid:26)(cid:3)(cid:10)(cid:25)(cid:8)G(cid:14)(cid:3)(cid:13)(cid:5)(cid:8)(cid:31)(cid:28)(cid:14))(cid:6))(cid:3)(cid:7)(cid:3)(cid:19)(cid:25)(cid:10)(cid:8)(cid:5)(cid:14)(cid:8)(cid:6)(cid:7)(cid:10)(cid:14)(cid:8) (cid:27)(cid:6)(cid:18)(cid:5)(cid:1...
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2 Fall 2012 This is the maximum attainable value! Lecture 1, Slide #19 e.g., Binary entropy function h( p) Heads (or C=1) with pr p obability Tails (or C=0) with probability 1− p h(p) 1.0 1.0 0.5 0 p 0 0.5 1.0 H(C) = −plog2 p − (1− p)log2(1− p) = h(p) 6.02 Fall 2012 Lecture 1, Slide #20 Connection to...
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fewer binary digits on average, the receiver will have some uncertainty about the outcome described by the message. If we send more binary digits on average, we’re wasting the capacity of the communications channel by sending binary digits we don’t have to. Achieving the entropy lower bound is the “gold standard” ...
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The (cid:1)natural(cid:2) fixed-length encoding uses two binary digits for each choice, so transmitting the results of 1000 choices requires 2000 binary digits. 6.02 Fall 2012 Lecture 1, Slide #24 Variable-length encodings (David Huffman, in term paper for MIT graduate class, 1951) Use shorter bit sequences for...
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've removed. Label the left branch with a “0”, and the right branch with a “1”. –  Add to S a new symbol that represents this new node. Assign this new symbol a probability equal to the sum of the probabilities of the two nodes it replaces. 6.02 Fall 2012 Lecture 1, Slide #26 Huffman Coding Example •  •  Initi...
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cid:17)(cid:10)(cid:13)(cid:21)(cid:3) (cid:12)(cid:8) 0(cid:8) (cid:15)(cid:8) (cid:2)(cid:8) ,(cid:8) -(cid:8) ,,(cid:8) ,-(cid:8) Why isn’t this a workable code? The expected length of an encoded message is (.333+.5)(1) + (.083 + .083)(2) = 1.22 bits which even beats the entropy bound ☺ 6.02 Fall 2012 Lecture ...
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6.824 2006 Lecture 2: I/O Concurrency Recall timeline [draw this time-line] Time-lines for CPU, disk, network How can we use the system's resources more efficiently? What we want is *I/O concurrency* Ability to overlap I/O wait with other useful work. In web server case, I/O wait mostly for net transfer to ...
https://ocw.mit.edu/courses/6-824-distributed-computer-systems-engineering-spring-2006/218dc61c38c02d26cb8dcb9193268a4c_lec2_concurrency.pdf
can be used to get CPU concurrency. Of course O/S designer had to work a lot harder... CPU concurrency much less important than I/O concurrency: 2x, not 100x In general, very hard to program to get good scaling. Usually easier to buy two separate computers, which we *will* talk about. Multiple process proble...
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. saved registers). Semantics: per-process resources: addr space, file descriptors per-thread resources: user stack, kernel stack, kernel state Kernel can schedule one thread per CPU This sounds like just what we want for our server BUT kernel threads are usually expensive, just like processes Kernel has ...
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1. non-blocking system calls 2. uniform event delivery mechanism Typical O/S provides only partial support for event notification yes: new TCP connections, arriving TCP/pipe/tty data no: file-system operation completion Similarly, not all system calls operations can be started w/o waiting yes: connect(), sock...
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register interest in events The point: this preserves the serial natures of the events Programmer sees events/functions occuring one at a time Cite as: Robert Morris, course materials for 6.824 Distributed Computer Systems Engineering, Spring 2006. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute...
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8.701 0. Introduction 0.4 Literature Introduction to Nuclear and Particle Physics Markus Klute - MIT 1 Recommended Books © Cambridge University Press. All rights reserved. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/fairuse. Introduction to High Energy P...
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Welcome back to 8.033! Image Courtesy of Wikipedia. Summary of last lecture: • Space/time unification • More 4-vectors: U, K • Doppler effect, aberration • Proper time, rest length, timelike, spacelike, null The Three Types of 4-Vectors: SPACELIKE NULL TIMELIKE Future Past DXth Dx > 0 |Dx| > cDt DXth Dx = 0 |Dx| ...
https://ocw.mit.edu/courses/8-033-relativity-fall-2006/21e5641958305fcb6e704fe8b59b404d_lecture7_kinem4.pdf
OCW. THE TWIN PARADOX IS IT RIGHT? Time dilaton examples: GRB’s SUPERNOVAE CLOCKS ON PLANES GPS GPS uses a constellation of 24 “NAVSTAR” satellites that are 11,000 miles above the earth's surface. How GPS receivers calculate your location: The positioning process: 1. Satellite 1 transmits a signal that contains...
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8.04: Quantum Mechanics Massachusetts Institute of Technology Professor Allan Adams 2013 February 7 Lecture 2 Experimental Facts of Life Assigned Reading: E&R 16,7, 21,2,3,4,5, 3all NOT 4all!!! 1all, 23,5,6 NOT 2-4!!! Li. 12,3,4 NOT 1-5!!! Ga. 3 Sh. We all know atoms are made of: • electrons – Cathode...
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being a little clever (but mostly obsessed) that spectral emission lines followed the formula λn ≈ (3646 angstrom) · 1 − −1 4 n2 for n ∈ {3, 4, 5, . . .} . (0.1) Rydberg and Ritz then found that λ−1 = R · (n1 − n2 ) for ni ∈ Z, n2 > n1 −2 −2 (0.2) where R is the Rydberg constant dependent on the particular e...
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Editor’s note: where did that come from? I mean, am I supposed to explain the reason for the spectra right here and now, or leave it as an open question for readers to ponder? 8.04: Lecture 2 3 Figure 2: Photoelectric effect experimental schematic Figure 3: Photoelectric dependence of I on V : expectation (top)...
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.03, and 8.033, the double-slit experiment seems fairly convincing! Figure 5: Schematic of double-slit interference and diffraction One thing to ask is, where is the light when it hits the wall? In fact, it is everywhere, as it is 8.04: Lecture 2 5 not localized. But the intensity shows an interference pattern. ...
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1(y) + e iθ2(y)). 6 8.04: Lecture 2 The intensity, up to a constant ensuring the correct dimensions, is |A(y)|2 = 2A2 0 · (1 + cos(θ1(y) − θ2(y))), which reduces to This yields maxima at |A(y)|2 ≈ 2A2 0 · (1 + cos( 2πfy λD )). fy = Dλn as expected. Note that maxima correspond to constructive interferenc...
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) (0.4) That’s enough about light for now. What about atoms? Well, we are not as confident that they exist, so let us stick with electrons for now. While the properties of color and hardness in electrons are disturbing, we should be able to agree that if electrons are truly particles, they would be localized and wo...
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low, the wavelength, which becomes the resolution of the electron’s spatial location, becomes too large to remain meaningful. Hence, determining through which slit an electron passes does away with the interference 8 8.04: Lecture 2 pattern. This means that every force must be quantized like E = hν and p = hλ−1,...
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3.032 Mechanical Behavior of Materials Fall 2007 Buckling: Long, thin beam under end-loaded compression P [N] δ max [m] r θ = dδ/dx Lecture 3 (09.10.07) 3.032 Mechanical Behavior of Materials Fall 2007 Beam bending: End-loaded cantilever Images removed due to copyright restrictions. Please see: Silva, Emilio C. C. M.,...
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3.46 PHOTONIC MATERIALS AND DEVICES Lecture 5: Waveguide Design—Optical Fiber and Planar Waveguides Lecture Fiber Optics Optical fiber ≡ core + cladding guided if n2 > n1 power loss to cladding if < n1 n ⎛ 1 ⎟⎟⎟⎟⎟ θ = sin−1⎜⎜⎜ ⎜n2⎝ n2 ⎠ c ⎞ JK K each mode travels with β, v , U (x,y), P, k single mode (small c...
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⎜⎜ ⎝ ⎢ ⎣ 1 ⎞2 ⎤ 2 n ⎥ 1 ⎟⎟⎟⎟ ⎥ n2 ⎠ ⎥ ⎦ 2 1 = (n2 − n1 ) 2 2 θ = sin−1(NA) a 1 NA = (n2 − n2 )2 ≈ n2(2Δ) 1 2 1 2 3.46 Photonic Materials and Devices Prof. Lionel C. Kimerling Lecture 5: Waveguide Design—Optical Fiber and Planar Waveguides Page 1 of 5 Lecture θ = acceptance ∠ for fiber a ≡ exit angle for ...
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= rate of change of u(r) in core γ = rate of U(r) in cladding kT 2 = n2 k0 2 2 − β2 2 2 2 2 γ = β −n k1 0 rate of decay high ⇒ low penetration 2 kT + γ 2 2 (n2 −n )k = (NA)2 2 0 2 1 2 k0 ↑ γ ↓⇒ penetration into cladding , k T kT > NA ⋅ 0 ⇒ γ imaginary, wave escape core 3.46 Photonic Materials and Devices Pro...
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= 1.452, Δ = 0.01, NA = 0.205 λ = 0.85 μm (GaAs) a (core) = 25 μm ⇒ V = 37.9, M = 585 0 remove cladding ⇒ n1 = 1, NA = 1 ⇒ V = 184.8, M > 13,800 3.46 Photonic Materials and Devices Prof. Lionel C. Kimerling Lecture 5: Waveguide Design—Optical Fiber and Planar Waveguides Page 3 of 5 Lecture Notes Group...
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est travel, slowest velocity ⇒ power low profile n1 n2 n ( ) 2 n r n= 2 2 ⎡ ⎢ − ⎜⎜ 1 2 ⎢ ⎢⎣ p ⎛ ⎞ r ⎜ ⎟⎟⎟⎟ ⎝ ⎠ a ⎤ ⎥Δ⎥ ⎥⎦ n = n2 @ r = 0 = n1 @ r = a r ≤ a p = 1 p = 2 p → ∞ n2(r) linear n2(r) quadratic n2(r) step function Δ = 2 2 n2 − n1 22n2 3.46 Photonic Materials and Devices Prof. Lionel ...
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Maslab Software Engineering January 5th, 2005 Yuran Lu Agenda  Getting Started On the Server Using the Documentation Design Sequence Tools The Maslab API  Design Principles  Threading in Java On the Server  Put these lines in your .environment:  add 6.186  add -f java_v1.5.0  setenv JAVA_HOME /...
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work together  Making sure one team member’s changes doesn’t break another team member’s code Modularity and the Design Process  Modular Design  Provides abstraction  Gives up fine-control abilities, but makes code much more manageable  The Design Process  Top-down vs. Bottom-up  Write out specification...
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, Ball-seeking, Obstacle Avoidance, etc.)  Higher-level strategic control Using Threading  Look in Sun’s Java tutorial, or Ed Faulkner’s Java reference  Look at the Java API:  Thread, Runnable, wait(), notify(), sleep(), yield()  Must take care to avoid deadlock Synchronization in Threading  Allows block...
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Matrices We have already defined what we mean by a matrix. In this section, we introduce algebraic operations into the set of matrices. Definition. If A and B are two matrices of the same size, say k by n, we define A t B to be the k by n matrix obtained by adding the corresponding entries of A and B, m d we defin...
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tak- ing the "dot product" of the 1-'th row of A with the j-th column of B . Schematically, This definition seems rather strange, but it is in fact e,xtremely useful. Motivation will come later! One important justification for this definition is the fact that this product operation satisfies some of the familar "...
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; thus their sum is also defined. The distributivity formula now follows from the equation n 1 s=l is SJ sj a ( b . + c ) = I a b n s=l is sj n + I a c s=l is sj' The other distributivity formula and the homogeneity formula are proved similarly. We leave them as exercises. Now let us verify associativity. ...
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. Then as s ranges from 1 to m, a1.l but one of the terms of this sunanation vanish. The only one that does not vanish is the one for which s = i, and in that case &. 0+. .. + 0 + TiQ l j + O+ . . a + 0= Q5. 1s c = ij = 1. We conclude that Ar, entirely similar proof shows that B * I = B if B has rn columns. m R...
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. Which of the field axioms (the algebraic axioms that the real numbers satisfy) hold for this set? (Such an algebraic object is called in modem algebra a "ring with identity.I ! ) Systems -of linear equations Given numbers a and given numbers cl,,c for i = 1 ij k , we wish to study the following, which is call...
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Representing elements of Vn as column matrices is so convenient that we will adopt it as a convention throughout this section, whenever we wish. Example 1. Consider the system [Here we use x, y, z for the unknowns instead of xl, x2, for convenience.] This system has no solution, since X 3 r the sum of the first t...
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shall need is stated in the following theorem: Tlieorem 2, Consider the system of equations A'X = C, where A is a k by' n matrix and C is a k by 1 matrix. Let B be the matrix obtained by applying an elementary row operation to A, arrd let C' be the matrix obtained by applying the same elementary row operation to C....
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wish to determine the dimension of this solution space, and to find a basis for it. Dflfinition. Let A be a matrix of size k by n. Let W be the row space of A; let r be the dimension of W. Then r equals the number of non-zero rows in the echelon form of A. It follows at once that r k. It is also true that r 5 n, ...
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