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19 Pages

### Cycle_Canceling

Course: EE 15.082, Spring 2010
School: Native American...
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Word Count: 309

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and 15.082 6.855J Cycle Canceling Algorithm 1 A minimum cost flow problem 0 2 30, \$7 25 1 25, \$5 20, \$6 3 0 20, \$2 20, \$1 5 -25 10, \$4 0 4 25, \$2 2 The Original Capacities and Feasible Flow 0 2 30,25 25 1 20,0 The feasible flow can be found by solving a max flow. 3 0 25,5 5 -25 25,15 20,10 20,20 10,10 0 4 3 Capacities on the Residual Network 2 5 1 25 15 10 20 3 10 10 20 5 20 10 4 5 4 Costs on the...

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and 15.082 6.855J Cycle Canceling Algorithm 1 A minimum cost flow problem 0 2 30, \$7 25 1 25, \$5 20, \$6 3 0 20, \$2 20, \$1 5 -25 10, \$4 0 4 25, \$2 2 The Original Capacities and Feasible Flow 0 2 30,25 25 1 20,0 The feasible flow can be found by solving a max flow. 3 0 25,5 5 -25 25,15 20,10 20,20 10,10 0 4 3 Capacities on the Residual Network 2 5 1 25 15 10 20 3 10 10 20 5 20 10 4 5 4 Costs on the Residual Network 2 7 1 -7 -5 6 3 Find a negative cost cycle, if there is one. 5 4 -4 2 -2 -1 2 -2 5 5 Send flow around the cycle 2 Send flow around the negative cost cycle The capacity of this cycle is 15. 1 25 15 20 3 5 4 Form the next residual network. 6 Capacities on the residual network 2 20 1 10 15 5 3 20 5 10 25 10 20 10 4 5 7 Costs on the residual network 2 -7 5 -6 6 3 Find a negative cost cycle, if there is one. 2 -2 5 -4 4 7 1 2 -2 -1 8 Send flow around the cycle 2 Send flow around the negative cost cycle The capacity of this cycle is 10. 1 4 10 20 3 20 5 Form the next residual network. 9 Capacities on the residual network 2 20 1 10 15 5 3 10 15 20 25 10 10 10 4 5 10 Costs in the residual network 2 -7 1 -6 6 3 Find a negative cost cycle, if is there one. 2 -2 5 5 -4 4 7 2 -1 1 11 Send Flow Around the Cycle 2 Send flow around the negative cost cycle The capacity of this cycle is 5. 1 10 10 20 4 5 3 5 Form the next residual network. 12 Capacities on the residual network 5 2 25 1 5 20 25 5 10 15 5 15 10 10 4 3 5 13 Costs in the residual network 4 7 1 -7 5 -6 3 Find a negative cost cycle, if there is one. 2 -2 5 2 -4 2 -2 -1 1 4 14 Send Flow Around the Cycle 2 Send flow around the negative cost cycle The capacity of this cycle is 5. 1 5 10 4 10 3 5 Form the next residual network. 15 Capacities on the residual network 5 2 25 1 5 20 5 20 25 5 20 5 15 4 3 5 16 Costs in the residual network 4 7 1 Find a negative cost cycle, if there is one. -7 5 -6 3 2 -2 5 2 -4 2 -1 1 4 There is no negative cost cycle. But what is the proof? 17 Compute shortest distances in the residual network 7 7 2 -4 -7 5 -6 3 2 -2 5 2 -1 1 4 11 4 0 1 Let d(j) be the shortest path distance from node 1 to node j. Next let (j) = -d(j) And compute c 10 12 18 Reduced costs in the residual network 7 2 0 -0 0 2 1 0 4 The reduced costs in G(x*) for the optimal flow x* are all non-negative. 3 0 0 5 0 0 11 4 0 1 10 12 19
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Native American Education Services - EE - 15.082
15.082 and 6.855JDepth First Search1Initialize2 1 1 pred(1) all Unmark= 0 nodes in N; next := 1 order(next) = Mark node s 1 LIST:= cfw_1 LIST 1 next 1248 75 3 69Select a node i in LIST2 1 1 In depth first search, i is the last node in LIST 3 5
Native American Education Services - EE - 15.082
15.082 and 6.855JDijkstra's Algorithm with simple buckets (also known as Dial's algorithm)1An ExampleInitialize distance labels Initialize buckets. 01 4 2 1 2 3 2 3 3 5244 2 6 Select the node with the minimum temporary distance label. 01 4 5 6
Native American Education Services - EE - 15.082
15.082 and 6.855JFlow Decomposition1begin Initialize while y do begin Select(s, y) Search(s, y) if a cycle C is found then do begin let = Capacity(C, y) Add Flow( , C) to cycle flows Subtract Flow( , C) from y. end if a path P is found then do begin le
Native American Education Services - EE - 15.082
15.082 and 6.855JThe Ford-Fulkerson Augmenting Path Algorithm for the Maximum Flow Problem1Ford-Fulkerson Max Flow23 2 3 1 4 1 151 2 2s4 3tThis is the original network, plus reversals of the arcs.2Ford-Fulkerson Max Flow23 2 3 1 4 1 151 2
Native American Education Services - EE - 15.082
15.082 and 6.855JLabel Correcting Algorithm1An Example 23 1 6 3 4 3 2 43 -2 3Initialize d(1) := 0; d(j) := for j 1015 3 -42 76 In next slides: the number inside the node will be d(j). Violating arcs will be in thick lines.2An Example2
Native American Education Services - EE - 15.082
15.082 and 6.855JMin Global Cut Animation1Initialize11 5 3 3 2 Saturate the arcs out of node 1. Update the residual network4 41 6 1 562Initialize1 5 4 3 2 1 0 4 6 3 5 2 We will never push from node 1 5 again or into node 1. 5 3 3 3 2 Compute di
Native American Education Services - EE - 15.082
15.082 and 6.855JModified Label Correcting Algorithm1The Modified Label Correcting Algorithm 23 1 6 3 4 3 2 53 -2 3Initialize d(1) := 0; d(j) := for j 1 LIST := cfw_1014 3 -42 76 In next slides: the number inside the node will be d(j).2A
Native American Education Services - EE - 15.082
15.082 and 6.855J Network Simplex Animations1Calculating A Spanning Tree Flow1 1 -6 2 1 3 5 3 6 -4 73A tree with supplies and demands. (Assume that all other arcs have a flow of 0) What is the flow in arc (4,3)?242Calculating A Spanning Tree Flow
Native American Education Services - EE - 15.082
15.082 and 6.855JThe Goldberg-Tarjan Preflow Push Algorithm for the Maximum Flow ProblemPreflow Push23 2 3 1 4 1 151 4 2s4 3tThis is the original network, plus reversals of the arcs.Preflow Push23 2 3 1 4 1 151 4 2s4 3tThis is the origi
Native American Education Services - EE - 15.082
15.082 and 6.855J February 25, 2003Radix Heap Animation1An Example from AMO (with a small change)Initialize distance labels Initialize buckets and their ranges. Insert nodes into buckets.2 0 1 3 4 71325 15420 10 20 9 868 15 16 31 32 63352
Native American Education Services - EE - 15.082
15.082 and 6.855JThe Shortest Augmenting Path Algorithm for the Maximum Flow Problem1Shortest Augmenting Path23 2 3 1 4 1 151 4 2s4 3tThis is the original network, plus reversals of the arcs.2Shortest Augmenting Path23 2 3 1 4 1 151 4 2s
Native American Education Services - EE - 15.082
15.082 and 6.855JSpanning Tree Algorithms1The Greedy Algorithm in Action352 2104 48 156 61 1402520 15305 521 11173 37 72The Greedy Algorithm in Action35 352 2 25 2510 104 4 30 30 5 5815 156 6 17 171 140 4020 203 31121 21
Native American Education Services - EE - 15.082
15.082J and 6.855JSuccessive Shortest Path Algorithm1The Original Costs and Node Potentials0 2 7 5 0 1 6 3 040 42125 02The Original Capacities and Supplies/Demands5 2 30 23 1 20 3 -7 25 20 20 10 -2 4255 -193Select a supply node and find t
Native American Education Services - EE - 15.082
15.082 and 6.855J Topological Ordering1Preliminary to Topological SortingLEMMA. If each node has at least one arc going out, then the first inadmissible arc of a depth first search determines a directed cycle.1 4 6 7 3COROLLARY 1. If G has no directe
University of Aarhus - BUS - 1332
Tecumseh (March 1768 October 5, 1813) also Tecumtha or Tekamthi, was a Native American leader ofthe Shawnee and a large tribal confederacy that opposed the United States during Tecumseh's War and the War of 1812. He grew up in the Ohio country during the
University of Aarhus - BUS - 1332
My First Semester in CollegeStarting my fist semester in college, I was nervous and scared to go into it. I thought to myself, &quot;that's it I slacked off enough in high school, and I barely studied, I need to get my act together and start off strong&quot;. Then
University of Aarhus - BUS - 1332
First YearFall Semester CHEM 1117. Chemistry for Engineers Laboratory CHEM 1372. Chemistry for Engineers ECE 1100. Introduction to Electrical and Computer Engineering ENGL 1303. Freshman Composition I HIST 1377. The United States to 1877 1 MATH 1431. Cal
University of Aarhus - BUS - 1332
Midterm Study Guide- 1377 Professor Hopkins MW 11:00-12:30, AH 540 See blackboard emails regarding your specific TA. If you need to contact any of us, please do so through blackboard- not personal email. READ THIS FIRSTThis midterm exam is a writing exam
University of Aarhus - BUS - 1332
Final Exam Study Guide- 1377 FA 09 MW Professor Hopkins READ THIS FIRSTThe Final Exam is Monday, December 14, from 5-8 pm, in the lecture auditorium. You must show your UH Student ID when turning in your finished exam, and sign the class roster. Like the
University of Aarhus - BUS - 1332
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Maryland - MATH - 464
Homework #1 Due: Tuesday, February 2, 20101. (2pt) Compute the following integrals:1 log( x)dx0where log(x) denotes the natural logarithm of x, and 1 x a dx 1 for a fixed parameter a&gt;1. What happens for a=1 ?2. (5pt) Consider the following function:
Maryland - MATH - 464
Homework #2 Due: Tuesday, February 9, 2010Compute the following integrals. Note: Use of Matlab (or any other software) is not permitted. 1 1. x 2 + 1 dx 02.- x21 dx , for a fixed real parameter a&gt;0 + a2 1 dx , for a fixed real parameter a&gt;0 + 1)( x
Maryland - MATH - 464
Homework #3 Due: Thursday, February 25, 2010Compute the Fourier coefficients, and expand in Fourier series the following 1-periodic functions. Note: Use of Matlab (or any other software) is not permitted. 1.2(1 - 2 x ), for 0 &lt; x &lt; 12.26(1 - 6 x +
Maryland - MATH - 464
Homework #4 Due: Thursday, March 4, 2010Note: Use of Matlab (or any other software) is not permitted. I. Compute the Fourier transform of the following functions (1-8): 1. 1 , 1 x 2 f ( x) = 0 , otherwise 0 , x &lt; -3 ( x - 3) / 2 , - 3 x &lt; -1 f ( x) = 1 ,
Maryland - MATH - 464
Homework #5 Due: Thursday, March 11, 2010Note: Use of Matlab (or any other software) is not permitted. I. (Exercise 3.23) Let f be a suitably regular function on R with the Fourier transform F. What can you deduce about F if you know that:1.- f ( x)d
Maryland - MATH - 464
Homework #6 Due: Tuesday, March 23, 20101. (2pts) Find the Fourier transform of the function f:RR1/ 2f ( x) =-1 / 2e- ( x -u ) 2du2. (2pts) Let f0,f1:RR be functions defined by 2 2 f 0 ( x) = e - x , f 1 ( x) = xe - x Compute the following convolu
Maryland - MATH - 464
Homework #7 Due: Thursday, April 8, 2010Note: Use of Matlab (or any other software) is not permitted. (3pts) Assume f:RR is a square-integrable function whose Fourier transform is supported in [0,2]. Assume we know the samples cfw_f(nT) , n=.,-2,-1,0,1,2
Maryland - MATH - 464
Homework #8 Due: Thursday, April 15, 2010Note: Use of Matlab (or any other software) is not permitted.I.Consider the box function for the window function g=. Compute the windowed Fourier transform Vgf of the following functions: a. (2pts) f ( x) = e 2i
Maryland - MATH - 464
Homework #9 Due: Thursday, April 22, 2010Note: Use of Matlab (or any other software) is not permitted. I. (see Exercise 3.9) Find the Fourier transform of each of the following functions: x a. (1pt) f ( x) = cos( x) b. (1pt) f ( x) = (2 x + 1) (2 x - 1)
Maryland - MATH - 464
Homework #10 Due: Thursday, April 29, 2010Note: Use of Matlab (or any other software) is not permitted. I. (see Exercise 4.2) Use Poisson's relation to find the Fourier series for each of the following p-periodic functions on R: a. (1pt) f ( x) =m = -
Maryland - MATH - 464
Homework #11 Due: Thursday, May 6, 2010Note: Use of Matlab (or any other software) is not permitted. I. (see Exercise 4.11) A 1-periodic function on R having the Fourier series:f ( x) =k = -c e k2 ikx(with F[k] replaced by ck) takes the values: 1 1
Maryland - MATH - 464
Quiz 1, MATH 464, 6 April 2010Your Name: _(1) [3 pts] A band-limited signal is sampled at its critical (Nyquist) sampling rate of 100 KHz. Thus in one-second interval 100,000 samples are collected. If only 1000 samples are used to compute the signal at
Maryland - MATH - 464
Quiz 1, MATH 464, 6 April 2010Your Name: _(1) [3 pts] A band-limited signal is sampled at its critical (Nyquist) sampling rate of 100 KHz. Thus in one-second interval 100,000 samples are collected. If only 1000 samples are used to compute the signal at
Maryland - MATH - 464
Solutions to Homework 1 1. (a) For 0 &lt; t &lt; 1 use integration by parts in:1 1 1log(x)dx = x log(x)|1 - tt tx(log(x) dx = -t log(t)-t1dx = -t log(t)-1+tThen1 1log(x)dx = lim0 t0log(x)dx = lim (-t log(t) + t) - 1 = -1t t 0Note:tlim t log(t) =
Maryland - MATH - 464
R. BalanHomework #2 SolutionsMATH 4641. 01 dx = 2+1 x 0/2 01 sec2 ()d = 2 tan () + 1 0/2 0sec2 () = sec2 () 0/2d = /202. -1 dx = 2 2 + a2 x1 2 dx = 2 2 + a2 x a1 2 dx = 2 /a2 + 1 x |a|y21 2 dy = /2 = +1 |a| a3. If a = 1, we have tha
Maryland - MATH - 464
R. BalanHomework #3 SolutionsMATH 4641. (1 - 2x) = 2 2. 2 (1 - 6x + 6x2 ) = 6 3. 3 (x - 3x2 + 2x3 ) = 3 4. 4 (1 - 30x2 + 60x3 - 30x4 ) = 90 5. f (x) = 1 + 8 i(-1)k (-1)k - 1 + 4k 4 2 k 2 e2ikx = 1 1 - 8 2 (-1)k 1 sin(2kx) - 2 k 1 cos(2(2p + 1)x) (2p +
Maryland - MATH - 464
R. BalanHomework #4 SolutionsMATH 4641. f1 = 1[1,2] , F1 () = e-3i 2. f2 (x) = 3. f3 (x) = e-a|x| , F3 () = 4. f4 (x) = Then replace by a: f (x) = 5. f5 (x) = sin(2x)f1 (x) = 1 2ix (e f1 (x) - e-2ix f1 (x) 2i x2 1 + a2 F () = -2a| e a 2 1 = F3 (x)|a=2
Maryland - MATH - 464
R. BalanHomework #5 SolutionsMATH 464I. 1. F (0) = 1 2. F (0) = -2i 3. F (1) + F (-1) = 0 4.sF (s)ds = 0-5. F (s) = F (s) , that means F (s) is real valued II. 6. We look for a function f : R C that satisfies the given equation. Since we have the fr
Maryland - MATH - 464
R. BalanHomework #6 SolutionsMATH 4641. Note the function f is the convolution between the box function and the unit Gaussian function , that is:f (x) =-(u)(x - u)du , (u) =1 0, |u| &lt; , |u| &gt;1 2 1 2, (u) = e-|u|2Thus the Fourier transform is t
Maryland - MATH - 464
R. BalanHomework #7 SolutionsMATH 4641. Method 1 2 Let g(x) = e-2ix f (x). Note G(s) = F (s + ) and thus G(s) = 0 for |s| &gt; . Hence g B (that is, g is -band limited). Shannon's sampling formula yields g(x) =ng(nT )sinc(x - nT TBut g(nT ) = e-in f (
Maryland - MATH - 464
R. BalanHomework #8 SolutionsMATH 464I. The window is g(x) = (x). a) f (x) = e2ix . The windowed Fourier transform isVg f (w, t) =-e-2iwx f (x)g(x - t)dx =e-2ix(w-1) (x - t)dx == e-2it(w-1)e-2iy(w-1) (y)dy = e-2it(w-1) sinc(w - 1)b) f (x) = e2ia
Maryland - MATH - 464
R. BalanHomework #9 SolutionsNote:MATH 464I.a. f (x) = cos(x)(x/).f (x) =Its Fourier transform is1 ix x e + e-ix 2 F (s) =1 2x 1 e-2isx (eix +e-ix )( )dx = 2 - y=x ,x 1 e-2i(s-1/2)x ( )dx+ 2 -x e-2i(s+1/2)x ( )dx -Change the integration v
Maryland - MATH - 464
R. BalanI.Homework #10 SolutionsMATH 464(a) Poisson's summation formula implies:f (x) =m=-e-(x-mp) =21 pe2ikx/p F (k/p)k=-2where Thus:F (s) = F(e-x )(s) = e-s 1 p2f (x) =e2ikx/p e-kk=-2/p2which is the Fourier series expansion of f (x)
Punjab Engineering College - ECON - ECON111
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North Alabama - MG - mg498
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Ohio State - ECE - 341
Untitled Normal PagePage 1 of 3http:/eewww.eng.ohio-state.edu/~kimju/ee432/hw2/hw2.htm1/26/01Untitled Normal PagePage 2 of 3http:/eewww.eng.ohio-state.edu/~kimju/ee432/hw2/hw2.htm1/26/01Untitled Normal PagePage 3 of 3http:/eewww.eng.ohio-state.e
Ohio State - ECE - 341
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Ohio State - ECE - 341
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Ohio State - ECE - 341
Ohio State - ECE - 331
ECE331 Homework #1 (Due Friday, April 04, 2008)1. Relative to electrons and electron states, what does each of the four quantum numbers specify? 2. Sodium Chloride (NaCl) exhibits predominantly ionic bonding. The Na and Cl ions have electron structures t
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ECE331 Homework #3 (Due Friday, April 25, 2008)1. (a) The position of an electron is determined to within 1 Ao. What is the minimum uncertainty in its momentum? (b) An electron's energy is measured with an uncertainty of 1 eV. What is the minimum uncerta