Converting a maximum preflow to a maximum flow. We define a maximum preflow X O as a preflow with the maximum possible flow into the sink. (a) Show that for a given maximum preflow xo, some maximum flow x* with the same flow value as xo, satisfies the condition that Xu :s x- for all arcs (i, j) E A. (Hint: Use flow decomposition.) (b) Sug gest a labeling algorithm that converts a maximum preflow into a maximum flow in at most n + m augmentations. (e) Suggest a variant of the shortest augmenting path algo rithm that would convert a maximum preflow into a maximum flow in O(nm) time. (Hi nt: Define distance labels from the source node and show that the algorithm will creat

Converting a maximum preflow to a maximum flow. We define a maximum preflow X O as a preflow with the maximum possible flow into the sink. (a) Show that for a given maximum preflow xo, some maximum flow x* with the same flow value as xo, satisfies the condition that Xu :s x- for all arcs (i, j) E A. (Hint: Use flow decomposition.) (b) Sug gest a labeling algorithm that converts a maximum preflow into a maximum flow in at most n + m augmentations. (e) Suggest a variant of the shortest augmenting path algo rithm that would convert a maximum preflow into a maximum flow in O(nm) time. (Hi nt: Define distance labels from the source node and show that the algorithm will creat

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Find the simplest POS expression using K- Maps for the following functions. For each, list all prime implicants found in the K-Map, then list only the essential prime implicants, and then list the simplest solution. i. f (x1, x2, x3) = II M (0, 1, 2, 4, 5) ii. f (x1,x2, x3, x4) = ПМ (0, 1, 2,4, 5) Em (1, 2,5, 6, 7,8, 10, 14)
If the only information given about variable x2 in an LP is x2 ≥3, what modifications need to be made to the LP to put it in Canonical Inequality Form? You may assume that the rest of the problem is already in Canonical Inequality Form. Be specific and show your work.
Simplify the following functions using K-maps, and then derive the corresponding simplified Boolean expressions. Note that d stands for don't-care cases. a) F(x, y, z) = (0,2,4,5) b) F(x, y, z) = (0,1,2,3,5) c) F(x, y, z) = Σ(1,2,3,6,7) d) F(x, y, z) = Σ(3,4,5,6,7) e) F(A,B,C,D) = Σ(4,6,7,15) f) F(w, x, y, z) = Σ(0,2,8,10, 12, 13, 14) g) F(w, x, y, z) = Σ(0, 2, 3, 5, 7, 8, 10, 11, 14, 15) h) F(A, B, C, D) = [(1,3,6,9,11,12,14) i) F(x, y, z) = Σ(0,4,5,6) and d(x, y, z) = Σ(2,3,7)
Let X = {1,2,..., 100} , and consider two functions f : X → R and g: X → R. The Chebyshev metric of f and g is given by: d(f,g) = max|f(x) – g(x)| IEX Write a function d (f,g) that calculates the Chebyshev metric of any two functions f and g over the values in X.
A soft to very soft silty clay layer 3 m thick exists between a layer of permeable clayey silty sand at top and a nearly impermeable layer of stiff silty clay at bottom. Oedometer tests were carried out on a sample of the soft clay 19 mn thick. From the plot of AH versus log time, the value of time fj00 (the time at 100% primary consolidation) wa 100 min. From the graph of e versus log o' the compression index C, was 0.32 and the void ratio e, at the end of the primary consolidation was 0.72. Estimate the secondary compression of the silty clay layer that would occur from time of completion of the primary consolidation to 30 years of the load application in the field.
(2) The long term steady state vector q is independent of the initial vector. To verify this statement, let's consider the same transition matrix P as in the weather example but use another initial vector of [0.5, 0.5] (the values must sum up to be 1). Follow the power method to approximate q. You may stop at X(7). Compare your answers with the following and select all the correct options. U U U U x(1) X(0) P = [0.7, 0.3] x(1) X(0) P = [0.6, 0.4] X(2) = x(1) p = [0.68, 0.32] X(2) = x(1) p = [0.78, 0.22] x(3) = x(2) p = [0.712, 0.288] X(3) X(2) P [0.812, 0.188] = = = X(4) X(3) P = [0.8248, 0.1752] = X(4)= x(3) P = [0.7248, 0.2752] X(5) X(4) P = [0.82992, 0.17008] X(5) X(4) P = [0.72992, 0.27008] = X(6) X(5) p = [0.831968, 0.168032] = X(6) = X(5) p = [0.7327872, 0.2672128] X(7) = x(6) p = [0.73311488, 0.26688512] X(7) X(6) p = [0.8327872, 0.1672128] =
Consider the flow network shown in the following figure (left), where the label next to each arc is its capacity, and the initial s-t flow on right. (b) Apply the Ford-Fulkerson algorithm to N starting from the flow f. The augmenting path used in each step of the algorithm must be given. (c) Give the cut that corresponds to the maximum flow obtained in (b), which is suggested in the Max-Flow-Min-Cut theorom.
Derive an exact closed form bound in terms of n on the number of times function Fcalled in each code segment.Use floor/ceilings as needed. Clearly explain your answer and show your work.
Problem 8. Consider the loss function in Problem 6. M J= Σ (Συ,ί (m) WiX +Συ - m=1 i=0 i=0 Give a gradient descent algorithm. I.e., write pseudocode (with loops, if-branches, etc.) with the derivative computed.
A steel Column is 10m long and Ø1.2m. It carries a load of 102MN. Given the modulus of Elasticity is 256GPa. Find the stress, Strain, anddetermine how much the column is compressed. {show all the work}
A manufacturer of programmable calculators is attempting to determine a reasonable free-service period for a model it will introduce shortly. The manager of product testing has indicated that the calculators have an expected life of 60 months. Assume product life can be described by an exponential distribution. T / MTBF -T / MTBF T| MTBF -т / МТВF T| MTBF -т / МТВF 0.10 0.9048 2.60 0.0743 5.10 0.0061 0.20 0.8187 2.70 0.0672 5.20 0.0055 0.30 0.7408 2.80 0.0608 5.30 0.0050 0.40 0.6703 2.90 0.0550 5.40 0.0045 0.50 0.6065 3.00 0.0498 5.50 0.0041 0.60 0.5488 3.10 0.0450 5.60 0.0037 0.70 0.4966 3.20 0.0408 5.70 0.0033 0.80 0.4493 3.30 0.0369 5.80 0.0030 0.90 0.4066 3.40 0.0334 5.90 0.0027 1.00 0.3679 3.50 0.0302 6.00 0.0025 1.10 0.3329 3.60 0.0273 6.10 0.0022 1.20 0.3012 3.70 0.0247 6.20 0.0020 1.30 0.2725 3.80 0.0224 6.30 0.0018 1.40 0.2466 3.90 0.0202 6.40 0.0017 1.50 0.2231 4.00 0.0183 6.50 0.0015 1.60 0.2019 4.10 0.0166 6.60 0.0014 1.70 0.1827 4.20 0.0150 6.70 0.0012 1.80 0.1653 4.30…