(4) Use the power method to approximate the PageRank values for the pages (i.e., the steady statevector q) in part (3). Again the choice of initial vector is unimportant. Let's just use [0.5, 0.5, 0]. i.e.,X(0) = [0.5, 0.5, 0]. Note that the values in a vector must sum up to be 1 because the surfer must bein one of the states (visiting one of the pages) at a time. You may stop at X(6). Compare your answerswith the following and select all the correct options.UUUrUUx(1) X(0) P = [0.26, 0.26, 0.48]x(1)=X(0) P = [0.16, 0.16, 0.68]x(2) x(1) p = [0.1952, 0.1952, 0.6096]=X(2) X(1) p = [0.2952, 0.2952, 0.4096]X(3) = X(2) p = [0.19003733, 0.19003733, 0.61992533]X(3) X(2) p = [0.29003733, 0.29003733, 0.41992533]=X(4) = x(3) P = [0.29079452, 0.29079452, 0.41841095]X(4)X(5)= x(4) p = [0.19068347, 0.19068347, 0.61863306]=X(3) P = [0.19079452, 0.19079452, 0.61841095]X(5) X(4) P = [0.29068347, 0.29068347, 0.41863306]X(6)X(5) p [0.19069976, 0.19069976, 0.61860048]=X(6) X(5) p = [0.29069976, 0.29069976, 0.41860048]=

The Long-Term Steady-State Vector (q)In a Markov chain, the long-term steady-state vector "q"…

(4) Use the power method to approximate the PageRank values for the pages (i.e., the steady state vector q) in part (3). Again the choice of initial vector is unimportant. Let's just use [0.5, 0.5, 0]. i.e. X(0) = [0.5, 0.5, 0]. Note that the values in a vector must sum up to be 1 because the surfer must b in one of the states (visiting one of the pages) at a time. You may stop at X(6). Compare your answ with the following and select all the correct options. U U U r U x(1) X(0) P = [0.26, 0.26, 0.48] x(1) X(0) P = [0.16, 0.16, 0.68] X(2) = x(1) p [0.1952, 0.1952, 0.6096] X(2)x(1) p = [0.2952, 0.2952, 0.4096] X(3) X(2) P = [0.19003733, 0.19003733, 0.61992533] X(3) X(2) p = [0.29003733, 0.29003733, 0.41992533] = X(4) X(3) P = [0.29079452, 0.29079452, 0.41841095] = X(4) X(5) = x(4) P = [0.19068347, 0.19068347, 0.61863306] X(5) X(4) P = [0.29068347, 0.29068347, 0.41863306] = X(3) P [0.19079452, 0.19079452, 0.61841095] = X(6) X(5) P= [0.19069976, 0.19069976, 0.61860048] = X(6) X(5) p = [0.29069976, 0.29069976, 0.41860048] =

(4) Use the power method to approximate the PageRank values for the pages (i.e., the steady state vector q) in part (3). Again the choice of initial vector is unimportant. Let's just use [0.5, 0.5, 0]. i.e. X(0) = [0.5, 0.5, 0]. Note that the values in a vector must sum up to be 1 because the surfer must b in one of the states (visiting one of the pages) at a time. You may stop at X(6). Compare your answ with the following and select all the correct options. U U U r U x(1) X(0) P = [0.26, 0.26, 0.48] x(1) X(0) P = [0.16, 0.16, 0.68] X(2) = x(1) p [0.1952, 0.1952, 0.6096] X(2)x(1) p = [0.2952, 0.2952, 0.4096] X(3) X(2) P = [0.19003733, 0.19003733, 0.61992533] X(3) X(2) p = [0.29003733, 0.29003733, 0.41992533] = X(4) X(3) P = [0.29079452, 0.29079452, 0.41841095] = X(4) X(5) = x(4) P = [0.19068347, 0.19068347, 0.61863306] X(5) X(4) P = [0.29068347, 0.29068347, 0.41863306] = X(3) P [0.19079452, 0.19079452, 0.61841095] = X(6) X(5) P= [0.19069976, 0.19069976, 0.61860048] = X(6) X(5) p = [0.29069976, 0.29069976, 0.41860048] =

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Question

(4) Use the power method to approximate the PageRank values for the pages (i.e., the steady state
vector q) in part (3). Again the choice of initial vector is unimportant. Let's just use [0.5, 0.5, 0]. i.e.,
X(0) = [0.5, 0.5, 0]. Note that the values in a vector must sum up to be 1 because the surfer must be
in one of the states (visiting one of the pages) at a time. You may stop at X(6). Compare your answers
with the following and select all the correct options.
U
U
U
r
U
U
x(1) X(0) P = [0.26, 0.26, 0.48]
x(1)
=
X(0) P = [0.16, 0.16, 0.68]
x(2) x(1) p = [0.1952, 0.1952, 0.6096]
=
X(2) X(1) p = [0.2952, 0.2952, 0.4096]
X(3) = X(2) p = [0.19003733, 0.19003733, 0.61992533]
X(3) X(2) p = [0.29003733, 0.29003733, 0.41992533]
=
X(4) = x(3) P = [0.29079452, 0.29079452, 0.41841095]
X(4)
X(5)
= x(4) p = [0.19068347, 0.19068347, 0.61863306]
=
X(3) P = [0.19079452, 0.19079452, 0.61841095]
X(5) X(4) P = [0.29068347, 0.29068347, 0.41863306]
X(6)
X(5) p [0.19069976, 0.19069976, 0.61860048]
=
X(6) X(5) p = [0.29069976, 0.29069976, 0.41860048]
=

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