6) This question uses the modulus function. If x + y, x – yl is the positive difference between x and y, e.g. |5 – 6.1| = 1.1. The algorithm described by the flow chart below is to be applied to the five pieces of data below. U(1) = 6.1, U(2) = 6.9, U(3) = 5.7, U(4) = 4.8, U(5) = 5.3 %3D %3D a Obtain the final output of the algorithm using the five values given for U(1) to U(5). b In general, for any set of values U(1) to U(5), explain what the algorithm achieves.
6) This question uses the modulus function. If x + y, x – yl is the positive difference between x and y, e.g. |5 – 6.1| = 1.1. The algorithm described by the flow chart below is to be applied to the five pieces of data below. U(1) = 6.1, U(2) = 6.9, U(3) = 5.7, U(4) = 4.8, U(5) = 5.3 %3D %3D a Obtain the final output of the algorithm using the five values given for U(1) to U(5). b In general, for any set of values U(1) to U(5), explain what the algorithm achieves.
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...
Related questions
Question
![6)
This question uses the modulus function. If x + y, \x – yl is the positive difference
between x and y, e.g. |5 – 6.1| = 1.1. The algorithm described by the flow chart below is to
be applied to the five pieces of data below.
U(1) = 6.1, U(2) = 6.9, U(3) = 5.7, U(4) = 4.8, U(5) = 5.3
%3D
a Obtain the final output of the algorithm using the five values given for U(1) to U(S).
b In general, for any set of values U(1) to U(5), explain what the algorithm achieves.
Start
I = 1, A = U(1),
Temp = 15 - U(1)|
Box 1
Вох 2
I = 1 + 1
Вох 3
M = |S – U()|
Is
Вох 4
M< Temp ?
Yes
A = U()
Temp = M
%3D
No
Воx 5
Is
Yes
Вох 6
I< 5?
No
Print A
Вох 7
Stop](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F944d3dc9-d1fd-460b-87fa-0a1b133990bc%2F094255a3-7b28-45b6-9a50-be26c2372999%2F8p4a1b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6)
This question uses the modulus function. If x + y, \x – yl is the positive difference
between x and y, e.g. |5 – 6.1| = 1.1. The algorithm described by the flow chart below is to
be applied to the five pieces of data below.
U(1) = 6.1, U(2) = 6.9, U(3) = 5.7, U(4) = 4.8, U(5) = 5.3
%3D
a Obtain the final output of the algorithm using the five values given for U(1) to U(S).
b In general, for any set of values U(1) to U(5), explain what the algorithm achieves.
Start
I = 1, A = U(1),
Temp = 15 - U(1)|
Box 1
Вох 2
I = 1 + 1
Вох 3
M = |S – U()|
Is
Вох 4
M< Temp ?
Yes
A = U()
Temp = M
%3D
No
Воx 5
Is
Yes
Вох 6
I< 5?
No
Print A
Вох 7
Stop
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