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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)
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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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
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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