Suppose Ethernet physical addresses are chosen at random (using true random bits). (a) What is the probability that on a 1024-host network, two addresses will be the same? (b) What is the probability that the above event will occur on one or more of 220 networks? (c) What is the probability that, of the 230 hosts in all the networks of (b), some pair has the same address? 1 Hint: The calculation for (a) and (c) is a variant of that used in solving the so-called Birthday Problem: Given N people, what is the probability that two of their birthdays (addresses) will be the same? The second person has probability 1 - 35 of having a different birthday from the first, the third has probability 1 - 365 of having a different birthday from the first two, and so on. The probability that all birthdays are different is thus 2 2 N (1-365) × (1-385) ××(1-¹) 365 which for smallish N is about 1 - 1+2+...+(N − 1) 365

Operations Research : Applications and Algorithms
4th Edition
ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Wayne L. Winston
Chapter20: Queuing Theory
Section20.4: The M/m/1/gd/∞/∞ Queuing System And The Queuing Formula L = Λw
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Suppose Ethernet physical addresses are chosen at random
(using true random bits).
(a) What is the probability that on a 1024-host network, two
addresses will be the same?
(b) What is the probability that the above event will occur on one
or more of 220 networks?
(c) What is the probability that, of the 230 hosts in all the
networks of (b), some pair has the same address?
Hint: The calculation for (a) and (c) is a variant of that used in
solving the so-called Birthday Problem: Given N people, what is
the probability that two of their birthdays (addresses) will be the
same? The second person has probability 1-3 of having a
365
different birthday from the first, the third has probability 1 365
of having a different birthday from the first two, and so on. The
probability that all birthdays are different is thus
2
(1-365)×(1-325) × × (1-N5¹)
X X
365
which for smallish N is about
1+2+
1
· + (N − 1)
365
...
Transcribed Image Text:Suppose Ethernet physical addresses are chosen at random (using true random bits). (a) What is the probability that on a 1024-host network, two addresses will be the same? (b) What is the probability that the above event will occur on one or more of 220 networks? (c) What is the probability that, of the 230 hosts in all the networks of (b), some pair has the same address? Hint: The calculation for (a) and (c) is a variant of that used in solving the so-called Birthday Problem: Given N people, what is the probability that two of their birthdays (addresses) will be the same? The second person has probability 1-3 of having a 365 different birthday from the first, the third has probability 1 365 of having a different birthday from the first two, and so on. The probability that all birthdays are different is thus 2 (1-365)×(1-325) × × (1-N5¹) X X 365 which for smallish N is about 1+2+ 1 · + (N − 1) 365 ...
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