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Counting Internet Addresses. In IPv4 (Version 4 of the Internet Protocol) an address is a string of 32 bits: network number (netid) + host number (hostid). Three forms of addresses: Class A: 0 + 7-bit netid + 24-bit hostid (largest network).

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counting internet addresses
Counting Internet Addresses
  • In IPv4(Version 4 of the Internet Protocol) an address is a string of 32 bits: network number (netid) + host number (hostid).
  • Three forms of addresses:
    • Class A: 0 + 7-bit netid + 24-bit hostid (largest network).
    • Class B: 10 +14-bit netid + 16-bit hostid (medium-sized network).
    • Class C: 110 +21-bit netid + 8-bit hostid (small network).
    • 1111111 is not available as netid of a class A network.
    • No hostids with all 0s or all 1s.
  • How many different IPv4 addresses are available?
slide5

Counting Internet Addresses

  • xa, xb and xc, denote the number of Class A, Class B and Class C addresses available, respectively.
  • Three forms of addresses:
    • Class A: 0 + 7-bit netid + 24-bit hostid (largest network)
    • Class B: 10 +14-bit netid + 16-bit hostid (medium-sized network)
    • Class C: 110 +21-bit netid + 8-bit hostid (small network)
  • xa= (27-1)*(224-2)=2,130,706,178.
  • xb=214*(216-2)=1,073,709,056.
  • xc=221*(28-2)=532,676,608.
  • Total=xa+xb+xc=3,737,091,842.
slide10

E1

E1∩ E2

E2

| E | = | E1 U E2 | = |E1| + |E2| - | E1∩ E2 |

slide12

How many bit strings of length eight either start with an 1 bit or end with the two bits 00?

E= { all such bit strings }

E1 = { bit strings start with 1} { 1 xxxxxxx }

E2 = { bit strings end with 00} {xxxxxx00}

E1∩ E2= {bits start with 1 and end with 00} {1xxxxx00}

|E1| = 27 |E2| = 26 | E1∩ E2 | = 25

| E | = | E1 U E2 | = |E1| + |E2| - | E1∩ E2 | = 128+64-32=160.

problem
Problem
  • How many positive integers between 100 and 999 inclusive
    • are divisible by 7?
    • are odd?
    • have the same three decimal digits?
    • are not divisible by 4?
    • are divisible by 3 or 4
    • are not divisible by either 3 or 4
    • are divisible by 3 but not by 4?
    • are divisible by 3 and 4?
problem17
Problem
  • How many functions are there from the set { 1, 2, . . ., n }, where n is a positive integer to the set { 0, 1 }

-- that are one-to-one?

    • that assign 0 to both 1 and n?
    • that assign 1 to exactly one of the positive integers less than n?
problem18
Problem
  • A Palindrome is a string whose reversal is identical to itself (e.g., madam, tenet) How many bit strings of length n are palindromes?

n is even,

n is odd

slide19

Pigeonhole

Pigeonhole

slide20

Pigeonhole

Pigeonhole

Pigeonhole

slide22

Pigeonhole

Pigeonhole

Pigeonhole

example application of pigeonhole principle
Example (Application of pigeonhole principle)

Among any n+1 positive integers not exceeding 2n, there must be an integer that divides one of the other integers.

Take n =5, n+1 =6, 2n=10

{1 4 7 8 9 10} 4 divides 8

Take n=10, n+1 =11, 2n=20

{1 3 4 6 8 9 11 15 17 19 20} 3 divides 6

slide25

Among any n+1 positive integers not exceeding 2n, there must be an integer that divides one of the other integers.

Solution: a1, a2, … an+1

Write aj = 2kj qj

qj are odd numbers less than 2n, and there are at most

n such qs.

So by pigeonhole principle, two of the qs must be equal (say qi = qj)

ai = 2ki qi

aj = 2kj qj

if ki < kj, then ai divides aj

slide26

1. What is the minimum number of students, each of whom comes from one of the 50 states, who must be enrolled in a university to guarantee that there are at least 100 who come from the same state?

2. How many numbers must be selected from the set {1, 2, 3, 4, 5, 6} to guarantee that at least one pair of these numbers add up to 7?

3. Show that among any group of five (not necessarily consecutive) integers, there are two with the same remainder when divided by 4.

announcement
Announcement
  • 2nd Midterm next Monday (any objection?).
  • Grade worksheet and extra credit assignment will be available next Monday.
  • One more chapter to cover (Chapter 9).

Today

6 7 8 9 10 11 12

13 14 15 16 17 18 19

20 21 22

282930

All grades except final

5th quiz

Final Exam

2nd Midterm

EC assignment due

slide28

3 Florida has 16 million people. At least N people in Florida have the same three initials and were born on the same day of the year (not necessarily in the same year). What is N?

4 Let (xi, yi), i= 1, 2, 3, 4, 5, be a set of five distinct points with integer coordinates in the xy plane. Show that the midpoint of the line joining at least one pair of these points has integer coordinates.

slide29

(3, 2)

(-2, 1)

(0.5, 1.5)

(2, 0.5)

(-0.5, 0)

(1,- 1)

Three midpoints all have fractional coordinates

5 3 permutations and combinations
5.3 Permutations and Combinations

Permutation (of a set of n distinct objects) is an orderedarrangement of these objects. (The number of all possible permutations is denoted as P(n).)

r-Permutation is an ordered arrangement of r elements of a set. (The number of r-permutations is denoted as P(n, r).)

The set (of 6 elements)

2 different permutations

2 different 3-permutations

5 3 permutations and combinations31
5.3 Permutations and Combinations

In how many ways can we select three students from a group of five students to stand in line for a picture? In how many ways can we arrange all five of these students in a line for a picture?

  • 5 ways to select the first student, 4 ways to select the second student, 3 ways to select the last student …… 5*4*3=60
  • 5 ways to select the first student, 4 ways to select the second student …… 5*4*3*2*1 = 5!
5 3 permutations and combinations32
5.3 Permutations and Combinations

Theorem: P(n, r) = n *(n-1)*(n-2) …. (n-r+1).

Corollary 1: P(n, r) = n! / (n-r)!.

Problem: How many ways are there to select a first-prize winner, a second-prize winner and a third-prize winner from 100 different people who have entered a contest?

Problem: How many permutations of the letters ABCDEFGH contain the string ABC?

5 3 permutations and combinations33
5.3 Permutations and Combinations

r-combination (of a set of n distinct elements) is an unordered

selection of r elements from the set. (The number of all

r-combinations is denoted as C(n, r).)

The set (of 6 elements)

3 different 3-permutations

2 different 3-combinations

Same selection (different ordering)

5 3 permutations and combinations34
5.3 Permutations and Combinations

For example C(4, 2)= 6

{a, b, c, d } => {a, b}, {a, c}, {a, d}, {b, d}, {b, c} {c, d}

6 = 4! / (2! (4-2)! )

There are 4*3 r-permutations: each combination gives 2 2-permutations

Theorem: C(n, r) = n! / (r ! (n-r) ! )

Proof: r-permutations of the set can be obtained by forming the C(n, r) r-combincations of the set and then ordering the elements in each r-Combination.

P(n, r)= C(n, r) * P(r, r)

5 3 permutations and combinations35
5.3 Permutations and Combinations

Corollary C(n, r) = C(n, n-r).

C(n, r) = n! /(r! (n-r)! ) = C(n, n-r).

Picking r-elements from the set is the same as picking the remaining n-r elements.

Example: How many poker hands of five cards can be dealt from a standard deck of 52 cards? Also, how many ways are there to select 47 cards from a standard deck of 52 cards?

C(52, 5) for both.

5 3 permutations and combinations36
5.3 Permutations and Combinations

How many bit strings of length n contain exactly r 1s?

(Each such bit string is determined by the positions of 1s in the bit string)

How many ways are there for 5 women and 3 men to stand in a line so that no two men stand next to each other?

P(5, 5)*P(6, 3)