Chapter04 - CHAPTER 4 IP Addresses: Classful Addressing...

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1 CHAPTER 4 IP Addresses: Classful Addressing Exercises 1. a. 2 8 = 256 b. 2 16 = 65536 c. 2 64 = 1.846744737 × 10 19 3. 3 10 = 59,049 5. a. 0x72220208 b. 0x810E0608 c. 0xD022360C d. 0xEE220201 e. 0xF1220208 7. a. 8/4 = 2 b. 16/4 = 4 c. 24/4 = 6 9. a. Class C b. Class D c. Class A d. Class B e. Class E
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SECTION 2 11. 13. This message must travel through a router because the netids of the two addresses are different (128.23 versus 128.45). 15. Network 8.0.0.0: Class A Network 131.45.0.0: Class B See Figure 4.1 . 17. Source address: 108.67.18.70 Destination address: 255.255.255.255 19. Source address: 123.27.19.24 Destination address: 0.67.89.56 21. An address such as x.y.z.t/32 means that the network is just one node. 23. Subtract the 2 addresses; the result is 146 + 96 × 256 = 24722. Subtracting 1, we get 24,721 addresses. 25. The first, second, and the fourth bytes can be easily found. We need to apply the AND operator to the third byte.
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This note was uploaded on 04/26/2010 for the course CSE CS501 taught by Professor Dmathur during the Winter '10 term at National Institute of Technology, Calicut.

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Chapter04 - CHAPTER 4 IP Addresses: Classful Addressing...

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