Question

comm & networks fund

12. (8pts) Change the following IP addresses from dotted- decimal notation to binary notation. a. 197.163.222.19 b. Show in details how to convert 197 into binary 13. (&pts) Change the following IP addresses from binary notation to dotted-decimal notation. a. 11011111 10110000 00011111 01011101 b. Show in details how to convert 01011101 into decimal 14. (6pts) Find the class of the following IP addresses (i.e., classful addressing) a. 11110111 11110011 10000111 11011101 b. 10101111 11000000 11110000 00011101 c. 11011111 10110000 00011111 01011101 15. (6pts) Find the netid and the hostid of the following IP addresses a. 114.34.2.8 b. 132.56.8.6 c. 208.34.54.12 16. (6pts) in a block of addresses, we know the IP address of one host is 25.34.12.56/18. a. What is the first address network address) in this block? b. What is the last address (limited broadcast address) in this block? c. How many addresses are there in this block of addresses? 17. (4pts) Which fields of the IPv4 header change from router to router? 18. (10pts) An IPv4 datagram has arrived with the following information in the header in hexadecimal): 4A 05 02 B411 03 30 A4 1B 06 58 50 7C 4E 03 02 BA 3C 13 72 a. What is the header size? b. Are there any options? c. What is the offset value ? d. What is the size of the data? e. How many more routers can the packet travel to? f. What is the identification number of the packet? g. What is the type of service? h. What is the destination IP address? i. Is this the first, middle or last fragment? Why?

Answer 1

192.163. 222 ) 9 197 in binary fan 2 | 19? 2 198 - ) 2 | 49 -0 d 4 4 4 4 Cd 2 12 4-1 / 197) 1 1 00 010 1), !! 2 - 0 ( 10 ) ordes Write ithe bits in reverse from bottom to top. (222 1l-0 55 - 81 -| 23 - ) 20 - 0 lg 10 O DI ) 10 4 170 Plo 2-1 6 - c\-|9s \S 19 2-1 -0 2 | 3 - 0 + / - / -) | | 16001 insbit (poo doo | 192.168 222 -19 15 24 (6 2 10 | St CScanned with CamScanner 00101 +10 10 0g/. 0| 11/0 0001/00 1 ,

11011111 10110000 00011111 01011101 11011111 conversion 1101//// 20 t 27 26 25 24 23 22 2 1827 +1+2° +0x25+ 1x2 + 1x2² +1x 2412 + x20 128+64 +0+16+8+4 +2+) 223. 10110000 = 1 x 2 + 1x 2 + 1 x 24 128 +32+16=176 0001111! 7 1678+4+2+1 =31 01011101-7X 267 64 +1% 2% +1x2+2x2°=64+1678+4 +1=93 So the answer is 223.176.31.93. following IP addresses 14. class of the Q: 11110111, 11110011 1000 011111011101 Boi 247243. 135, 221.. Class A 71-126 class B 128-11 247 Belongs to Clan C- 192-223 class E R clano=224 to 239 Clan E²2.40 to 255 CScanned with CamScanner

b TUTUI!! TUUUUUU MITTUU on lol 128 +32 +15. 128+64 1 28 +64 +32 46 . 32+16+841 175. 192.240.57 125 - B«Jamp to cуам в 0 Dose a olare 10 lat 1100 lard 111on C. 11011111.10110000 COLLVO 11101 . class c Belongs to class c 15. Net id I a. 114.34.2.8. and host id class A U class A Netid → 114.0.0.0 Hostid != 0. 34.2.8 class Air One byte retid and three bestes bastid - Clar Bi- Two bytes netid and two by terhesie class 3: Three byte netice and One byte hast it b. 132.56.8-6 Net id = 132.56.0-0 Hostin= 0.0.8-6 class c. 208..34:54:12 Nenid = 208-34-540 hostid - 0.0.0.12 CScanned with CamScanner

. know the IP address of 16. In a - pre host block of address, are is 25-34.12-56118 . This is a class A address : 2 catubali frist addsen (network address Default subnet mask 255.0.0.0 ved Addresses are Hello New sub net mask 180 255.dodendbbed 1000000-OVOOO O meni 8. Netida 255.255. 19200 Hestia total no of bits borrowed Sombost Total address (R indicates 2n (36= 20 = 1024 No of hosts in each network 214 -2 bits) (14 > no of remarning bad First - 16382 address :-- mask is anded with ZP. 25.34.12.56 AND. 225.225. 192. o 25. 3400.0 Last Address :- Limited Broadcast = 25.34.00 111111111111111 = 25.34.63.255 CScanned with CamScanner