Question

6. Review: IP Addresses and Masks In this problem we are going to review IP addresses and bit masks. An IP address has 32 bits (4 bytes) and is written in dotted decimal form. This means that there are 4 decimal numbers, each of which is followed by a decimal point, such as 172.39.168.1 A mask can also be written in dotted decimal form. This problem will show you how to perform bitwise AND of an IP address and a mask. Remember that for each decimal number you will need 8 binary digits so that your binary version of the IP address has 32 bits. IP Address is 172.39.168.1 Mask is 255.255.0.0 a. Write the IP address in binary. Keep a few spaces between each byte to make it easier to read. b. Write the mask in binary. Keep a few spaces between each byte to make it easier to read. c. Perform the bitwise AND of the binary IP address and the binary mask: d. Write the result found in (c) in dotted decimal format: e. Compare the dotted decimal IP Address and mask. Does it surprise you to see the Os in your answer in part (d)? Think about why the Os are in the position they are in. f. What does the mask 255.255.0.0 tell you? g. Thinking about your answer in step (e), try to perform the bitwise AND of the following sets WITHOUT making any conversions to binary and write the result in dotted decimal format: a. 172.39.168.1 AND 255.255.255.255 b. 172.39.168.1 AND 255.255.255.0 c. 172.39.168.1 AND 255.255.0.0 d. 172.39.168.1 AND 255.0.0.0

Answer 1

01 "72:39.168.1 IP Address Maska Toollooloo oololol = 0030 Mask 255.255.0.0 10 10 1200 (9) 172 in binasy = = 0 0 1 0 0 11 in =10101000 - 00000001 265 0 = 11111112101 = 0000 00.00. 1 Now ulice IP Address in binasy ssrin 6 s 101010000100111101010000000000|| osa ③ Mask in bin any = llllllllllllllllllL 0000000000000000 © Bitwise AND do IP an B mask= 10101100 oolool11 10101000 oool oool Till li lillllll DODDOO 00 0.0000 DO AND 101011000010011000OOOOo Oooooooo

Now the bitwise And of IP and mask = 10101100 00/0011) OODOD000 00000000 Now asitting it in to decimal 10101100 172 is ! no 100111 39 = OD000000-00 Now in decimal format 172.39:0901 - © Given Ip was 172.39.168.1 you aolo Given Given mask » 255.255.0.0 Bitwise AND of 172.39:00 IP and mask o o since A subnet mash is usef to tell the network devices about locae) addresses and remote addresses .. Here o means in subh ef mask that bitwire AND will be zero resulting no network addresses. lt o no netu dan 10

f mask = 255.255.0.0 This is the default subnet mask which is used by class 'B' networks. in class address so total number of B = 256X256 = 65,536 lei masx 255.255.0.0 means that . 16 bits are use B for network... Address while grest 16 bits are used for Host Address. I Bitwise AND Bitwise and can be pe tormed using operator (&). AND 255.255.255.255 a) 172.39.168.1 Approach. I since 255 a llill112 so performing Bitwise And with IP (Any decimal) and 255 coill results equal to decimali ed: 172 ~ 10101100 Puct 255.2 IIIII!! 10101100 = 172. AND with 255 Give the so performing same

ach-I use Bitweise And obecator. ega 172 & 255 Approach = 172 Approgor awestion coith now solving rest -I. @ 17239.168.1 Cc. 255 same as IP because subnet mask=255 for AND 255.250 Past egoh P Egoh 17.39.168.1 6 172.39.168.1 and 255255.255.0 d o QUA su becall QUAS mask=0 - 172.39. 168.04 - © 172. 39.168.1 AND 255.255.0.0 172,39.0.0 0 172- 39.168.1 AND 255.0.0.0 I 172.0.0.0

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