Question

Given the class B network address 128.10.0.0 will be divided into subnets. Compute: a. The minimum number of bits necessary to obtain the number of subnets required. b. The maximum number of hosts in each subnet. c. The subnet mask. d. The dotted decimal IP address of the specified host on the specified subnet. (b) (c) Max # Of Subnets (a) Min # Of Subnet Bits (d) Dotted Decimal Max # of Hosts per Subnet Subnet Mask Subnet Host # IP Address 1,900 1,900 1,489 1,757 828 799 45 19 25 55 49 950 950 60 469

Answer 1

Class B 128.10.0.0 In class B, the first 16 bits forms network Id bits. 1) Max# of subuet = 1900. Flog,1900)!. .: a) Min # of subnet bits="!. 32f0+16) b) Max no. of host /subuet 2 -2 = = 2* 2:30 . 32- Y c) Subnet mask: 11111111 111111111111111 11100000 (make first 27 , 255.255.255.224. bits 1) d) Subnet #21489 Host # 19 HTRA 10111010 00010011 10000000 00001010 128.10.186.19. Max # of subuet = 1900 a) Min # of subuet bits = 11 b) Max # of host/subuet = 30 c) Subuet mask = 255.255.255.224. d) Subnet # : 1757 host #225 1000 0000 00001010 11011011 10111001 128.10.219. 185.

3) Max # of subuet =950 . a) Min # of subnet bits = log29507-10. 5) Max no. of host /subnet - 2** 22= 62. (10+16) 111000000 c) subnet mask: 1111111111111 255.255.255.192 Subnet # - 828 Host # = 55 10000000 0000101011001110 128.10.206.247. 1111011 4 Max # of subuet = 950 a) Min # of subnet bits = log2950] = 10 b) Mase no. of host/subnet - 62. 5) Subnet mask : 255.255.255.192. Subnet # 2799 Host # • 49 1000 0000 0000 1010 11000 111 10110001 128.10.199.177. 5 Max # of subnet = 60. a) Min # of subnet bits = log,607-6. b) Max. no. of host /subnet = 232-(1676) 32-22 =Q -210-2 =1022 c) Subnet mask: 11111111 m do 00000000 255.255.252.0. Subnet #:45 , host #4469 1000 000 00001010 10110 001 11010101 128.10. 177.2213.
Please upvote... Thanks...