1) Assume a company has purchased a class C network using the IP base address of

Question

1) Assume a company has purchased a class C network using the IP base address of 192.168.111.0. The company has five departments containing 20, 25, 30, 35 and 40 user computers each. The class C network needs to be divided into five subnetworks, one for each department.

(a) Design how you would divide up your IP address range for these five subnetworks. For each subnet give the particular addresses the user computers would occupy, the likely address of the subnet's router, and the broadcast address.

(b) Assume a single router will direct traffic between the computers in these subnets and the internet. Write down the IP address/netmask combinations you would see in this router's forwarding table to handle this traffic.

Explanation / Answer

given ip is 192.168.111.0 this is class c so default subnet mask is 255.255.255.0

given number of hosts required =20, 25, 30, 35 and 40 =150

number of subnets required =5

so

2^s-2 = 5 (s- number of binary 1's in the requred subnet mask's host id.)

we get s=3

so subnet mask is 255.255.255.224 (255.255.255.11100000 host id has three 1's (11100000 -224 in binary))

number of hosts per subnet

2^h -2 (h- number of binary 0's in host id (11100000) so h=5)

number of hosts per subnet =30

but our requirement has 35,40 hosts per subnet which is greater than our hosts.so,we faild in allocating those hosts using subnet mask 255.255.255.224

there is a simple solution for this we can rather divide the network in to 4 subnets first and then afterwards we can divide it further

dividing into 4 subnets --> class c generally has 254 hosts (256-2 (that 2 corresponds to broadcast address and the initial address that is allocated to subnet))

so we get 62+62+62+62 (normally it would be 64 each but we subtract 2 for broadcast and network address)

for this we can use the mask 255.255.255.192(/26)

now we can divide it further

62+62+30+30+30+30 in which we need only 62+62+30+30+30

for this we can use the subnet mask 255.255.255.224(/27)

below table gives all the details of the subnets--

this table is arranged according to the size of allocation (ascending order)

the allocation of ip address starts with E and then followed by D and then C,B,A .

IP address/netmask combinations used for routing table---

Subnet Name Needed Size Allocated Size Address Mask Dec Mask Assignable Range Broadcast A 20 30 192.168.111.192 /27 255.255.255.224 192.168.111.193 - 192.168.111.222 192.168.111.223 B 25 30 192.168.111.160 /27 255.255.255.224 192.168.111.161 - 192.168.111.190 192.168.111.191 C 30 30 192.168.111.128 /27 255.255.255.224 192.168.111.129 - 192.168.111.158 192.168.111.159 D 35 62 192.168.111.64 /26 255.255.255.192 192.168.111.65 - 192.168.111.126 192.168.111.127 E 40 62 192.168.111.0 /26 255.255.255.192 192.168.111.1 - 192.168.111.62 192.168.111.63

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