Consider a subnet with the block 125.119.40.192/26. Write out the IPv4 address 1
ID: 3807282 • Letter: C
Question
Consider a subnet with the block 125.119.40.192/26. Write out the IPv4 address 125.119.40.192 in binary. Write out the mask for this block in binary. What are the first and last addresses in this block? Write this in dotted decimal notation only. Suppose an ISP wants to make four subnets from this block, with each block having the same number of lP addresses. Give the blocks (using the a.b.c.d/x notation) for these four subnets. What field in the IP header can be used to to ensure that a packet is forwarded through no more than N routers? If a router has k interfaces, how many IP addresses does it have? Does IPv6 have a fixed or variable length header? Do machines within the same subnet communicate to each other through a router? Must a new machine joining subnet S get its IP address from a DHCP server from inside the same subnet S?Explanation / Answer
a>
125 --> 01111101
119 --> 01110111
40 --> 00101000
192 --> 11000000
Therefore binary representation is concatenation of the binary data:
01111101011101110010100011000000
b> mask : 11111111111111111111111111000000 (26 ones followed by 32 -26 zeros)
c>
first address is
01111101011101110010100011000000 or 125.119.40.192
last address is
01111101011101110010100011111111 or 125.119.40.255
d>
Since block uses /26 masking value, we have 6 unmasked values. Using this 6 bits we can represent 64 different IPs
If we divide into 4 equal subgroups, each will have 16 IPs, which need 4bits, so the mask is /28 (32-4)
The address of 4 blocks would be
01111101011101110010100011000000 or 125.119.40.192/28
01111101011101110010100011010000 or 125.119.40.208/28
(192 + 16 = 208)
01111101011101110010100011100000 or 125.119.40.224/28
(208 + 16 = 224)
01111101011101110010100011110000 or 125.119.40.240/28
(224 +16 = 240)