An ISP has CIDR block 156.94.0.0/16. The ISP wants to create 18 sub-blocks from

Question

An ISP has CIDR block 156.94.0.0/16. The ISP wants to create 18 sub-blocks from this block.
a. How many bits should the ISP add to the mask for the sub-blocks?
b. What are the addresses (include slash notation) for each sub-block?
c. What is the maximum number of hosts in each sub-block?

Assume that NJIT has a CIDR address space 128.235.204.0/19. Which of the following addresses could be host addresses within that space?
a. 135.235.160.0
b. 135.235.204.117
c. 136.235.205.1
d. 135.235.204.255
e. 135.236.207.223


We wish to assign a CIDR block of 1024 addresses (including all 0’s and all 1’s), starting at 128.211.192.0.
a. What is the maximum address in this block? [Hint: figure out how many bits are required to represent 1024 hosts. The maximum address is when all those bits are 1]
b. How many bits are in the host id?
c. How many bits are in the net id?
d. What is the mask in binary, dotted decimal, and slash notation?

Explanation / Answer

a. In order to create an additional 18 blocks you would need to allocate another 5 bits to the mask, which would actually give you 32 more sub-blocks. b. 156.94.0.0 156.94.8.0 156.94.16.0 156.94.24.0 156.94.32.0 156.94.40.0 156.94.48.0 156.94.56.0 156.94.64.0 156.94.72.0 156.94.80.0 156.94.88.0 156.94.96.0 156.94.104.0 156.94.112.0 156.94.120.0 156.94.128.0 156.94.136.0 c. Each network can support up to 2046 hosts. 2. a. The maximum address in this block is 128.211.195.255, which is a broadcast address. b. There are 10 bits in the host ID. c. There are 22 bits in the net id. d. The mask is 11111111111111111111110000000000 11111111.11111111.11111100.00000000 /22

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