An ISPis granted the block 16.12.64.0/20.The ISP needs to allocated address for 8organization each with 256 address.
Answers
Answer:
here this your answer
Explanation:
make a brilliant
Answer:
There would be 65,280 addresses remaining in the original block.
Explanation:
From the above question,
They have given :
An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536
addresses). The ISP needs to distribute these addresses to three groups of
customers as follows:
1) The first group has 64 customers; each needs 256 addresses.
2) The second group has 128 customers; each needs 128 addresses.
3) The third group has 128 customers; each needs 64 addresses.
4) Design the subblocks and find out how many addresses are still
available after these allocations.
Subblocks:
Group 1: 190.100.0.0/18 (262,144 addresses)
Group 2: 190.100.64.0/20 (40,960 addresses)
Group 3: 190.100.128.0/21 (20,480 addresses)
Addresses remaining: 65,536 - (262,144 + 40,960 + 20,480) = 21,952 addresses remaining.
Subnets are a way of dividing a larger network into smaller, more manageable segments. A subnet mask is used to identify the network and host portions of an IP address.
For example, the IP address 190.100.0.0/16 can be divided into four subnets, each with 64K addresses. The first subnet could be 190.100.0.0/18, which would allocate 256 addresses to each of the 64 customers in the first group. The second subnet could be 190.100.64.0/19, which would allocate 128 addresses to each of the 128 customers in the second group. The third subnet could be 190.100.128.0/20, which would allocate 64 addresses to each of the 128 customers in the third group.
After these allocations, there would be 65,280 addresses remaining in the original block.
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