as i already have the solution to the first part, i only need the answers to e f
ID: 3703769 • Letter: A
Question
as i already have the solution to the first part, i only need the answers to e f g and h please!
(20pts) Suppose you are an ISP that owns a /17 address CIDR block starting at 241.172.128.0.
Answer the following questions to allocate address blocks to 8 customers who want to pay for the smallest CIDR blocks to support 31 (valid) users and you want to keep the remaining addresses in one large contiguous block.
What is the new submask in slash notation for each customer? /20
How many valid hosts can each customer actually allocate? 2^12-2=4094
Show start and end addresses of each sub-block?
in order from subnet 1 to subnet 8.
241.172.0.0 to 241.172.15.254
241.172.16.0 to 241.172.31.254
241.172.32.0 to 241.172.47.254
241.172.48.0 to 241.172.63.254
241.172.64.0 to 241.172.79.254
241.172.80.0 to 241.172.95.254
241.172.96.0 to 241.172.111.254
241.172.112.0 to 241.172.117.254
How big is the remaining CIDR block? 20 bits long
Now divide the remaining contiguous large block for 5 more customers that want to pay for the smallest CIDR block to support 14 users and you want to keep the remaining address in one large contiguous block.
e. What is the new submask in slash notation for each customer?
f. How many customers can actually be added using this contiguous large block?
g. Show starting and ending addresses of each subblock?
h. How big is the remaining CIDR block after the 5 customers have been allocated their blocks?
Explanation / Answer
As per your wanting
for each customer new CIDR block slash notation will be /20.Since to assign the new subnet to 8 customers we need
The 3 bits will be taken from the host part of the /17.
so now the new slash will become /20.
After this 12 bits are there that means the no. of valid hosts can be allocate are
2 hosts are subtracted because they are the subnet ID and broadcast address of the network.
Start and end of each sub-block are given below
In order from subnet 1 to subnet 8.
241.172.48.0 to 241.172.63.254
241.172.64.0 to 241.172.79.254
241.172.80.0 to 241.172.95.254
241.172.96.0 to 241.172.111.254
241.172.0.0 to 241.172.15.254
241.172.16.0 to 241.172.31.254
241.172.32.0 to 241.172.47.254
241.172.112.0 to 241.172.117.254
The remaining CIDR block is 20 bits long.
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