An isp is granted a block of addresses starting with 120.60.4.0/20. the isp wants to distribute these blocks to 100 organizations with each organization receiving 8 addresses only. design the subblocks and give the slash notation for each subblock. find out how many addresses are still available after these allocations.

Answer

896 addresses

Explanation

Let's assume that the addresses means that the broadcast and the network ID in the allocated block can be included.

Also, note that there are /20.

First, we'll calculate the numbers of bits left by:.

2^(32-20)

= 2^12.

So, out of the remaining 12 bits left, we need to provide 100 subnets.

To solve this,

there are ceil(log 100) (logarithm in base 2).

Ceil(log 100)

= Ceil(6.64)

= 7.

The above points to the fact that 7 bits out of the 12 bits remaining, are for the Subnet ID part and remaining 5 (12 - 7) bits can be used for addresses allocated per subnet i.e. 2^5=32

But since we're using 100 Subnet IDs, that means there are (128 - 100)

= 28

28 IDs left in spare because we're making use of 7 bits.

This means that we can address a total of 128 networks but we are using just 100.

Subnet ID each having 32 addresses. So 28×32=896 addresses are left with the ISP.

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