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.
