Now we are left with 128-100 =28 Subnet ID each having 8 addresses. So 28×8=

224addresses are left with the ISP.

@Abbas2131

Didn't got this, It will be really appreciated if you can elaborate.

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An ISP is granted a block of addresses starting with 120.60.4.0/22. The ISP wants to distribute these blocks to 100 organizations with each organization receiving just eight addresses. Design the subblocks and give the slash notation for each subblock. Find out how many addresses are still available after these allocations.

+1 vote

Assuming that by addresses we mean we can include the broadcast and the network ID in the block we allocate.

We have /22 means we have 10 bits in spare and 22 bits from the network ID part.

We need 100 subnets each of equal size, so SUBNET ID = ceil(100) = 7 bits and remaining 3 bits can be used for addresses allocated per subnet i.e. 2^3=8

So CIDR of first block = 120.60.4.0/**29**

Now we are left with 128-100 =28 Subnet ID each having 8 addresses. So 28×8=**224** addresses are left with the ISP.

0

Now we are left with 128-100 =28 Subnet ID each having 8 addresses. So 28×8=

224addresses are left with the ISP.

@Abbas2131

Didn't got this, It will be really appreciated if you can elaborate.

0

@Sumit Singh Chauhan

we have a CIDR of /22 so how many address are left ?

i.e. 2^{(32-22)} = 2^{10 }

out of the 10 bits remaining we need to provide 100 Subnets

so we need Ceil(log 100) i.e. 7 bits from the remaining 10 bits for the subnet ID part, but since we use only 100 subnet IDs we are left with 28 IDs in spare becuase using 7 bits we can address 128 Networks but we only used 100,

and since 28 networks are remaining and each network (subnet) has 8 hosts asssociated, we still have 28*8=224 hosts left.

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