Number of Hosts?

Dubuku57

Hi everyone,

I have a little VLSM problem..there is this qn i was attempting, let me try to describe the scenario..

3 switches, A - 7 hosts, B - 60 hosts, C- 66 hosts
They all connected to 1 router A, B, C respectively.

The Network to be subnetted is
192.168.151.0
ip subnet-zero configured and use RIP v2 as the routing
protocol.must address the network and at the same time converse unused addresses for future growth

Need to choose 4 adds out of the 6 below for the 3 switches and one serial link on Router B. The bolded are the correct answers. BUT i got 192.168.151.249/28 instead of .239/29 for the Switch A with 7 hosts. Doesnt /29 mean 8 hosts and thus only 8-2=6 usable ones?

192.168.151.249/28
192.168.151.224/30
192.168.151.239/29 - F0/0 on Router A linked to Switch A(7 hosts)
192.168.151.30/25 - F0/0 on Router C linked to Switch C(66 hosts)
192.168.151.225/30 - S0/1 on Router B linked to RouterA (WANlink)
192.168.151.149/26 - F0/0 on Router B linked to Switch B(60 hosts)

Pls helpings! Thank you all so much!

sandman748

You're correct. Not to mention 192.168.151.239/29 is a broadcast address and unusable.

blackngold4877

You are right. the /28 mask gives increments of 16, that network would be 192.168.151.240 /28, with .255 the broadcast and 14 usable addresses. As was stated, the .239 /29 address given the increments of 8 would put that in the 192.168.151.32 /29 network which would make .39 the broadcast address and unusable.

Dubuku57

Yay! Thank you so much!