Answer Posted / jitendera kumar sinha

172.17.32.0/23
mask of that ip is
sunated in the
172.17.32.0/27
sunating of 4 bit
total number of n/w
2^4-2=14(according to cisco)
now taol nomber of valabe bit for host is 16-11=5
so taotal nomber of avilable host is 2^5-2=30
note**(why eleven is redused from 16.ans becoz this is a
xlass b adress and it have 16 bit for n/w and 16 bit for
host but we have the mask of 27 then we have taken 11 bit
from host for n/w.so we redused that eleven bit from the 16)
now
now hat is the mask of that ip
255.255.128+64+32+16+8+4+2+1.128+64+32
i.e
mask is 255.255.255.224/27
so the block size is 256-224=32
so the n/w is
172.17.32.0---------------------.31
.32------------------------------.63 and so on

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