subnneting doubts

Alzuaak12 · ‎10-05-2017

helloo from Spain, ive got this cuestion

Which of these hosts adresses could be assigned to a host of the subnet 192.168.15.19/28. Explain your answer...

(A) 192.168.15.17

(B) 192.168.15.14

(C) 192.168.15.29

(D) 192.168.15.15

(C) 192.168.15.31

is it possible the question is not correct written? i think it must to be 192.168.15.0/28 isnt it? please help

GRANT3779 · ‎10-05-2017

Only making an assumption but it may be a typo and actual question should read

192.168.15.16/28

Answer would be A and C

for the /28 your networks would be -

192.168.15.0

192.168.15.16

192.168.15.32

....... and so so forth.

A and C fit within the network specified in question.

Host range would be 192.168.15.17 through to 192.168.15.30

192.168.15.31 would not as it would be the broadcast address of that subnet.

Alzuaak12 · ‎10-05-2017

thank about your answer, could you explain me how do you obtain this
conclusion? i don understand in the question adress typed as a
192.168.15.19 is a host adress? or a subnetwork adress? and which is the
procedure you have employed...thanks you..

Bill Longman · ‎10-09-2017

@Alzuaak12, if you take the IP address that they give you, 192.168.15.19 and place it into a /28 subnet, you must arrive at the conclusion that the network there is 192.168.15.16 because /28 uses 4 bits for the hosts leaving 28 for the full network bits.

Address:   192.168.15.19        11000000.10101000.00001111.0001 0011
Netmask:   255.255.255.240 = 28 11111111.11111111.11111111.1111 0000
Wildcard:  0.0.0.15             00000000.00000000.00000000.0000 1111
Network:   192.168.15.16/28     11000000.10101000.00001111.0001 0000
Broadcast: 192.168.15.31        11000000.10101000.00001111.0001 1111

HostMin:   192.168.15.17        11000000.10101000.00001111.0001 0001
HostMax:   192.168.15.30        11000000.10101000.00001111.0001 1110

From here you can see that any IP address from .17 to .30 are valid as host IP addresses in this subnet.

If you count up from /24 to /28 to can remember how many hosts there are. You know that /24 is 8 bits (256) so /25 is 7 bits (128), /26 is 6 bits (64), /27 is 5 bits (32), and /28 is 4 bits (16). Which leaves you a network, a broadcast and 14 hosts.

Bill Longman · ‎10-09-2017

@Alzuaak12, if you take the IP address that they give you, 192.168.15.19 and place it into a /28 subnet, you must arrive at the conclusion that the network there is 192.168.15.16 because /28 uses 4 bits for the hosts leaving 28 for the full network bits.

Address:   192.168.15.19        11000000.10101000.00001111.0001 0011
Netmask:   255.255.255.240 = 28 11111111.11111111.11111111.1111 0000
Wildcard:  0.0.0.15             00000000.00000000.00000000.0000 1111
Network:   192.168.15.16/28     11000000.10101000.00001111.0001 0000
Broadcast: 192.168.15.31        11000000.10101000.00001111.0001 1111

HostMin:   192.168.15.17        11000000.10101000.00001111.0001 0001
HostMax:   192.168.15.30        11000000.10101000.00001111.0001 1110

From here you can see that any IP address from .17 to .30 are valid as host IP addresses in this subnet.

If you count up from /24 to /28 you can determine how many hosts there are. You know that /24 is 8 bits (256) so /25 is 7 bits (128), /26 is 6 bits (64), /27 is 5 bits (32), and /28 is 4 bits (16). Which leaves you a network, a broadcast and 14 hosts.