Exam 200-301 topic 1 question 16 discussion

For /28 network, There (2^4)=16 Subnets with each having (2^4-2)=14 host (14 +1 Network ID+ 1Broadcast ID)=16 Subnets are 192.168.16.0 192.168.16.16 ..... 192.168.16.128 192.168.16.144 (Above this network ID there will be address 192.168.16.143 which is a broadcast ID of Network 192.168.16.128

For anyone still confused, I break it down a bit easier: grab 192.168.16.143/28 10001111 is 143 in binary (the last octet is all we're worried about since it's /28). Draw your line at /28, 1000 | 1111. You remember how to get your network/broadcast, first/last? Notice all 1's at the right of the line. That usually means that's your broadcast right? And all 0's is your network. So we can see that this is actually a broadcast.

x.x.x.143 is the last address of the subnet range, It is reseved for broadcast purpose.

The network ID marks the beginning of a subnet and is always even. The broadcast ID marks the end of a subnet and is always odd.

D is correct. 192.168.16.143/28 is a broadcast IP for 192.168.16.128/28 subnet.

143=10001111 ,240=11110000 1111=last address in the subnet=broadcast

1. Find the network address of 192.168.16.143 /28 - 240 = (128 64 32 16) 8 4 2 1 = 128 + 64 +32 +16 -> binary of 255.255.255.240: 1111 0000 (240) -> binary of 192.168.16.143: 1000 1111 (143 = 128 + 8 + 4 +2 +1) -> network address in binary: 1000 0000 (128) -> network address: 192.168.16.128 2. Find the range: - /28: 240 in subnet -> 255 - 240 = 15 hosts - range of hosts: 192.168.16.128 to 192.168.16.143 - first usable: 192.168.16.129 and last usable: 192.168.16.142 -> 192.168.16.143 is a broadcast -> bad mask error ----- Now practice together guys, please solve this 10.10.10.130 /27 is the bad mask 1. find the subnet mask 2. find the network address (find binary of 10.10.10.130 and /27) 3. find range, first usable, and last usable

during the test do we have to do this calculation in our head or will they give us calculator.

Selected D

subnet address=192.168.16.128 Host address range = 192.168.16.129-192.168.16.142 broadcast address=192.168.16.143

D is correct Network: 192.168.16.128/28 Broadcast: 192.168.16.143 Usable IPs: 192.168.16.129 - 192.168.16.142

Not A: it's possible to configure a private IP address on an interface Not B: /28 is a prefix not a mask, and all routers support them Not C: network addresses are always even numbers (host part all 0's) D: broadcast addresses are always odd numbers (host part all 1's)

According to the mask 255.255.255.240 we take the last octet as reference and subtract to see the subnet increment: 256-240= 16. Then a multiple of 16 close to 143 is searched, in this case it is 144 which would be the following address of network and therefore 143 would be a broadcast address.

D is correct

D is definitely correct.

D is correct because the next subnet address is 192.168.16.144

zumy insegnami la via