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Question 16

Refer to the exhibit. Which statement explains the configuration error message that is received?

    Correct Answer: D

    The configuration error message appears because the IP address 192.168.16.143 with a subnet mask of 255.255.255.240 (/28) is a broadcast address. In a /28 network, each subnet has 16 IP addresses. The subnet that includes 192.168.16.143 has an IP range from 192.168.16.128 to 192.168.16.143, where 192.168.16.143 is the broadcast address and cannot be assigned to a device.

Discussion
ZUMYOption: D

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

ZUMY

Shortcut to find 1.First calculate subnets (barrowed 4 bits 2^4=16 subnets) or 256-240 (16) 2. Then do a math (256/16)=16 subnets, if so (144/16)=9 subnets so 144 is a subnet address and 143 is a broadcast address of previous network ID (128) it means (128+16)144

CustomexitOption: D

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.

pud

thx, that's really smart thinking. when all host bits are 1s means broadcast.

__sbOption: D

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)

Micah_TENGWAOption: D

D is correct because the next subnet address is 192.168.16.144

tonyisabelOption: D

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

marcojmnezOption: D

255.255.255.240 -->/28 Block size=256-240=16 Usable IPs. las part of the IP is 143 and there are 144 IPs to 0 to 143. 144/16=9 and hence 192.168.16.143 is a broadcast IP. Explained by Samitha

DatBroNZOption: D

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

Shaz313Option: D

D is definitely correct.

AlsaherOption: D

D is correct

tannucOption: D

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

LazizS

- 224 = (128+64+32)16 8 4 2= 128+64+32 - binary of 255.255.255.224: 1110 0000(224) - binari of 10.10.10.130: 1000 0010 - network address 128 -255-224 = 31 hosts - range of hosts - 10.10.10.128 to 10.10.10.158 - 10.10.10.129 first usable address -10.10.10.159 broadcast address

Giuseppe_001

zumy insegnami la via

ZUMYOption: D

D is correct. If list out the subnet for /28 It will be like 192.168.16.0 192.168.16.16 .. 192.168. 16.128 -------> here the last IP is 192.168. 16.143 is a broadcast 192.168.16.144 192.168.16.160.. Last 192.168. 16.240

ZUMY

Dear moderator Please remove this comment.Thx

kalistroOption: D

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.

aman87Option: D

D is correct

MattschwimOption: D

D is the correct choice Contact me for full questions <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="deb3bfaaaab7aab7ad9eb1abaab2b1b1b5f0bdb1b3">[email protected]</a>

Starlord2535

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

MSTAHIROption: D

Selected D