Ip addressing

MikdillyMikdilly Posts: 309Member
Was taking the ccna prep tests on Cisco's website, the 3rd test has a question that goes something like. 'You are using an address range of 172.18.0.0/23. What are the valid host addresses if subnet zero is usable?' (Choose three)

A) 172.18.0.0
B) 172.18.0.174
C) 172.18.1.0
D) 172.18.1.236
E) 172.18.1.255
F) 172.18.2.182

They claim the answers are B, C, and D, which does look correct. My question is wouldn't F be correct as well?

Comments

  • georgemcgeorgemc Posts: 429Member
    No, F would not be correct.

    The valid range of usable host addresses with subnet 0 enabled would be:

    172.18.0.1 - 172.18.1.254 with 172.18.0.0/23 being the subnet address and 172.18.1.255 being the broadcast. 172.18.2.182 falls into a different subnet.
    WGU BS: Business - Information Technology Management
    Start Date: 01 October 2012
    QFT1,PFIT in progress.
    TRANSFERRED/COMPLETED: AGC1,BBC1,LAE1,QBT1,LUT1,QLC1,QMC1,QLT1,IWC1,INC1,INT1,BVC1,CLC1,MGC1, CWV1 BNC1, LIT1,LWC1,QAT1,WFV1,EST1,EGC1,EGT1,IWT1,MKC1,MKT1,RWT1,FNT1,FNC1, BDC1,TPV1 REQUIRED:
  • redgoblinredgoblin Posts: 57Member ■■□□□□□□□□
    F would not be valid because using 172.18.2.0 network is only possible with a /22 address and they're specifically asking you to use a /23
  • markzabmarkzab Posts: 619Member
    For visualization purposes a /23 mask would look like:

    11111111.11111111.11111110.00000000 = /23

    The last bit in the 3rd octet has a value of 1, so therefor you could not have a 2 in the 3rd octet set for a host. A 2 in the 3rd octet would actually be the next available subnet after subnet 0.

    172.18.0.0
    172.18.2.0
    172.18.4.0
    172.18.6.0
    etc...
    "You, me, or nobody is gonna hit as hard as life. But it ain't how hard you hit; it's about how hard you can get hit, and keep moving forward. How much you can take, and keep moving forward. That's how winning is done!" - Rocky
  • malwethmalweth Posts: 42Member ■■□□□□□□□□
    I'm guessing that "subnet zero" being permitted confused you...

    The question is checking that you can subnet, but also that you understand "subnet zero." They want to see that you know answer "A" is not valid even with subnet zero. Unfortunately, specifying this may also seem to indicate that you're to iterate this over all subnets.

    What's implied by the network in question is that you are finding the first subnet, not all subnets in the class. Typically this would be specified (or a higher network to be subnetted will be).

    F is a host in the 2nd subnet.

    --- if "subnet zero" were disallowed answer "F" would be the only correct option.
    128  64  32  16  |   8   4   2   1
    128 192 224 240  | 248 252 254 255
     25  26  27  28  |  29  30  31  32
    
  • MikdillyMikdilly Posts: 309Member
    I was under the impression that since /23 is mask 255.255.254 and 256 minus 254 is 2, I would increment each subnet by 2 and any valid host address that fell into those increments could be used. The wording of the question asked for valid host addresses with subnet zero being used, it doesn't say just in the first subnet. Valid subnets would be 18.0 all the way up to 18.252.
  • markzabmarkzab Posts: 619Member
    Mikdilly wrote:
    I was under the impression that since /23 is mask 255.255.254 and 256 minus 254 is 2, I would increment each subnet by 2 and any valid host address that fell into those increments could be used. The wording of the question asked for valid host addresses with subnet zero being used, it doesn't say just in the first subnet. Valid subnets would be 18.0 all the way up to 18.252.

    Your subnets are going up by 2's, but you just didn't interpret the question properly. The first stament noted: "You are using an address range of 172.18.0.0/23. What are the valid host addresses if subnet zero is usable?"

    They are basically saying in the first sentence that the subnet they want you to focus on is the 172.18.0.0 subnet. The second sentence is supposed to reaffirm that subnet zero is being used so you can use the 0.0 subnet to find its hosts.
    "You, me, or nobody is gonna hit as hard as life. But it ain't how hard you hit; it's about how hard you can get hit, and keep moving forward. How much you can take, and keep moving forward. That's how winning is done!" - Rocky
  • MikdillyMikdilly Posts: 309Member
    Thanks for all of the replies, if i see address range on the exam i'll just concentrate on the network specified.
  • vjsousavjsousa Posts: 24Member ■□□□□□□□□□
    How can C be correct. That is not a valid Host IP address, it is a network ID.
    Vinny
  • markzabmarkzab Posts: 619Member
    vjsousa wrote:
    How can C be correct. That is not a valid Host IP address, it is a network ID.

    Your subnets jump by 2...

    172.18.0.0
    172.18.2.0
    172.18.4.0
    172.18.6.0

    1.0 is a host address.
    "You, me, or nobody is gonna hit as hard as life. But it ain't how hard you hit; it's about how hard you can get hit, and keep moving forward. How much you can take, and keep moving forward. That's how winning is done!" - Rocky
  • vjsousavjsousa Posts: 24Member ■□□□□□□□□□
    Please explain how C would be correct. The range of IP addresses for the first subnet is 172.18.0.1 - 172.18.1.254. Answer C doesn't make any sense.
    Vinny
  • widjerdwidjerd Posts: 17Member ■□□□□□□□□□
    vjsousa wrote:
    Please explain how C would be correct. The range of IP addresses for the first subnet is 172.18.0.1 - 172.18.1.254. Answer C doesn't make any sense.

    /23 mask lets you use the 8th bit in octlet 3 as a host address.
    The range you specified actually includes 172.18.1.0
  • NetstudentNetstudent Posts: 1,694Member
    If this were a class C address, then yes that would be a subnet address. This is class B.

    172.18.0.0 is the subnet, 0.1, 0.2, 0.3, 0.4
    > 0.254 then 0.255 are all hosts

    After 0.255 the third Octet will increment up to 1 and the fourth octet will start all over again starting with 172.18.1.0 ,1.1, 1.2, 1.3

    These are all valid hosts within 172.18.0.0/23
    There is no place like 127.0.0.1 BUT 209.62.5.3 is my 127.0.0.1 away from 127.0.0.1!
  • waruwaru Posts: 41Member ■■□□□□□□□□
    vjsousa wrote:
    Please explain how C would be correct. The range of IP addresses for the first subnet is 172.18.0.1 - 172.18.1.254. Answer C doesn't make any sense.

    You kind of explained it yourself. Going by the valid range of 172.18.0.1 - 172.18.1.254, 172.18.0.0 has to be the network address and 172.18.1.255 has to be the broadcast for the subnet. That being the case how could 172.18.1.0 be anything other than a valid host address? It is not possible to have 2 network addresses or 2 broadcast addresses in 1 subnet.

    Just because an address ends in a 0 does not automatically make it a network address.
Sign In or Register to comment.