Training test question

jekylljekyll Member Posts: 6 ■■□□□□□□□□
Which one of the following IP addresses is NOT assignable to a machine in a network?
a) 193.158.20.254
b) 54.1.0.0
c) 137.4.0.0
d) 192.178.0.254

Answer: C)

Why? What are the steps to come to that conclusion?

Comments

  • JDMurrayJDMurray MSIT InfoSec, CISSP, SSCP, GSEC, EnCE, C|EH, CySA+, PenTest+, CASP+, Security+ Surf City, USAAdmin Posts: 11,675 Admin
    Maybe the C answer option is supposed to read "127.4.0.0". No 127.0.0.0/8 IP address should be directly assignable to an IP interface.
  • EANxEANx Member Posts: 1,078 ■■■■■■■■□□
    C is a class B address where the first two octets identify the network. Since none of the remaining bits are set, the all 0s indicates the network address for that subnet mask.
  • kaijukaiju Member Posts: 420 ■■■■■■□□□□
    **Beware** It seems like that is a VCE **** question. I just corrected a VCE for a friend because I felt sorry for him. 25% of the questions were poorly worded, contained typos that made the question not answerable or the supposed correct answer actually was not correct.

    If you have the time and money I will suggest getting the Pearson test prep or Boson.

    Back to the question. I agree with JDMurray. 127.0.0.0/8 or 127.0.0.0 ~ 127.255.255.255 is the special used IPV4 loopback address network that will never be assigned to an interface.


    Work smarter NOT harder! Semper Gumby!
  • jekylljekyll Member Posts: 6 ■■□□□□□□□□
    edited July 2019
    EANx said:
    C is a class B address where the first two octets identify the network. Since none of the remaining bits are set, the all 0s indicates the network address for that subnet mask.
    I think you are right.
    In that case, and if I understood correctly, the following IP addresses wouldn't be valid either because they are also network addresses:
    54.0.0.0 (Class A)
    137.192.0.0 (Class B)
    192.192.192.0 (Class C)
    I made up these three addresses just to provide an example of what you are saying. This type of question is just about counting the octets of the default subnet mask and check if the IP addresses have an extra number outside the subnet mask octets or not. 

    Thank you and thanks to the others for participating also.
  • kaijukaiju Member Posts: 420 ■■■■■■□□□□
    I made the "trick quick question assumption w/o really looking at the question. Now that I look at it, if you cannot visualize the answer a glance it is best to apply the base /16 subnet mask for class B and then do the math:

    100001000.00000100.00000000.00000000 binary for 137.4.0.0
    11111111.11111111.00000000.00000000 binary for /16
    all bits in the address after the 16th are flipped to "0" to ascertain the network address and the same bits after the 16th are flipped to "1" to get the broadcast address. 

    provides ip space for
    100001000.000001000.00000000.00000000 - 100001000.00000111.11111111.11111111
    137.4.0.0 - 137.7.255.255




    Work smarter NOT harder! Semper Gumby!
  • lwwarnerlwwarner Member Posts: 147 ■■■□□□□□□□
    IMO, the question is bogus. Given an appropriate mask, 137.4.0.0 is a perfectly valid IP address. For example, 137.4.0.0/13 works just fine:

    2620-TS-FRS#sh run


    Bills-MacBook-Pro:~ lww$ ping 137.4.0.0


    As you can see, 137.4.0.0 can, in fact, be assigned "to a machine in a network." FYI, the output above is from a real router (137.4.0.0) and a real laptop (137.4.0.1). This is just one of those things that works as expected in the real world, but doesn't work in PT.

    Really, CIDR has been around since 1992, can't we just forget about IP address classes already?

  • JDMurrayJDMurray MSIT InfoSec, CISSP, SSCP, GSEC, EnCE, C|EH, CySA+, PenTest+, CASP+, Security+ Surf City, USAAdmin Posts: 11,675 Admin
    lwwarner said:
    IMO, the question is bogus. Given an appropriate mask, 137.4.0.0 is a perfectly valid IP address. For example, 137.4.0.0/13 works just fine:

    ...

    Really, CIDR has been around since 1992, can't we just forget about IP address classes already?

    This was my reasoning in thinking that the 137.4.0.0 answer option was a typo. Almost any IP address can be valid to assign to an interface given the ability to change the netmask. However, no netmask is given in this exam item, so we must assume the default (class-based) mask for each IP address given.

    There is some excellent thought happening in this discussion. Keep it up, people! :D
  • lwwarnerlwwarner Member Posts: 147 ■■■□□□□□□□
    Hi JD, I'm glad to see some familiar names are still here! I haven't been active here in a long time, but I'm gearing up to recertify before February so I'm back. Hopefully I can learn & contribute along the way!

    I'm not sure what I think about being tagged as an "Ancient Member" though... :o
  • JDMurrayJDMurray MSIT InfoSec, CISSP, SSCP, GSEC, EnCE, C|EH, CySA+, PenTest+, CASP+, Security+ Surf City, USAAdmin Posts: 11,675 Admin
    Dude, you and I are both "ancient" on this board!  :s
  • kaijukaiju Member Posts: 420 ■■■■■■□□□□
    I do agree that the absence of a mask requires the use of assumption to derive an answer. Totally unnecessary question. 
    Work smarter NOT harder! Semper Gumby!
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