Can someone show me this

protocol167protocol167 Posts: 20Member ■□□□□□□□□□
I know this is a easy one for you guys but was wondering if someone can explain how they came about with the broadcast ID and Network IP address.

Thanks

Of the following choices, which IP address should be assign to the pc host?

A. 192.168.5.5
B. 192.168.5.32
C. 192.168.5.40
D. 192.168.5.63
E. 192.168.5.75

Answer: C

Explanation:
The subnet mask used on this Ethernet segment is /27, which translate to
255.255.255.24. Valid hosts on the 192.168.5.33/27 subnet are 192.168.5.33 thru
192.168.5.62, with 192.168.5.32 used as the network IP address and 192.168.5.63
used as the broadcast IP address. Therefore, only choice C falls within the
usable IP range.

Comments

  • milliampmilliamp Posts: 135Member
    Of the following choices, which IP address should be assign to the pc host?

    A. 192.168.5.5
    B. 192.168.5.32
    C. 192.168.5.40
    D. 192.168.5.63
    E. 192.168.5.75

    The question cannot be answered on this information alone.
  • protocol167protocol167 Posts: 20Member ■□□□□□□□□□
    Here is the actual questions

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  • milliampmilliamp Posts: 135Member
    Subnetting is just working in binary.

    IP's are 32 bits. /27 means you have 5 host bits.

    Since we start counting at 0, 5 bits is 32 possible IP addresses.

    This means every 32 IP addresses begings a new subnet.

    Here are the boundries for /27.

    0-31
    32-63
    64-95
    96-127
    128-159
    160-193
    192-223
    224-255

    192.168.5.33/27 means it falls in the 32-63 range.

    It can't be .32 or 63 becasue they are the network and broadcast addresses.

    The only IP left that falls in that range is 192.168.5.40.
  • StoticStotic Posts: 248Member
    /27 = 255.255.255.224

    Take the 224 and subtract it from 256
    256-224=32

    32 is your block size.

    The first network starts at 32. Add your block size again (32) and you get 64 which will be your next network. Therefore your network is between .32 - .63. .32 is the network address and .63 is your broadcast address so an assignable host address has to be between .33 and .62. .40 is the only answer that fulfills this.
  • protocol167protocol167 Posts: 20Member ■□□□□□□□□□
    Awesome. Thanks guys for explaining it. I just wasn't sure where and how they got the .33 from. But now I understand it.

    Thanks!!!!!!
  • iDShaDoWiDShaDoW Posts: 67Member ■■□□□□□□□□
    I gotta agree with milliamp, with the original question stated that way, there's not enough information to get your answer.

    Did you just not bother to type the part in the original question about it using a /27 subnet mask?

    If they only give you the /27 subnet mask in the explanation after giving you the answer then the question is no good.
  • mastercormmastercorm Posts: 64Member ■■□□□□□□□□
    yea, i don't care for that question either. should be more information. the explanations are all good though.
    Working towards MCSE w/Security, then CCNA, then CCSP, and, eventually CISSP
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