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Thread: Help please

  1. Junior Member Registered Member
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    #1

    Default Help please

    [IMG]file:///C:/DOCUME%7E1/Pranab/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]
    IN this exhibit, two router was connected by wan . first router has 14 host and 10 servers and 2nd router has 33 hosts and 23 host.
    Refer to the exhibit. A network administrator has to develop an IP addressing scheme that uses the 192.168.1.0 /24 address space. The network that contains the serial link has already been addressed out of a separate range. Each network will be allocated the same number of host addresses. Which network mask will be appropriate to address the remaining networks?

    answer is 255.255.255.192, but i could not understand i t. please are you help me.???
    Last edited by Pranab Roy; 04-07-2011 at 08:34 AM. Reason: some word are need to add
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    #2
    Quote Originally Posted by Pranab Roy View Post
    IN this exhibit, two router was connected by wan . first router has 14 host and 10 servers and 2nd router has 33 hosts and 23 host.
    Refer to the exhibit. A network administrator has to develop an IP addressing scheme that uses the 192.168.1.0 /24 address space. The network that contains the serial link has already been addressed out of a separate range. Each network will be allocated the same number of host addresses. Which network mask will be appropriate to address the remaining networks?
    answer is 255.255.255.192, but i could not understand i t. please are you help me.???
    My understanding:

    Minimum IP addresses needed by first router = 14 hosts + 10 servers = 24 host addresses (the IP addresses used by the servers are also considered as host address)

    Minimum IP addresses needed by second router = 33 hosts + 23 host = 56 host addresses (hosts and host?)

    Condition 1: WAN IP addresses have been catered for, so we should not worry about it.

    Condition 2: Each network will be allocated the same number of host addresses

    Given: 192.168.1.0 /24 (/24 = 255.255.255.0)

    11000000 . 10101000 . 00000001 . 00000000 = 192.168.1.0
    11111111 . 11111111 . 11111111 . 00000000 = 255.255.255.0

    Host bits that we can borrow to create subnet are highlighted in blue and there are a total 8 of them.


    Condition 2 states that each network (network A & network B) will have the same host addresses, so we need to use the larger one out of the two calculated minimum host numbers ===> 24 & 56. Hence we will take 56 as the minimum host addresses for network A and network B.

    In order to cater for 56 hosts, we need to retain the appropriate host bits. Let's try 2^5 and the answer is 32, so since it is smaller than 56, it is not suitable. Let's try for 2^6, and the answer is 64. Are 64 host addresses arge enough to cater for the requirement of 56 minimum host addresses? Absolutely. (Usable hosts are [2^6] - 2 = 62].

    Now come back to visit our default and given subnet mask:

    11111111 . 11111111 . 11111111 . 00000000 = 255.255.255.0

    Since we are to retain 6 host bits for our 64 host addressses, we will be left with two host bits to borrow from to create new subnets, by turning 0 into 1:

    11111111 . 11111111 . 11111111 . 11000000 = 255.255.255.192

    And there you go, you've got your answer.
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