Help please

Pranab Roy

[IMG]file:///C:/DOCUME%7E1/Pranab/LOCALS%7E1/Temp/moz-screenshot.png[/IMG] [FONT=&quot]
[/FONT][FONT=&quot][/FONT][FONT=&quot] 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?[/FONT]
answer is [FONT=&quot]255.255.255.192, but i could not understand i t. please are you help me.???
[/FONT]

chopsticks

Pranab Roy wrote: »

[FONT=&quot]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.[/FONT]
[FONT=&quot]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?[/FONT]
answer is [FONT=&quot]255.255.255.192, but i could not understand i t. please are you help me.???[/FONT]

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 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.