Thank You

Congratulation guys passed CCNA. I hope you can share new watson vce files. Please send it to my email khoaanh1989@gmail.com

Thanks.

A maximum of 50 hosts per subnet is a maximum of 50 hosts per subnet !

hpaul602@gmail.com

tnx

10.0.00000000.0
10.0.00000001.0 – 1
10.0.00000010.0 – 2
10.0.00000011.0 – 3

1st octet – 8 bits
2nd octet – 8 bits
3rd octet – 6 bits that are the same

All in all 22 bits that are all the same.

Could anyone explain following. Why it shouldn’t be /16 because only first 2 octets are not changed.

we can guess the initial subnet mask is /24 because 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0 belong to different networks.

Q1 explanation:

60 hosts = 2^6 which means 6 bits is assigned for host bit. This equals /26 subnet mask.
Now we’ve two choices: 192.168.1.64 /26 or 192.168.1.56 /26.
Question is asking for subnet address, so we need to find the subnet for option 1:
- last octet .64 in binary = 0100 0000, subnet mask is /26 so the last 6 bits are 0s which
makes it equals 0100 0000 = .64 (192.168.1.64)
- broadcast address (the last 6 bits are 1s) = 0111 1111 = 127 (192.168.1.127)

Option 2:
- last octet .56 = 0011 1000, network address (last 6 bits are 0s) = 0000 0000 (192.168.1.0)
- broadcast address = 0011 1111 = 63 (192.168.1.63)

From the results: options 1 is clearly the network address, while option 2 is one of the usable address. Hence, the answer is option 1.

Question asked for 7 LANs (7 subnets) = 2^3 which means 3 bits is assigned for subnet. It’s using class C address, which has by default 24 bits for network.
So we’ve 8 bits left to be used by the subnet and host (3 already being assigned for subnet), so we’ve 5 bits left for host = 2^5 = 32 hosts. Remember, first and last available IP are used for network and broadcast, which left us 30 – 2 = 30 available usable IP addresses.

Since the question asked for the “most efficient” summarized route, you need to pick the answer that includes all the routes listed, but limits any additional routes (as much as possible). Answer “C” offers all the routes listed in the picture and advertises no additional routes to R2.

In addition to the routes shown in the picture, answer “B” would advertise 172.1.0.0 – 172.1.3.255 to R2 and any packets for those network(s) would be routed toward R1. Since R1 does not have those network(s), all those packets would either be dropped by R1 or be caught in a routing loop if R1 had a default gateway pointing to R2.

Hope that helps.

Thanks.

for the following IP adress 130.22.40.1 and 194.174.3.2

a) maximum hosts is 60 for network( subnet) ?

b) Bits number of hosts?

c) Bits numbers of network ( subnet)?

d) Broadcast and network (subnet) ID?

To summarise you have to include all subnets (it’s like subnetting a subnet)

The octet position and block size (increment) of combined networks now become important.

Now the third octet has to be subnetted because it’s a summary of all networks.

A block or increment of 4 in the third octet is a /22 (252)

For Q8 this covers 172.1 .[4-8].0

Answer: C

Explanation

Network 172.1.4.0/25 and network 172.1.4.128/25 can be grouped to a single network 172.1.4.0/24

Network 172.1.4.0/24 + Network 172.1.5.0/24 + Network 172.1.6.0/24 + Network 172.1.7.0/24 can be grouped to a single network 172.1.4.0/22 because we have all 4 subnetworks so we can move left 2 bits (22=4).
++++++++++++++++++++++++++++++

why did it become /22????? if the 4 networkds are subnetted to /24

By the way, as it seems a lot of people are not too familiar with the ip subnet-zero command, it applies to the network and not the hosts. You always minus 2 for host calculation. ip subnet-zero (which is enabled by default) allows the use of the first and last subnets.

Please update when you get a chance. Is 9tut enough ?

Traditionally, it was strongly recommended that subnet zero and the all-ones subnet not be used for addressing. According to RFC 950 leavingcisco.com, “It is useful to preserve and extend the interpretation of these special (network and broadcast) addresses in subnetted networks. This means the values of all zeros and all ones in the subnet field should not be assigned to actual (physical) subnets.” This is the reason why network engineers required to calculate the number of subnets obtained by borrowing three bits would calculate 23-2 (6) and not 23 (8). The -2 takes into account that subnet zero and the all-ones subnet are not used traditionally.”

“RIP version 1 is the only routing protocol in use on the network and subnet 0 is not being used, ” are we safe to assume that the router has the no ip suibnet zero command configured? If so then are we not counting the first LAN 0 Subnet and the last broadcast subnet?

Last Octect – 00000000 – Typically if we are looking for 7 subnets then we would need to take 3 bits with 2^3= 8 subnets- 11100000. Total of 8 subnets which meets the LAN requirements but we can’t use the first and last subnets if the no ip subnet zero is applied where the new subnet formula becomes 2^3-2 = 6 subnets which does not meet the LAN requirements.

So I was thinking could the answer really be C. 14 since if we borrow 4 bits then it would 2^4-2 = 14 subnets. Or if it’s not a true no ip subnet zero formula and we are just ignoring the very first 0 subnet then 2^3 = 8 – 1 subnet = 7. Where now we have exactly 7 subnets not counting only the 0 subnet.

This question has really confused me lol, if someone can shed some further light it would be appreciated.

Thank you.

It would be appreciate .