You have an IP address of 172.16.7.160 and a subnet mask of 255.255.255.192

What is the address of the last host?

A. 172.16.7.190
B. 172.16.7.191
C. 172.16.7.254
D. 172.16.7.127

I say it's C. 254 Is that right?

The subnet mask is 255.255.255.192 Class B. That means you borrowed 10 bits. 192 = 128 and 64, 2 bits. 256-192=64. There is your first Network ID.
You start at 64.

N ID Usable B ID
64 65-126 127
128 129-190 191
192 193-254 255

As you can see from this table, 254 is the LAST usable host from this IP address.

Is this correct?

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Hello all

I think that Wrong!

Because 255-192=63 =>

IP: 172.16.7.160+(next 63 IPs)=

172.16.7.223



We have 63 Ip from 172.16.7.160 to 172.16.7.223

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Subnet mask of 255.255.255.192

This generates a total of 2,048 subnets, each with 64 addresses only 62 are usable, 2,046 subnets usable or 2,047 if you use ip subnet-zero on all your interfaces (remeber that this is a subnet mask applied to a class 'B' address, 172.16.0.0). Only the 10 leftmost bits are used in the netmask like this

11111111.11111111.11111111.11000000

Cool hunh?

Your networks will range from 2 to the nth power with the usable hosts only being the answer minus 2 ((2 to the n) minus 2) since you cannot use the first and the last addresses since they are the network number and the broadcast address address for the sub-network. So what are the ranges? Since they gave you an addres of 172.16.7.160, lets stay inside the 172.16.7.0 subnetwork.

0-63 is the first
64-127 is the second
128-191 is the third
192-255 is the fourth

How can I tell? Check out the last octet in binary (remeber that all zeros in the host portion mean network number and all 1's mean broadcast)

00000000 - 00111111 (0-63)
01000000 - 01111111 (64-127)
10000000 - 10111111 (128-191)
11000000 - 11111111 (192-255)

Now put your IP Host address in the ranges and it falls between 128-191. Since 191 is all ones in the host portion, that cannot be used as it is the broadcast address for this network. The address just before it or 190 is the last usable host in these ranges.

A) 172.16.7.190

So how do you determine this on the test quickly? There are tons of tricks but I found that just memorizing the values associated with the bits will always give quick results-

10000000 - 128
11000000 - 192
11100000 - 224
11110000 - 240
11111000 - 248
11111100 - 252
11111110 - 254
11111111 - 255

You also must know your powers of 2, this is true for the real world also. For masks, there won't be any other values because they have to be continous from left to right in order to work at all. By the time you take the test, you should be able to roll these numbers off your tongue.

Sorry for being so verbose, there were issues with other answers that I have given and I want to make sure that I am very clear on where I got the answer. Also helps me to insure that I didn't miss anything.

Hope that helps and good luck!!

__________________
A Black Hole is God dividing by zero.

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Subnet mask of 255.255.255.192 Class B

This generates a total of 2,048 subnets, each with 64 addresses only 62 are usable, 2,046 subnets usable or 2,047 if you use ip subnet-zero on all your interfaces (remeber that this is a subnet mask applied to a class 'B' address, 172.16.0.0). Only the 10 leftmost bits are used in the netmask like this

11111111.11111111.11111111.11000000

Are you sure it's 2048 subnets?
10 bits borrowed right? Well, 2^10-2=1022
I still don't see how you got 190 when 254 is the last usable host number.

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quote:

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Originally posted by username
Are you sure it's 2048 subnets?
10 bits borrowed right? Well, 2^10-2=1022

---

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Thank you Sir. I'll keep reading your post and practicing subnetting until it becomes clear.

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(*What you have to realize is that the question really asks "What is the address of the last host in the subnet to which 172.16.7.160 belongs?"*)

Thank you!!!!! Now I can see it. 160 fits between 129 and 190. Since 190 is the LAST usable host for that subnet, 190 is the answer they are looking for. YES!

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Here'another one...hope it helps..

Given:
172.16.7.160
255.255.255.192

Find the last available host?

Get the last octet of the subnetmask and subtract it from 256.

256 - 192 = 64 (no. of subnets)

Keep on adding to reach the fourth octet of 192
64 + 64 + 64 = 192 (stop here you reach the last subnet)

Now heres the rules:
1. The last number before the next available subnet is the broadcast address of that subnet.
2. The number before the broadcast of that subnet is the last available host for that subnet.

so... 172.16.7.191 is the broadcast
and...172.16.7.190 is the last host here...

hope you get it....

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Thank you! I went through the academy but subnetting was always fuzzy. My instructor's method was not very effective for a lot of the students. Many students in the end still couldn't get it.

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Your example is very easy. If you read my webpage on subnetting Todd Lammle is great. I put your example in my webpage and tried to explain the answer. The correct is (A) 172.16.7.190. My exaplanation is easier to understand. See my webpage.

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