Subnetting Practice Questions: Test 1

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Subnetting Practice Work:

1. Find the Subnet
2. Find the 1st Usable Host
3. Find the Last Usable Host
4. Find the Broadcast Address

OR

You will be asked if the IP Address is a valid Host address or a valid Host address for a particular Subnet.

For Example:

124.59.10.1 /10

1. Find the Subnet = 124.0.0.0
2. Find the 1st Usable Host = 124.0.0.1
3. Find the Last Usable Host = 124.63.255.254
4. Find the Broadcast Address = 124.63.255.255

Please Note I will post the answers tomorrow for you to check your work.


1.) 55.110.67.205 /16





2.) 88.248.235.250 255.255.255.248





3.) 56.58.128.76 255.128.0.0





4.) 198.13.70.25 255.255.255.192





5.) 150.75.222.94 /18





6.) 198.134.190.70 /28





7.) 47.165.237.34 /21





8.) Is Ip address 79.246.255.1 with a subnet mask 255.224.0.0 a valid host ip address?





9.) 107.30.205.80 255.255.255.240





10.) Is 193.43.65.127 with a 255.255.255.192 mask a vaild Host IP address?

 

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Subnetting Practice Work:

1. Find the Subnet
2. Find the 1st Usable Host
3. Find the Last Usable Host
4. Find the Broadcast Address

OR

You will be asked if the IP Address is a valid Host address or a valid Host address for a particular Subnet.

For Example:

124.59.10.1 /10

1. Find the Subnet = 124.0.0.0
2. Find the 1st Usable Host = 124.0.0.1
3. Find the Last Usable Host = 124.63.255.254
4. Find the Broadcast Address = 124.63.255.255

1.) 157.129.136.141 255.255.255.224

2.) 44.204.221.73 /27

3.) 181.39.102.219 255.255.128.0

4.) 145.177.71.142 255.255.254.0

5.) Host 215.83.56.220 /27 can't ping Host 215.83.56.225 /27. The newly hired Network Admin can't figure out why since they both have the same /27 subnet mask. Whats wrong?

 

Subnetting Practice Work:

1. Find the Subnet
2. Find the 1st Usable Host
3. Find the Last Usable Host
4. Find the Broadcast Address

OR

You will be asked if the IP Address is a valid Host address or a valid Host address for a particular Subnet.

For Example:

124.59.10.1 /10

1. Find the Subnet = 124.0.0.0
2. Find the 1st Usable Host = 124.0.0.1
3. Find the Last Usable Host = 124.63.255.254
4. Find the Broadcast Address = 124.63.255.255

1.) 17.46.23.96 255.248.0.0





2.) 139.7.98.15 255.255.128.0






3.) 199.180.150.119 /26 can't communicate with 199.180.150.128 /26 what is the problem?






4.) 205.39.228.185 255.255.255.224






5.) 15.19.11.2 /15

 

Thanks.

I'll have a crack at those and see how I make out.

 

So the other night I got an alert on an IP Address. We are implementing some new management software with the goal of having 1 alerting system for our entire enterprise. I am not going to comment on this aspect, but...so I got an alert on an IP Address. I attempted to find the IP. Tracert, NSLookup, Doing a search on our Visio Maps...Tracert is very nice obviously since it shows you all the layer 3 hops. However in our environment now most of the time it stops at our COLOs before it hits the Provider Cloud since we are no longer using
P2P links. I tried a last ditch effort by checking a piece of software that contains all our IP Scope Information.

Nothing...Nadda...Zilch. BUT I had a High Priority Auto Ticket on this device that I had to do something with. I could close it. But then someone might ask why I didn't act on a High Priority issue.

Whos rules are these anyways? Man...

Ok..so I started to look at the IP Address and then it hit me. It was the Network Address. Not an address actually assigned to an interface. It was a /30. /30 has two usable host ips. This was the network addy. So I incremented it by 1 and found the device. Why it alerted on that? I have no idea.

Of course with IP Subnet Zero you can use the network ip I believe to assign it to an interface, however I haven't played with that and I really don't see a huge point in doing so. You just add a layer of complexity to troubleshooting. And trust me..there are all ready layers of complexity.

So the moral of the story is...Knowing Subnetting is a good thing. Plus I feel it is good to practice a couple problems on a regular basis. I do 5 a night every night I work. Probably overkill, but it keeps it in my mind.

 

NetEyeBall,

At least in Cisco world, it also helps to know that you can configure a router to use or not use the Subnet Zero and the default setting based on Cisco IOS version. According to Cisco...

Link:

• Subnet Zero and the All-Ones Subnet [IP Addressing Services] - Cisco Systems - http://www.cisco.com/en/US/tech/tk648/tk361/technologies_tech_note09186a0080093f18.shtml

 

I hear you.

 

Okay, I've been playing with some of those subnets and this is what I came up with (for better or worse).


1.) 55.110.67.205 /16

55.110.0.0 | 55.110.0.1 | 55.110.255.254 | 55.110.255.255


2.) 88.248.235.250 255.255.255.248

88.248.235.248 | 88.248.235.249 | 88.248.235.254 | 88.248.235.255



3.) 56.58.128.76 255.128.0.0

56.0.0.0 | 56.0.0.1 | 56.127.255.254 | 56.127.255.255


4.) 198.13.70.25 255.255.255.192


192.13.70.0 | 192.13.70.1 | 192.13.70.62 | 192.13.70.63


5.) 150.75.222.94 /18


150.75.192.0 | 150.75.192.1 | 150.75.255.254 | 150.75.255.255


6.) 198.134.190.70 /28

198.134.190.64 | 198.134.190.65 | 198.134.190.78| 198.134.190.79


7.) 47.165.237.34 /21

47.165.232.0 | 47.165.232.1 | 47.165.239.254 | 47.165.239.255


8.) Is Ip address 79.246.255.1 with a subnet mask 255.224.0.0 a valid host ip address?

Not sure about this one, but I think yes.

79. 224.0.0 | 79.224.0.1 | 79.255.255.254 | 79.255.255.255.


9.) 107.30.205.80 255.255.255.240


107.30.205.80 | 107.30.205.81 | 107.30.205.94 | 107.30.205.95


10.) Is 193.43.65.127 with a 255.255.255.192 mask a vaild Host IP address?


Answer: No, it's a broadcast address.


193.43.65.64 | 193.43.65.65 | 193.43.65.126| 193.43.65.127

 

Thanks R.H.Lee for the link. I knew as of release 12.0 that use of subnet zero could be done. I had forgotten that you could do it with all 1s as well.

 

Answers:

1.) 55.110.67.205 /16

1. Find the Subnet = 55.110.0.0
2. Find the 1st Usable Host = 55.110.0.1
3. Find the Last Usable Host = 55.110.255.254
4. Find the Broadcast Address = 55.110.255.255

2.) 88.248.235.250 255.255.255.248

1. Find the Subnet = 88.248.235.248
2. Find the 1st Usable Host = 88.248.235.249
3. Find the Last Usable Host = 88.248.235.254
4. Find the Broadcast Address = 88.248.235.255

3.) 56.58.128.76 255.128.0.0

1. Find the Subnet = 56.0.0.0
2. Find the 1st Usable Host = 56.0.0.1
3. Find the Last Usable Host = 56.127.255.254
4. Find the Broadcast Address = 56.127.255.255

4.) 198.13.70.25 255.255.255.192

1. Find the Subnet = 198.13.70.0
2. Find the 1st Usable Host = 198.13.70.1
3. Find the Last Usable Host = 198.13.70.62
4. Find the Broadcast Address = 198.13.70.63

5.) 150.75.222.94 /18

1. Find the Subnet = 150.75.192.0
2. Find the 1st Usable Host = 150.75.192.1
3. Find the Last Usable Host = 150.75.255.254
4. Find the Broadcast Address = 150.75.255.255

6.) 198.134.190.70 /28

1. Find the Subnet = 198.134.190.64
2. Find the 1st Usable Host = 198.134.190.65
3. Find the Last Usable Host = 198.134.190.78
4. Find the Broadcast Address = 198.134.190.79

7.) 47.165.237.34 /21

1. Find the Subnet = 47.165.232.0
2. Find the 1st Usable Host = 47.165.232.1
3. Find the Last Usable Host = 47.165.239.254
4. Find the Broadcast Address = 47.165.239.255

8.) Is Ip address 79.246.255.1 with a subnet mask 255.224.0.0 a valid host ip address?

Yes. It falls in the 79.224.0.0 subnet and is a valid host ip address.

9.) 107.30.205.80 255.255.255.240

1. Find the Subnet = 107.30.205.80
2. Find the 1st Usable Host = 107.30.205.81
3. Find the Last Usable Host = 107.30.205.94
4. Find the Broadcast Address = 107.30.205.95

10.) Is 193.43.65.127 with a 255.255.255.192 mask a vaild Host IP address?

No. 193.43.65.127 is the broadcast address for the 193.43.65.64 subnet.

 

Thanks Lee. I just re-read your link. I had the concept incorrect. It has nothing to do with host ips, but rather using the all zeros or all ones SUBNET.

Damn...this stuff never seems to get easy.

 

subnetting practice test 1

question 6:

6.) 198.134.190.70 /28




Your answer:

6.) 198.134.190.70 /28

1. Find the Subnet = 198.134.190.64
2. Find the 1st Usable Host = 198.134.190.65
3. Find the Last Usable Host = 198.134.190.78
4. Find the Broadcast Address = 198.134.190.79


I just can't see this!!
My answer:
196.134.190.1 - first address
196.134.190.14 - last
196.134.190.15 Broadcast

What am I doing wrong???

 

198.134.190.70 /28
look at the last octet in binary
01000110
subnet mask is
11110000
that means the network portion of the last octect is
01000000=64 (all host bits set to 0)
network address is 196.134.190.64

 

Oh what an arse I am....I see it now...there's 16 subnets
2^4.....

many thanks for your response.