Subnetting

Which command will assign the last usable IP address from the 192.168.32.128/28 subnetwork to
a router interface?
A. Router(config-if)# ip address 192.168.32.142 255.255.255.240
B. Router(config-if)# ip address 192.168.32.143 255.255.255.240
C. Router(config-if)# ip address 192.168.32.158 255.255.255.240
D. Router(config-if)# ip address 192.168.32.142 255.255.255.248
E. Router(config-if)# ip address 192.168.32.144 255.255.255.224
F. Router(config-if)# ip address 192.168.32.158 255.255.255.224

So
/28=255.255.255.240 (11111111.11111111.11111111.1111.0000)
subnet: 2^4 = 16
host-per subnet: (2^4)-2=14
valid subnet: 256-240=16 (increment number)

0,16,32,64,80,96,112,128,144,160,176,192,208,224,240,256

broadcast address: subnet #9 : 192.168.32.143
valid host: 192.168.32.129 - 192.168.32.142

broadcast address: subnet #10: 192.168.159
valid host: 192.168.32.145 - 192.168.32.158

I can understand: D,E,F are incorrect.

B is the broadcast address.

Why the answer is A? not C? Please explain.

Thank you,

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Comments

FrankGuthrie

yengiang wrote: »

Which command will assign the last usable IP address from the 192.168.32.128/28 subnetwork to
a router interface?
A. Router(config-if)# ip address 192.168.32.142 255.255.255.240
B. Router(config-if)# ip address 192.168.32.143 255.255.255.240
C. Router(config-if)# ip address 192.168.32.158 255.255.255.240
D. Router(config-if)# ip address 192.168.32.142 255.255.255.248
E. Router(config-if)# ip address 192.168.32.144 255.255.255.224
F. Router(config-if)# ip address 192.168.32.158 255.255.255.224

So
/28=255.255.255.240 (11111111.11111111.11111111.1111.0000)
subnet: 2^4 = 16
host-per subnet: (2^4)-2=14
valid subnet: 256-240=16 (increment number)

0,16,32,64,80,96,112,128,144,160,176,192,208,224,240,256

broadcast address: subnet #9 : 192.168.32.143
valid host: 192.168.32.129 - 192.168.32.142

broadcast address: subnet #10: 192.168.159
valid host: 192.168.32.145 - 192.168.32.158

I can understand: D,E,F are incorrect.

B is the broadcast address.

Why the answer is A? not C? Please explain.

Thank you,

Well, you kinda answered the question yourself. With the 192.168.32.128/28 this means the subnet mask is 255.255.255.240 (This already rules out awnsers D, E and F). This means the subnet has 16 IP addresses from which only 14 can be used as the first is the network address, and the last address is the broadcast addresses.

In the 192.168.32.128/28 network, the last address is 192.168.32.143, because 192.168.32.144 is the start of the next IP subnet. Seeing that the last IP address is used for broadcast, which is 143, the last usable address which you can assign to a host is .142.

Looking at the answers, only A is correct. The reason answer C is not correct is that 192.168.32.158 is the last usable IP address of the next IP subnet, which begins with 192.168.32.144 - 192.168.32.159. The next subnet is 192.168.32.160 - 192.168.32.176 and so on. You already stated that you knew that answer B is the broadcast address.

1. By the determining that the /28 subnet mask is 255.255.255.240, you could immediately rule out D, E & F
2. By using the 192.168.32.128 address and the 255.255.255.240 subnet mask, you know that the subnet has 16 addresses, 192.168.32.128 - 192.168.32.143. This rules out answer C
3. The last IP address in a subnet is used for broadcast. with bullet number 2, you have determined the last IP address is 192.168.32.143, so this is the broadcast and thus you eliminate answer B.
4. The only answer left is A

Hope this is clear.

yengiang

Thank you. I just need to focus on what was being asked in the questions.:)

FrankGuthrie

yengiang wrote: »

Thank you. I just need to focus on what was being asked in the questions.:)

No problem, good luck with your studies.

yengiang

The internetwork is using subnets of the address 192.168.1.0 with a subnet mask of
255.255.255.224. The routing protocol in use is RIP version 1. Which address could be assigned to the FastEthernet interface on RouterA which has 13 hosts?

A. 192.168.1.31
B. 192.168.1.64
C. 192.168.1.127
D. 192.168.1.190
E. 192.168.1.192

subnet: 2^3=8
host-per subnet: (2^5)-2=30
valid subnet: 256-224=32
increment: 0,32,64,96,128,160,192

Answer: not A,C, or E
why D but not B? what did I do wrong?

Thanks

SecurityThroughObscurity

B is a network address and couldnt be assigned to the interface.

FrankGuthrie

Remember! You cannot use the network address, which is the first usable IP address in a subnet. Also you can't use the last IP address in a subnet, as it is used for broadcast address within an ip subnet.

A. 192.168.1.31 - This is the broadcast address of the network 192.168.1.0 - 192.168.1.31, because the next address is the IP of the next subnet 192.168.1.32)
B. 192.168.1.64 - This is the next address of the next subnet Network 1: 192.168.1.0 - 192.168.1.31, Network 2: 192.168.1.32 - 192.168.1.63, Network 3: 192.168.1.64 - 192.168.1.95
C. 192.168.1.127 - Using the logic above and further writing out the subnets: Network 4: 192.168.1.96 - 192.168.1.127, So the last address is the broadcast address and cannot be assigned.
D. 192.168.1.190 - Using the logic above and even further writing out the subnets: Network 5:192.168.1.128 - 192.168.1.159, Network 6:192.168.1.160 - 192.168.1.191. Network 7:192.168.1.192 - 192.168.1.223. That immediately eliminates Answer E as answer E is a network address. The only answer left is D. To verify this, we look at network 6, 191 is the broadcast address, so the address before it, .190, can be given out to a host/Interface.
E. 192.168.1.192

yengiang

Great explanation! Thanks...

yengiang

Please help. I can't do this in my head within a minute.

A new LAN segment is allocated the network number 172.16.0.0/25. What range of addresses are
available for hosts on that network?
A. 172.16.0.1 through 172.16.0.254
B. 172.16.0.1 through 172.16.0.126
C. 172.16.0.129 through 172.16.0.254
D. 172.16.0.1 through 172.16.1.254
E. 172.16.1.1 through 172.16.1.126
F. 172.16.1.1 through 172.16.1.254

jvrlopez

Don't worry about speed at the moment. That will will come with time and practice.

The /25 means that there are 25 out of a possible 32 bits "on" (1s) in the network: all 8 bits in the first 3 octets and 1 more in the last octet. As the last octet has 1 bit on, this reads as 10000000 which equals 128.

Since the first 3 octets have all of their bits "on", that leaves no room for hosts on those octets, so we are only concerned about the last octet. Since our first 3 octets will not be changing, this rules out answers D, E, and F on the spot.

In our last octet, subtracting 128 from 256 gives you a block size of 128, so every network is 128 IPs long (this is not your range, though!).

Bear in mind that the first IP in the subnet is reserved for its network's ID and the last IP is for its broadcast, so each subnet will have 126 IP addresses available for hosts (this is your range).

As your block size is 128, start at 0 and count by 128s until you max out, which each increment of 128 being a new subnet. So your valid subnets would be 172.16.0.0 and 172.16.0.128.

172.16.0.0 is broken down as follows:

Network ID (first IP in subnet) - 172.16.0.0
Range - 172.16.0.1 - 172.16.0.126
Broadcast (last IP in subnet) - 172.16.0.127

172.16.0.128 is broken down as follows:

Network ID (first IP in subnet) - 172.16.0.128
Range - 172.16.0.129 - 172.16.0.254
Broadcast (last IP in subnet) - 172.16.0.255

Does this question allow for more than one correct answer? If that is the case, then B and C would be valid given the network number 172.16.0.0 and /25 ( 255.255.255.128 ) mask.

A is not correct because 172.16.0.1 through 172.16.0.254 does not take into account the IPs 172.16.0.127 and 172.16.0.128 which are reserved for the first subnet's broadcast and second subnet's network ID, respectively.

yengiang

B is the only answer. Per Todd Lammle, it's one of the hardest subnetmask. I just need to re-read this section again to tackle this kind of question. Thanks for breaking it down.

jvrlopez

If B is the only answer allowed, I'm guessing it's the "best answer" because it is the first correct subnet and range.

Keep at it! Practice daily, learn from your mistakes, and you'll be surprised at how quick it comes.

I try and do at least 50 subetting questions (net ID, range, mask, networks/hosts per mask, etc) throughout the day on my phone and at work.

RouteMyPacket

yengiang wrote: »

B is the only answer. Per Todd Lammle, it's one of the hardest subnetmask. I just need to re-read this section again to tackle this kind of question. Thanks for breaking it down.

There is nothing hard about it...you need to remember CIDR notation and what they actually mean

RIGHT NOW, grab some paper and write this down and post it at your desk

CIDR Notation Subnet Addresses

/32 255.255.255.255 1
/31 255.255.255.254 2
/30 255.255.255.252 4
/29 255.255.255.248 8
/28 255.255.255.240 16
/27 255.255.255.224 32
/26 255.255.255.192 64
/25 255.255.255.128 128
/24 255.255.255.0 256
/23 255.255.254.0 512
/22 255.255.252.0 1,024
/21 255.255.248.0 2,048
/20 255.255.240.0 4,096
/19 255.255.224.0 8,192
/18 255.255.192.0 16,384
/17 255.255.128.0 32,768
/16 255.255.0.0 65,536

Remember you cannot use the "Network" or "Broadcast" address so take "2" away from every number above to know how many "usable" addresses there are.

Now with CIDR, it's easy they have already told you the amount of bits used in the mask. Remember we only have 32 bits to work with in the first place so

What does /24 mean? Let's look at it in binary

11111111.11111111.11111111.00000000

Ok, all bits are used in the first three octets so that translates to

255.255.255.0 or /24

How many "1's" do you see above? 24? Bingo...

/23

11111111.11111111.11111110.00000000

255.255.254.0

So how did we come up with 254 in the third octet? Remember

128 64 32 16 8 4 2 1

We used 7 bits in the third octet to get our /23 so add them up

128+64+32+16+8+4+2 = 254

/28

11111111.11111111.11111111.11110000

128+64+32+16 = 240

255.255.255.240

/26

11111111.11111111.11111111.11000000

128+64 = 192

255.255.255.192

Subnetting is easy trust me, it's about experience and exposure which you are now getting. This forum doesn't allow me to write the addresses above right, it pushes them together so you will have to write them down or copy/paste and stretch them out

yengiang

jvrlopez: I'm still struggling with the same question everyday but my speed is getting better.

routemypacket: Thanks for showing a good way to learn CIDR.

Here's another question:

The command ip route 192.168.100.160 255.255.255.224 192.168.10.2 was issued on a router.
No routing protocols or other static routes are configured on the router. Which statement is true
about this command?

A. The interface with IP address 192.168.10.2 is on this router.
B. The command sets a gateway of last resort for the router.
C. Packets that are destined for host 192.168.100.190 will be sent to 192.168.10.2.
D. The command creates a static route for all IP traffic with the source address 192.168.100.160.

Why is C the answer? Thanks

yengiang

Which command excludes seven host addresses from a DHCP pool that has a 255.255.255.192
mask?
A. router(config)#ip dhcp excluded-address 192.168.7.0 192.168.7.6
B. router(config)#ip dhcp excluded-address 192.168.7.121 192.168.7.127
C. router(config)#ip dhcp excluded-address 192.168.7.159 192.168.7.165
D. router(config)#ip dhcp excluded-address 192.168.7.193 192.168.7.198

valid subnet: 256-192=64

increment: 0,64,128,192

rule out D because it excludes only 6 hosts.
Not A: because it's included the 192.168.7.0 (Network address)
Not B: because 192.168.7.127 (broadcast address)
C is the answer...please confirm.. thanks

RouteMyPacket

yengiang wrote: »

jvrlopez: I'm still struggling with the same question everyday but my speed is getting better.

routemypacket: Thanks for showing a good way to learn CIDR.

Here's another question:

The command ip route 192.168.100.160 255.255.255.224 192.168.10.2 was issued on a router.
No routing protocols or other static routes are configured on the router. Which statement is true
about this command?

A. The interface with IP address 192.168.10.2 is on this router.
B. The command sets a gateway of last resort for the router.
C. Packets that are destined for host 192.168.100.190 will be sent to 192.168.10.2.
D. The command creates a static route for all IP traffic with the source address 192.168.100.160.

Why is C the answer? Thanks

ip route 192.168.100.160 255.255.255.224 192.168.10.2

What is that? A static route that's what, but what does it mean?

192.168.100.160/27

That's the network we have, so what IP's fit in that segment? We have 32 addresses in a /27 network so out network is

192.168.100.160 - 192.168.100.191

C is correct simply because 192.168.100.190 is in our network segment.

RouteMyPacket

yengiang wrote: »

Which command excludes seven host addresses from a DHCP pool that has a 255.255.255.192
mask?
A. router(config)#ip dhcp excluded-address 192.168.7.0 192.168.7.6
B. router(config)#ip dhcp excluded-address 192.168.7.121 192.168.7.127
C. router(config)#ip dhcp excluded-address 192.168.7.159 192.168.7.165
D. router(config)#ip dhcp excluded-address 192.168.7.193 192.168.7.198

valid subnet: 256-192=64

increment: 0,64,128,192

rule out D because it excludes only 6 hosts.
Not A: because it's included the 192.168.7.0 (Network address)
Not B: because 192.168.7.127 (broadcast address)
C is the answer...please confirm.. thanks

So write out each of your networks and tell us why you believe C is correct

192.168.7.0
192.168.7.64
192.168.7.128
192.168.7.192

There are you networks, so what's the answer?

yengiang

Network 3: 192.168.7.128

valid hosts: 192.168.7.129 thru 192.168.7.190
broadcast: 192.168.7.191

C. router(config)#ip dhcp excluded-address 192.168.7.159 192.168.7.165

Thanks for clearing that up.

yengiang

Please help. I'm having a hard time doing Class B.

A network administrator has subnetted the 172.16.0.0 network using a subnet mask of
255.255.255.192. A duplicate IP address of 172.16.2.120 has accidentally been configured on a
workstation in the network. The technician must assign this workstation a new IP address within
that same subnetwork. Which address should be assigned to the workstation?
A. 172.16.1.80
B. 172.16.2.80
C. 172.16.1.64
D. 172.16.2.64
E. 172.16.2.127
F. 172.16.2.128

draught

Kinda a trick question. What you are really answering for is what address to use in the 172.168.2.120 /28 subnetwork.
The class B part is just there to throw you off because regardless the subnet mask you are dealing with is class C.

Start at 172.168.2.0 with a 255.255.255.192 subnetmask your prefix is /26 and the interval is 64.

Keep adding up 64 to see what subnet the workstation is located in.

0,64,128

You can see your range is 64 -127 and so the answer is B.

Aidan Aiken

Nice Information Thanks

steveyeung

192.168.32.128/28=192.168.32.1000|0000
the last usable ip address means the host bit part should be 1110
=192.168.32.1000|1110
=192.168.32.142 255.255.255.240

SysnetNotes

yengiang wrote: »

Please help. I'm having a hard time doing Class B.

A network administrator has subnetted the 172.16.0.0 network using a subnet mask of
255.255.255.192. A duplicate IP address of 172.16.2.120 has accidentally been configured on a
workstation in the network. The technician must assign this workstation a new IP address within
that same subnetwork. Which address should be assigned to the workstation?
A. 172.16.1.80
B. 172.16.2.80
C. 172.16.1.64
D. 172.16.2.64
E. 172.16.2.127
F. 172.16.2.128

Here the given subnet mask is 255.255.255.192 so block size will be 64 and change will be taking place in last octet only
As per the given information, range will be from 172.16.2.64 - 172.16.2.127

Answer B (172.16.2.80) is the only one coming in that range

Answer A - will be a different network (wrong)
Answer C- will be a different network (wrong)
Answer D - it will be the network address (wrong)
Answer E - it will be the broadcast address (wrong)
Answer F - It will be different network (wrong)

Hope this helps

yengiang

Thank you all. I got freaked out sometime just looking at the answer. I need to review more about Class B.

yengiang

Which command would you use to configure a static route on Router1 to network
192.168.202.0/24 with a nondefault administrative distance?

A. router1(config)#ip route 1 192.168.201.1 255.255.255.0 192.168.201.2
B. router1(config)#ip route 192.168.202.0 255.255.255.0 192.168.201.21
C. router1(config)#ip route 5 192.168.202.0 255.255.255.0 192.168.201.2
D. router1(config)#ip route 192.168.202.0 255.255.255.0 192.168.201.25

Rule out A and C. It's also a Class C default subnet. I'm stuck...please help.

Increment: ??

jvrlopez

There is no block size/increment on a /24 on a class C network. All the bits are "off" (0s) and are for hosts. Its just one big network with all 256 addresses, 254 of which are usable.

192.168.202.0 - Net ID
192.168.202.1 - First usable IP
192.168.202.254 - Last usable IP
192.168.202.255 - Broadcast

Static routes are ip route [destination address] [subnet mask] [next hop address or exit interface] [administrative distance]

If you want to forward all traffic destined for a certain network, in our case, 192.168.202.0, use the network ID as your destination address, use the correct mask, 255.255.255.0 in our case, and then the address (or interface) all traffic for that network is being forwarded to.

To use the default administrative distance, just leave it blank. The question asks for a non-default admin distance so you could use any number between 0 and 255, with the lowest being the best, 0 being directly connected and 255 being unreachable.

bbarrick

Actually I believe the answer is C. But I would rather have someone else explain it. My reason is, the key wording is "nondefault administrative distance". Neither B nor D change that. A technically doesn't either since the default is 1.

jvrlopez

Can you even set the AD before the addresses?

Never seen that before and just opened up an IOS session and it wouldn't recognize the command with the AD written in the space that the answer choices give.

bbarrick

I'm not sure. It just seemed the most logical to me but when I looked the command up on Cisco it show's the AD in the middle.

yengiang

"To use the default administrative distance, just leave it blank. The question asks for a non-default admin distance so you could use any number between 0 and 255, with the lowest being the best, 0 being directly connected and 255 being unreachable".

B. router1(config)#ip route 192.168.202.0 255.255.255.0 192.168.201.21
D. router1(config)#ip route 192.168.202.0 255.255.255.0 192.168.201.25

The answer is D but I don't understand why?

bbarrick

Is this from Mr. Odoms book? If so, send him a message on Facebook. He usually gets back to me within a day or so.