Subnetting

yengiangyengiang Posts: 29Member ■■□□□□□□□□
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

  • FrankGuthrieFrankGuthrie Posts: 245Member
    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.
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    Thank you. I just need to focus on what was being asked in the questions.:)
  • FrankGuthrieFrankGuthrie Posts: 245Member
    yengiang wrote: »
    Thank you. I just need to focus on what was being asked in the questions.:)

    No problem, good luck with your studies.
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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
  • B is a network address and couldnt be assigned to the interface.
  • FrankGuthrieFrankGuthrie Posts: 245Member
    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
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    Great explanation! Thanks...
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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
  • jvrlopezjvrlopez Posts: 910Member
    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.
    And so you touch this limit, something happens and you suddenly can go a little bit further. With your mind power, your determination, your instinct, and the experience as well, you can fly very high. ~Ayrton Senna
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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.
  • jvrlopezjvrlopez Posts: 910Member
    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.
    And so you touch this limit, something happens and you suddenly can go a little bit further. With your mind power, your determination, your instinct, and the experience as well, you can fly very high. ~Ayrton Senna
  • RouteMyPacketRouteMyPacket Posts: 1,104Member
    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
    Modularity and Design Simplicity:

    Think of the 2:00 a.m. test—if you were awakened in the
    middle of the night because of a network problem and had to figure out the
    traffic flows in your network while you were half asleep, could you do it?
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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
  • RouteMyPacketRouteMyPacket Posts: 1,104Member
    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.
    Modularity and Design Simplicity:

    Think of the 2:00 a.m. test—if you were awakened in the
    middle of the night because of a network problem and had to figure out the
    traffic flows in your network while you were half asleep, could you do it?
  • RouteMyPacketRouteMyPacket Posts: 1,104Member
    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?
    Modularity and Design Simplicity:

    Think of the 2:00 a.m. test—if you were awakened in the
    middle of the night because of a network problem and had to figure out the
    traffic flows in your network while you were half asleep, could you do it?
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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.
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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
  • draughtdraught Posts: 218Member ■■■■□□□□□□
    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 AikenAidan Aiken Posts: 3Registered Users ■□□□□□□□□□
    Nice Information Thanks
  • steveyeungsteveyeung Posts: 44Member ■■□□□□□□□□
    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
  • SysnetNotesSysnetNotes Posts: 45Member ■■□□□□□□□□
    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
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    Thank you all. I got freaked out sometime just looking at the answer. I need to review more about Class B. icon_sad.gif
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    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: ??
  • jvrlopezjvrlopez Posts: 910Member
    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.
    And so you touch this limit, something happens and you suddenly can go a little bit further. With your mind power, your determination, your instinct, and the experience as well, you can fly very high. ~Ayrton Senna
  • bbarrickbbarrick Posts: 241Member
    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.
  • jvrlopezjvrlopez Posts: 910Member
    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.

    icon_confused.gif:
    And so you touch this limit, something happens and you suddenly can go a little bit further. With your mind power, your determination, your instinct, and the experience as well, you can fly very high. ~Ayrton Senna
  • bbarrickbbarrick Posts: 241Member
    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.
  • yengiangyengiang Posts: 29Member ■■□□□□□□□□
    "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?
  • bbarrickbbarrick Posts: 241Member
    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.
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