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Subnetting Test

Discussion in 'General Cisco Certifications' started by NetEyeBall, Nov 29, 2006.

  1. NetEyeBall

    NetEyeBall Kilobyte Poster

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    I am writting a subnetting practice test for one of my co-workers everyday till he kills me or passes the CCNA.

    Here is part 1. I will post the answers tomorrow.

    --------------------------------------------------------------

    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.) 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?
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  2. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    OK - I'm probably about to make a fool of myself....

    [HIDE]55.110.0.0
    55.110.0.1
    55.110.255.254
    55.110.255.255[/HIDE]
    [HIDE]88.248.235.248
    88.248.235.249
    88.248.235.254
    88.248.235.255[/HIDE]
    [HIDE]56.0.0.0
    56.0.0.1
    56.127.255.254
    56.127.255.255[/HIDE]
    [HIDE]198.13.70.0
    198.13.70.1
    198.13.70.62
    198.13.70.63[/HIDE]
    [HIDE]150.75.192.0
    150.75.192.1
    150.75.255.254
    150.75.255.255[/HIDE]
    [HIDE]198.134.190.64
    198.134.190.65
    198.134.190.78
    198.134.190.79[/HIDE]
    [HIDE]47.165.232.0
    47.165.232.1
    47.165.239.254
    47.165.239.255[/HIDE]
    [HIDE]yes[/HIDE]
    [HIDE]107.30.205.0.80
    107.30.205.0.81
    107.30.205.94
    107.30.205.95[/HIDE]
    [HIDE]No - it is the broadcast address for that net[/HIDE]

    Whew - serious care needed there!

    I have assumed current CIDR rules.

    I wonder how many I messed up... :biggrin

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  3. NetEyeBall

    NetEyeBall Kilobyte Poster

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    You did good! One answer had a slight typo...but I know what you ment!
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  4. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Oops - I'd be interested to know what I got wrong. I went through the questions several times!

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  5. NetEyeBall

    NetEyeBall Kilobyte Poster

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    You didn't get anything wrong! Here is the typo...

    107.30.205.0.80
    107.30.205.0.81
    107.30.205.94
    107.30.205.95


    lol...this is one of those special 5 octet ip addresses! ;)

    So what did you think? A good test?

    (I think I wasn't excited enough when you took it or at least it didn't show in my post above. I am sorry! I work nights and during the day I tend to be sleep deprived...but lets start over!!! Congrats!!!! You got 100%!!!!!) :)
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  6. NetEyeBall

    NetEyeBall Kilobyte Poster

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    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.
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  7. NetEyeBall

    NetEyeBall Kilobyte Poster

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    Here is Test 2. A bit smaller since I didn't have too much time to write a larger one..Answers posted tomorrow...:D

    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?
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  8. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Oops! :oops:

    Yes - very good. Made you think hard, and there were some nice 'misdirections'!

    Thanks!

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  9. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    [HIDE]157.129.136.128
    157.129.136.129
    157.129.136.158
    157.129.136.159[/HIDE]
    [HIDE]44.204.221.64
    44.204.221.65
    44.204.221.94
    44.204.221.95[/HIDE]
    [HIDE]181.39.0.0
    181.39.0.1
    181.39.127.254
    181.39.127.255[/HIDE]
    [HIDE]145.177.70.0
    145.177.70.1
    145.177.71.254
    145.177.71.255[/HIDE]
    [HIDE]Sack him. They are on different networks. 215.83.56.192 and 215.83.56.224[/HIDE]

    Shouldn't be any 5 byte IPs there <giggle>

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  10. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Come on everybody - I don't want to be the only one trying these questions! :biggrin :p

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  11. NetEyeBall

    NetEyeBall Kilobyte Poster

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    Got got it 100% again!
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  12. NetEyeBall

    NetEyeBall Kilobyte Poster

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    You have Router1, Router2, and Router3 all connected with point to point T1s. Each Router has its own Lan connecting to F0. Router1's Lan needs to support 100 hosts. Router2's Lan needs to support 20 hosts. Router3's Lan needs to support 20 hosts. The network administrator has decided to subnet 192.168.10.0 in the most efficient way to ensure connectivity between all the devices, but needs help doing it.

    Design this network using VLSM in the 192.168.10.0 range.

    (I need more practice on this design topic myself so if I screw it up let me know! :) ) My answer will be posted later today.
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  13. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    I'm sitting in a hotel in Brum, so am not really in the best position to attempt this!

    Your reference to T1s made me check where in the world you were!
    [HIDE]
    I'll guess at:
    router1 192.160.10.0/25
    router2 192.160.10.128/27
    router3 192.160.10.160/27
    [/HIDE]
    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  14. NetEyeBall

    NetEyeBall Kilobyte Poster

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    You have Router1, Router2, and Router3 all connected with point to point T1s. Each Router has its own Lan connecting to F0. Router1's Lan needs to support 100 hosts. Router2's Lan needs to support 20 hosts. Router3's Lan needs to support 20 hosts. The network administrator has decided to subnet 192.168.10.0 in the most efficient way to ensure connectivity between all the devices, but needs help doing it.

    Design this network using VLSM in the 192.168.10.0 range.

    Router 1 F0 192.168.10.1 /25

    Subnet Mask= /25
    Find the Subnet = 192.168.10.0
    Find the 1st Usable Host = 192.168.10.1
    Find the Last Usable Host = 192.168.10.126
    Find the Broadcast Address = 192.168.10.127

    Router 1 S0 192.168.10.129 /30 (Connected to Router 2)

    Subnet Mask= /30
    Find the Subnet = 192.168.10.128
    Find the 1st Usable Host = 192.168.10.129
    Find the Last Usable Host = 192.168.10.130
    Find the Broadcast Address = 192.168.10.131

    Router 2 S0 192.168.10.130 /30 (Connected to Router 1)

    Subnet Mask= /30
    Find the Subnet = 192.168.10.128
    Find the 1st Usable Host = 192.168.10.129
    Find the Last Usable Host = 192.168.10.130
    Find the Broadcast Address = 192.168.10.131

    Router 2 F0 192.168.10.161 /27

    Subnet Mask= /27
    Find the Subnet = 192.168.10.160
    Find the 1st Usable Host = 192.168.10.161
    Find the Last Usable Host = 192.168.10.190
    Find the Broadcast Address = 192.168.10.191

    Router 2 S1 192.168.10.133 /30 (Connected to Router 3)

    Subnet Mask= /30
    Find the Subnet = 192.168.10.132
    Find the 1st Usable Host = 192.168.10.133
    Find the Last Usable Host = 192.168.10.134
    Find the Broadcast Address = 192.168.10.135

    Router 3 S0 192.168.10.134 /30 (Connected to Router 2)

    Subnet Mask= /30
    Find the Subnet = 192.168.10.132
    Find the 1st Usable Host = 192.168.10.133
    Find the Last Usable Host = 192.168.10.134
    Find the Broadcast Address = 192.168.10.135

    Router 3 F0 192.168.10.193 /27

    Subnet Mask= /27
    Find the Subnet = 192.168.10.192
    Find the 1st Usable Host = 192.168.10.193
    Find the Last Usable Host = 192.168.10.222
    Find the Broadcast Address = 192.168.10.223
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  15. techtr0n1c

    techtr0n1c Bit Poster

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    i rekon you or we(cf) should create this thing as a website

    testmysubnetting.co.uk is available to buy, what do people think

    it may be a dumb idea but its another to the list of ideas ;)
     
    Certifications: HNC Computer Networking, Network+
    WIP: A+, 70-290, 70-270,CCNA
  16. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Hm - this must be a Cisco oddity then.

    I assume that the routers are connected with something like PPP to run over the T1s. Usualy you don't have to allocate IPs on the WAN side of the interface, normal routing will take care of things, but I have seen a case where such a thing was insisted on.

    Most of the docs I've seen suggest you only need IPs on the WAN when special considerations are needed.

    So is this the 'Cisco way' for normal routing?

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  17. NetEyeBall

    NetEyeBall Kilobyte Poster

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    Yes, you would need to have an ip address and mask on wan interfaces. I am not sure if this is a Cisco Oddity since I have only worked on Cisco equipment. How would it know how to route unless you have an ip address on the Wan interface? Good discussion!!!!
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  18. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    Simple - it is a point-to-point connection - so what you stick in one end will come out of the other!

    Strictly - if you adopted this view for everything then *all* router/modems would need such an allocation. Take the typical one, there *should* be a IP addy on each interface, so there should be one on *both* the LAN side (which is the normal one set up) *and* the WAN side, which in the normal world is almost always ignored.

    Look at the usual routing table - (I'm being non-Cisco/ generic here). Assume that you have two interfaces - 'eth0' and 'wan0'. Then the routing is easy, anything for the local network gets sent to eth0, and anything else gets sent to wan0.

    The few docs I have (mostly Lucent) suggest that you only need to assign addresses to the WAN side when things get really complicated.

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  19. NetEyeBall

    NetEyeBall Kilobyte Poster

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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.) 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
     
    Certifications: CCNA, A+, N+, MCSE 4.0, CCA
    WIP: CCDA, CCNP, Cisco Firewall
  20. Spice_Weasel

    Spice_Weasel Kilobyte Poster

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    Actually, any device that has an ip stack loaded requires an ip address for full functionality. A router, for example, will have an ip address on the LAN side (typically private) and one on the WAN side (typically public). The cable/dsl modem will not usually have an ip address because it is operating primarily at layer 2. PPP is a layer 2 protocol that can be used to encapsulate many L3 protocols, e.g. ppp uses ipcp to carry ip traffic. PPP itself does not require any L3 addressing, but whatever is being encapsulated (IP, IPX, etc.) does need some kind of L3 addressing. For example, a home router has a private ip on the LAN side, and on the WAN side (dsl) uses PPP to negotiate a connection to your ISP. The dsl modem is functioning as a L2 bridge, connecting the Ethernet router WAN interface to the ATM dsl ISP side. Your router negotiates a PPPoE or PPPoA session with the DSLAM or aggregation point at the ISP, with the dsl modem acting as a bridge.

    What is often done to save addresses is to use the "ip unnumbered" command on point to point links. This allows a L3 interface to send traffic over the link without needing to assign a unique ip to the interface. Since it can only send frames to the device on the other side it can "borrow" an ip address from another interface on the router. For example, you might have assigned ip add 10.10.35.60 / 28 to ether0/0. On s0/2 (a point to point link) you could issue the command "ip unnumbered eth0/0" and the serial line will use the same ip as eth0/0, saving you an address. Note that usually you will borrow the ip address from a loopback interface, not a physical interface.

    Loopback interfaces are invaluable, always configure one on your routers ;)


    Spice_Weasel
     
    Certifications: CCNA, CCNP, CCIP, JNCIA-ER, JNCIS-ER,MCP
    WIP: CCIE

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