Alcatel-lucent Scalable IP Networks Exam - B

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  • 1/35 Questions

    1. When a frame with an unknown destination MAC address enters a switch, the switch will forward it out which ports?

    • A. None
    • B. All
    • C. All unicast ports
    • D. All except the port that received the frame
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About This Quiz

Preparing for the Network Routing Specialist I (NRS I) Certifications Exam (4A0-100)

Alcatel-lucent Scalable IP Networks Exam - B - Quiz

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  • 2. 

    2. The primary difference in the way Ethernet hubs and Ethernet switches handle traffic is ______.

    • A. Hubs forward broadcast traffic out every port, switches do not.

    • B. Switches eliminate the need for thicknet cabling.

    • C. Switches support multiple physical connections to hosts.

    • D. Switches forward unicast traffic only to a specific destination port.

    Correct Answer
    A. D. Switches forward unicast traffic only to a specific destination port.
    Explanation
    This is opposed to hubs that act like a “wire in a box” and forward frames out
    every port. Both hubs and switches forward broadcast traffic out all ports.

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  • 3. 

    3. Which of the following is not true about Link Aggregation Groups?

    • A. They protect against single or multiple link failures.

    • B. They can contain up to eight physical links.

    • C. They can protect against a switch failure by calculating multiple paths to the root.

    • D. They can be configured to enter a down state if a certain number of links in the bundle fail.

    Correct Answer
    A. C. They can protect against a switch failure by calculating multiple paths to the root.
    Explanation
    This is a description of STP, not LAG.

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  • 4. 

    4. Which of the following is not true of the STP protocol?

    • A. It calculates a root bridge.

    • B. It uses a cost value on each port to determine the path to the root bridge.

    • C. It ensures a loop-free topology.

    • D. It provides load-sharing capability.

    Correct Answer
    A. D. It provides load-sharing capability.
    Explanation
    STP cannot provide load-sharing because only a single path exists between each
    segment and the root bridge to avoid loops.

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  • 5. 

    5. The advantage of using VLANs is ______.

    • A. They can increase the security of your network.

    • B. They can interconnect multiple broadcast domains.

    • C. They can limit the amount of broadcast traffic between groups of devices.

    • D. A and C but not B

    Correct Answer
    A. D. A and C but not B
    Explanation
    VLANs do not provide routing between broadcast domains although they do
    create separate broadcast domains.

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  • 6. 

    6. Which of the following statements is false?

    • A. Routers provide broadcast domain separation.

    • B. Hubs provide collision domain separation.

    • C. VLANs provide broadcast domain separation.

    • D. Switches provide collision domain separation.

    Correct Answer
    A. B. Hubs provide collision domain separation.
    Explanation
    Hubs simply forward all frames out all ports and so do not provide collision
    domain separation.

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  • 7. 

    7. The method that LAGs use to provide load balancing is best described as:

    • A. Aggregates all source/destination conversations into a single conversation equally across all links

    • B. Uses the same physical link for each source/destination conversation

    • C. Statistically balances conversations based on the source MAC address

    • D. Distributes egress frames equally across all links in the bundle

    Correct Answer
    A. B. Uses the same physical link for each source/destination conversation
    Explanation
    This method is used to ensure that there is no frame reordering as required by the
    802.3ad standard.

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  • 8. 

    8. Given the following code: Config> lag 1 Config>lag# description “LAG from PE1 to PE2” Config>lag# port 1/1/1 1/1/2 1/1/3 1/1/4 1/1/5 1/1/6 Config>lag# port-threshold 2 action down Config>lag# dynamic-cost Config>lag# no shutdown Which answer correctly describes what happens when Ports 1/1/5 and 1/1/6 fail?

    • A. Nothing because the port threshold of 2 active links has not been reached

    • B. The LAG begins using equal costing across all links because of the dynamiccost parameter.

    • C. The LAG updates its BPDUs and recalculates STP.

    • D. The LAG changes its OSPF cost for the bundle but takes no other action.

    Correct Answer
    A. D. The LAG changes its OSPF cost for the bundle but takes no other action.
    Explanation
    This is because the “dynamic-cost” option is configured on the bundle so that
    each time a link fails, the OSPF cost is updated. No other action is taken because
    the threshold is 2 and there are still four out of six active links functioning.

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  • 9. 

    9. What is the primary reason that Ethernet switched networks require STP?

    • A. STP provides for link backup between switches.

    • B. A loop-free topology is more efficient.

    • C. Redundant paths can lead to broadcast storms and FDB instability.

    • D. STP updates the OSPF routing protocol cost upon link failure.

    Correct Answer
    A. C. Redundant paths can lead to broadcast storms and FDB instability.
    Explanation
    This describes the problem of loops in a switched network, which is the primary
    reason to use STP.

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  • 10. 

    10. The mechanism that STP uses to prevent loops in an Ethernet switched network is ______.

    • A. STP elects a root and selectively blocks higher cost paths to the root from each bridge.

    • B. STP blocks ports on all bridges that are not the root bridge.

    • C. STP proactively changes all paths to the root bridge so that they are equal cost.

    • D. STP uses BPDUs to set up a virtual path between each source and destination pair.

    Correct Answer
    A. A. STP elects a root and selectively blocks higher cost paths to the root from each bridge.
    Explanation
    This is an accurate description of the way STP functions to prevent loops.

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  • 11. 

    11. What determines how the root bridge is elected?  

    • A. The bridge priority

    • B. The MAC address of the lowest switch port

    • C. The bridge priority unless there is a tie, and then the lowest MAC address

    • D. The BID unless there are multiple bridge priorities that are equal

    Correct Answer
    A. C. The bridge priority unless there is a tie, and then the lowest MAC address
    Explanation
    This is an accurate description of how a root bridge is selected. Answer D is wrong

    because the BID is always used to determine the root bridge (note the unless

    qualifier). In fact, answer C describes what the BID is: bridge priority plus MAC

    address.

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  • 12. 

    12. What distinguishes an alternate port from a back-up port in STP?

    • A. The alternate port has a higher path to the root.

    • B. The back-up port has a lower priority.

    • C. The back-up port is used only when the alternate port fails.

    • D. The back-up port is on the same switch as the designated switch.

    Correct Answer
    A. D. The back-up port is on the same switch as the designated switch.
    Explanation
    The alternate port is on a non-designated switch on the same segment.

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  • 13. 

    13. Which of the following is false regarding VLANs?

    • A. They provide for broadcast domain separation.

    • B. A single VLAN can exist on multiple switches.

    • C. They require a separate physical connection per VLAN for interswitch links.

    • D. They use a 12-bit VLAN ID to identify each VLAN.

    Correct Answer
    A. C. They require a separate physical connection per VLAN for interswitch links.
    Explanation
    This is not true if a VLAN trunk is used between the switches.

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  • 14. 

    14. Which STP port state is characterized by the port accepting and recording MAC address information, but not forwarding any frames out the port?

    • A. Blocking

    • B. Forwarding

    • C. Listening

    • D. Learning

    Correct Answer
    A. D. Learning
    Explanation
    In this state, the bridge learns information about MAC addresses but does not yet
    forward frames.

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  • 15. 

    15. The technology that allows multiple customers with the same VLANs to use the same provider backbone for their Ethernet traffic is known as ______.

    • A. VLAN trunking

    • B. VLAN tunneling

    • C. VLAN stacking

    • D. IEEE 802.1p

    Correct Answer
    A. C. VLAN stacking
    Explanation
    VLAN stacking allows a provider to stack its own VLAN information in front of
    the customer’s VLAN information to support customers with overlapping VLANs.

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  • 16. 

    Which of the following is not a reason networks built on Ethernet alone cannot scale to a global?

    • A. Excessive broadcasts would make the network unusable.

    • B. Ethernet lacks hierarchical addressing.

    • C. Ethernet switches cannot build forwarding tables.

    • D. Ethernet cables can only be of a limited length.

    Correct Answer
    A. C. Ethernet switches cannot build forwarding tables.
    Explanation
    Ethernet switches, in fact, do build FDB tables of MAC addresses. The other
    answers are all factors that prevent Ethernet from creating global networks.

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  • 17. 

    Which of the following is true about Layer 3 addressing?

    • A. It is embedded in the device’s firmware.

    • B. It provides for a logical hierarchy.

    • C. It allows for duplicate addresses on the Internet.

    • D. Addresses are not required to be registered if they are used on the Internet.

    Correct Answer
    A. B. It provides for a logical hierarchy.
    Explanation
    This is one of the main reasons that IP can support global networks—the use of
    hierarchical addressing.

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  • 18. 

    Which of the following is not true about an IP packet?        

    • A. The TTL field ensures that IP packets have a limited lifetime.

    • B. The maximum size is 65,535 octets.

    • C. The total length field includes the IP header.

    • D. The current version is IPv5.

    Correct Answer
    A. D. The current version is IPv5.
    Explanation
    The current version is IPv4, and the next version is IPv6.

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  • 19. 

    Which of the following is a valid host IP address?

    • A. 192.168.300.4

    • B. 255.70.1.1

    • C. 224.0.0.1

    • D. 10.254.1.1

    Correct Answer
    A. D. 10.254.1.1
    Explanation
    10.254.1.1 is the only valid host address. The other addresses are either out of supported
    IP address ranges, are in the broadcast range (255), or are multicast (224).

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  • 20. 

     An IP address has a first octet represented in binary as 11000001; the equivalent in decimal is ______.

    • A. 190

    • B. 193

    • C. 192

    • D. 11,000,001

    Correct Answer
    A. B. 193
    Explanation
    Calculate this as 28 + 27 + 20 = 128 + 64 + 1 = 193.

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  • 21. 

     The address 224.100.1.1 under traditional classful addressing would be ______.

    • A. Class A

    • B. Class B

    • C. Class C

    • D. None of the above

    Correct Answer
    A. D. None of the above
    Explanation
    This is a Class D or multicast address. You can tell this by examining the first
    octet in binary: 224 = 128 + 64 + 32 = 11100000. The first 3 bits are 1, so this
    indicates that it is a Class D address.

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  • 22. 

    Which of the following is not a private address?

    • A. 172.18.20.4

    • B. 10.0.1.1

    • C. 200.1.1.254

    • D. 192.168.0.1

    Correct Answer
    A. C. 200.1.1.254
    Explanation
    The private address ranges are 10.0.0.0/8, 172.16.0.0/12, and 192.168.0.0/16.

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  • 23. 

     Which of the following is not a reason that subnetting is superior to class-based addressing? .

    • A. It reduces the Internet routing table size.

    • B. You can identify the host portion of the address without the need for a mask.

    • C. It creates greater internal address flexibility.

    • D. It allows for more efficient use of address space

    Correct Answer
    A. B. You can identify the host portion of the address without the need for a mask.
    Explanation
    This is true for classful addressing, but not for classless addressing. Using classless
    addressing, you need the subnet mask to determine how many bits are used for the
    subnet and how many for the host.

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  • 24. 

     Given a network address of 192.168.100.0/24, what is the maximum number of subnets you can create if each subnet must support at least seven hosts?

    • A. 16

    • B. 32

    • C. 4

    • D. 8

    Correct Answer
    A. A. 16
    Explanation
    If you need seven hosts per subnet, this means that you need to leave at least 4 bits
    for each subnet because 24 – 2 = 14 available hosts. This leaves 4 bits (8 – 4 = 4)
    for the subnet, which means you can have 24 = 16 subnets. 3 bits for each subnet is
    not sufficient because two addresses (all zeros and all ones) cannot be used.

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  • 25. 

    If your network address is 10.1.0.0/16 and you have subnetworks that all support at least 300 hosts, how many subnets do you have?

    • A. 255

    • B. 64

    • C. 100

    • D. 128

    Correct Answer
    A. D. 128
    Explanation
    The trick here is recognizing that if you need 300 hosts per subnet, then you need
    to use more than 8 bits of the IP address for hosts because you can only have 255
    max hosts for an 8-bit octet. If you borrow one of the bits from the third octet,
    this gives you a maximum of 510 hosts, and leaves 7 bits of the third octet for subnets.
    27 = 128, so you can have 128 subnets.

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  • 26. 

    Which of the following is the correct representation of mask 255.192.0.0?

    • A. /8

    • B. /11

    • C. /10

    • D. /16

    Correct Answer
    A. C. /10
    Explanation
    The key here is to remember that 255 means “all 1s,” so you know that the first
    octet has all the bits used and therefore the mask is at least /8. 192 translates into
    the highest 2 bits of the next octet: 27 + 26 = 128 + 64 + 192, so this means that
    10 bits are used so the subnet mask is /10.

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  • 27. 

    A network with a /30 mask allows you to have how many usable host addresses?

    • A. 4

    • B. 2

    • C. 6

    • D. 0

    Correct Answer
    A. B. 2
    Explanation
    The key here is “usable” host addresses. A /30 leaves 2 bits for hosts, which would
    translate into 22 = 4. However, the 0 host is the subnet address and the “all 1s”
    host is the broadcast, so those addresses cannot be used. This leaves only two
    usable host addresses.

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  • 28. 

    Given the address 10.1.1.0/24, the most correct description of 10.1.1.0 is ______.

    • A. Host 0 on the 10.1.1.0 subnet

    • B. Network 10.1.1.0

    • C. Illegal because 10.0.0.0 is a Class A

    • D. Subnet 10.1.1.0

    Correct Answer
    A. D. Subnet 10.1.1.0
    Explanation
    B is not as correct because technically 10.1.1.0 is a subnet of network 10.0.0.0.

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  • 29. 

    The concept of allowing a single route entry to represent many network addresses is known as ______.

    • A. CIDR

    • B. Route aggregation

    • C. VLSM

    • D. Classless addressing

    Correct Answer
    A. B. Route aggregation
    Explanation
    CIDR is a technology that implements route aggregation, but the question asks
    what the “concept” is called, and the concept is route aggregation.

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  • 30. 

    How many subnets can be created from network 10.0.0.0/8 if each subnet must support at least 31 hosts?

    • A. 216

    • B. 218

    • C. 219

    • D. 224

    Correct Answer
    A. B. 218
    Explanation
    The key is to determine how many bits are available for subnetting. We know we
    need 31 hosts, so we know we need at least 5 bits because 25 = 32. However, we
    lose 2 bits for each subnet for the subnet and broadcast address, so we actually
    need 6 bits for the hosts because using 5 bits would only give us 30 hosts per subnet.
    Since there are 24 total available bits for subnets on network 10.0.0.0/8,
    24 – 6 host bits = 18 subnet bits, and 218 = 262,144 available subnets.

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  • 31. 

    Given network 175.100.0.0/16, if you create four subnets, how many addresses are available on each subnet?

    • A. 16,384

    • B. 4,096

    • C. 16,382

    • D. 4,094

    Correct Answer
    A. C. 16,382
    Explanation
    If there are four subnets, that means we have used 2 of the available host bits
    because 22 = 4. This leaves 16 – 2 = 14 bits for subnets, and 214 – 2 = 16,382.

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  • 32. 

    What is the correct “all hosts” broadcast address for subnet 10.15.0.0/17?

    • A. 10.15.255.255

    • B. 10.15.0.255

    • C. 10.15.127.255

    • D. 10.15.128.255

    Correct Answer
    A. C. 10.15.127.255
    Explanation
    Using a /17 means that we have used 1 bit for the subnet, leaving 7 bits in the
    third octet. The “all 1s” means to set all of those bits to 1 and all the bits in the
    last octet to 1. This means that the last octet is 255 (8 bits set to 1) and the 3rd
    octet is 127 (highest bit is 27 = 128, so 255 – 128 = 127).

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  • 33. 

    Which of the following is not allowed?  

    • A. subnet 10.0.0.0/16

    • B. subnet 10.255.0.0/16

    • C. subnet 10.10.10.0/16

    • D. host 10.10.10.0/32

    Correct Answer
    A. C. subnet 10.10.10.0/16
    Explanation
    This value cannot be used as a subnet address since the host portion is not all
    zeros. The other network addresses given all have the host portion all zeros. Any
    valid address can be used as a host address.

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  • 34. 

    Given network 135.100.0.0/16, you need nine subnets, and of these nine, one subnet needs to be split into 13 additional subnets. Choose the most likely masks you would create for this.

    • A. /20 for the first eight subnets, /23 for the remaining 13

    • B. /20 for the first eight subnets, /24 for the remaining 13

    • C. /24 for all subnets

    • D. /19 for the first eight subnets, /24 for the remaining 13

    Correct Answer
    A. B. /20 for the first eight subnets, /24 for the remaining 13
    Explanation
    Because nine subnets are needed, we need to use 4 bits of the host address in the
    third octet because 24 = 16 possible subnets (3 bits would only yield eight subnets).
    Then we need 13 more subnets, which means we need to take another 4 bits to
    give us 16 more possible subnets (24 = 16). This means we need 16 + 4 = /20 for
    the first eight subnets and 16 + 4 + 4 = /24 for the remaining 13 subnets.

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  • 35. 

    Given network 176.200.0.0/16 and a subnet that supports 4,387 hosts, what is the most likely mask for the subnet?

    • A. /20

    • B. /17

    • C. /21

    • D. /19

    Correct Answer
    A. D. /19
    Explanation
    We need to determine how many bits are needed to support 4,387 hosts. We know
    that using 8 bits gives us 255 hosts (all 1s for a single octet). If we double this by
    using an additional bit, that gives us 510 hosts; double again gives 1,020; double
    again gives us 2,040; double again gives us 4,080; and double again gives us 8,160.
    We had to double five times to get the desired number of hosts, which means we
    had to take 5 bits + original 8 bits = 13 total bits for hosts from the available 16
    (/16 was the original network). This leaves 3 bits left for subnetting (16 – 13 = 3),
    so /16 + 3 bits = /19 subnet mask.

    Rate this question:

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Jul 14, 2011
    Quiz Created by
    Temi

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