Juniper Junos, Associate (JNCIA-Junos) (JN0-106)

✅ Option A (Correct)

Reasoning: Layer 2 switches are susceptible to traffic loops if a loop prevention mechanism like Spanning Tree Protocol (STP) is not properly configured or fails. Loops can lead to broadcast storms, MAC address table instability, and network instability.

✅ Option B (Correct)

Reasoning: When a switch receives a broadcast frame, it floods the frame out of all ports within the same VLAN, except the port on which it was received. This is a fundamental characteristic of how switches handle broadcast traffic.

❌ Why the other choices are incorrect:

• Option C is incorrect: Switches actively learn MAC addresses by inspecting the source MAC address of incoming frames and storing them in a MAC address table (also known as a CAM table). This learning process allows switches to make intelligent forwarding decisions.
• Option D is incorrect: Each port on a Layer 2 switch typically operates as its own collision domain. This significantly reduces collisions compared to a hub, which creates a single collision domain for all connected devices.

A /24 subnet mask indicates that the first 24 bits are reserved for the network portion of the IP address, leaving the remaining 8 bits (32 - 24 = 8) for host addresses. The total number of possible addresses in a subnet is calculated as 2ⁿ, where 'n' is the number of host bits. In this case, 2⁸ equals 256 possible IP addresses.

However, two of these addresses are reserved and cannot be assigned to hosts: the network address (all host bits set to 0) and the broadcast address (all host bits set to 1). Therefore, the number of usable host addresses is the total number of addresses minus 2. So, 256 - 2 = 254 usable hosts.

Reference: https://www.juniper.net/documentation/us/en/software/junos/junos-qfx/topics/concept/ip-address-overview.html

✅ Option A (Correct)

Reasoning: LC (Lucent Connector or Little Connector) is a common fiber-optic connector known for its small form factor and push-pull latching mechanism. It is widely used in high-density applications and for SFP/SFP+ transceivers.

✅ Option C (Correct)

Reasoning: SC (Standard Connector or Subscriber Connector) is a fiber-optic connector with a square profile and a push-pull mechanism. It's known for its robust performance and is often used in general networking and FTTx applications.

❌ Why the other choices are incorrect:

• Option B is incorrect: RJ-45 is a common connector for twisted-pair copper Ethernet cables, not fiber optics.
• Option D is incorrect: BNC (Bayonet Neill-Concelman) is a quick-connect/disconnect radio frequency connector used for coaxial cables, not fiber optics.

Correct Options Explained:

• Option C: TCP uses a three-way handshake to establish a connection.

  This is a fundamental characteristic of TCP. Before any data is exchanged, TCP establishes a connection using a three-step process: SYN (synchronize), SYN-ACK (synchronize-acknowledgment), and ACK (acknowledgment). This handshake ensures that both hosts are ready to communicate and agree on initial sequence numbers.

• Option D: TCP provides reliable, ordered delivery using sequence numbers.

  TCP is designed for reliable data transfer. It uses sequence numbers to ensure that data segments are reassembled in the correct order at the destination, and acknowledgments (ACKs) along with retransmission timers to guarantee delivery. If a segment is lost, the sender will retransmit it until an ACK is received.

Incorrect Options Explained:

• Option A: TCP guarantees fixed latency for application flows.

  TCP does not guarantee fixed latency. While it manages flow control and congestion control to optimize network usage, latency is an end-to-end network characteristic influenced by many factors (e.g., distance, network congestion, processing delays) that TCP itself cannot guarantee to fix or control.

• Option B: TCP broadcasts datagrams by default on Ethernet.

  TCP operates at the Transport Layer (Layer 4) and communicates between specific hosts using IP addresses and port numbers, resulting in unicast communication. Broadcasting is typically a Layer 2 (Ethernet) or Layer 3 (IP) mechanism used for discovery or specific protocols, not for standard TCP data transfer. TCP uses "segments," not "datagrams" (which are typically associated with UDP).

A router (R1) operates at Layer 3 (Network Layer) to forward IP packets and Layer 2 (Data Link Layer) for frame encapsulation. When R1 receives a packet from User A destined for User B, it performs a routing lookup using User B's IP address to determine the next hop, which is R2. While the source and destination IP addresses within the packet's Layer 3 header remain unchanged (unless NAT is involved), the Layer 2 frame header is modified. R1 removes the incoming Layer 2 header and constructs a new Layer 2 header for the egress interface. The destination MAC address of this new Layer 2 frame will be the MAC address of the next hop device, which is Router R2's interface directly connected to R1. This allows the frame to be delivered across the local segment between R1 and R2.

Reference: https://www.juniper.net/documentation/us/en/software/junos/junos-overview/topics/topic-map/understanding-junos-os.html

To determine the fourth usable host address for the 172.16.40.128/29 subnet, we first analyze the subnet mask.

A /29 subnet mask indicates that 3 bits are available for host addresses (32 - 29 = 3). This allows for 2³ = 8 total addresses. In any subnet, the first address is the network address, and the last is the broadcast address, leaving 6 usable host addresses.

For 172.16.40.128/29:

• Network address: 172.16.40.128
• 1st usable host: 172.16.40.129
• 2nd usable host: 172.16.40.130
• 3rd usable host: 172.16.40.131
• 4th usable host: 172.16.40.132
• 5th usable host: 172.16.40.133
• 6th usable host: 172.16.40.134
• Broadcast address: 172.16.40.135

Therefore, the fourth usable host address is 172.16.40.132.

To support 50 hosts, we need a subnet mask that provides at least 50 usable host IP addresses. The number of usable hosts in a subnet is calculated using the formula 2^h - 2, where 'h' is the number of host bits. We subtract 2 for the network address and the broadcast address, which are not assignable to hosts.

• A: /26 corresponds to 26 network bits. This leaves 32 - 26 = 6 host bits. The number of usable hosts is 2^6 - 2 = 64 - 2 = 62. This satisfies the requirement of 50 hosts.
• B: /27 leaves 5 host bits (32 usable hosts, 30 usable).
• C: /28 leaves 4 host bits (16 usable hosts, 14 usable).
• D: /29 leaves 3 host bits (8 usable hosts, 6 usable).

Therefore, /26 is the smallest subnet mask that can accommodate 50 hosts.

For a point-to-point link, only two usable IP addresses are required: one for each end of the link. In IPv4 subnetting, the number of usable host addresses is calculated as 2^{(32 - subnet_mask)} - 2 (subtracting the network address and broadcast address).

• A: 192.168.1.0/30
  A /30 subnet mask leaves 2 bits for host addresses (32 - 30 = 2). This results in 2² = 4 total addresses. After deducting the network address and broadcast address, there are 4 - 2 = 2 usable host addresses. This perfectly fits the requirement for a point-to-point link.
• B: 192.168.1.0/29
  A /29 subnet provides 3 bits for hosts (32 - 29 = 3), yielding 2³ = 8 total addresses. This allows for 8 - 2 = 6 usable host addresses.
• C: 192.168.1.0/24
  A /24 subnet provides 8 bits for hosts (32 - 24 = 8), yielding 2⁸ = 256 total addresses. This allows for 256 - 2 = 254 usable host addresses.
• D: 192.168.1.0/28
  A /28 subnet provides 4 bits for hosts (32 - 28 = 4), yielding 2⁴ = 16 total addresses. This allows for 16 - 2 = 14 usable host addresses.

Therefore, 192.168.1.0/30 is the correct combination for a point-to-point link requiring only two usable host addresses.

To determine the maximum number of assignable IP addresses for hosts in a network, we first identify the subnet mask. A /24 network signifies that the first 24 bits are used for the network portion, leaving the remaining 8 bits for the host portion of an IPv4 address (32 total bits - 24 network bits = 8 host bits). The total number of possible IP addresses in a subnet is calculated as 2 to the power of the number of host bits (2^8 = 256). However, two addresses are reserved and cannot be assigned to hosts: the network address (where all host bits are zero, e.g., 192.168.1.0) and the broadcast address (where all host bits are one, e.g., 192.168.1.255). Therefore, the number of usable IP addresses for hosts is 256 - 2 = 254.

The IPv6 address is 128 bits long, a significant increase from IPv4's 32 bits, which provides an exceptionally large address space. This fundamental characteristic makes option A correct.

The IPv6 header explicitly removed the checksum field present in IPv4. Error detection is instead handled by upper-layer protocols (like TCP and UDP) or at the data link layer to improve performance. Therefore, option B is incorrect.

IPv6 does not use ARP (Address Resolution Protocol) for neighbor discovery. Instead, it utilizes the Neighbor Discovery Protocol (NDP), which is part of ICMPv6, to resolve IP addresses to link-layer addresses and discover neighboring devices. Thus, option C is incorrect.

IPv6 addresses that begin with fe80::/10 are classified as Link-Local addresses. These addresses are intended for communication only on the local network segment (link) and are not routed by routers to other network segments. Therefore, option D is correct.