Network Fundamentals

An interface with MAC address 00:1C:2D:AA:BB:CC builds its 64-bit interface identifier using modified EUI-64. Which interface identifier results?

• A001C:2DFF:FEAA:BBCC, splitting the MAC and inserting FFFE with the first octet unchanged
• B021C:2DFF:FEAA:BBCC, splitting the MAC, inserting FFFE, and flipping the seventh bit of the first octetcheck_circle Correct
• C001C:2DFE:FFAA:BBCC, inserting the bytes FEFF in the middle and leaving the first octet unchanged
• D031C:2DFF:FEAA:BBCC, inserting FFFE and flipping the eighth bit of the first octet

Why A is wrong: This correctly splits the MAC and inserts FFFE in the middle but skips flipping the universal/local bit, leaving the first octet as 00 instead of 02; that omitted bit-flip is the single most common modified EUI-64 mistake.

Why B is correct: Modified EUI-64 splits the OUI from the device portion, inserts FFFE between them, and inverts the seventh bit of the first octet (00 becomes 02), producing exactly 021C:2DFF:FEAA:BBCC.

Why C is wrong: The inserted value is FFFE, not FEFF; reversing those two bytes is a plausible slip but yields an invalid interface identifier, and this answer also fails to flip the universal/local bit.

Why D is wrong: The bit that modified EUI-64 inverts is the seventh bit, not the eighth; flipping the wrong bit turns 00 into 01 rather than 02, so 031C is wrong and so is the stated bit position.

SW2 has the following SVI configured: interface Vlan10 ip address 10.0.0.1 255.255.255.192 no shutdown A junior engineer needs to know how many addresses on this subnet can be assigned to end devices such as printers and access points. How many usable host addresses does the subnet provide?

interface Vlan20
 ip address 10.4.20.1 255.255.255.192

• A62 usable host addressescheck_circle Correct
• B64 usable host addresses
• C63 usable host addresses
• D30 usable host addresses

Why A is correct: 255.255.255.192 is a /26 with 6 host bits, giving 2 to the power 6 (64) total addresses minus 2 reserved, which leaves 62 addresses for end devices.

Why B is wrong: 64 is the total number of addresses in a /26 block, but it counts the network and broadcast addresses, which cannot be assigned to devices.

Why C is wrong: Subtracting only the network address (or only the broadcast) gives 63, an off-by-one error that forgets one of the two reserved addresses.

Why D is wrong: 30 is the usable count for a /27 (255.255.255.224), so this answer applies the wrong mask and the wrong block size to the interface.

A network engineer is choosing between TCP and UDP for a real-time voice application that can tolerate occasional lost packets but cannot tolerate the delay caused by retransmission. Which characteristic of UDP makes it the better fit for this traffic?

• AUDP is connectionless and does not retransmit lost segments, so delivery proceeds without the latency of acknowledgements and resends.check_circle Correct
• BUDP guarantees in-order delivery using sequence numbers while still avoiding any retransmission delay.
• CUDP applies flow control with a sliding window so the receiver is never overwhelmed by voice data.
• DUDP establishes a connection first but then disables retransmission to keep latency low for media streams.

Why A is correct: UDP adds no connection setup, acknowledgements, or retransmission, so a lost voice packet is simply skipped rather than resent, keeping latency low for real-time media.

Why B is wrong: UDP does not provide sequencing or ordered delivery; that is a TCP feature. The claim of guaranteed ordering makes this option incorrect despite sounding reassuring.

Why C is wrong: Windowing and flow control belong to TCP, not UDP. UDP sends without negotiating a window, so attributing this mechanism to UDP is wrong.

Why D is wrong: UDP is connectionless and never establishes a session, so describing it as establishing a connection contradicts its core design and is incorrect.

On SW2, host H1 (MAC 00aa.bbcc.00a1) connects to Gi0/4 in VLAN 20 and is learned dynamically. The default MAC address-table ageing time is unchanged, and H1 then goes silent, sending no frames at all. After how long, with no further traffic from H1, will SW2 remove the dynamic entry for 00aa.bbcc.00a1?

SW2# show mac address-table aging-time vlan 20
Vlan    Aging Time
----    ----------
  20    300

• AAfter 300 seconds, the default ageing time shown for VLAN 20, because no traffic refreshed the entry.check_circle Correct
• BAfter 30 seconds, matching the default forward-delay timer used while the port settles.
• CAfter 120 seconds, the default ageing interval for dynamically learned addresses.
• DNever, because dynamic entries persist until the switch is reloaded or the port goes down.

Why A is correct: The default ageing time is 300 seconds, and because each frame from a source resets that timer, a host that stays completely silent has its entry removed once 300 seconds elapse.

Why B is wrong: 30 seconds is the combined spanning-tree listening and learning delay, not the MAC ageing interval, so it does not govern when a learned entry expires.

Why C is wrong: 120 seconds is a plausible-sounding guess and is close to other protocol timers, but the default MAC ageing time on Cisco switches is not 120 seconds.

Why D is wrong: That describes static or sticky entries; dynamic entries are aged out after the ageing timer expires when no frames refresh them, so they do not persist indefinitely.

A junior engineer claims that an interface assigned 172.32.5.10/24 is using RFC 1918 private space and therefore cannot be reached from the internet without NAT. Which assessment of this address is accurate?

• AIt is private, because the entire 172.16.0.0 to 172.31.255.255 block is reserved by RFC 1918 and 172.32 falls inside it.
• BIt is private, because any address in the 172.0.0.0/8 range is reserved for internal use under RFC 1918.
• CIt is public, because the RFC 1918 range is 172.16.0.0/12, spanning only 172.16.0.0 to 172.31.255.255, and 172.32.5.10 is outside it.check_circle Correct
• DIt is public, because RFC 1918 reserves only 172.16.0.0/16, so any second octet other than 16 is globally routable.

Why A is wrong: The RFC 1918 block in this range is 172.16.0.0/12, which spans 172.16.0.0 to 172.31.255.255; 172.32 is one step beyond the upper bound, so the stated range is incorrect.

Why B is wrong: RFC 1918 reserves 172.16.0.0/12, not the whole 172.0.0.0/8; treating the entire first octet 172 as private wrongly enlarges the reservation by sixteen times.

Why C is correct: RFC 1918 reserves 172.16.0.0/12, whose 12-bit mask covers the second octet from 16 through 31; 172.32 exceeds 31, so 172.32.5.10 is globally routable public space, not private.

Why D is wrong: The conclusion that 172.32 is public is right, but the reasoning is wrong: RFC 1918 reserves a /12, not a /16, so second octets 16 through 31 are all private.