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Ruankao Network Administrator (Junior) practice questions

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Question 1Computer architecture, logic and data representation

In 软考 网络管理员 computer-system teaching, which description matches the classic Von Neumann architecture?

  1. Instructions and data use completely separate, non-shared memories with no common bus
  2. A single main memory holds both instructions and data, fetched over a shared bus Correct answer.
  3. The ALU stores the entire operating system in ROM and never reads RAM
  4. I/O devices execute machine instructions instead of the CPU

Explanation

Von Neumann architecture stores programs and data in the same main memory and uses a shared bus, so instruction fetch and data access compete for that path (the Von Neumann bottleneck). Harvard architecture, by contrast, uses separate instruction and data memories. The ALU does not replace RAM as the OS store, and I/O devices do not execute the instruction stream instead of the CPU.

Question 2Computer architecture, logic and data representation

Why do 网络管理员 textbooks place a small cache between the CPU and main memory?

  1. To hold recently used instructions and data so average access time is lower than DRAM alone Correct answer.
  2. To replace the ALU for floating-point arithmetic
  3. To assign public IPv4 addresses to hosts
  4. To convert analog voice to Ethernet frames

Explanation

Cache is a small, fast SRAM hierarchy (L1/L2/L3) that exploits locality: recently or nearby referenced blocks are reused, so the CPU waits on DRAM less often. Cache does not perform ALU math, does not allocate IP addresses, and is not a voice codec or Ethernet MAC.

Question 3Computer architecture, logic and data representation

A 网络管理员 is asked to protect a file server against a single disk failure with the simplest redundant RAID. Which level mirrors entire disks?

  1. RAID 0 striping with no redundancy
  2. RAID 4 with a single dedicated parity disk only, without mirroring
  3. RAID 1 mirroring (disk mirroring) Correct answer.
  4. JBOD spanning with no redundancy

Explanation

RAID 1 writes the same data to two (or more) disks so one disk can fail without data loss. RAID 0 stripes for speed with zero redundancy. RAID 5/6 use distributed parity rather than full mirrors. JBOD concatenates disks with no RAID protection.

Question 4Computer architecture, logic and data representation

What is the 8-bit binary representation of decimal 255, a value 网络管理员 candidates must convert when reading subnet masks?

  1. 01111111
  2. 10000000
  3. 11111110
  4. 11111111 Correct answer.

Explanation

2^8 − 1 = 255, so eight 1-bits are 11111111, which is also 0xFF. That pattern appears in 255.255.255.0 and similar masks. 01111111 is 127, 10000000 is 128, and 11111110 is 254.

Question 5Computer architecture, logic and data representation

A data-communication item uses even parity on the 7-bit word 1101010. What parity bit is appended so the total number of 1-bits is even?

  1. 1, because the word already has an odd number of 1-bits
  2. 0, because the word already has an even number of 1-bits Correct answer.
  3. 1, because even parity always appends 1
  4. The CRC-32 remainder 0x04C11DB7

Explanation

1101010 contains four 1-bits (positions with 1: four of them), which is even. Even parity therefore appends 0 so the 8-bit frame still has an even count of 1s. Parity is a 1-bit check, not a CRC polynomial remainder. Even parity does not always append 1.

Question 6Computer architecture, logic and data representation

Nyquist sampling of a 4 kHz analog voice channel (standard PCM teaching in 软考 网络管理员) requires a sampling frequency of at least:

  1. 2 kHz
  2. 4 kHz
  3. 8 kHz Correct answer.
  4. 64 kHz

Explanation

The sampling theorem requires fs > 2fmax. For a 4 kHz voice band, the Nyquist rate is 8 kHz, which is why G.711 PCM uses 8000 samples/s. 2 kHz and 4 kHz undersample; 64 kHz is a bit-rate figure for 8-bit PCM at 8 kHz (8 × 8000 = 64 kbit/s), not the sampling frequency.

Question 7Computer architecture, logic and data representation

Shannon’s noisy-channel capacity used in 网络管理员 data-communication items is which formula (C in bit/s, B in Hz, S/N linear power ratio)?

  1. C = B log2(1 + S/N) Correct answer.
  2. C = 2B log2(M) with no noise term
  3. C = B × (S/N) with no logarithm
  4. C = 2B only, independent of SNR and signalling levels

Explanation

The Shannon–Hartley theorem gives C = B log2(1 + S/N) as the AWGN capacity bound. Nyquist’s noiseless formula C = 2B log2(M) uses discrete signalling levels M and does not include SNR. Capacity is not B times SNR without a log, and it is not simply 2B.

Question 8Computer architecture, logic and data representation

Compared with a Hamming code that can correct single-bit errors, what is the usual role of a CRC on an Ethernet frame?

  1. CRC always corrects every bit error in the payload
  2. CRC encrypts the payload with AES
  3. CRC assigns the IPv4 address of the sender
  4. CRC detects (but does not by itself correct) many bit errors via a polynomial remainder, as in the Ethernet FCS Correct answer.

Explanation

Cyclic redundancy checks detect accidental bit errors by comparing a polynomial remainder (Ethernet’s 32-bit FCS). They do not encrypt, do not assign IP addresses, and typical CRC usage is detection plus discard/retransmit, not Hamming-style single-bit correction in the NIC.

Question 9Computer architecture, logic and data representation

A switched Ethernet link that sends and receives at the same time is operating in which data-communication mode?

  1. Simplex
  2. Full duplex Correct answer.
  3. Half duplex with CSMA/CD collisions required
  4. Broadcast radio only

Explanation

Full duplex allows simultaneous transmit and receive, which is how modern switched Ethernet typically runs, so CSMA/CD is not needed on that link. Simplex is one direction only. Half duplex shares the medium and uses CSMA/CD on classic shared Ethernet. Ordinary copper Ethernet is not 'broadcast radio only'.

Question 10Computer architecture, logic and data representation

Why does the Internet (and 网络管理员 TCP/IP teaching) use packet switching rather than dedicating a circuit for every conversation?

  1. Every packet is guaranteed a reserved 64 kbit/s circuit like PCM voice
  2. Packet switching forbids sharing any router output link
  3. Packets share links statistically, improving utilization when traffic is bursty Correct answer.
  4. Packet switching eliminates the need for IP addresses

Explanation

Packet switching multiplexes independent datagrams on shared links. Bursty data uses capacity when other flows are idle, which is more efficient than a dedicated circuit per session. IP still needs addresses. Circuits (classic PSTN) reserve capacity whether used or not.

Question 11Computer architecture, logic and data representation

A noiseless channel has bandwidth B = 4000 Hz and uses 8 distinct signalling levels (M = 8). Using Nyquist’s formula C = 2B log2(M), the maximum bit rate is:

  1. 4000 bit/s
  2. 24000 bit/s Correct answer.
  3. 8000 bit/s
  4. 32000 bit/s

Explanation

log2(8) = 3, so C = 2 × 4000 × 3 = 24000 bit/s. 4000 bit/s ignores the 2B and log2(M) factors. 8000 bit/s is 2B with M = 2 (binary). 32000 would correspond to a different M or B.

Question 12Computer architecture, logic and data representation

A noisy channel has B = 3000 Hz and S/N = 1023 (linear). Using C = B log2(1 + S/N), the Shannon capacity is:

  1. 30000 bit/s Correct answer.
  2. 3000 bit/s
  3. 1023 bit/s
  4. 6000 bit/s

Explanation

1 + S/N = 1024 = 2^10, so log2(1024) = 10 and C = 3000 × 10 = 30000 bit/s. This is a standard 软考-style Shannon drill. Do not confuse B with C, or S/N with C, or Nyquist 2B = 6000.

Question 13Computer Network Fundamentals and Architecture

How many layers does the ISO OSI reference model used in 软考 网络管理员 architecture items have?

  1. 3
  2. 4
  3. 7 Correct answer.
  4. 12

Explanation

OSI has seven layers: Physical, Data Link, Network, Transport, Session, Presentation, Application. TCP/IP is commonly taught as four layers (or five if Link is split). There is no 12-layer OSI model.

Question 14Computer Network Fundamentals and Architecture

In the four-layer TCP/IP model taught for 网络管理员, which layer contains IP?

  1. Internet (network) layer Correct answer.
  2. Application layer
  3. Physical layer only, with no IP header
  4. Presentation layer of OSI as a required TCP/IP layer

Explanation

IP is the Internet-layer (OSI Network layer) protocol that forwards datagrams between networks. HTTP/FTP/DNS sit at Application. TCP/IP does not use OSI Presentation as its own required layer. The Physical layer carries bits, not the IP header as its protocol.

Question 15Computer Network Fundamentals and Architecture

The TCP/IP Link (network-access) layer corresponds most closely to which OSI layers?

  1. OSI Session and Presentation only
  2. OSI Transport only
  3. OSI Application only
  4. OSI Physical and Data Link Correct answer.

Explanation

TCP/IP Link covers framing, MAC, and physical signalling - OSI layers 1 and 2. Transport is TCP/UDP (OSI 4). Application in TCP/IP covers OSI 5–7. Session/Presentation are not a separate TCP/IP layer.

Question 16Computer Network Fundamentals and Architecture

Which protocol belongs at the TCP/IP application layer in 网络管理员 Internet-application teaching?

  1. IPv4
  2. HTTP Correct answer.
  3. Ethernet CSMA/CD
  4. The HDLC flag byte only

Explanation

HTTP is an application protocol for the Web. IPv4 is Internet layer. Ethernet/CSMA/CD is Link/data-link. HDLC flags are data-link framing, not WWW.

Question 17Computer Network Fundamentals and Architecture

Which statement correctly contrasts TCP and UDP for 网络管理员 candidates?

  1. UDP provides a three-way handshake and retransmission by default
  2. TCP is connectionless and never uses port numbers
  3. TCP is connection-oriented and can retransmit lost segments; UDP is connectionless with no retransmission Correct answer.
  4. Both TCP and UDP operate at the OSI Network layer instead of Transport

Explanation

TCP establishes a connection (three-way handshake), numbers segments, and retransmits. UDP sends datagrams without a connection or built-in retransmission. Both are Transport-layer protocols and both use 16-bit ports.

Question 18Computer Network Fundamentals and Architecture

A DNS query from a 网络管理员 workstation uses which well-known port on the server (UDP/TCP)?

  1. 25
  2. 80
  3. 53 Correct answer.
  4. 110

Explanation

IANA well-known port 53 is DNS (UDP for typical queries, TCP for large responses and zone transfers). 25 is SMTP, 80 is HTTP, 110 is POP3.

Question 19Computer Network Fundamentals and Architecture

The default TCP port for HTTPS (HTTP over TLS) is:

  1. 80
  2. 443 Correct answer.
  3. 21
  4. 23

Explanation

HTTPS uses TCP 443. HTTP cleartext uses 80. FTP control is 21. Telnet is 23. 网络管理员 service questions expect these defaults even when administrators later change them.

Question 20Computer Network Fundamentals and Architecture

Match the defaults: FTP control, SMTP, and POP3. Which triple is correct?

  1. FTP 23, SMTP 53, POP3 80
  2. FTP 21, SMTP 25, POP3 110 Correct answer.
  3. FTP 80, SMTP 110, POP3 25
  4. FTP 443, SMTP 21, POP3 53

Explanation

FTP control is TCP 21 (data 20 in active mode), SMTP is 25, POP3 is 110. Mixing those with Telnet 23, DNS 53, HTTP 80, or HTTPS 443 is a common exam trap.

Question 21Computer Network Fundamentals and Architecture

In TCP’s three-way handshake, which sequence is correct?

  1. ACK, then FIN, then SYN
  2. SYN, then SYN-ACK, then ACK Correct answer.
  3. FIN, then FIN-ACK, then RST only
  4. ARP request, then ICMP redirect, then DHCP ACK

Explanation

The client sends SYN, the server replies SYN-ACK, the client sends ACK. FIN is used to close. ARP and DHCP are not the TCP handshake.

Question 22Computer Network Fundamentals and Architecture

ARP on an Ethernet LAN is used to:

  1. Map an IPv4 address to a 48-bit MAC address Correct answer.
  2. Map a MAC address to a DNS domain name
  3. Encrypt TCP payloads
  4. Assign public AS numbers for BGP

Explanation

Address Resolution Protocol resolves a next-hop IPv4 address to a MAC so a frame can be sent on the LAN. DNS maps names to IP. ARP does not encrypt and is not BGP.

Question 23Computer Network Fundamentals and Architecture

The ping diagnostic used daily by a 网络管理员 sends which ICMP messages by default on IPv4?

  1. TCP SYN to port 80
  2. UDP to port 53 only
  3. ICMP Echo Request, expecting Echo Reply Correct answer.
  4. SMTP MAIL FROM probes

Explanation

Classic ping is ICMP Echo Request/Reply (Type 8/0 on IPv4). Some tools can use other probes, but the 软考 default is ICMP echo. TCP SYN is a port scan/connect test, not ping. DNS uses 53. SMTP is mail.

Question 24Computer Network Fundamentals and Architecture

A host’s IPv4 default gateway is used when:

  1. The destination IP is on the same subnet as the host, so ARP to the gateway is always required first
  2. The destination IP is not on a locally connected subnet, so the packet is sent to the configured router Correct answer.
  3. DNS fails, so all packets are sent to 255.255.255.255
  4. The host has no IP address at all

Explanation

The default route (0.0.0.0/0) sends off-subnet traffic to the gateway router. Same-subnet destinations are delivered directly via ARP to the target MAC, not via the gateway. Broadcast 255.255.255.255 is not a substitute for a gateway. A host without an IP cannot use a unicast gateway path.

Question 25Computer Network Fundamentals and Architecture

Each router that forwards an IPv4 datagram decrements which header field, dropping the packet when it reaches 0?

  1. Source port
  2. TTL (Time to Live) Correct answer.
  3. Checksum of the Ethernet FCS only
  4. The DNS serial number

Explanation

IPv4 TTL (IPv6 Hop Limit) is decremented at each hop to prevent routing loops. When TTL hits 0, the router discards the datagram and may send ICMP Time Exceeded (the basis of traceroute). Ports are transport fields. Ethernet FCS is link-layer. DNS SOA serial is not in the IP header.

Question 26Computer Network Fundamentals and Architecture

DHCP address assignment on IPv4 uses the DORA sequence. Which order is correct?

  1. ACK, Request, Offer, Discover
  2. Discover, Offer, Request, ACK Correct answer.
  3. Offer, Discover, ACK, Request
  4. Request, Discover, ACK, Offer

Explanation

The client broadcasts Discover, servers Offer, the client Requests one offer, the server ACKs. That is DORA. Reversing the letters is a common trap. DHCP server UDP port 67 and client 68 carry these messages.

Question 27Computer Network Fundamentals and Architecture

PAT (NAPT) in a small office 网络管理员 design typically:

  1. Maps many private RFC 1918 addresses to one (or a few) public IPv4 address(es) using port numbers Correct answer.
  2. Replaces Ethernet with Token Ring
  3. Assigns a globally unique public IPv4 address to every internal printer without sharing
  4. Disables all DNS so names cannot leak

Explanation

Port Address Translation multiplexes many inside private addresses onto a shared public address by rewriting source ports. That conserves public IPv4. It does not change LAN MAC technology, does not give every printer a unique public IP, and does not require disabling DNS.

Question 28Computer Network Fundamentals and Architecture

A 网络管理员 must pick TCP versus UDP. Which choice matches typical Internet practice?

  1. Use UDP for a bank file copy that must not lose a single byte, with no application retry
  2. Use TCP for DNS zone transfers and HTTP page downloads; UDP is common for short DNS queries Correct answer.
  3. Use TCP because UDP cannot use port numbers
  4. Use UDP for SMTP message transfer between MTAs as the RFC-required default

Explanation

HTTP and SMTP use TCP. DNS queries are usually UDP/53; zone transfers (AXFR) use TCP. File copies that need reliability use TCP (or an application that retries). UDP does use ports. SMTP between mail servers is TCP 25, not UDP.

Question 29Computer Network Fundamentals and Architecture

IANA well-known (system) TCP/UDP ports used by privileged Internet services occupy which range?

  1. 1024–49151 only
  2. 0–1023 Correct answer.
  3. 49152–65535 only
  4. Only port 8080

Explanation

Ports 0–1023 are well-known/system ports (HTTP 80, HTTPS 443, DNS 53, etc.). 1024–49151 are registered. 49152–65535 are dynamic/private. 8080 is a common alternative HTTP port in the registered range, not the whole well-known space.

Question 30Computer Network Fundamentals and Architecture

IMAP’s default TCP port, which 网络管理员 mail items contrast with POP3 110, is:

  1. 25
  2. 143 Correct answer.
  3. 53
  4. 67

Explanation

IMAP uses TCP 143 (IMAPS 993). POP3 is 110 (POP3S 995). SMTP is 25. DNS is 53. DHCP server is UDP 67. IMAP keeps mail on the server for multi-device access; POP3 traditionally downloads.

Question 31Data Link Layer and LAN Technologies

IEEE 802.3 is the family of standards for which LAN technology that 网络管理员 must master?

  1. Ethernet Correct answer.
  2. ATM cells only
  3. Bluetooth PAN only
  4. Frame Relay PVC only

Explanation

IEEE 802.3 specifies Ethernet (historically including CSMA/CD). IEEE 802.11 is wireless LAN. ATM, Bluetooth, and Frame Relay are different technologies.

Question 32Data Link Layer and LAN Technologies

Classic shared-media half-duplex Ethernet uses which media-access method?

  1. Token passing as on IEEE 802.5 Token Ring only
  2. CSMA/CA as the Ethernet-wired default
  3. CSMA/CD (Carrier Sense Multiple Access with Collision Detection) Correct answer.
  4. TDMA cellular slots only

Explanation

Half-duplex Ethernet on a shared coaxial or hub collision domain uses CSMA/CD. Wi-Fi (802.11) uses CSMA/CA. Token Ring is 802.5. Full-duplex switched Ethernet does not need CD on that link.

Question 33Data Link Layer and LAN Technologies

A 网络管理员 comparing wired Ethernet with office Wi-Fi should remember that IEEE 802.11 typically uses:

  1. CSMA/CD on the radio medium
  2. CSMA/CA (collision avoidance) rather than Ethernet’s CD Correct answer.
  3. Token Ring beacons on 2.4 GHz
  4. ATM signalling for every smartphone

Explanation

Wireless stations cannot always detect collisions the way wired CSMA/CD does, so 802.11 uses CSMA/CA (and often RTS/CTS). Do not apply CSMA/CD as the 802.11 access method. Token Ring and ATM are different families.

Question 34Data Link Layer and LAN Technologies

A multi-port Ethernet hub (repeater) operates primarily at which OSI layer and has which collision-domain effect?

  1. Layer 3, creating a separate collision domain per port
  2. Layer 7, terminating HTTP
  3. Layer 2 switching with a CAM table
  4. Layer 1; all ports share one collision domain Correct answer.

Explanation

A hub is a multi-port repeater at the Physical layer: it floods bits to every port, so all attached hosts share one collision domain. A switch is Layer 2 with per-port collision domains. Routers are Layer 3. Hubs do not terminate HTTP.

Question 35Data Link Layer and LAN Technologies

An Ethernet switch forwards frames using which addresses and primarily which OSI layer?

  1. IPv6 addresses at Layer 4
  2. MAC addresses at Layer 2 (data link) Correct answer.
  3. UDP ports at Layer 1
  4. AS numbers at Layer 1

Explanation

A transparent bridge/switch learns source MACs and forwards/filters on destination MAC (Layer 2). IP routing is Layer 3. Ports at Layer 4 are firewall/load-balancer functions, not basic switching. AS numbers are BGP.

Question 36Data Link Layer and LAN Technologies

A router interconnecting two IPv4 subnets primarily operates at which layer and has which broadcast-domain effect?

  1. Layer 1 repeater; broadcasts flood both sides unchanged
  2. Layer 2 bridge; a single broadcast domain spans both LAN segments
  3. Layer 3; it does not forward Ethernet broadcasts between interfaces Correct answer.
  4. Layer 7 only; it cannot forward IP packets

Explanation

Routers forward IP packets between networks and terminate Layer-2 broadcast domains: an Ethernet broadcast on one interface is not flooded out the other. Hubs/repeaters are L1. Switches/bridges extend a broadcast domain (unless VLANs). Routers absolutely forward IP.

Question 37Data Link Layer and LAN Technologies

An Ethernet MAC address as used in 网络管理员 LAN teaching is typically:

  1. 48 bits (6 octets), burned-in or locally administered Correct answer.
  2. 32 bits like IPv4
  3. 128 bits like IPv6
  4. 16 bits like a TCP port

Explanation

IEEE 48-bit MAC addresses (EUI-48) identify Ethernet/Wi-Fi interfaces. IPv4 is 32-bit, IPv6 is 128-bit, TCP ports are 16-bit. Do not confuse the lengths.

Question 38Data Link Layer and LAN Technologies

IEEE 100BASE-TX Fast Ethernet over copper typically uses which media and distance?

  1. One pair of Cat3, 2 km
  2. Eight pairs of Cat8, 10 km
  3. Two pairs of Category 5 (or better) UTP, about 100 m Correct answer.
  4. Single-mode fiber only, 80 km mandatory

Explanation

100BASE-TX uses two pairs of Cat5 (or better) twisted pair with a 100 m channel, the same copper distance class as 10BASE-T. It is not a 2 km Cat3 one-pair scheme, not Cat8 10 km, and not mandatory long-haul SMF (that would be optical PHYs such as 100BASE-FX/LX variants with different reaches).

Question 39Data Link Layer and LAN Technologies

IEEE 1000BASE-T Gigabit Ethernet over copper uses which typical cabling?

  1. One pair of telephone Cat3, 100 m
  2. Coaxial 10BASE5 thicknet only
  3. Two pairs of Cat5, same as 100BASE-TX
  4. Four pairs of Category 5 or better UTP (commonly Cat5e), about 100 m Correct answer.

Explanation

1000BASE-T (IEEE 802.3ab) transmits on all four pairs of Cat5/Cat5e (or better) UTP for a 100 m channel. 100BASE-TX uses only two pairs. 10BASE5 is obsolete thick coaxial. Cat3 one-pair cannot carry 1000BASE-T.

Question 40Data Link Layer and LAN Technologies

The conventional Ethernet payload MTU used in 网络管理员 IP/LAN items is:

  1. 64 bytes of IP header only
  2. 1500 bytes Correct answer.
  3. 64 kilobytes of TCP window only
  4. 9000 bytes as the mandatory standard MTU on all office switches

Explanation

Standard Ethernet allows a 1500-byte payload (IP packet) in the frame; the Layer-2 frame is larger once headers and FCS are added. Jumbo 9000-byte MTUs are optional and not universal. TCP window size is not the Ethernet MTU. 64 bytes relates to minimum frame size, not the IP MTU.

Question 41Data Link Layer and LAN Technologies

How many usable IPv4 host addresses are in a /24 LAN (usable hosts = 2^h − 2, excluding network and broadcast)?

  1. 254 Correct answer.
  2. 256
  3. 255
  4. 252

Explanation

A /24 has 8 host bits, so 2^8 = 256 addresses. Subtract the network identifier and the directed broadcast: 256 − 2 = 254 usable hosts. Counting 256 or 255 forgets those two reserved addresses; 252 is 2^8 − 4 or a /30 confusion.

Question 42Data Link Layer and LAN Technologies

The dotted-decimal mask 255.255.255.192 corresponds to which prefix length?

  1. /24
  2. /25
  3. /26 Correct answer.
  4. /30

Explanation

192 in the last octet is 11000000, so two extra network bits beyond /24, which is /26. /24 is 255.255.255.0, /25 is 255.255.255.128, /30 is 255.255.255.252.

Question 43Data Link Layer and LAN Technologies

A /26 IPv4 subnet has how many usable host addresses (2^h − 2)?

  1. 64
  2. 62 Correct answer.
  3. 32
  4. 30

Explanation

A /26 has 6 host bits: 2^6 = 64 addresses, minus network and broadcast leaves 62 usable hosts. 64 forgets −2. 32 and 30 belong to /27 (2^5 − 0 or −2).

Question 44Data Link Layer and LAN Technologies

Host 192.168.10.130/26 belongs to which network (subnet ID)?

  1. 192.168.10.0/26
  2. 192.168.10.64/26
  3. 192.168.10.192/26
  4. 192.168.10.128/26 Correct answer.

Explanation

A /26 block size is 64 in the last octet: subnets 0, 64, 128, 192. 130 falls in 128–191, so the network is 192.168.10.128. 0, 64, and 192 are the other /26s on that /24.

Question 45Data Link Layer and LAN Technologies

For 192.168.10.130/26, the directed broadcast address is:

  1. 192.168.10.191 Correct answer.
  2. 192.168.10.255
  3. 192.168.10.129
  4. 192.168.10.128

Explanation

The 128/26 range is 192.168.10.128–191. The last address 191 is the subnet broadcast. 255 would be the /24 broadcast. 128 is the network ID. 129 is the first usable host.

Question 46Data Link Layer and LAN Technologies

Host 172.16.5.67/27: what is the subnet network address?

  1. 172.16.5.0/27
  2. 172.16.5.32/27
  3. 172.16.5.64/27 Correct answer.
  4. 172.16.5.96/27

Explanation

A /27 block size is 32: last-octet subnets 0, 32, 64, 96, … 67 is in 64–95, so the network is 172.16.5.64/27. Broadcast would be 172.16.5.95; usable hosts 65–94.

Question 47Data Link Layer and LAN Technologies

A /27 subnet has how many usable IPv4 hosts?

  1. 32
  2. 30 Correct answer.
  3. 62
  4. 14

Explanation

Host bits = 5, so 2^5 − 2 = 30. 32 is the block size. 62 is /26. 14 is /28 (2^4 − 2).

Question 48Data Link Layer and LAN Technologies

Which address is an RFC 1918 private IPv4 address suitable for a 网络管理员 small LAN?

  1. 8.8.8.8
  2. 172.32.0.1
  3. 192.0.2.1
  4. 192.168.10.5 Correct answer.

Explanation

RFC 1918 private blocks are 10.0.0.0/8, 172.16.0.0/12 (172.16.0.0–172.31.255.255), and 192.168.0.0/16. 192.168.10.5 is in 192.168.0.0/16. 8.8.8.8 is public DNS. 172.32.0.1 is just outside 172.16/12. 192.0.2.0/24 is documentation (TEST-NET-1), not RFC 1918.

Question 49Data Link Layer and LAN Technologies

A 网络管理员 must pick the longest mask (smallest subnet) that still fits 60 usable hosts. Which prefix works, given usable = 2^h − 2?

  1. /27 (30 usable) - enough because 30 > 60/2
  2. /26 (62 usable) Correct answer.
  3. /30 (2 usable)
  4. /28 (14 usable)

Explanation

Need at least 60 usable addresses: /27 gives 30 (too small), /26 gives 62 (fits), /25 gives 126 (fits but is larger). /28 and /30 are far too small. The longest mask that still fits is /26.

Question 50Data Link Layer and LAN Technologies

Host 10.1.8.200/22: what is the IPv4 subnet network address?

  1. 10.1.0.0/22
  2. 10.1.4.0/22
  3. 10.1.8.0/22 Correct answer.
  4. 10.1.8.200/22 as the network ID

Explanation

A /22 mask is 255.255.252.0; the third octet increments in blocks of 4: 0, 4, 8, 12, … 8.200 sits in 10.1.8.0–10.1.11.255, so the network is 10.1.8.0/22. The host address 10.1.8.200 is not the subnet ID. 10.1.0.0/22 and 10.1.4.0/22 are earlier blocks.