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Ruankao Network Planning Designer (Senior) study materials
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Ruankao Network Planning Designer (Senior) practice questions

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Question 1Computer architecture, logic and data representation

In 软考 网络规划设计师 computer-system teaching, which description matches classic Von Neumann architecture?

  1. Instructions and data use completely separate memories with no shared bus
  2. A single main memory holds both instructions and data on a shared bus Correct answer.
  3. The ALU stores the operating system in ROM and never reads RAM
  4. I/O devices fetch and execute the machine-instruction stream 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 (the Von Neumann bottleneck). Harvard architecture splits instruction and data memories. The ALU does not replace RAM as the OS store, and I/O devices do not execute the instruction stream in place of the CPU.

Question 2Computer architecture, logic and data representation

A disk array must copy a large backup image into RAM with minimal per-byte CPU involvement. Which I/O method is designed for that transfer?

  1. Programmed I/O that busy-waits on every byte in the CPU
  2. DMA (direct memory access) by a controller into main memory Correct answer.
  3. Disabling interrupts and having the CPU poll the controller status register
  4. Replacing the memory bus with an RS-232 serial console

Explanation

DMA lets an I/O controller move blocks between a device and main memory while the CPU continues other work, with an interrupt typically only at completion. Programmed I/O/polling burns CPU cycles on every transfer. Keyboard polling and RS-232 consoles are unrelated to bulk disk-to-RAM copies.

Question 3Computer architecture, logic and data representation

A file server needs to survive any single disk failure and keep most of the raw capacity. Which RAID level stripes data and distributed parity (not a full mirror)?

  1. RAID 0 striping with no redundancy
  2. RAID 1 mirroring of entire disks
  3. RAID 5 striped data with distributed parity Correct answer.
  4. JBOD concatenation with no RAID protection

Explanation

RAID 5 stripes data and parity across disks so any one disk can fail and capacity is about (N−1) disks. RAID 0 has no redundancy. RAID 1 mirrors whole disks and uses 50% of raw capacity on a pair. JBOD is not a redundant RAID level.

Question 4Computer architecture, logic and data representation

Why do 网络规划设计师 textbooks place SRAM cache between the CPU and DRAM?

  1. To hold recently used blocks so average access time is lower than DRAM alone Correct answer.
  2. To assign public IPv4 addresses to campus hosts
  3. To terminate BGP sessions on the processor die
  4. To convert analog voice to Ethernet frames

Explanation

Cache exploits temporal and spatial locality: recently or nearby referenced blocks are reused in fast SRAM, so the CPU waits on DRAM less often. Cache does not allocate IP addresses, speak BGP, or act as a voice codec or NIC.

Question 5Computer architecture, logic and data representation

A classic telephone voice channel is band-limited to 4 kHz and sampled at the Nyquist rate with 8-bit linear PCM. What is the resulting bit rate?

  1. 32 kbps
  2. 64 kbps Correct answer.
  3. 128 kbps
  4. 1.544 Mbps

Explanation

Nyquist sampling of a 4 kHz analog voice signal requires at least 8 kHz. With 8 bits per sample, 8000 × 8 = 64 000 bit/s, the DS0/E0 64 kbps timeslot used in PCM hierarchies. 32 kbps is a common ADPCM rate, 128 kbps is two such channels, and 1.544 Mbps is a T1 line rate, not a single voice PCM stream.

Question 6Computer architecture, logic and data representation

Nyquist’s noiseless-channel formula is C = 2B log2(M). For B = 4 kHz and 16-level signaling (M = 16), what is C?

  1. 8 kbps
  2. 16 kbps
  3. 32 kbps Correct answer.
  4. 64 kbps

Explanation

log2(16) = 4, so C = 2 × 4000 × 4 = 32 000 bit/s. The factor 2B is the maximum symbol rate; each symbol carries 4 bits when M = 16. 8 kbps would be 2B with binary signaling forgotten; 16 kbps is B log2(M) without the 2; 64 kbps is PCM voice, not this Nyquist example.

Question 7Computer architecture, logic and data representation

Ethernet frames carry a 32-bit FCS. Which error-detection method is that FCS?

  1. Even parity over each byte only
  2. A simple Internet checksum as in IPv4 headers
  3. CRC (cyclic redundancy check) using a generator polynomial Correct answer.
  4. A Hamming SECDED code that corrects two-bit errors in hardware

Explanation

IEEE 802.3 uses a CRC-32 frame check sequence computed from a generator polynomial; it detects burst errors far better than per-byte parity or the IPv4 header checksum. Ethernet FCS is detection, not a Hamming corrector for two-bit errors on the wire.

Question 8Computer architecture, logic and data representation

DWDM packs many optical channels onto one fiber by assigning each channel a different wavelength. That is an example of which multiplexing method?

  1. TDM (time-division multiplexing) of 64 kbps timeslots only
  2. FDM/WDM (frequency/wavelength-division multiplexing) Correct answer.
  3. Statistical multiplexing of Ethernet collision domains
  4. CDMA spreading codes on a copper access loop

Explanation

WDM (including DWDM/CWDM) is optical frequency-division multiplexing: channels occupy different wavelengths on one fiber. TDM shares time slots on one wavelength or electrical stream. Ethernet statistical multiplexing and CDMA are different sharing methods and are not how DWDM grids are built.

Question 9Computer architecture, logic and data representation

Classic 10BASE5/10BASE-T Ethernet used Manchester line coding. What is the characteristic of Manchester encoding?

  1. A mid-bit transition represents the bit and also supplies a clock edge Correct answer.
  2. A long string of 1s produces no transitions, so the receiver loses clock
  3. Each bit is sent as five baud using 4B/5B as on FDDI only
  4. The line is analog QAM with no binary mid-bit edge

Explanation

Manchester encoding places a guaranteed transition in the middle of each bit interval, encoding the bit value by the direction of that edge and giving the receiver a clock. NRZ can have long runs without transitions. 4B/5B is a block code used on 100BASE-X/FDDI, not Manchester. QAM is a passband analog scheme.

Question 10Computer architecture, logic and data representation

Two codewords differ in 3 bit positions (Hamming distance 3). How many bit errors can this code always detect, and how many can it always correct?

  1. Detect 3 and correct 3
  2. Detect 2 and correct 1 Correct answer.
  3. Detect 1 and correct 2
  4. Detect 0 and correct 3

Explanation

For minimum distance d, a code can detect up to d−1 errors and correct up to floor((d−1)/2) errors. With d = 3 that is detect 2 and correct 1. It cannot reliably correct 2-bit errors or detect 3-bit errors as a guaranteed bound.

Question 11Computer architecture, logic and data representation

Shannon’s noisy-channel capacity is C = B log2(1 + S/N). For B = 3 kHz and SNR = 30 dB (S/N = 1000), which value is the best approximation of C?

  1. 3 kbps
  2. 6 kbps
  3. 30 kbps Correct answer.
  4. 3 Mbps

Explanation

30 dB means S/N = 10^(30/10) = 1000. Then C = 3000 × log2(1001) ≈ 3000 × 9.97 ≈ 29.9 kbps, about 30 kbps. 6 kbps is roughly Nyquist binary 2B; 3 kbps is B itself; 3 Mbps off by 100× from a unit error (kHz vs MHz).

Question 12Computer architecture, logic and data representation

On a high delay-bandwidth satellite link, which ARQ strategy retransmits only the lost frames and buffers out-of-order arrivals at the receiver?

  1. Stop-and-wait with window size 1
  2. Go-back-N, which discards out-of-order frames and rewinds the sender
  3. Selective-repeat ARQ Correct answer.
  4. Unacknowledged UDP with no retransmission

Explanation

Selective Repeat retransmits only missing sequence numbers and lets the receiver reassemble from a window of out-of-order frames, which uses the satellite pipe more efficiently than Go-back-N rewinds. Stop-and-wait sends one frame at a time. UDP has no ARQ.

Question 13Network Layer Protocols and Routing

A campus access switch that forwards frames using destination MAC addresses (no IP routing) operates primarily at which OSI layer?

  1. Layer 1 physical only, as a shared-media hub
  2. Layer 2 data link (transparent bridging/switching) Correct answer.
  3. Layer 3 network (IP routing)
  4. Layer 7 application (HTTP proxy)

Explanation

A MAC-learning Ethernet switch is an OSI Layer 2 (data-link) device: it forwards frames by destination MAC and CAM table. A hub is Layer 1. An IP router or Layer 3 switch routing between VLANs is Layer 3. An HTTP proxy is application-layer.

Question 14Network Layer Protocols and Routing

Which transport protocol provides a connection-oriented, reliable byte stream with congestion control, as used by HTTPS?

  1. UDP
  2. ICMP
  3. TCP Correct answer.
  4. ARP

Explanation

TCP is connection-oriented, acknowledges data, retransmits loss, and implements congestion control; TLS/HTTPS runs on TCP. UDP is datagram and unreliable. ICMP is a network-layer diagnostic/control protocol. ARP maps IPv4 to MAC on a LAN.

Question 15Network Layer Protocols and Routing

Which block is one of the RFC 1918 private IPv4 ranges used on enterprise campuses?

  1. 172.16.0.0/12 Correct answer.
  2. 11.0.0.0/8
  3. 192.0.2.0/24
  4. 224.0.0.0/4

Explanation

RFC 1918 private space is 10.0.0.0/8, 172.16.0.0/12, and 192.168.0.0/16. 11.0.0.0/8 is public. 192.0.2.0/24 is TEST-NET-1 documentation space. 224.0.0.0/4 is multicast.

Question 16Network Layer Protocols and Routing

What is the correct order of segments in a TCP three-way handshake?

  1. SYN, then ACK, then FIN
  2. SYN, then SYN-ACK, then ACK Correct answer.
  3. FIN, then FIN-ACK, then ACK
  4. RST, then SYN, then ACK

Explanation

The client sends SYN, the server replies SYN-ACK, and the client completes with ACK. FIN sequence is connection teardown, not setup. RST aborts. A lone ACK after SYN without SYN-ACK is not the three-way open.

Question 17Network Layer Protocols and Routing

Which TCP/UDP well-known port pair is correct for HTTPS versus unencrypted HTTP?

  1. HTTPS 80, HTTP 443
  2. HTTPS 443, HTTP 80 Correct answer.
  3. HTTPS 22, HTTP 23
  4. HTTPS 53, HTTP 25

Explanation

IANA well-known ports: HTTP is TCP 80 and HTTPS (HTTP over TLS) is TCP 443. 22 is SSH, 23 is Telnet, 53 is DNS, and 25 is SMTP.

Question 18Network Layer Protocols and Routing

What is the standard maximum transmission unit (MTU) of Ethernet v2 payload (IP packet) without jumbo frames or 802.1Q?

  1. 576 bytes
  2. 1280 bytes
  3. 1500 bytes Correct answer.
  4. 9000 bytes

Explanation

Classic Ethernet MTU is 1500 bytes of payload. 576 is the IPv4 minimum reassembly size hosts must handle. 1280 is the IPv6 minimum link MTU. 9000 is a common jumbo-frame size, not the IEEE default.

Question 19Network Layer Protocols and Routing

IEEE 802.11 WLANs use CSMA/CA rather than Ethernet CSMA/CD. Why is collision detection not used on Wi-Fi?

  1. Wi-Fi stations cannot reliably hear all other transmitters (hidden terminal) while sending Correct answer.
  2. Wi-Fi has no MAC layer
  3. CSMA/CD requires IPv6
  4. Access points forbid any carrier sensing

Explanation

A transmitting Wi-Fi radio cannot listen for collisions the way a shared Ethernet NIC can, and hidden terminals may not hear each other. 802.11 therefore uses CSMA/CA (and optional RTS/CTS) to avoid collisions. Wi-Fi has a MAC; CSMA/CD is not tied to IPv6; APs still use carrier sensing.

Question 20Network Layer Protocols and Routing

How many usable IPv4 host addresses are in a /26 LAN (usable = 2^h − 2)?

  1. 30
  2. 62 Correct answer.
  3. 64
  4. 126

Explanation

A /26 has 6 host bits: 2^6 = 64 addresses. Subtract network and directed broadcast: 64 − 2 = 62 usable hosts. 30 is /27, 64 forgets the two reserved addresses, and 126 is /25.

Question 21Network Layer Protocols and Routing

Host 192.168.10.45/27 sits on which IPv4 network, and what is the directed broadcast?

  1. Network 192.168.10.0, broadcast 192.168.10.255
  2. Network 192.168.10.32, broadcast 192.168.10.63 Correct answer.
  3. Network 192.168.10.45, broadcast 192.168.10.46
  4. Network 192.168.10.64, broadcast 192.168.10.95

Explanation

A /27 has a 32-address block. 45 div 32 = 1 remainder 13, so the block starts at 192.168.10.32 and ends at .63. Usable hosts are .33–.62; .45 is a valid host. /24 would be .0/.255. The host address is not the network. The next /27 is .64/.95.

Question 22Network Layer Protocols and Routing

From 172.16.0.0/16 you must create subnets that each support 500 hosts. What prefix do you use, and how many such subnets fit in the /16?

  1. /24 with 256 subnets
  2. /23 with 128 subnets Correct answer.
  3. /22 with 64 subnets
  4. /25 with 512 subnets

Explanation

500 hosts need 9 host bits because 2^8 − 2 = 254 is too small and 2^9 − 2 = 510 ≥ 500, so /23. Borrowing 7 bits from /16 to /23 yields 2^7 = 128 subnets. /24 has only 254 usable hosts. /22 has 1022 hosts (oversized). /25 has only 126 usable hosts.

Question 23Network Layer Protocols and Routing

IEEE 802.1Q inserts a 4-byte VLAN tag. How many bits identify the VLAN ID, and what is the usual usable VID range?

  1. 8 bits, VIDs 1–255
  2. 12 bits, VIDs 1–4094 (0 and 4095 reserved) Correct answer.
  3. 16 bits, VIDs 1–65534
  4. 24 bits, VIDs 1–16 million as in VXLAN

Explanation

The 802.1Q TCI contains a 12-bit VID (0–4095). VID 0 means priority-tagged with no VLAN, and 4095 is reserved, so user VLANs are 1–4094. VXLAN uses a 24-bit VNI, which is a different encapsulation.

Question 24Network Layer Protocols and Routing

In multi-area OSPFv2, which statement is required for a correct backbone design?

  1. Every area must attach to area 0 (backbone), directly or via a virtual link Correct answer.
  2. Area 0 is forbidden; all areas must be totally stubby
  3. OSPF cannot run on Ethernet broadcast links
  4. Each ABR may belong to at most one area

Explanation

OSPF requires a contiguous backbone (area 0). Non-backbone areas connect to area 0 on ABRs; a virtual link is the textbook repair when area 0 is discontiguous. OSPF does run on broadcast Ethernet (DR/BDR). An ABR sits on area 0 plus at least one other area.

Question 25Network Layer Protocols and Routing

BGP is a path-vector protocol. Which well-known mandatory attribute lists the ASes a route has traversed and is used to prevent AS-level loops?

  1. MED (MULTI_EXIT_DISC)
  2. LOCAL_PREF
  3. AS_PATH Correct answer.
  4. Origin code IGP/EGP/Incomplete only, with no AS list

Explanation

AS_PATH is the sequence of AS numbers; a router rejects an update that already contains its own AS, which prevents loops. MED influences inbound choice among multiple exits. LOCAL_PREF is used inside an AS for outbound preference. ORIGIN is a separate attribute and does not list ASes.

Question 26Network Layer Protocols and Routing

Which IPv6 prefix is the link-local unicast range that every interface configures (typically fe80::/64 on the link)?

  1. 2000::/3 global unicast
  2. ff00::/8 multicast
  3. fe80::/10 link-local unicast Correct answer.
  4. fc00::/7 unique-local as the only address an interface needs

Explanation

IPv6 link-local unicast is fe80::/10 (commonly used as fe80::/64 per link). 2000::/3 is global unicast. ff00::/8 is multicast. Unique-local fc00::/7 (typically fd00::/8 locally assigned) is optional private space, not the mandatory link-local.

Question 27Network Layer Protocols and Routing

IPv4 DHCP address assignment uses DORA. What is the correct four-step order?

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

Explanation

The client broadcasts DHCPDISCOVER, a server unicasts or broadcasts DHCPOFFER, the client DHCPREQUESTs a lease, and the server DHCPACKs. Other orders reverse client and server roles.

Question 28Network Layer Protocols and Routing

Many campus hosts share one public IPv4 address using different transport port numbers. What NAT mode is that?

  1. Static one-to-one NAT only
  2. PAT / NAT overload (NAPT) Correct answer.
  3. NAT64 of IPv6-only hosts without a stateful table
  4. Destination NAT for a single published server with no port mapping

Explanation

Port Address Translation (NAT overload/NAPT) maps many private tuples (IP, port) onto one (or a few) public IP addresses distinguished by ports. Static NAT is 1:1. NAT64 is IPv6/IPv4 translation. A single published server uses DNAT/port forwarding, not many-to-one source overload.

Question 29Network Layer Protocols and Routing

In IEEE 802.1D STP, which switch becomes the root bridge?

  1. The switch with the highest numeric Bridge ID
  2. The switch with the lowest Bridge ID (priority + MAC) Correct answer.
  3. The switch with the most 10 Gbps ports regardless of Bridge ID
  4. The first switch to boot, permanently

Explanation

STP elects the lowest Bridge ID, which is 16-bit priority concatenated with the 48-bit MAC. Operators lower priority on the intended core. Port speed affects root-path cost, not the root election key. Boot order is not the tie-breaker defined by 802.1D.

Question 30Network Layer Protocols and Routing

On Ethernet with a 1500-byte IPv4 MTU, what TCP MSS should endpoints typically advertise (20-byte IP + 20-byte TCP, no options)?

  1. 1500 bytes
  2. 1480 bytes
  3. 1460 bytes Correct answer.
  4. 1400 bytes as a mandatory IEEE value

Explanation

MSS is the maximum TCP payload: MTU − IP header − TCP header = 1500 − 20 − 20 = 1460 bytes when headers have no options. 1500 would double-count headers. 1480 subtracts only IP. 1400 is a common conservative clamp for tunnels, not the IEEE Ethernet default MSS.

Question 31Network Layer Protocols and Routing

From 192.168.10.0/24 you VLSM for LANs of 100, 50, 25, and 10 hosts, assigning largest-first. Which prefix is correct for the 50-host LAN?

  1. /25 (126 usable), overlapping the 100-host /25
  2. /26 (62 usable) as the second block after a /25 Correct answer.
  3. /29 (6 usable), too small for 50 hosts
  4. /24 for every LAN, ignoring VLSM

Explanation

100 hosts need /25 (126 usable) on 192.168.10.0/25. Next, 50 hosts need /26 (62 usable), placed on 192.168.10.128/26. A second /25 would overlap. /29 has only 6 usable hosts. Keeping a single /24 wastes addressing and is not VLSM.

Question 32Network Layer Protocols and Routing

An OSPF network statement should match 172.16.8.0/22. Which wildcard mask is correct?

  1. 0.0.0.255
  2. 0.0.3.255 Correct answer.
  3. 0.0.7.255
  4. 255.255.252.0

Explanation

A /22 mask is 255.255.252.0, so the inverse wildcard is 0.0.3.255. 172.16.8.0 is aligned on /22 (blocks of 4 in the third octet: 0, 4, 8, 12, …). 0.0.0.255 is /24. 0.0.7.255 is /21. 255.255.252.0 is a subnet mask, not an OSPF wildcard.

Question 33Network Layer Protocols and Routing

What is the line rate of a standard E1 circuit used on Chinese WAN access, and how is it structured?

  1. 1.544 Mbps as 24 × 64 kbps (T1)
  2. 2.048 Mbps as 32 × 64 kbps timeslots Correct answer.
  3. 10 Mbps as a single Ethernet collision domain
  4. 155.52 Mbps as STM-1 only

Explanation

E1 (CEPT G.704) is 2.048 Mbps: 32 timeslots of 64 kbps. T1 is 1.544 Mbps (24 DS0s plus framing). Fast Ethernet and STM-1 are different hierarchies. Chinese 软考 WAN items use E1, not T1, as the primary PDH example.

Question 34Network Layer Protocols and Routing

A 1 Gbps path has 40 ms RTT. What is the bandwidth-delay product in bytes (BDP = bandwidth × RTT)?

  1. 5 MB (40 Mbit) Correct answer.
  2. 40 MB
  3. 125 kB
  4. 1 GB

Explanation

1 × 10^9 bit/s × 0.040 s = 4 × 10^7 bits = 40 Mbit. In bytes: 40 000 000 / 8 = 5 000 000 bytes = 5 MB. That is the in-flight data a TCP window must cover to fill the pipe. 40 MB forgets bits-to-bytes. 125 kB is 1 Mb in 1 ms, not this product. 1 GB is the link rate with seconds ignored.

Question 35Network Interconnection and Device Configuration

Which device floods every incoming Ethernet frame out all other ports and does not learn MAC addresses?

  1. Layer 2 switch
  2. Layer 3 router
  3. Repeater hub Correct answer.
  4. Stateful firewall

Explanation

A hub is a Layer 1 repeater: one collision domain, no MAC learning. A switch learns MACs and forwards selectively. A router forwards IP packets between networks. A stateful firewall inspects sessions, not hub-style flooding.

Question 36Network Interconnection and Device Configuration

Where is a perimeter stateful firewall typically placed in a campus with a DMZ?

  1. Only on every PC as a host firewall, with no border device
  2. At the Internet/WAN edge, with a DMZ interface for published servers Correct answer.
  3. In place of the core switch fabric for all east-west traffic only
  4. As a replacement for interior OSPF routers

Explanation

Campus designs put a stateful firewall (or NGFW) at the Internet edge, often with a DMZ leg for externally reachable servers, while interior switches/routers handle campus forwarding. Host firewalls complement but do not replace the border. The core should not be replaced by a firewall chokepoint for all east-west flows unless that is an explicit zero-trust fabric.

Question 37Network Interconnection and Device Configuration

A DNS A record maps which pair?

  1. A hostname to an IPv4 address Correct answer.
  2. A hostname to an IPv6 address
  3. A domain to a mail exchanger host
  4. An IPv4 address to a hostname (reverse)

Explanation

A records map names to IPv4 addresses. AAAA maps to IPv6. MX maps a domain to mail servers. PTR is reverse mapping from address to name.

Question 38Network Interconnection and Device Configuration

An enterprise WLAN of many thin APs managed by a controller is the Fit AP + AC model. What does the AC typically centralize?

  1. RF/SSID policy, roaming, and AP management, while APs still radio-serve clients Correct answer.
  2. Only PoE power, with no WLAN control plane
  3. Replacement of all campus core switches
  4. Termination of every OSPF adjacency from access PCs

Explanation

In AC+Fit AP (thin AP) designs, the wireless controller holds configuration, roaming, and often encryption/forwarding policy while APs provide RF. PoE is a switch function. The AC does not replace the wired core or speak OSPF to user PCs.

Question 39Network Interconnection and Device Configuration

A reverse proxy that distributes HTTPS sessions across several web servers by URL or cookie is acting as which product?

  1. A Layer 4 load balancer using only TCP 5-tuples, never HTTP
  2. A Layer 7 (application) load balancer / ADC Correct answer.
  3. An unmanaged Layer 2 hub
  4. A DWDM transponder

Explanation

URL and cookie switching are application-layer (L7) load-balancer/ADC functions. L4 balancers hash IP/port only. Hubs and DWDM optics do not parse HTTPS applications.

Question 40Network Interconnection and Device Configuration

What distinguishes a Layer 3 campus switch from a pure Layer 2 switch in 软考 product selection?

  1. It can route between VLANs (SVI/inter-VLAN routing) in hardware Correct answer.
  2. It cannot learn MAC addresses
  3. It must run BGP and cannot run OSPF
  4. It replaces the need for any default gateway on hosts

Explanation

A Layer 3 switch still switches MACs inside a VLAN but also routes IP between SVIs or routed ports, typically in ASIC. It still learns MACs. It may run OSPF or static routing; BGP is optional. Hosts still need a default gateway (the SVI VIP).

Question 41Network Interconnection and Device Configuration

IEEE 802.1AX/802.3ad LACP is used between two aggregation switches. What does a working LAG provide?

  1. One logical link from multiple physical links, with member failure surviving Correct answer.
  2. A spanning-tree loop that must keep all member ports blocking forever
  3. Layer 7 HTTP cookie persistence
  4. Replacement of fiber with analog microwave only

Explanation

LACP bundles ports into one EtherChannel/LAG so STP/IP see a single link and traffic hashes across members; if one member fails, remaining members carry traffic. LAG is not an HTTP ADC feature and is not a microwave PHY.

Question 42Network Interconnection and Device Configuration

IEEE 802.3af PoE at the PSE is specified to deliver up to how much power per port for a Type 1 / Class 3 PD?

  1. 4.0 W
  2. 15.4 W Correct answer.
  3. 60 W
  4. 90 W

Explanation

802.3af Type 1 PSE budget is 15.4 W per port (about 12.95 W at the PD after cable loss). 60 W and 90 W are 802.3bt Type 3/4 levels. 4 W is below Class 3.

Question 43Network Interconnection and Device Configuration

Which storage access method presents block devices over TCP (port 3260) rather than a file-share protocol?

  1. NFS file sharing
  2. SMB/CIFS file sharing
  3. iSCSI Correct answer.
  4. FTP of individual files only

Explanation

iSCSI carries SCSI block commands over TCP (commonly port 3260), so servers see disks. NFS and SMB are file-system (NAS) protocols. FTP copies files, it is not a block SAN.

Question 44Network Interconnection and Device Configuration

SNMP polling of a router uses which default UDP port on the agent, and which version adds USM user-based security instead of a community string?

  1. Port 80 with SNMPv1 communities only
  2. Port 161 with SNMPv3 USM (v2c still uses communities) Correct answer.
  3. Port 23 Telnet with SNMPv3
  4. Port 443 HTTPS with SNMPv1

Explanation

SNMP agents listen on UDP 161; traps often use 162. SNMPv3 introduces User-based Security Model (authentication/privacy). v1/v2c still use community strings. HTTP/Telnet/HTTPS ports are not the SNMP agent port.

Question 45Network Interconnection and Device Configuration

A product that inspects HTTP/HTTPS for SQL injection and XSS, sitting in front of web servers, is primarily a:

  1. WAF (web application firewall) Correct answer.
  2. Mail transfer agent
  3. OTDR optical tester
  4. Unicast PIM-SM RP

Explanation

A WAF understands HTTP(S) application attacks (injection, XSS, OWASP classes) beyond IP/port filtering. An MTA handles mail. An OTDR tests fiber. PIM-SM is multicast routing.

Question 46Network Interconnection and Device Configuration

A CDN caches static objects near users. What is the main network-planning benefit?

  1. It eliminates the need for any origin DNS
  2. It reduces origin load and WAN/Internet traffic for cacheable content Correct answer.
  3. It replaces OSPF inside the campus core
  4. It converts all TCP sessions into ATM PVCs

Explanation

CDNs keep copies of cacheable objects at edge nodes, so many hits never reach the origin, cutting origin CPU and backbone/Internet traffic. DNS is still required. CDNs do not replace campus IGP or ATM.

Question 47Network Interconnection and Device Configuration

Twenty 802.3af Class 3 APs each draw 15.4 W from the PSE. What minimum PoE budget must the access switch reserve for those APs alone?

  1. 154 W
  2. 308 W Correct answer.
  3. 370 W
  4. 15.4 W total for all twenty

Explanation

20 × 15.4 W = 308 W at the PSE if every AP is Class 3 802.3af. 154 W is 10 APs. 370 W is a typical switch budget number, not this calculation. 15.4 W would power only one AP.

Question 48Network Interconnection and Device Configuration

A data-center fabric uses spine-leaf (every leaf dual-homed to all spines, ECMP). How does that differ from classic campus three-tier (core–汇聚–接入)?

  1. Spine-leaf is optimized for east-west any-to-any with predictable hop count; three-tier is optimized for north-south campus aggregation Correct answer.
  2. Spine-leaf forbids ECMP while three-tier requires it
  3. Three-tier cannot use Ethernet; spine-leaf cannot use IP
  4. They are identical topologies with different names only

Explanation

Spine-leaf gives a leaf-to-leaf path of two hops via any spine and hashes on ECMP, which suits server-to-server (east-west) traffic. Campus three-tier aggregates users north-south through 汇聚 and core. Both can run Ethernet/IP; ECMP is typical on spines, not forbidden.

Question 49Network Interconnection and Device Configuration

In SDN with OpenFlow, which plane makes forwarding decisions centrally, and which plane is the switch ASIC that matches flow tables?

  1. Control plane on the controller; data plane on the switches Correct answer.
  2. Data plane on the controller; control plane only on each PC NIC
  3. Management plane replaces both and OpenFlow is unused
  4. Both planes must run as BGP route reflectors

Explanation

SDN separates the control plane (controller computes and installs flows, often via OpenFlow southbound) from the data plane (switch/router hardware forwards). NICs are not the SDN control plane. Management (NETCONF/CLI) is a third plane. BGP RR is a specific routing design, not the definition of OpenFlow SDN.

Question 50Network boundaries and protected communication

Encrypting a backup file so unauthorized readers cannot recover plaintext primarily protects which CIA property?

  1. Integrity
  2. Availability
  3. Confidentiality Correct answer.
  4. Physical tamper-evidence of the disk tray

Explanation

Confidentiality (机密性) is hiding content from unauthorized subjects; encryption is the classic control. Integrity uses hashes/MACs/signatures. Availability uses redundancy and DDoS defense. A sealed tray is physical security, not encryption’s primary CIA mapping.