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NSSCAS Computer Science (8231) study materials
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NSSCAS Computer Science (8231) practice questions

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Question 1Data representation

A register stores the 8-bit binary value 10110110. What is its denary equivalent?

  1. 182 Correct answer.
  2. 204
  3. 166
  4. 218

Explanation

The place values for an 8-bit binary number are 128, 64, 32, 16, 8, 4, 2 and 1. Setting the bits at positions 128, 32, 16, 4 and 2 gives 128 + 32 + 16 + 4 + 2 = 182.

Question 2Data representation

A video file has a size of 3 MiB. How many KiB is this, using binary prefixes?

  1. 24576
  2. 3145728 Correct answer.
  3. 3072
  4. 3000

Explanation

A mebibyte (MiB) is 1024 kibibytes (KiB), so the conversion is 3 × 1024 KiB = 3072 KiB. This uses binary prefixes rather than the decimal prefix approximation of 1000.

Question 3Data representation

What is the 8-bit two's complement representation of the denary value -37?

  1. 11011011 Correct answer.
  2. 10100101
  3. 11011010
  4. 00100101

Explanation

The positive value +37 in 8 bits is 00100101. Inverting the bits gives 11011010, and adding 1 produces 11011011, which is the two's complement form of -37.

Question 4Data representation

A text file is encoded in UTF-8. The character 'é' has Unicode code point U+00E9. How many bytes does this character occupy in the UTF-8 encoding?

  1. 4
  2. 1 Correct answer.
  3. 3
  4. 2

Explanation

UTF-8 uses one byte for code points U+0000 to U+007F and two bytes for U+0080 to U+07FF. Since U+00E9 lies in the two-byte range, it is stored as the byte sequence C3 A9.

Question 5Data representation

A 30-second stereo audio recording is sampled at 44.1 kHz with 16-bit resolution per sample. Using 1 MiB = 1024 × 1024 bytes, what is the approximate file size of the uncompressed PCM data?

  1. 40.4 MiB
  2. 10.1 MiB
  3. 2.5 MiB
  4. 5.0 MiB Correct answer.

Explanation

The number of samples is 44,100 × 30 = 1,323,000. At 16 bits (2 bytes) per sample and two channels, the data size is 1,323,000 × 2 × 2 = 5,292,000 bytes. Dividing by 1,048,576 bytes per MiB gives approximately 5.0 MiB.

Question 6Data representation

The denary value 190 is to be stored as a single hexadecimal byte. Which hexadecimal value represents 190?

  1. CB
  2. BE Correct answer.
  3. BC
  4. BD

Explanation

Divide 190 by 16: the quotient is 11 with remainder 14. In hexadecimal, 11 is B and 14 is E, so the byte is BE.

Question 7Data representation

In packed binary-coded decimal (BCD), how is the denary value 58 represented in an 8-bit code?

  1. 0101 1000 Correct answer.
  2. 0101 0011
  3. 0011 1010
  4. 1000 0101

Explanation

Packed BCD stores each denary digit in its own 4-bit nibble. The digit 5 is 0101 and the digit 8 is 1000, giving 0101 1000.

Question 8Data representation

What is the range of denary integers that can be represented in an 8-bit sign-and-magnitude representation?

  1. −128 to +127
  2. −127 to +127 Correct answer.
  3. −255 to +255
  4. 0 to +255

Explanation

Sign-and-magnitude uses one bit for the sign and seven bits for the magnitude. The largest magnitude is 1111111 in binary, which is 127, so the range is −127 to +127; zero has two representations.

Question 9Data representation

An uncompressed bitmap image is 800 pixels wide and 600 pixels high. Each pixel uses a 24-bit colour depth. What is the approximate file size in MiB? (Use 1 MiB = 1024 × 1024 bytes.)

  1. 0.92 MiB
  2. 2.75 MiB
  3. 1.37 MiB Correct answer.
  4. 0.46 MiB

Explanation

The image contains 800 × 600 = 480,000 pixels. At 24 bits per pixel it needs 11,520,000 bits, which is 1,440,000 bytes, or about 1.37 MiB.

Question 10Data representation

Two unsigned 8-bit binary values, 0110 1100 and 1011 0101, are added. What is the 8-bit result and the carry-out from the most significant bit?

  1. 0010 0001, carry 1 Correct answer.
  2. 0010 0001, carry 0
  3. 0001 0001, carry 1
  4. 1101 0001, carry 1

Explanation

0110 1100 is 108 and 1011 0101 is 181; their sum is 289. In binary, 289 is 1 0010 0001, so the lower eight bits are 0010 0001 and the carry-out is 1.

Question 11Data representation

Which statement correctly distinguishes lossy compression from lossless compression in data representation?

  1. Lossy compression permanently removes some data, while lossless compression allows exact reconstruction. Correct answer.
  2. Neither method removes data, but lossy compression stores the file using fewer bits.
  3. Lossy compression allows exact reconstruction, while lossless compression permanently removes some data.
  4. Both methods permanently remove data, but lossy compression removes a larger proportion.

Explanation

Lossless compression preserves all original data so the file can be restored exactly, whereas lossy compression discards some information that cannot be recovered. The key distinction is exact reconstruction, not simply the amount of storage saved.

Question 12Data representation

Standard 7-bit ASCII defines a character set. How many distinct character codes can this set represent?

  1. 512
  2. 64
  3. 128 Correct answer.
  4. 256

Explanation

A 7-bit code has 2^7 possible bit patterns, which equals 128. Standard ASCII uses these 128 codes for control characters and printable characters.

Question 13Data representation

An unsigned 8-bit register holds 0011 0100. It is logically shifted two places to the left, with no bits lost. What is the new 8-bit value and the effect on the denary value?

  1. 1101 0000; the value is multiplied by 4. Correct answer.
  2. 1001 0000; the value is multiplied by 4.
  3. 1101 0000; the value is multiplied by 2.
  4. 0000 1101; the value is divided by 4.

Explanation

Each logical left shift moves bits towards the most significant end and multiplies an unsigned value by 2. Shifting two places therefore multiplies by 4, and 0011 0100 (52) becomes 1101 0000 (208).

Question 14Data representation

A system adds an odd parity bit to the left of a 7-bit ASCII character. The data bits are 1001011. Which parity bit should be added, and what 8-bit value is transmitted?

  1. 1; 01001011
  2. 0; 01001011
  3. 1; 11001011 Correct answer.
  4. 0; 11001011

Explanation

Odd parity requires the total number of 1 bits in the transmitted byte to be odd. The data 1001011 contains four 1s, so a parity bit of 1 is needed, giving 11001011 when placed on the left.

Question 15Data representation

In packed BCD, the 8-bit values 0011 1000 (38) and 0010 0101 (25) are added using binary addition. What intermediate result is produced, and what correction is needed to obtain valid packed BCD?

  1. 0101 1101; add 0110 to the upper nibble, giving 1011 1101.
  2. 0101 1101; add 0110 to the lower nibble, giving 0110 0011. Correct answer.
  3. 0110 0011; no correction is needed because the result is already valid BCD.
  4. 0101 1101; subtract 0110 from the lower nibble, giving 0101 0111.

Explanation

Binary addition gives 0101 1101, but the lower nibble 1101 represents 13, which is not a valid BCD digit. Adding 0110 to that nibble corrects it to 0011 and produces the BCD result 0110 0011, representing 63.

Question 16Data representation

In data representation, a hexadecimal digit is often used as a compact shorthand for a group of binary bits. How many binary bits does one hexadecimal digit represent?

  1. Four bits Correct answer.
  2. Sixteen bits
  3. Two bits
  4. Eight bits

Explanation

Each hexadecimal digit has one of 16 values (0–F), and 16 = 2^4, so it maps to a unique 4-bit binary pattern. Thus one hexadecimal digit corresponds to one nibble.

Question 17Data representation

A database must store characters from many different writing systems, including characters that are not defined in ASCII. Which property of Unicode makes it suitable for this task?

  1. It uses exactly seven bits for every character code.
  2. It stores every character as a fixed-length bitmap.
  3. It assigns code points to many scripts. Correct answer.
  4. It guarantees that every character is compressed losslessly.

Explanation

Unicode defines a large set of code points, allowing it to represent characters from many scripts rather than only the 128 codes of standard ASCII. It is a character-encoding standard, not a bitmap format or a compression method.

Question 18Data representation

A sound is digitised at the same sampling rate twice: once with 8-bit resolution and once with 16-bit resolution. What is the main effect of using 16-bit resolution instead of 8-bit resolution?

  1. The number of samples taken per second is doubled.
  2. The sound is converted from analogue to digital without sampling.
  3. Each sample can use more amplitude levels, improving accuracy. Correct answer.
  4. The file size is halved because fewer samples are needed.

Explanation

Resolution, or bit depth, sets how many discrete amplitude levels each sample can store. Moving from 8-bit to 16-bit resolution increases the levels from 256 to 65,536, so each sample can be more accurate, while the sampling rate stays the same.

Question 19Data representation

An uncompressed bitmap image is changed from 24-bit colour depth to 8-bit colour depth, with the same width and height. What is the most direct effect on the stored image data?

  1. The image becomes lossless because fewer bits are used per pixel.
  2. The image resolution increases because more pixels are stored.
  3. The image is converted to a vector graphic with scalable shapes.
  4. The image has fewer distinct colours and a smaller file size. Correct answer.

Explanation

Colour depth is the number of bits used to store each pixel's colour; 8-bit depth allows 256 colours per pixel, while 24-bit depth allows millions. With the same width and height, reducing bits per pixel reduces both the number of available colours and the stored data size.

Question 20Data representation

Which pair of 8-bit two's complement values produces arithmetic overflow when added in an 8-bit register?

  1. 1111 1111 and 0000 0001
  2. 1000 0000 and 0111 1111
  3. 0100 0000 and 0011 1111
  4. 0111 1111 and 0000 0001 Correct answer.

Explanation

In 8-bit two's complement, the range is −128 to +127. Adding 0111 1111 (+127) and 0000 0001 (+1) gives +128, which cannot be represented; the 8-bit result is 1000 0000, whose sign bit indicates a negative value, so overflow has occurred.

Question 21Data representation

An 8-bit two's complement representation is used. Which denary range can be represented?

  1. -128 to +127 Correct answer.
  2. -127 to +127
  3. 0 to 255
  4. -255 to +255

Explanation

In 8-bit two's complement, the most significant bit has place value -128 and the remaining seven bits have place values +64 down to +1. The smallest value is therefore -128 and the largest is 127.

Question 22Data representation

An image uses a palette of exactly 64 colours. What is the minimum colour depth per pixel needed to store an index for every colour?

  1. 6 bits Correct answer.
  2. 8 bits
  3. 64 bits
  4. 4 bits

Explanation

Colour depth is the number of bits used to store each pixel. With 6 bits per pixel there are 2^6 = 64 possible bit patterns, so a palette of 64 colours requires at least 6 bits.

Question 23Data representation

A mono audio recording is sampled at 8 kHz with 8-bit resolution. Using 1 kbit/s = 1000 bits/s, what is the bit rate?

  1. 16 kbit/s
  2. 32 kbit/s
  3. 64 kbit/s Correct answer.
  4. 8 kbit/s

Explanation

The bit rate equals the number of samples per second multiplied by the number of bits per sample. Here 8,000 samples/s × 8 bits = 64,000 bits/s, which is 64 kbit/s using the stated conversion.

Question 24Data representation

An 8-bit two's complement register contains 1101 1000. It is arithmetic right shifted two places. What is the resulting 8-bit pattern?

  1. 1110 1100
  2. 0011 0110
  3. 1111 0110 Correct answer.
  4. 1111 1100

Explanation

An arithmetic right shift copies the sign bit into the vacated positions on the left and shifts the other bits right. For 1101 1000, shifting two places gives 1111 0110.

Question 25Communication and network technologies

In a star topology, what is the immediate effect when the central switch fails?

  1. Attached devices lose connectivity with one another. Correct answer.
  2. Only the device on the failed port loses connectivity.
  3. Each device continues sending directly to every other device.
  4. The network falls back to a bus topology automatically.

Explanation

In a star topology, every attached device reaches the rest of the network through the central switch. If that switch fails, the devices have no shared path and cannot exchange data unless a replacement or redundant central device takes over.

Question 26Communication and network technologies

A file-transfer application requires every byte to arrive in order and needs lost segments retransmitted. Which transport-layer protocol should it use?

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

Explanation

TCP provides connection-oriented, reliable delivery using sequence numbers, acknowledgements, and retransmission of lost segments. UDP is connectionless and does not guarantee delivery or ordering, while ICMP and ARP serve other network functions.

Question 27Communication and network technologies

A subnet uses the IPv4 prefix 192.168.10.0/27. How many usable host addresses does each subnet provide?

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

Explanation

A /27 leaves five bits for host addresses, so the block contains 2^5 = 32 addresses. The all-zero network address and all-ones broadcast address cannot be assigned to hosts, leaving 30 usable addresses.

Question 28Communication and network technologies

A host on 192.168.1.0/24 sends an IPv4 packet to a server on 10.0.0.0/8. The host has a default gateway. Which MAC address is used as the destination in the Ethernet frame sent from the host?

  1. The MAC address of the default gateway Correct answer.
  2. The MAC address of the remote server
  3. The broadcast MAC address FF:FF:FF:FF:FF:FF
  4. The MAC address of the sending host's network interface

Explanation

For an IPv4 destination outside the host's local subnet, the host sends the frame to its default gateway. It uses ARP to learn the gateway's MAC address, so the destination MAC is the gateway's, not the remote server's.

Question 29Communication and network technologies

A router receives a packet destined for 172.16.5.130. Its routing table contains 172.16.0.0/16 via router A, 172.16.5.128/25 via router B, and 0.0.0.0/0 via router C. Which route is chosen and why?

  1. The route via A, because it was added before the more specific route.
  2. The route via C, because the default route is used whenever the destination is not an exact network address.
  3. The route via B, because 172.16.5.128/25 is the longest prefix that matches the destination. Correct answer.
  4. The route via A, because its /16 prefix covers a larger address range.

Explanation

A router chooses the longest prefix that matches the destination address. The /25 route 172.16.5.128/25 covers 172.16.5.128 to 172.16.5.255, which includes 172.16.5.130, and it is more specific than the /16 and default routes.

Question 30Communication and network technologies

A router receives a frame, removes the link-layer header, and chooses an outbound interface by looking up the packet's destination IPv4 address. At which OSI layer is this forwarding decision primarily made?

  1. Physical layer
  2. Network layer Correct answer.
  3. Transport layer
  4. Data-link layer

Explanation

IPv4 addressing and route selection are network-layer functions, so a router uses the destination IP address to choose the next hop. The data-link layer handles local frame delivery, while the transport and physical layers do not make IP forwarding decisions.

Question 31Communication and network technologies

A client knows the domain name of a web server but not its IPv4 address. Which protocol is used to resolve the domain name into an IP address before the client can connect?

  1. ARP
  2. DNS Correct answer.
  3. DHCP
  4. ICMP

Explanation

DNS is the name-resolution system that maps domain names such as www.example.com to IP addresses. DHCP assigns host configuration, ARP resolves IPv4 addresses to MAC addresses on a local link, and ICMP carries control and error messages.

Question 32Communication and network technologies

Two 802.11 stations are associated with the same access point but are too far apart to hear each other's transmissions. When both transmit at the same time, their signals can collide at the access point. Which mechanism is designed to reduce this hidden-node problem?

  1. CSMA/CD collision detection
  2. VLAN tagging on the wireless frame
  3. RTS/CTS exchange before data transmission Correct answer.
  4. NAT translation of the source address

Explanation

The hidden-node problem occurs when stations cannot sense each other, so carrier sensing alone is insufficient. An RTS/CTS exchange reserves the medium and informs nearby stations, including hidden ones, to defer transmission.

Question 33Communication and network technologies

A small office has one public IPv4 address and several internal hosts using private addresses in 192.168.1.0/24. Which technique lets all internal hosts make simultaneous outbound TCP connections through the single public address?

  1. Proxy ARP with duplicate public addresses
  2. Port address translation using source port numbers Correct answer.
  3. DHCP address pooling on the public link
  4. Static one-to-one NAT for every internal host

Explanation

Port address translation, also called NAT overload or PAT, maps multiple private IP addresses to one public address by tracking unique source port numbers. The port mapping lets returning packets be forwarded to the correct internal host.

Question 34Communication and network technologies

Two BGP routes to the same external prefix have equal weight, local preference, origin type, AS_PATH length, and MED. One route was learned from an eBGP peer and the other from an iBGP peer. According to the standard BGP best-path selection process, which route is preferred?

  1. The route with the higher BGP router ID, because router ID is checked first
  2. The eBGP route, because eBGP is preferred after earlier attributes tie Correct answer.
  3. The route with the lower MED, because MED is checked before peer type
  4. The iBGP route, because iBGP has a lower administrative distance

Explanation

After weight, local preference, locally originated status, AS_PATH length, origin type, and MED are equal, BGP prefers the route learned from an eBGP peer over one learned from an iBGP peer. This preference helps select routes learned from outside the local autonomous system.

Question 35Communication and network technologies

A switch that forwards frames using MAC addresses operates primarily at which OSI model layer?

  1. Physical layer
  2. Data link layer Correct answer.
  3. Network layer
  4. Transport layer

Explanation

A switch forwarding frames by MAC address is a Layer 2 device, so it operates at the data link layer. The network layer handles logical addressing and routing, the transport layer handles end-to-end delivery, and the physical layer transmits bits.

Question 36Communication and network technologies

On a half-duplex Ethernet segment using CSMA/CD, two stations transmit at the same time and a collision occurs. What should each station do next?

  1. Change the frame to full-duplex mode and resend it.
  2. Retransmit immediately after a fixed one-slot delay.
  3. Continue sending the frame and rely on TCP to detect the collision.
  4. Stop, send a jam signal, then wait a random backoff. Correct answer.

Explanation

CSMA/CD stations detect a collision, stop transmitting, send a jam signal so all stations notice the collision, and then use binary exponential backoff before retrying. TCP retransmission is irrelevant because the collision occurs at the link layer.

Question 37Communication and network technologies

An Ethernet trunk between two switches carries frames for VLANs 10 and 20 using IEEE 802.1Q. How does the receiving switch identify the VLAN of an arriving frame?

  1. It reads the VLAN identifier from the IP header.
  2. It reads the 12-bit VLAN ID in the 802.1Q tag. Correct answer.
  3. It uses the ARP cache entry for the source host.
  4. It maps the destination MAC address to a VLAN.

Explanation

An IEEE 802.1Q trunk inserts a tag after the source MAC address. The tag contains a 12-bit VLAN identifier, so the receiving switch uses that field to associate the frame with a VLAN.

Question 38Communication and network technologies

A client opens a TCP connection to a server. Which sequence of control flags is used to establish the connection?

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

Explanation

TCP uses a three-way handshake: the client sends SYN, the server replies with SYN-ACK, and the client completes with ACK. This synchronises sequence numbers before data transfer.

Question 39Communication and network technologies

A router has four static routes for 172.16.4.0/24, 172.16.5.0/24, 172.16.6.0/24, and 172.16.7.0/24. Which single summary route covers exactly these four subnets?

  1. 172.16.0.0/21
  2. 172.16.4.0/23
  3. 172.16.4.0/22 Correct answer.
  4. 172.16.4.0/24

Explanation

The four /24 networks 172.16.4.0 through 172.16.7.0 occupy the address range 172.16.4.0–172.16.7.255. A /22 mask provides four contiguous /24 blocks with the network address 172.16.4.0, so 172.16.4.0/22 is the exact summary.

Question 40Communication and network technologies

A technician connects a new workstation to a LAN. The workstation has no manually configured IPv4 settings but can reach the internet after startup. Which protocol supplied the workstation's IP address, subnet mask, and default gateway?

  1. DHCP Correct answer.
  2. DNS
  3. ARP
  4. ICMP

Explanation

DHCP uses a client-server exchange to lease an IPv4 address, subnet mask, default gateway, and optional DNS server addresses when a host starts up. DNS resolves names, ARP maps IPv4 addresses to MAC addresses on the local link, and ICMP carries error and diagnostic messages. None of those protocols assigns a complete IP configuration to a new workstation.

Question 41Communication and network technologies

Which wireless security protocol uses AES-CCMP for both confidentiality and integrity?

  1. Open system authentication
  2. WPA2 Correct answer.
  3. WEP
  4. WPA with TKIP

Explanation

WPA2 is based on IEEE 802.11i and uses AES-CCMP to provide confidentiality and integrity for wireless frames. WEP uses RC4, WPA with TKIP uses the Temporal Key Integrity Protocol, and open system authentication provides no encryption. Therefore WPA2 is the protocol in the list that matches AES-CCMP.

Question 42Communication and network technologies

A network administrator is assigned the IPv4 block 10.20.0.0/16 and must divide it into subnets that each support at least 500 usable host addresses. What is the longest prefix length that satisfies this requirement?

  1. /24
  2. /25
  3. /23 Correct answer.
  4. /22

Explanation

A /23 prefix leaves 32 - 23 = 9 host bits, so it provides 2^9 - 2 = 510 usable IPv4 addresses, which meets the 500-host requirement. A /24 provides only 254 usable addresses, so it is too small. A /22 would also meet the requirement, but it is a shorter prefix and therefore not the longest prefix that works.

Question 43Communication and network technologies

An ISP assigns an organization the IPv6 prefix 2001:db8:acad::/48. The organization plans to use a uniform /64 prefix for every subnet. How many /64 subnets can the organization create from this assignment?

  1. 65,536 Correct answer.
  2. 4,096
  3. 4,294,967,296
  4. 256

Explanation

The /48 prefix reserves 48 bits for the network, and a /64 subnet uses 64 network bits, leaving 64 - 48 = 16 bits for subnet identifiers. Therefore the organization can create 2^16 = 65,536 distinct /64 subnets. The other values correspond to 8, 12, or 32 subnet bits, not the 16 bits available here.

Question 44Communication and network technologies

A TCP sender notices that its retransmission timer expires and responds by reducing its congestion window. Later, the receiver advertises a window size of zero, and the sender pauses transmission. Which pair of mechanisms is illustrated?

  1. Congestion control for the timeout; flow control for the zero window Correct answer.
  2. Congestion control for both the timeout and the zero window
  3. Error control for the timeout; flow control for the zero window
  4. Flow control for the timeout; congestion control for the zero window

Explanation

TCP congestion control responds to network congestion, such as a retransmission timeout, by reducing the congestion window. TCP flow control uses the receiver's advertised window to ensure the sender does not overflow the receiver's buffer; a zero window pauses transmission until the receiver advertises space again. The two mechanisms operate independently, so a timeout is not flow control and a zero window is not congestion control.

Question 45Communication and network technologies

A network application sends short status messages where occasional loss is acceptable and no connection setup is required. Which transport-layer protocol matches these requirements?

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

Explanation

UDP is a connectionless transport protocol and does not retransmit lost segments, so it suits short messages that tolerate occasional loss. TCP establishes a connection and recovers lost data, while ICMP and ARP operate below the transport layer.

Question 46Communication and network technologies

A switched LAN has redundant links between three switches. Which function does IEEE 802.1D Spanning Tree Protocol perform?

  1. It automatically assigns VLAN membership to every access port on each switch.
  2. It encrypts all frames carried over inter-switch trunk links.
  3. It blocks selected ports to create a loop-free logical topology. Correct answer.
  4. It balances traffic across every redundant link at the same time.

Explanation

STP prevents Layer 2 forwarding loops by electing a root bridge and placing redundant ports into a blocking or discarding state. It does not assign VLANs, encrypt trunk frames, or use all redundant links simultaneously for a single topology.

Question 47Communication and network technologies

An employee needs to read the same mailbox from a laptop and a smartphone, and wants the read/unread state to stay synchronized between both devices. Which application-layer protocol should the mail client use?

  1. IMAP Correct answer.
  2. SMTP
  3. POP3
  4. SNMP

Explanation

IMAP keeps messages on the mail server and synchronizes mailbox state, including read/unread flags, across multiple clients. SMTP is used to send or relay mail, POP3 typically downloads messages to one device, and SNMP is for network management.

Question 48Communication and network technologies

Two branch routers establish an IPsec VPN over the Internet. The company requires the original source and destination IP addresses of every packet to be hidden from intermediate routers. Which IPsec mode meets this requirement?

  1. Aggressive mode, because it establishes the tunnel more quickly than main mode.
  2. Tunnel mode, because it encapsulates the entire original packet in a new IP packet. Correct answer.
  3. Transport mode, because it encrypts only the upper-layer payload and leaves the original IP header visible.
  4. Main mode, because it negotiates security associations before any user data is sent.

Explanation

IPsec tunnel mode encapsulates the complete original IP packet inside a new IP packet, so intermediate routers see only the new outer header. Transport mode leaves the original IP header visible, and main and aggressive modes are IKE negotiation modes rather than data encapsulation modes.

Question 49Hardware and computer architecture

In the von Neumann architecture, what is the function of the program counter (PC)?

  1. It counts the number of instructions executed since the processor was reset.
  2. It decodes the opcode of the instruction currently in the pipeline.
  3. It holds the address of the next instruction to be fetched. Correct answer.
  4. It stores the data operand most recently fetched from main memory.

Explanation

The program counter is a processor register that holds the memory address of the next instruction to be fetched. After an instruction is fetched, the PC is normally incremented or updated by a branch so that the fetch unit can locate the following instruction.

Question 50Hardware and computer architecture

In a computer system, which property is characteristic of volatile main memory such as DRAM?

  1. It keeps its stored contents permanently after power is removed.
  2. It loses its stored contents when the power supply is removed. Correct answer.
  3. It stores only the bootstrap loader and no program data.
  4. It can be written only once during manufacture.

Explanation

Volatile memory requires continuous power to retain stored information, so DRAM contents are lost when power is removed. Non-volatile memory such as ROM or flash retains data without power.