Cambridge A Level Computer Science 9608 — 2021 May/June Paper 3 · Variant 3

1 In a particular computer system, two real numbers, A and B, are stored using floating-point representation with: • 12 bits for the mantissa • 4 bits for the exponent • two’s complement form for both mantissa and exponent. Number A Mantissa Exponent 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 Number B Mantissa Exponent 0 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 (a) (i) Identify whether each number is positive or negative. Justify your answer. Number A ......................................................................................................................... ........................................................................................................................................... Number B ......................................................................................................................... ........................................................................................................................................... [2] (ii) Convert the binary values of the mantissa and the exponent for each number to their separate denary values. A mantissa ........................................................................................................................ ........................................................................................................................................... A exponent ........................................................................................................................ ........................................................................................................................................... B mantissa ........................................................................................................................ ........................................................................................................................................... B exponent ........................................................................................................................ ........................................................................................................................................... [4] (iii) Calculate the denary value of each floating-point number using your values from part (a)(ii). Number A ......................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... Number B ......................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... [2] (b) State which number, A or B, is stored in normalised floating-point form. Justify your answer. Number ..................................................................................................................................... Justification ............................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... [3]

Mark scheme: Question Answer Marks 1(a)(i) A – negative, mantissa starts with a one 2 B – positive mantissa starts with a zero 1(a)(ii) 1 4 A mantissa: −0.5 // – 2 A exponent: 2 B mantissa: 0.875 // 7/8 B exponent: −1 1(a)(iii) A: −2 2 B: 0.4375 // 7/16 1(b) Number: B 3 Justification: Using the mantissa The first two bits are different // first bit 0 second bit 1

2 The TCP/IP protocol suite can be viewed as a stack with four layers. (a) Write the correct descriptions for the two layers and the correct layers for the two descriptions given in the following table. Layer Description Application Handles forwarding of packets Internet / Network Handles how data is physically sent [4] (b) (i) Explain why communication protocols are necessary. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Identify and describe one other communication protocol. State its purpose. Protocol ............................................................................................................................. Description ........................................................................................................................ ........................................................................................................................................... Purpose ............................................................................................................................. ........................................................................................................................................... [3]

Mark scheme: 2(a) One mark for each correct row 4 Layer Description Handles access to services // Application manages data exchange // defines protocols used Transport Handles the forwarding of packets Handles transmission of data /routing Internet / Network / IP addressing Network Access / Interface // Handles how data is physically sent (Data) Link // Physical 2(b)(i) One mark for each point max two 2 • All data is (sent and received) using the same rules • All data is (sent and received) using the same formats • Allows communications between devices operating on different platforms • The communication is independent of the software used • The communication is independent of the hardware used 2(b)(ii) One mark for protocol, one mark for description, one mark for use 3 For example: Protocol: FTP// File Transfer Protocol Purpose: To directly transfer data between two computers over a network Use: upload and download files over the Internet Protocol: SMTP // Simple Mail Transfer Protocol Purpose: protocol for sending email Use: used by mail servers to forward email messages Protocol: POP3 //Post Office Protocol 3 // IMAP // Internet Message Access Protocol Purpose: mail is held for you by a remote server until you download it Use: to receive e-mail Protocol: BitTorrent Purpose: protocol for peer-to-peer file sharing Use: decentralised distribution of data Protocol: Ethernet Purpose: To send/receive data along a cable Use: for local area networks Protocol: HTTP(S)//Hypertext Transfer Protocol (Secure) Purpose: transfer of web pages/hypertext Use: browsing websites Protocol: WiMAX // Worldwide Interoperability for Microwave Access (AXess) Purpose: to provide wireless broadband Use: where there is no wired infrastructure

3 Describe, with the aid of a diagram for each one, the bus and star network topologies. Bus Description ....................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... Star Description ....................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... [6]

Mark scheme: 3 One mark for each point to Max 6. Max 4 for Bus, max 4 for Star 6 Bus Max 2: • Diagram showing bus topology • Diagram with correct labels (e.g. Terminator, Workstation, Backbone) Max 2: • All nodes connected to a single cable … • … with a terminator at each end • Uses half duplex Star Max 2: • Diagram showing star topology • Diagram with correct labels (e.g. Server / Central Device, Workstation, Individual Connection) Max 2: • All nodes connected to a central device … • … each node has its own connection • All data is transferred via the central device • .. using a bi-directional connection

4 (a) The truth table for a logic circuit with four inputs is shown. INPUT OUTPUT P Q R S X 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 (i) Write the Boolean expression for the truth table as a sum-of-products. X = .............................................................................................................................. [2] (ii) Complete the Karnaugh Map (K-map) for the truth table. PQ 00 01 11 10 00 01 RS 11 10 [2] (iii) The K-map can be used to simplify the expression in part (a)(i). Draw loops around appropriate groups of 1s in the table in part (a)(ii) to produce an optimal sum-of-products. [2] (iv) Write the simplified sum-of-products expression for your answer to part (a)(iii). X = ............................................................................................................................... [2] (b) Simplify your expression for X in part (a)(i) using Boolean algebra. Show your working. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2]

Mark scheme: 4(a)(i) One mark for 2 correct, two marks for 4 correct and no other terms 2 X = P.Q.R.S + P.Q.R.S + P.Q.R.S + P.Q.R.S 4(a)(ii) Two marks for fully correct K-map 2 One mark for a K-map with one error Zero marks for a K-map with two or more errors PQ 00 01 11 10 00 1 0 0 0 01 0 0 1 0 RS 11 0 0 1 0 10 1 0 0 0 4(a)(iii) One mark for each correct loop max two 2 PQ 00 01 11 10 00 1 0 0 0 01 0 0 1 0 RS 11 0 0 1 0 10 1 0 0 0 4(a)(iv) One mark per point 2 • P.Q.S • +P.Q.S X = P.Q.S + P.Q.S 4(b) One mark for correct use of distributive law 2 One mark for correct use of complementary law One mark for correct use of redundancy law One mark for correct use of idempotent law Max two X = P.Q.R.S + P.Q.R.S + P.Q.R.S + P.Q.R.S X = P.Q.S ( R + R ) + P.Q.S ( R + R ) X = P.Q.S + P.Q.S // P.Q.S (1) + P.Q.S (1)

5 The following syntax diagrams for a programming language show the syntax of: • an assignment statement • a variable • an unsigned integer • a digit • a letter • an operator assignment statement variable = variable operator variable unsigned integer variable letter letter unsigned integer digit digit letter operator digit X + 1 Y 2 – Z * 3 (a) Give reasons why each of these statements is invalid. X = XY + 21 ................................................................................................................................................... YZ := YZ * 3 ................................................................................................................................................... XY = XY − 5 ................................................................................................................................................... [3] (b) Complete the Backus-Naur Form (BNF) for the syntax diagrams shown. <letter> has been completed for you. <letter> ::= X|Y|Z <assignment_statement> ::= ................................................................................................................................................... <variable> ::= ................................................................................................................................................... <digit> ::= ................................................................................................................................................... <unsigned_integer> ::= ................................................................................................................................................... <operator> ::= ................................................................................................................................................... [5] (c) The syntax of a variable is changed to allow one or two letters followed by zero, one or two digits. (i) Draw an updated syntax diagram for the variable. [3] (ii) Give the BNF for the revised variable. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3]

Mark scheme: 5(a) • X is not a variable 3 • := should be = for an assignment statement • 5 is not a valid digit 5(b) <assignment_statement> ::= 5 <variable> = <variable><operator><variable> 1 |<variable> = <variable><operator><unsigned_integer> 1 <variable> ::= <letter><letter> 1 <unsigned_integer> ::= <digit>|<digit><digit> 1 <digit>::= 1 | 2 | 3 <operator> ::= + | - | * 1 5(c)(i) 3 or unsigned integer • two letters and two digits / one unsigned integer and arrows in and out seen • allows for one or two letters at start • zero, one or two digits // zero or one unsigned integer at end 5(c)(ii) Three marks for completely correct 3 Two marks for four alternatives correct One mark for three alternatives correct <variable> ::= <letter> | <letter><digit> | <letter><digit><digit> | <letter><letter> | <letter><letter><digit> | <letter><letter><digit><digit> Or Three marks for completely correct Two marks for three alternatives correct One mark for two alternatives correct <variable> ::= <letter> | <letter><unsigned integer> | <letter><letter> | <letter><letter><unsigned integer>

6 Encryption is used to provide security when messages are transferred over a communication link. (a) (i) Explain the way in which asymmetric key cryptography is used to encrypt a message being sent from one computer user to another over the Internet. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .................................................................................................................................... [4] (ii) State two benefits of using asymmetric key cryptography. 1 ........................................................................................................................................ ........................................................................................................................................... 2 ........................................................................................................................................ ........................................................................................................................................... [2] (b) (i) Explain the way in which Transport Layer Security (TLS) provides communication security over a computer network. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [4] (ii) State two situations where the use of TLS would be appropriate. 1 ........................................................................................................................................ ........................................................................................................................................... 2 ........................................................................................................................................ ........................................................................................................................................... [2]

Mark scheme: 6(a)(i) Any four from: 4 • Asymmetric encryption / cryptography uses a matching pair of keys • A public key (available to everyone) • … receiver’s public key used for encrypting the message before it is sent • A private key (only known to the owner of the keys) • … receiver’s private key for decrypting the message after it has been received 6(a)(ii) Any two from: 2 • Increased message security as one key is private • Allows message authentication • Allows non-repudiation • Detects tampering 6(b)(i) Any four from: 4 • A protocol with two layers • …Handshake and Record • A TLS/digital/public key certificate is used for authentication • Handshake uses asymmetric cryptography • … to generate agreed parameters • … establish a shared session key • The shared session key provides symmetric cryptography for • … sending and receiving data (record layer) • At end of session all parameters, keys, etc. erased 6(b)(ii) Any two from: 2 • Browsers accessing secure websites e.g. bank transactions • VPNs • Email • VOIP

7 Four shipping containers are used to store goods on the dockside at a port. The temperature inside each container should be kept between 5 and 8 degrees Celsius inclusive. Each container has a temperature sensor. A computer system is programmed to control each container’s temperature by: • turning on the heater and turning off the air conditioning unit when the temperature falls below 5 degrees • turning off the heater and turning on the air conditioning unit when the temperature rises above 8 degrees. (a) (i) State the name given to the type of system described. ..................................................................................................................................... [1] (ii) Justify your answer to part (i). ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) The computer system stores the temperature readings for the four sensors in two’s complement form and in four eight-bit memory locations with addresses 301 to 304. 301 0 0 0 0 1 0 0 1 Container 1 302 0 0 0 0 0 1 1 1 Container 2 303 0 0 0 0 0 1 1 0 Container 3 304 1 1 1 1 1 1 1 0 Container 4 State the container number(s) where the temperature is out of range and give the value(s) of these temperature(s) in denary. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) The status of the heaters and the air conditioning units is shown at location 300. A value of 1 means that the device is on and a value of 0 (zero) means that the device is off. The status of the heaters is shown in the most significant four bits; the status of the air conditioning units is shown in the least significant four bits. The pattern of bits at location 300 shows that the heater for container 4 is on and the air conditioning unit for container 1 is on. Container number 1 2 3 4 1 2 3 4 300 0 0 0 1 1 0 0 0 Heater Air conditioning Show the pattern of bits when the heater is on for containers 1 and 2 and no air conditioning units are on. 300 [1] (d) The following table shows assembly language instructions for the container computer system that has one general purpose register, the Accumulator (ACC). Instruction Explanation Label Op code Operand LDM &n Load the hexadecimal number n to ACC Load the contents of the location at the given LDD <address> address to ACC Store the contents of ACC at the given STO <address> address Bitwise AND operation of the contents of ACC AND &n with the hexadecimal number n Bits in ACC are shifted denary number n LSL #n places to the left. Zeros are introduced at the right hand end Compare the contents of ACC with the CMP &n hexadecimal number n Following a compare instruction, jump to JPE <address> <address> or <label> if the compare was True <label>: <op code> <operand> Labels an instruction If the bit for a container’s heater and the bit for the same container’s air conditioning unit are both set to 1, a routine at label ERROR is executed. This routine has not been provided. (i) These assembly language instructions check for an error in the container 1 system. LDD 300 AND &88 CMP &88 JPE ERROR Explain the purpose of each instruction. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [4]

Mark scheme: 7(a)(i) Control 1 7(a)(ii) Any two from: 2 • Automatically controls devices / heaters / air conditioning units • … using actuators • With the use of feedback // output affects the values that are input • To maintain the required temperature range 7(b) (Container) 1: 9 degrees 2 (Container) 4: –2 degrees 7(c) 1 300 1 1 0 0 0 0 0 0 7(d)(i) One mark per point 4 • Load the accumulator with status of heaters and air conditioning units // Load the accumulator with the contents of address 300 / • Mask out the bits for container 1 // Mask out 4th and 8th bit • See whether both heater is on and the air conditioning is on // Compare the contents of the accumulator/previous result with B10001000 / (&88) • IF the heater is on and the air conditioning is on / jump to ERROR routine // jump to ERROR routine if bit patterns are equal 7(d)(ii) One mark keeping instructions 1 and 4 the same 3 One mark AND &11 One mark CMP &11 LDD 300 AND &11 // AND B00010001 CMP &11 // CMP B00010001 JPE ERROR