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CAMBRIDGE INTERNATIONAL EXAMINATIONS
9691 COMPUTING
9691/12
Paper 1 (Written), maximum raw mark 75
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
om
.c
MARK SCHEME for the May/June 2014 series
s
er
GCE Advanced Subsidiary Level and GCE Advanced Level
Page 2
1
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
12
(a)
A
B
C
D
primary
memory
storage
magnetic
secondary
storage
optical
secondary
storage
solid state
secondary
storage
DVD – RAM
ROM
hard disk
flash memory
1 mark for each correct tick
[4]
(b) (i) A
[1]
(ii) B
[1]
(iii) C / D (If both ringed here then still award a mark)
[1]
(c) 1 mark for a benefit and 1 mark for a drawback
benefit
– faster start up speed (no “spin-up” required)
– no moving parts (so more robust)
– very fast read write seek (latency) times
– doesn’t require additional read/write hardware devices (just plugs into USB)
– can store data indefinitely provided that it is periodically refreshed
drawback
– vulnerable to magnetic fields and electrostatic charges
– limited write cycles
– more vulnerable to corruption if used as primary source of saving files
– because it is small, very easy to lose
© Cambridge International Examinations 2014
[2]
Page 3
2
Mark Scheme
GCE AS/A LEVEL – May/June 2014
Syllabus
9691
Paper
12
(i) Any two from:
circuit switching
– path decided on before data transmission starts
– system decides in which route to follow …
–
… and transmission goes according to this path/route
– for whole length of communication session, route is dedicated exclusively
– route only released when data transmission stops
packet switching
– data is broken up into packets
– packets are reassembled at destination
– packets are sent towards destination independent of each other
– each packet has to find its own route to destination
– decision as to which path to take is decided when each node is reached
– nodes are switches, routers, etc.
– each packet finds its way based on information it carries
[2]
(ii) Any two from:
baseband
– data sent as digital signals …
– … through the media as a single channel
– … that uses entire bandwidth of the media
– it is bi-directional
– (frequency-division) multiplexing is not possible
broadband
– data sent in form of analogue signals
– each transmission is assigned to a portion of the bandwidth …
– … thus multiple transmissions are possible at the same time
– communication is uni-directional
– to send and receive needs two pathways …
– … either by assigning a frequency for sending and a different frequency for receiving
– … or by using different communication paths
– multiplexing is possible using this method
[2]
(iii) Any two from:
ring topology
–
–
–
–
–
–
or if diagram described
faulty connections can cause whole network to fail
it is difficult to expand this type of network
works well under heavy loading
possible to form very large networks
no server
less secure (because data passes through all computers)
© Cambridge International Examinations 2014
Pa
age
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Ma
ark
k Sch
S hem
me
e
GCE
G E AS
AS/A LE
EV
VEL
L – Ma
ay//Ju
une
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2 14
Syllab
bus
s
96
691
1
Pa
ape
er
12
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star
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–
–
–
–
–
–
3
or if dia
d agram
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d
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itt iss ea
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s in
n th
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asier to exxpand th
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es conta
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r aso
ons
© Ca
amb
brid
dge
e In
nterrnattion
nal Ex
xam
mina
atio
ons
s 20
014
4
[2