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9618 · 14.2

Circuit switching, packet switching — practice questions

Practice and worked examples for 9618 Circuit switching, packet switching. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A real-time voice-over-IP (VoIP) call is being made over a busy corporate network. The network uses packet switching. Explain two potential problems that could degrade the quality of the call, linking them to the principles of packet switching.

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  1. Jitter: Because the network uses packet switching, each voice data packet is routed independently. Due to varying levels of congestion at different routers, packets may take different paths or experience different queuing delays. This results in variable arrival times at the destination, a phenomenon known as jitter. For a voice call, this can cause the audio to sound choppy or distorted as the receiving device struggles to play the audio smoothly. [1 mark for identifying jitter, 1 mark for explanation linked to packet switching principles].
  2. Packet Loss: During periods of high network congestion, routers may become overwhelmed with more packets than they can process. When a router's internal buffer (queue) is full, it may be forced to drop incoming packets. If voice packets are lost, there will be gaps in the audio stream, leading to words or syllables being missed by the listener, severely degrading call quality. [1 mark for identifying packet loss, 1 mark for explanation linked to congestion].

Worked example 2

A 1.2 MB file is transmitted over a packet-switched network. The file is divided into packets where each packet has a maximum payload size of 1500 bytes. Each packet header is 40 bytes. Calculate: (i) The number of packets required. (ii) The total size of all transmitted headers. (iii) The total amount of data transmitted across the network. (iv) The efficiency of the transmission. (Assume 1 MB = 1,000,000 bytes)

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First, convert the file size to bytes: 1.2 MB = 1.2 * 1,000,000 = 1,200,000 bytes.

(i) Number of packets: Number of packets = Total file size / Payload size per packet = 1,200,000 / 1500 = 800 packets. [1 mark]

(ii) Total size of headers: Total header size = Number of packets * Header size per packet = 800 * 40 bytes = 32,000 bytes (or 32 KB). [1 mark]

(iii) Total data transmitted: Total data = Original file size + Total header size = 1,200,000 bytes + 32,000 bytes = 1,232,000 bytes (or 1.232 MB). [1 mark]

(iv) Transmission efficiency: Efficiency = (Useful data / Total data transmitted) * 100 = (1,200,000 / 1,232,000) * 100 = 0.9740 * 100 ≈ 97.4%. [1 mark for correct calculation and answer]