Lab7 PDF

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Generated: 48 kbps

Chunk size: 16ms

Question 1: What is the rate at which data is generated at the sender (in bytes)? 48kbps = 6000 bytes pr second

Question 2: What is the size of the IP datagrams sent? You must clearly show the steps in how you calculated your answer and what elements the datagram consists of. Each chunk has 6000 B/s * 0.016 s= 96 bytes pr packet worth of sender generated data. Hader: Payload type: 7 bits Sequence number: 16 bits Timestamp: 32 bits Synchronization source identifier: 32 bits Header sum: 87 bits = 11 bytes Total 107 bytes.

Question 3: Explain how an arbitrary RTP packet in the application will look like. Use actual values from the application when possible. Include the size of the fields and elements of the packet The only given value we have is the payload type part of the packet. The payload number will be 9 as the format is g.722. Other that this it will include sequence number of 16 bits (what order the packet is in the stream, used for sorting and loss detection), timestamp of 32 bit (when the packet was sent, can be used to remove jitter), Synchronization source identifier of 32 bit (identifies the source of the stream)

Loss recovery schemes can be used to preserve acceptable audio quality in the presence of packet loss. We will focus on three types of loss anticipation schemes: forward error correction (FEC) with redundant encoded chunks, FEC with redundant lower-resolution audio stream, and interleaving. Question 4: For each of the three schemes listed in the previous paragraph, show how much additional bandwidth each of them require of our application. For the first type of FEC, suppose a redundant chunk is generated for every five original chunks. For the other type, suppose GSM is used for the low-bit rate encoded stream. For each scheme, include both the new transmission rate and the percentage increase. FEC redundant encoded chunks: With one redundant chunk generated every 5 chunks (n=5) an 1/5 = 20% additional bandwidth is needed wich is a total of 6687.5*1.2 = 8025 Bps. FEC redundant low-bit rate encoded stream: GSM rate: 13 kbps= 1625 Bps wich is about 20%. The total bandwidth needed is 9650 Bps Interleaving: Interleaving does not increase the bandwidth.

Question 5: For each of the three schemes, explain what happens if the first packet is lost in every group of five packets. Which scheme will have the better audio quality? FEC redundant encoded chunks: If only the first packet is lost the user can reconstruct the missing packet using the redundant encoded packet. This is because the redundant chunk is created by XORing all 5 packets in the chunk, and if only one packet is missing this can be constructed by reversing the operation on the redundant chunk and the remaining 4 packets. FEC redundant low-bit rate encoded stream: When a packet is missing the stream is constructed with the lower resolution strem instead. This leads to small drop in audio quality until the original packets starts arriving. Interleaving: The stream will be built without the missing packet, but due to interleaving weaving the packets before sending, the missing audio will be spread out over a longer period of time and the audio will only lack miniscule chunks. This helps the over all flow and not much real information is lost. With only one packet missing the best quality comes from FEC with redundant encoded chunks as the entire packet can be reconstructed.

Question 6: Make an illustration of the scenario in question 5 for FEC with redundant encoded chunks. Use the same n value as before. The figure should be made such that it’s clear that you understand how the scheme worksSimplification (n = 2) Packet 1: Packet 2: Packet 3: Packet 4: Packet 5: Redundant (1 XOR 2…):

01010100101  Lost 00011011111  …. …. …. 01001111010 

00011011111 …. …. …. 01001111010

The redundant packet should be XOR of all 5 values, but XOR with 5 values is not fun ☹. Reconstruction of 1 (1 XOR Redundant)

01010100101

The system is the same with a higher n but the XOR operations gets a bit cumbersome with 5 packets. The core of the system is the fact that the XOR operation ads 1 degree of freedom and the redundant chunk can be used to detect what values 1 missing packet has by preforming an XOR on the redundant chunk and the chunks that are not missing.

Question 7: For each of the three schemes, explain what happens if the first packet is lost in every group of two packets. Which scheme will have the better audio quality? FEC redundant encoded chunks: There are to many missing packets to perfectly construct the missing one and a approximation has to be made. This lowers quality FEC redundant low-bit rate encoded stream: Same as Q5, but with a lot more lower quality audio in the stream. Interleaving: Same as Q5 but the amount of holes in the stream will increase.

The best audio quality will come from scheme 2 where the missing audio will be concealed by the lower quality stream.

Question 8: Make an illustration of the scenario in question 7 for FEC with redundant lowerresolution audio stream. The figure should be made such that it’s clear that the student understands how the scheme works.

Packet numbers | HQ LQ Lost packet Original stream:

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Low quality:

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Constructed audio:

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Here we can see the missing packets being swapped out with the lower quality GSM packets from the redundant lower resolution stream.

Question 9: For each of the FEC schemes, how much playback delay does each scheme add? What can be said about the delay of the interleaving scheme? With FEC redundant chunks all n chunks has to be received before playback. With a chunk size of 16ms and a n = 5 the delay is 80ms. With FEC redundant audio only has to receive 2 packets before playback and therefore only has a 32ms playback delay....