Question

Given a link with a maximum transmission rate of 11.7 Mbps. Only two computers, X and Y, wish to transmit starting at time t = 0 seconds. Computer X sends fileX (14 MiB) and computer Y sends fileY (272 KiB), both starting at time t = 0. Computer X gets the transmission medium first, so Computer Y must wait. For the following calculations, assume maximum transmission rate during transmission. Suppose that entire files are sent as a stream (no packets, no multiplexing). At what time (t = ?) would File Y finish transmitting? Give answer in seconds, without units, and round to two decimal places.

Answer 1

Missing Details of Question

Statistical multiplexing is used, with details as follows o

- Packet Payload Size = 1000 Bytes

- Packet Header Size = 24 Bytes (overhead)

- Ignore Processing and Queueing delays

- Assume partial packets (packets consisting of less than 1000 Bytes of data) are padded so that they are the same size as full packets.

- Assume continuous alternating-packet transmission.

Answer:

Time = 10.47

Explanation:

Given

Initial Time, t = 0

File X = 14MiB = 14 * 1024 * 1024 = 14,680,064 bytes

File Y = 272KiB = 272 * 1024 = 278528 bytes

Transmission Rate = 11.7Mbps

Total Packet Length is calculated as (1024 * 8)/(11.7Mbps)

Total Packet Length = 8192 ÷ 11700

Total Packet Length = 0.700171 ms/packets

Number of packets

File X = 14,680,064 bytes ÷ 1000 bytes

File X = 14,680.064

File X = 14,680 ---- Approximated

File Y = 278528 bytes ÷ 1000 bytes

File Y = 278.528

File Y = 279 ------- Approximated

Number of Packets = 279 + 14,680

= 14959 packets

The implication of the above is that, File X after a total of 20202 packets are transmitted

File Y will finish transmitting at:

14959 * 0.700171

= 10473.86 ms

= 10.47386 s

= 10.47 s