Suppose host A wants to send a 500 MByte (1 MByte = 210 KByte = 1024 KByte, 1 KB
ID: 3561281 • Letter: S
Question
Suppose host A wants to send a 500 MByte (1 MByte = 210 KByte = 1024 KByte, 1 KByte = 210 Byte = 1024 Byte) file to host D. The path from host A to host D passes through the two routers B and C. The links on the path have the following bandwiths: Rab = I Mbps (link between A and B), Rbc = 1 Gbps (link between B and C), and Rcd = 1200Kbps (link between C and D). Assuming there is no other traffic in the network, what is the throughput for the file transfer in bps? How many minutes will it take to transfer the whole file from A to D (accurate to at least one decimal place)?Explanation / Answer
Part A
1 megabyte = 1,048,576 500mb x (1,048,576 bytes/1mb) = 524,288,000 bytes
500 MB = 500x(524,288,000) = 272144000000 bytes
Throughput = 272144000000 bytes / 917.1 sec
= 280243023.4 bytes / sec
Part B
To get from point A - > B
500 MBs / 1 MBs = 500 seconds
For B --> C
500 MB / 1000 MBs (since 1GB = 1000MBs) = 1/2 second
To get from C - D
500 Mb / 1200 KBs
= 500,000 KB / 1200 KBs = 416.6 second
Total Time = 416.6 sec + .5 sec + 500 sec = 917.1 sec = 15 min 17.1 seconds This is reasonable because 500 Mb is a big file.