Computer science · Grades 9–12
Ideal payload transfer time
Assume no protocol overhead or contention. MB and Mbps use decimal prefixes, not binary mebibytes.
Method and conditions
Assume no protocol overhead or contention. MB and Mbps use decimal prefixes, not binary mebibytes.
time = payload bits / bit rate
Worked examples
A 173 MB payload is sent over a constant 185 Mbps link. Use decimal MB and ignore overhead. Find transfer time. Round the final result to 4 decimal places if needed. Answer in s.
The payload is 1384 megabits. Divide by 185 megabits per second.
7.4811 s
A 251 MB payload is sent over a constant 45 Mbps link. Use decimal MB and ignore overhead. Find transfer time. Round the final result to 4 decimal places if needed. Answer in s.
The payload is 2008 megabits. Divide by 45 megabits per second.
44.6222 s
Try it
A 767 MB payload is sent over a constant 185 Mbps link. Use decimal MB and ignore overhead. Find transfer time. Round the final result to 4 decimal places if needed. Answer in s.
A 125 MB payload is sent over a constant 195 Mbps link. Use decimal MB and ignore overhead. Find transfer time. Round the final result to 4 decimal places if needed. Answer in s.
Check your work
Answers and marking notes
- 33.1676 s
The payload is 6136 megabits. Divide by 185 megabits per second.
- 5.1282 s
The payload is 1000 megabits. Divide by 195 megabits per second.
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