You are building a communication network based on RFC 1149 "IP over avian carrier". You attach two 128GB flash drives to your pigeon. You can write data to the flash drive at the rate of 20 MB/s and you can read data from the flash drive at the rate of 50MB/s. The flying speed of a pigeon is 80km/h and pigeons do not need to stop for food or rest before reaching destination. What is the total time needed for you to send and your friend who lives in Brazil (Rio is 13600 km from Adelaide) to receive a 4 ∗ 10^9 bit file of your favourite home movies? Solve for the number of seconds delay. Show your work.
Try an answer before revealing the guidance below.
Key Concepts
- Bits to bytes
- Write time (transmission)
- Flight time (propagation)
- Read time
- Unit conversion
Answer Approach
- Convert the file size from bits to MB.
- Compute the time to write the file onto the flash drive.
- Compute the pigeon’s flight time in seconds.
- Compute the time to read the file at the destination, then add all three.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
File size = 4 × 10^9 bits / 8 = 5 × 10^8 bytes = 500 MB (fits on one drive). Write: 500 MB / 20 MB/s = 25 s. Flight: 13,600 km / 80 km/h = 170 h = 170 × 3600 = 612,000 s. Read: 500 MB / 50 MB/s = 10 s. Total = 25 + 612,000 + 10 = 612,035 s (about 7.08 days). (Uses decimal MB; with binary MB the write and read times are slightly shorter.)
Total = write time + flight time + read timeFlight time = distance / speedShortcuts: K concepts · A approach · F answer · R reviewed · B bookmark · ← / → previous / next