IPA IP 2020 Question 95
A question IPA Japan published in 令和2年度 10月 ITパスポート試験 公開問題 問95, translated into English. ITPEC is based on the same exams.
On a communication line where the transmission speed is 20 Mbps (bits/second) and the transmission efficiency is 80%, how many seconds does it take to transmit 1 Gbyte of data? Here, 1 Gbyte = 103 Mbytes.
Explanation
500 seconds — convert the data to bits, then divide by the speed the line actually delivers.
- •Data: 1 Gbyte = 103 Mbytes = 1000 × 8 = 8000 Mbits
- •Effective speed: 20 × 0.8 = 16 Mbit/s
- •Time: 8000 ÷ 16 = 500 seconds
Answer (d)
Why not others:
- •(c) 62.5 — divides 1000 Mbytes by 16 Mbit/s, mixing bytes and bits: the factor 8 is missing
- •(b) 50 — leaves out both the factor 8 and the efficiency: 1000 ÷ 20
- •(a) 0.625 — would move 1 Gbyte in well under a second, far beyond what a 20 Mbit/s line can carry
Key rule: Transmission time = data in bits ÷ (line speed × transmission efficiency); a line speed in bps needs the bytes multiplied by 8 first.
Networks · AI-written, may contain errors
This question comes from the past questions IPA Japan publishes, translated into English by ITPEC Practice. ITPEC Practice is an independent study tool and is not affiliated with IPA or ITPEC. See the original paper (PDF) or Report an issue.