An 8 Mbps token ring uses a token holding timer of 10 ms. Ignoring header bits, what is the maximum frame size that a station can transmit during one token hold?

Introduction / Context:
In token ring networks, a station may transmit only while it holds the token, bounded by a token holding timer. The maximum frame length is limited by link rate and token hold time.

Given Data / Assumptions:

• Bit rate = 8 Mbps (8 x 10^6 bits per second).
• Token holding time = 10 ms (0.01 s).
• Header/overhead negligible for this calculation.

Concept / Approach:
Maximum bits sendable = data_rate * time_window. Convert bits to bytes by dividing by 8 to get frame size in bytes.

Step-by-Step Solution:
1) Compute bits in 10 ms: bits = 8 x 10^6 * 0.01 = 80,000 bits.2) Convert to bytes: 80,000 / 8 = 10,000 bytes.3) Therefore, the longest frame size is 10,000 B (ignoring headers).

Verification / Alternative check:
Cross-check units: Mbps is bits/s; multiplying by seconds yields bits; dividing by 8 yields bytes. No other constraints given, so 10,000 B stands.

Why Other Options Are Wrong:
8000 B: Corresponds to 64,000 bits, which would require only 8 ms at 8 Mbps; underestimates the maximum.

80,000 B: Equals 640,000 bits, exceeding what can be sent in 10 ms at 8 Mbps.

8 x 10^5 bit frame: 800,000 bits; also too large.

1000 B: Only 8,000 bits; far below the limit.

Common Pitfalls:
Confusing Mbps (bits) with MBps (bytes) and forgetting to divide by 8; mixing milliseconds and seconds.

Final Answer:
10,000 B frame