{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz103key

# quiz103key - Time/cycle = TC = TT RTT = 4 50 = 54 msec...

This preview shows page 1. Sign up to view the full content.

Quiz 3 CNT 5106C 11 February 2010 Instructions: Write your last name, then first initial in upper righthand corner of answer sheet, and write “Quiz 3” centered at top. Answer all questions. 1. Consider a path with round trip time 50 ms. and data rate 1 Mbps. Suppose you send segments of length 500 bytes to a receiver, using 1-bit ARQ. What is the effective data rate of the connection? 2. How many bits of sequence number are required to achieve the maximum data rate using SR-ARQ? Key 1. Length of segment L = 500 B x 8 bits/B = 4,000 bits Transmission time TT = L/R = 4,000/1,000,000 sec = 4 msec.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Time/cycle = TC = TT + RTT = 4 + 50 = 54 msec. Data/cycle = L = 4,000 bits (since 1-bit ARQ) Effective data rate = L/TC = 4,000/.054 = 74 kbps 2. For SR-ARQ, the efficiency is k/(1+2a), where k = send window size and 2a = normalized round trip time = RTT/TT = 12.5. For maximum data rate, k >= 1+2a = 1 + 12.5 = 13.5, so k >= 14. The number of bits needed to express k by itself if log2(14) = 4 bits, but in order to avoid confusing the receiver when the sequence number wrap around, there have to be at least 2k sequence numbers. Hence, sequence numbers must have at least 5 bits....
View Full Document

• Spring '09
• Helmy
• Round-trip delay time, Transmission Control Protocol, Fibonacci number, maximum data rate, data rate, effective data rate

{[ snackBarMessage ]}