Consider the bandwidth as 50 Kbps, one way transit time-240 msec and the segment size is 1000 bit. Consider the event of a segment transmission and the corresponding ACK reception. Find the maximum number of segments that can be outstanding during this duration.
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There are 23 segments that can be outstanding during this duration.
Explanation:
- Bandwidth is defined as the number of frequencies in a specific signal.
- Segment transmission is used by networks related to communication.
- This is a data package. It is segregated into small units. These units are used for transmission.
- ACK is abbreviated as acknowledgment code. These are a set of signals that any computer sends for transmission of data.
Answered by
1
Answer:
25
Explanation:
Maximum number of outstanding segments = 2*BDP +1
BDP in bandwidth delay product = 50*.24 = 12Kbit
since we have 1000 bit segment 12 data segments will be outstanding and
12 ACK will be outstanding and +1 for one ACK which has just reached the sender.
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