Which of the following is/are true about windowing in TCP/IP?
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For a window size of 3 the ack4 signal says send the 4th packet.
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For a window size of three, the packets will be sent in three different routes and acknowledgment is expected for each.
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For a window size of 3 the acknowledgement is always ack3 to notify THAT the 3 packets were received.
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None of the above.
In TCP windowing, a window size of 3 means 3 segments can be sent before requiring acknowledgment. The acknowledgment number (ACK4) indicates the next expected byte, not packet count. Option A correctly describes that ACK4 means 'send the 4th packet' as the next expected. Option B is wrong because windowing doesn't use multiple routes. Option C is wrong because the ACK number is cumulative, indicating the next expected sequence number, not just counting packets.
TCP's sliding window uses cumulative, next-expected-byte/segment acknowledgments: if packets 1-3 are sent in a window of size 3 and all arrive correctly, the receiver sends ACK 4, meaning "I've received everything through packet 3; send packet 4 next." That matches option one. TCP does not route different packets of the same window over "three different routes" needing individual acks per route — routing is a network-layer (IP) concern, and normally packets from one flow follow the same path. The idea that the ack is always just "ack3 to say 3 packets were received" misstates TCP's acknowledgment semantics, which reference the next expected sequence, not a count of packets received. So the first statement is the accurate one.