[Computer Network] Sliding window General doubt Tanenbaum p-333 closed

Question

To find an appropriate value for w we need to know how many frames can fit
inside the channel as they propagate from sender to receiver. This capacity is determined
by the bandwidth in bits/sec multiplied by the one-way transit time, or
the bandwidth-delay product of the link. We can divide this quantity by the
number of bits in a frame to express it as a number of frames. Call this quantity
BD. Then w should be set to 2BD + 1. Twice the bandwidth-delay is the number
of frames that can be outstanding if the sender continuously sends frames when
the round-trip time to receive an acknowledgement is considered. The ‘‘+1’’ is
because an acknowledgement frame will not be sent until after a complete frame
is received.

Can anyone tell why we add one to the max. window size?Does bandwidthn delay product means that time it takes first  bit to reach receiver to the first frame to reach receiver?Please help

Comments

Say bandwidth is 10bits/sec and delay is 25 sec,then bandwidth delay product is 25*10=250 bits,now in this 25 seconds , 250 bits can be transmitted from source to destination.

If frame size is 10 bits,means 25 frames will be transmitted.

So once 25 frames are dont,then sender will start the ack and in next 25 frames we will get first ack,why do we need +1 to that.In this case why it is 51 instead of 50?

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when you send a frame and waiting for its acknowledge,in that waiting time when last bit of frame reached to receiver take delay time 25 sec( so sender send more 25 frame(25*10=250bits, 250/10=25 frame) in that time).Then receiver send acknowledge that take 25sec due to propogation delay (in this duration sender send next 25 frame).so total frame send by sender are 1+25+25 until not get acknowledge of 1st frame. now you can clearly understand why 51 not 50.