Sliding Window Protocol - Window Size, Bits Required Efficiency, Throughput | UGC NET 2022 PYQs

How to Calculate
1. transmission and propagation delay.
2. The sender window size to get the maximum efficiency.
3. The minimum number of bits required in the sequence number field of the packet.
4. If only 6 bits are reserved for sequence number field, then the efficiency of the system is
5. The maximum achievable throughput is
A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.

[@Happy26500]

this was an amazing video cleared my lot of confusion , thankyou mam

[@TarjBaxi]

Great Work Mam!!

[@dr.bnagarajan9481]

Thanks 👍 Ji.. very useful and explained clearly thanks Ji

[@sorajsahu5136]

Very nice mam. Mujhe bht help hua

[@kasaiop]

liked for detail explaination

[@yaminisahu8110]

Mam ,propagation speed=distance/speed but why we multiply distance * speed
I have a doubt

[@qasidmubashir]

its a typo , its actually 1km / 6micro seconds .........
so propagation delay = 3000 km / (1km/6micro sec). = 3000 x 6 micro sec = 18000 u s

[@sudhakothari5941]

Thank mam . Nice video❤

[@mr.luxxyandreneo5168]

thanks , nice video

[@ugcnet8015]

Thank you so much Mam

[@vanshvansh3484]

Thnku mam ❤

[@trendyindia100]

Thanks mam