3) (8 pts) A hypothetical network has an end to end length of 1000 meters with a propagation speed of 2x108 m/s, and a single store-and-forward switch which introduces a queuing delay of 2.5 S to transmissions in both directions. The bandwidth of the link between transmitter and switch is 500 Mbps (500,000,000 bps) while the bandwidth of the link between switch and receiver is 200 Mbps (200,000,000 bps). The maximum frame size for transmission on this network is 5000 bits.

b) (2 pts) What is the one way (from one end to the other) propagation delay for the network? (Assume this calculation does not include any queuing delays)

d) (2 pts) Assume zero-size jamming frame which incurs no queueing delay and transmission time, but only propagation time. Suppose CSMA/CD (carrier sense multiple access with collision detection) is used on this network. Using the queuing delay provided and the propagation delay from part b, what is the minimum number of bits in a frame that can be used such that a transmitting node is guaranteed of detecting a collision?

Please help for part d. Thanks

5001 m. 500 Mbps 200Mbps. Between Transmitter and switch Transmission time = Tt - 1 500 500*106 10lls Propogation Time Tb. d = 500g = 250*10 = 2.5 Uls. Total delay o els+ 2.5ls : 12.5 els Between switch and receives Transmission time = 1. sooo 200*106 = 25ls. Tp= 2.54s. Total delay. 25+2.5 lowo .27.5 . Total delay from one end to other -12.5+27.5 Us = 40 bls. 5001 m. 500 Mbps 200Mbps. Between Transmitter and switch Transmission time = Tt - 1 500 500*106 10lls Propogation Time Tb. d = 500g = 250*10 = 2.5 Uls. Total delay o els+ 2.5ls : 12.5 els Between switch and receives Transmission time = 1. sooo 200*106 = 25ls. Tp= 2.54s. Total delay. 25+2.5 lowo .27.5 . Total delay from one end to other -12.5+27.5 Us = 40 bls