1 votes 1 votes The network consists of 4 hosts distributed as shown below: Assume this network uses CSMA/CD . And signal travels at $3$ $X$ $10^{5} \text{km/sec}$ If sender sends at $1 \text{Mbps}$. What could be the minimum size of packet? (a).$600$ $\text{bits}$ (b). $400$ $\text{bits}$ (c). $6000$ $\text{bits}$ (d). $1500$ $\text{bits}$ Computer Networks computer-networks + – Shefali asked Nov 7, 2015 Shefali 2.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 0 votes 0 votes Cannot solve this question without knowing sender and receiver. Question needs clarity. Let us assume A is a sender and D is receiver. Algorithm to calculate distance should be specified in GATE question Use the following equation to solve answer once you know the distance L>= B*2*d/v pC answered Jul 12, 2016 selected Jul 12, 2016 by pC pC comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes The minimum size of packet in CSMA/CD is TT>=2*TP. Length of packet=2*(distance/Speed of signal)*Bandwidth. Here all terms are common b/w all nodes except distance. So minimum the distance, minimum will be length of packet. Here minimum distance=40km L=2*(40/3*105)*106=800/3=266.66 bits Avdhesh Singh Rana answered Nov 8, 2015 Avdhesh Singh Rana comment Share Follow See all 2 Comments See all 2 2 Comments reply Himanshu1 commented Nov 8, 2015 reply Follow Share Not in options.. 0 votes 0 votes akash commented Nov 8, 2015 reply Follow Share should we consider 40 or 111(AB+ BD)? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes in CSMA/CD- L/B = 2(D/V) SO "L " IS DIRECTLY PROPORTIONAL TO "D" =>MINIMUM SIZE=MINIMUM DISTANCE,, AND WITH MINIMUM DISTANCE OF 90KM WE WILL GET 600BITS ..WITH DISTANCE OF 40,51,54,57 WE WILL GET <400 BITS.... SO 90 KM SATISFIES asu answered Nov 16, 2015 asu comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Suppose u get a value of the frame size by considering 40 km. There u will get frame size to be 266.67 bits. Now if u consider the frame size to be 266.67 bits and if a collision takes place between A and D, it will not suffice with the value of frame size as 266.67 bits. Because when u put the value in Tt ( transmission time ) and compare it with propagation time ( Tp ), then it will not satisfy the collision detection equation in CSMA / CD that is Tt > = 2*Tp. That is why you have to consider the longest path of 90 km to get the minimum frame size which is coming out to be 600 bits. Option A. user2525 answered Jul 10, 2019 user2525 comment Share Follow See all 0 reply Please log in or register to add a comment.