Hw01 hint

ComputerNetworkHWHw01 hintp.06(a)Propagation delay is equal tod / swheredis the distance andsis thewave propagation speed.(b)Transmition Delay is equal to L/R ,where L is the bits to be transmitted and R is the data transmition rate.( c )End to end delay can be obtained by sum Propagation delay and Transmition delay.(d)To transmit all bit in packet it must at least use up time equal to tranmition delay time.p.06(e & f)If time a means the time to place P use up a time unit and time b > a means the time to the same place P use up b time unit , then if at the same time a and b start propagating to place P(with a use up a time and b use up b time),then a must first arrive but b not yet at the time that a arrives at place P.(g) to solve transmition delay equals to propagetion delay function(L/R = d/s).

p.07To get the answer , you must sum the time that data transforming to packet and the time of transmition and the time of propagation .p.08(a)circuit switching used up all resources equally devided to all users.(b)according to the course slide , each user are idled in 90% of time.(c )you can use binomial distribution to represent it.(d)you can use basic probability theory and binomial distribution to represent it.p.09(a)all users will equally share the transmition rate when using circuit switching.(b)you can use sigma and binomial distribution to represent it.p.10 & p.11(10)to sum up the all three end to end delay of three links and proc delays.(11)proc = 0 to calculate p.10,and it does not introduce a transmission delay , so transmission delay is only counted at host A.p.12Queuing delay =(the total bits in queue have not been transmitted ) / tranmit rate.p.13(a)to sum up queueing delay for every packet without considering queueing delay for first packet.(b)consider that when one batch have been transmitted,then the next batch comes.p.14(a)To simplify the summation of transmition delay and queueing delay in this question(b) Let x = L/R Total delay = x/(1-ax) then you can plot them in 2D diagram (note:please note what y axis and x axis represent for)p.15Represent (L/R)/1-I using and a annd simplify itp.16Using N = a * d ,noted that 1.d = queueing delay+transmition delay 2.If there are N packet in queue ,then the first packet is transmitting , other packets are queueingp.17Try using sigma to generalize itp.20Consider that faster link must wait for slower link to transmit datap.21Max throughput on one link must be the max rate between all RateMax throughput one multiple link can sum up Rates of all linkp.22You can use probability basic theory and geomertic random variable to calculate(generalize) itp.25(a)dprop = d / s R * dprop is the answer , noted must correct the unit!(b)As big as the bandwidth(c )to describe what is bandwidth-delay product(d) the width of a bit = length of link / bandwidth-delay product(e )width = propagation speed / transmition ratep.26To solve S/R = length to determine Rp.27(a)same hint as p.25(a)(b)same hint as p.25(b)(c )same hint as p.25(e )p.28(a)to calculate total send time without acknowledgement(b)to calculate total send time with each packet considering acknowleddement time.(c )compare the cause of difference between (a) and (b)p.29(a)to calculate the propagation time(b)to calculate bandwidth-delay product(see p.25(a)hint(c )to calculate in one minute you must send all the photo data with this bandwidth,then find xp.33First calculate sending time of each packet,then find whole file delay,finally find out what S causes delay become minimum(probably need diffrenciation)