Exercicios Redes Solução

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Lucas Henrique Silva
CS436 – Chapter 1 – HW1
5 questions, each 20 points – total is 100
P17.
a. Generalize the end-to-end formula in section 1.4.3 section 1.4.3 for heterogeneous processing rates, and
propagation delays.
Dend−to−end =
∑N
j=1Dtransj + Dprop +
∑N
k=1Dprock
b. Repeat (a), now also suppose that there is an average queuing delay of dqueue at each node.
Dend−to−end =
∑N
i=1Dqueuei +
∑N
j=1Dtransj + Dprop +
∑N
k=1Dprock
P24.Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R
= 2Mbps. Suppose the propagation speed over the link is 2.5 ∗ 108 m/s.
a. Calculate the bandwidth-delay product, R* dqueue.
dprop =
20000∗103
2.5∗108 = 0.08s.
Bandwidth-delay =0.08 ∗ 2 ∗ 106 = 160, 000bits
b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as
one large message. What is the maximum number of bits that will be in the link at any given time ?
160,000bits
c. Provide an interpretation of the bandwidth- delay product.
Maximum number of bits that will be in the link at any given time.
d. What is the width ( in meters) of a bit in the link ? Is it longer than a foot-ball field?.
Width of a bit = lengthoflinkbandwidth−delay−product =
20∗106
160∗103 = 125
A football field have around 109 m so is bigger than a football field.
e. BitSize = LBandwidth−delay−product =
L
R∗(d s)
P25. Referring to problem P24 , suppose we can modify R. For what value of R is the width of a bit as long
as the length of the link?
Maximum number of bits that will be in the link at any given time.
P26. Referring to problem P24 bot now with a link of R = 1Gbps.
a. Calculate the bandwidth-delay product, R * D prop.
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Bandwidth-delay = (1∗10
9)∗(20∗106)
2.5∗108 = 80, 000, 000bit
b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as
one large message. What is the maximum number of bits that will be in the link at any given time ?
800,000bits
c. What is the width ( in meters) of a bit in the link ?
Width of a bit =D/delay = 20∗10
6
80∗106 = 0.25m
P32. Consider sending a large file of F bits from Host A to Host B. There are three links ( and two switches)
between A and B, and the links are uncongested ( that is, no queuing delays). Host A segments the file into
segments of S bits each and adds 80 bits of header to each segment, forming packets of L = 80+S bits. Each
link has a transmission rate of R bps. Find the value of S that minimizes the delay of moving the file from
Host A to Host B. Disregard propagation delay.
R is the link transmission rate. If S is smaller then R, so S could be sent as one packet otherwise, to
reduce the delay the size of the packet should be as close to the rate R as possible, so, to obtain the
minimum delay each pecket must be the size of R-80 this way S is S = R-80 bits , this way there is less
packets to send and less delay.
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