Review the Car-caravan Analogy in Section 14 Assume a Propagation Speed of 100 Km/hour
P2.
Equation 1.i gives a formula for the cease-to-cease delay of sending one packet of length L over Due north links of transmission rate R . Generalize this formula for sending P such packets dorsum-to-back over the N links.
The general case of sending one packet from source to destination over a path consisting of N links each of rates R. so,
stop-to-terminate delay (d) = N(L/R)
1 parcel = the end-to-terminate delay (d) = N(50/R)
P packets = d1
one packet x d1 = d ten p
i x d1 = [N(L/R)] x P
so as a result, the end-to-cease delay is
d1 = [N(L/R)] x P
P5.
Review the motorcar-caravan analogy in Section 1.4. Assume a propagation speed of 100 km/60 minutes.
a. Suppose the caravan travels 150 km, first in front of one tollbooth, passing through a 2d tollbooth, and finishing just later a third tollbooth. What is the end-to-end delay?
ans: full distance = 150 kms
the speed = 100 km/hr
so the total transmission filibuster = 150/100 = i.5 hrs
the time taken by each toll booth to attain a single car = 12 sec
then the time taken to reach 10 cars is = 120 sec or two min
the time taken by 3 toll booth to achieve 10 cars = 6 min
the end-to-end delay is = 1.5 hrs + half-dozen min = one hr 36 min
b. Repeat (a), now assuming that there are viii cars in the caravan instead of ten.
answer : the full transmission filibuster = 150/100 = 1.5 hrs
the fourth dimension taken by each toll booth to accomplish a unmarried car = 12 sec
the time taken to reach 8 cars = eight x 12 = 96 sec
the time taken by three toll booth to reach 8 cars = 3 x 96 = 288 sec
cease-to-terminate delay is = one.v hrs + 4 min 48 sec
so,the for finish to end delay for 8 cars is = 1 hour 34 min 48 sec
P6.
This elementary problem begins to explore propagation delay and manual
filibuster, two central concepts in information networking. Consider two hosts, A
and B, connected past a single link of rate R bps. Suppose that the two hosts
are separated by thou meters, and suppose the propagation speed forth the link
is s meters/sec. Host A is to transport a packet of size L bits to Host B.
a) Express the propagation filibuster, in terms of 1000 and s .
The distance between the two hosts A and B = yard meters
The propagation speed along the link = s meter/sec
So the propagation delay is = 1000/southward or distance/s
b) Determine the transmission time of the packet, in terms of L and R .
The size of the parcel = L bits
And the manual charge per unit of the link is R bps
The transmission of the packet = L/R sec
c) Ignoring processing and queuing delays, obtain an expression for the cease-to-end delay.
The end to end filibuster = d(proc) + d(queue) + d(trans) + d(prop)
Which would consequence in = d(trans) + d(prop)
d(trans)= L/R and d(prop)=m/s
The stop to end delay = (Fifty/R)+(m/south)
d) Suppose Host A begins to transmit the packet at time t = 0. At time t = d(trans) ,
where is the concluding fleck of the packet?
transmission delay is time it takes for the host to button the packet out.
then, at t = d(trans) the last bit of the packet has been pushed out or it was transmitted.
eastward) Suppose dprop is greater than dtrans . At time t = d(trans) , where is the first scrap of
the bundle?
the commencement bit is on the first bundle
f)Suppose d(prop) is less than d(trans) . At time t = d(trans) , where is the first bit of the packet?
if d(prop) < d(trans)
at time t = d(trans) the showtime flake has reached destination B.
1000) Suppose s = ii.v · ten^8 , Fifty = 120 $.25, and R = 56 kbps. Observe the distance yard so that d(prop) equals d(trans) .
s = 2.5 x 10^viii seconds
L = 120 bits R = 56 kbps
d(trans) = d(prop)
so, (L/R)=(k/s)
distance m = sL/R
= (2.5×10^8×120)/(56×1000)
m (distance) = 535.7 km
P7.
In this problem, we consider sending real-time voice from Host A to Host B
over a package-switched network (VoIP). Host A converts analog voice to a
digital 64 kbps bit stream on the fly. Host A and then groups the bits into 56-byte
packets. In that location is i link between Hosts A and B; its transmission charge per unit is 2
Mbps and its propagation delay is ten msec. As before long equally Host A gathers a
package, it sends it to Host B. Every bit soon as Host B receives an entire packet, it
converts the packet's $.25 to an analog signal. How much fourth dimension elapses from
the time a bit is created (from the original analog signal at Host A) until the
flake is decoded (as part of the analog indicate at Host B)?
host A coverts analog voice to digital chip stream = 64 kbps
host A grouping the $.25 = 56 byte packets
transmission rate = 2 Mbps
propagation delay = .01 sec
consider that start but in a packet. before this bit can be transmitted, all of the bits in the packet must be generated. ways,
= (56 ten 8)/ ( 64×10^3)sec = .007 sec
The fourth dimension required to transmit this packet = (56 x 8) /(2 x 10^half-dozen) = 0.000224 sec
time elapses from the fourth dimension a chip is created until the scrap is decoded
= .007 sec + .000224 sec + .01 sec
= .017224 sec
P9.
Consider the word in Section 1.3 of packet switching versus excursion
switching in which an example is provided with a one Mbps link. Users are
generating data at a charge per unit of 100 kbps when busy, but are busy generating
information but with probability p = 0.1. Suppose that the i Mbps link is replaced
past a 1 Gbps link.
a. What is N, the maximum number of users that tin can exist supported
simultaneously under circuit switching?
The total manual rate = 1 Gbps
Thedata generation rate of each user = 100 kbps
So the maximum number of possible users = 1 Gbps/ 100 Kbps = 10000
b. Now consider package switching and a user population of M users. Give a
formula (in terms of p , Chiliad , Due north ) for the probability that more than than N users are
sending data.
full user population = 1000
number of users transmitting = N
data generation probability of each user , P = 0.1
And then y'all use the binomial distribution formula
= M (sigma) northward=N+1 [MP^north(ane-p)^Thou-n]
P10.
Consider a packet of length 50 which begins at end system A and travels over
iii links to a destination finish organization. These iii links are connected by
2 packet switches. Let di, si, and Ri announce the length, propagation speed,
and the manual rate of link i, for i = 1, 2, 3. The packet switch delays
each packet by dproc. Assuming no queuing delays, in terms of di, si, Ri,
(i = ane,two,iii), and Fifty, what is the total terminate-to-end delay for the parcel? Suppose
at present the packet is ane,500 bytes, the propagation speed on all three links is ii.v ·
108 m/s, the transmission rates of all three links are ii Mbps, the package switch
processing filibuster is iii msec, the length of the first link is 5,000 km, the length
of the second link is 4,000 km, and the length of the terminal link is 1,000 km. For
these values, what is the terminate-to-end delay?
packet length = L
link i Length = di
propagation speed = si
manual rate = Ri
The first finish system requires to transmit the packet onto the start link = L/R1
The first cease system requires to transmit the packet onto the 2d link = L/R2
The first end arrangement requires to transmit the packet onto the third link = 50/R3
The packet propagates over the showtime link = d1/s1
The packet propagates over the 2d link = d2/s2
The bundle propagates over the 3rd link = d3/s3
end to end delay = L/R1+L/R2+L/R3+d1/s1+d2/s2+d3/s3+d(proc)+ d(pro)
packet size = 1500 bytes
The propagation speed on both links = 2.five 10 10^8
manual charge per unit of all three links = 2 Mbps
bundle switch processing filibuster = iii msec
length of the first link = 5000 km
length of the 2d link = 4000km
length of the last link = grand km
the first end system requires to transmit the packet onto the starting time link = L/R1 = .006 sec
packet propagates over the starting time link = .02 sec
the bundle switch requires to transmit the package onto the first link = 50/R1 = .006 sec
packet propagates over the second link = .016 sec
the package switch requires to transmit the packet onto the tertiary link = L/R1 = .006 sec
packet propagates over the third link = .004 sec
end to end delay = .006+.006+.006+.02+.016+.004+.003+.003
= 0.064 sec
P11.
In the above problem, suppose R1 = R2 = R3 = R and dproc = 0. Further suppose
the packet switch does non store-and-forward packets but instead immediately
transmits each bit information technology receives before waiting for the entire bundle to
go far. What is the end-to-end delay?
transmission charge per unit = R1=R2=R3
propagation delay = 0
package =1500 bytes = 1500 x 8 bits
propagation speed = 2.5 x 10^viii m/s
the terminate system requires to transmit the packet onto the link =L/R = (1500×eight)/(2×ten^6)= half dozen ms
length of the starting time link = 5000 km
the packet propagates over the 1st link delay = 5000×10^3 / 2.5×ten^8 = 20 ms
length of 2d link = 4000 km
the package propagates over the 2d link filibuster = 4000×10^three / two.5×10^8 = xvi ms
length of last link = 1000 km
the packet propagates over the third link delay = 1000×10^3 / 2.five×ten^8 = 4 ms
the terminate to stop delay = 6+20+16+4 = 46 ms
P13.
(a) Suppose N packets make it simultaneously to a link at which no packets
are currently being transmitted or queued. Each packet is of length L
and the link has manual rate R. What is the boilerplate queuing delay
for the N packets?
(b) Now suppose that North such packets arrive to the link every LN/R seconds.
What is the average queuing filibuster of a bundle?
P14.
Consider the queuing delay in a router buffer. Let I denote traffic intensity;
that is, I = La/R . Suppose that the queuing delay takes the form IL/R (1 – I )
for I < 1.
a. Provide a formula for the total filibuster, that is, the queuing filibuster plus the
manual filibuster.
b. Plot the total filibuster as a function of L/R .
P15.
Allow a denote the charge per unit of packets arriving at a link in packets/sec, and let μ
denote the link's manual rate in packets/sec. Based on the formula for
the full filibuster (i.due east., the queuing delay plus the transmission filibuster) derived in
the previous problem, derive a formula for the total delay in terms of a and μ .
(L/R)/(1-I)=(L/R)/(1-aL/R)
= (1/μ)/(one-a/μ)
=one/(μ-a)
P18.
Perform a Traceroute betwixt source and destination on the aforementioned continent
at iii unlike hours of the twenty-four hours.
a. Find the boilerplate and standard divergence of the circular-trip delays at each of
the three hours.
three trials the round trip delay between source and the router was
delay at 1st 60 minutes = 1.03 ms
delay at 2nd 60 minutes = .48 ms
delay at 3rd 60 minutes = .45 ms
-boilerplate = (one.03+.48+.45)/3 = .65 ms
-standard deviation = sq rt ((ane/3)[(1.02-.65)^2+(.48-.65)^ii+(.45-.65)^ii])
= sq rt (.0711) = 0.267 ms
b. Find the number of routers in the path at each of the three hours. Did the
paths change during any of the hours?
-the number of routers is nine between source and destination. the paths may accept changed over the corporeality of time.
c. Try to identify the number of Isp networks that the Traceroute packets pass
through from source to destination. Routers with similar names and/or similar
IP addresses should be considered equally part of the same ISP. In your experiments,
do the largest delays occur at the peering interfaces between adjacent ISPs?
-the number of Internet access provider networks are seven
-the largest delays are about probable to occur between the adjacent ISPs.
d. Repeat the above for a source and destination on different continents.
Compare the intra-continent and inter-continent results.
P19.
(a) Visit the site http://www.traceroute.org and perform traceroutes from two different
cities in France to the same destination host in the United states of america. How many
links are the same in the two traceroutes? Is the transatlantic link the same?
(b) Echo (a) just this time cull 1 city in France and another city in
Deutschland.
(c) Pick a city in the United States, and perform traceroutes to two hosts, each
in a different urban center in China. How many links are common in the ii
traceroutes? Do the 2 traceroutes diverge before reaching Cathay?
P20.
Consider the throughput example corresponding to Effigy 1.20(b). Now
suppose that there are G client-server pairs rather than 10. Denote Rs , Rc , and
R for the rates of the server links, customer links, and network link. Assume all
other links have abundant chapters and that at that place is no other traffic in the
network besides the traffic generated by the Yard customer-server pairs. Derive a
general expression for throughput in terms of Rs , Rc , R , and M .
P22.
Consider Figure 1.xix(b). Suppose that each link between the server and the
client has a package loss probability p, and the bundle loss probabilities for
these links are independent. What is the probability that a packet (sent past the
server) is successfully received by the receiver? If a packet is lost in the path
from the server to the client, then the server volition re-transmit the packet. On
boilerplate, how many times will the server re-transmit the packet in guild for
the client to successfully receive the packet?
effigy: server—R1–0–R2–0——-0–RN–client where 0=links
each link between the server and the client has a bundle loss probability=p.
probability of the packet is successfully received by the receiver(p1) = (one-p)^n
so, the server will transmit the packet in order for the client to successfully receive the packet on average = 1/(p1)
the server will re transmit the packet in order for the client to succefully receive the packet on average = (1/(p1))-1
P23.
Consider Effigy 1.19(a). Presume that we know the clogging link along the
path from the server to the client is the outset link with rate Rs bits/sec. Suppose
nosotros transport a pair of packets back to back from the server to the client, and in that location
is no other traffic on this path. Assume each packet of size L bits, and both
links take the same propagation delay dprop.
a. What is the packet inter-arrival time at the destination? That is, how much
time elapses from when the last bit of the first parcel arrives until the terminal
bit of the second packet arrives?
b. Now assume that the 2nd link is the bottleneck link (i.due east., Rc < Rs ). Is it
possible that the 2nd parcel queues at the input queue of the second
link? Explain. Now suppose that the server sends the second packet T seconds
after sending the first bundle. How large must T be to ensure no
queuing before the 2nd link? Explain.
suppose the first packets every bit A and the second package as B.
If the bottleneck link is the first link of the packet A, and the packet B is queued at the kickoff link. Since it is waiting for the transmission of package A. Then, the package inflow time at the destination is Fifty/Rs.
Now we will assume that the second link is the bottleneck link (exampleR c <R s ). Since both packets are sent dorsum to dorsum, it's possible that the 2nd parcel volition get in at the input queue of the second link earlier the second link finishes the transmission of the get-go bundle, which is:
L/Rs+50/Rs+d(prop) < 50/Rs+d(prop)+L/Rc …….(1)
The left side of the equation higher up shows the fourth dimension that is needed by the second packet to arrive at the input queue of the second link. And the right side of the equation shows the time needed by the first package of end its manual onto the 2d link transmission onto the 2d link.
The equation 1 is possible every bit Rc < Rs and it'southward clear that the second packet has to have a queuing delay at the input queue of the second link. If we transport the second packet T sec later, then the delay for the second bundle at the 2d link, and then
(Fifty/Rs)+(50/Rs)+(d(prop))+T < (L/Rs)+d(prop)+(L/Rc)
P24.
Suppose you lot would like to urgently deliver 40 terabytes data from Boston to
Los Angeles. You accept available a 100 Mbps dedicated link for information transfer.
Would you prefer to transmit the data via this link or instead use FedEx overnight
delivery? Explain.
40 terabytes of data is a very large amount of data. But I am given a 100 Mbps link to transfer. So it takes a lot time to transfer the information through the link provided. Also there may be some chances of missing information during this huge manual. So if the information is sent to exist urgently than it would exist better to do FedEx overnight delivery.
P25.
Suppose two hosts, A and B, are separated by xx,000 kilometers and are
connected by a direct link of R = ii Mbps. Suppose the propagation speed
over the link is 2.5 108 meters/sec.
a. Summate the bandwidth-delay product, R d(prop).
propagation filibuster = Distance/speed
= two x 10^7 / 2.five ten 10^8 = 0.08 sec
The transmission rate = two Mbps
So, bandwidth delay production = R x dprop = 2 x x^6 x .08 = 16 x 10^4 $.25
b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose
The file is sent continuously as 1 large bulletin. What is the maximum
number of bits that volition exist in the link at any given time?
file size = 8e5 bits
If the file is being sent continuously equally one message, a link can have the maximum number of bits at the same as ring width filibuster product.
The maximum number of bits at a given time will be sixteen x 10^4 bits
c. Provide an interpretation of the bandwidth-delay product.
Bandwidth delay product is the number of $.25 transmitted per/ sec when propagation delay is one second.
d. What is the width (in meters) of a scrap in the link? Is it longer than a football
field?
propagation speed over the said link = 2.v×10^8 m/s
so 1 fleck takes = (2.five×10^viii)/(ii×10^6) =125 m/flake
i fleck is 125 meters. and so its longer than a football field.
e. Derive a general expression for the width of a bit in terms of the propagation
speed s, the transmission rate R, and the length of the link 1000 .
The width of a bit is related with the propagation speed per bandwidth
then the general expression for the width of bit is = s/R
P27.
Consider trouble P25 but now with a link of R = 1 Gbps.
a. Calculate the bandwidth-delay production, R dprop .
Bandwidth delay product = R x dprop = 10^nine x 0.08 = 80000000 $.25
b. Consider sending a file of 800,000 $.25 from Host A to Host B. Suppose
the file is sent continuously as one large message. What is the maximum
number of $.25 that will be in the link at any given time?
The max number of bits in the link at any given time
= min (bandwidth filibuster production,bundle size)
=(80000000,800000)
and so, eight×10^5 bits can be sent as continuous transmission.
c. What is the width (in meters) of a bit in the link?
width of a bit in the link = due south/R
= 2.5e8/x^9 = 0.25 meters
P31.
In modern packet-switched networks, including the Internet, the source host
segments long, awarding-layer messages (for example, an paradigm or a music
file) into smaller packets and sends the packets into the network. The receiver
so reassembles the packets back into the original message. We refer to this
process as message segmentation . Effigy 1.27 illustrates the end-to-terminate
ship of a message with and without bulletin partitioning. Consider a
message that is 8 · 106 $.25 long that is to be sent from source to destination in
Figure 1.27. Suppose each link in the effigy is ii Mbps. Ignore propagation,
queuing, and processing delays.
figure i.27End-to-end message transport: (a) without message partition
= source—-packet switch(message)—-packet switch—-destination
figure 1.27(b) with message segmentation
= source(package)—-packet switch—-parcel switch—destination
a. Consider sending the message from source to destination without message
segmentation. How long does information technology take to motion the bulletin from the source
host to the first packet switch? Keeping in mind that each switch uses
shop-and-forrard packet switching, what is the total time to move the
message from source host to destination host?
message sent from source ti destination = 8e6
transmission rate = 2 Mbps
time to transport message from the source host to first packet switch = 8e6/2e6 = iv sec
with shop and forrad switching, the total fourth dimension to move message from source host to destination host for 3 links = 4 sec x 3 hops = 12 sec
b. At present suppose that the bulletin is segmented into 800 packets, with each
packet being 10,000 $.25 long. How long does it accept to motility the first
packet from source host to the first switch? When the first packet is being
sent from the starting time switch to the 2nd switch, the 2nd packet is being
sent from the source host to the first switch. At what time will the 2d
parcel be fully received at the beginning switch?
time to send offset parcel from source host to first bundle switch
= (x×10^3)/(2×10^half-dozen) = 0.005 seconds
c. How long does information technology accept to motion the file from source host to destination host
when message division is used? Compare this outcome with your
respond in function (a) and comment.
time at which first packet is received at the destination host = .005 10 3 link = .0015 sec
the fourth dimension at which last packet is received
= .0015 sec + 799×0.001 sec = 4.01 sec
the result is significantly less than it.
d. In addition to reducing filibuster, what are reasons to use message division?
In that location are a lot of skilful reasons to use message segmentation. 1 is that if any failure of delivery message, simply a small portion has to be retransmit, only not the whole bulletin. and other affair is that segmented packet can be transmitting different routes depending on congestion.
e. Discuss the drawbacks of bulletin partitioning.
Drawbacks of message segmentation
-if one segmented packet is missing, then the overall file cannot exist read at all.
-more than bandwidth of overhead
-the full corporeality of header bytes is more
-packets have to be put in sequence at the destination
Source: https://culbertsonj.wordpress.com/homework-1/
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