Problems
1. For a 5 Kbytes packets sent over a 10 Mbps
transmission link, what is the transmission time of the
packet?
(5K x 8) / 10M = 4 msec
2. A 600-byte packet is sent over a 20 Kbps point-to-point
link whose propagation delay is 10 msec. After how much
delay will the packet reach the destination?
10 msec + (600 x 8) / 20K = 250 msec
3. Given a 1 Gbps point to point copper wire (propagation
speed 200m/μs) with a length of 2km. Ignoring the
acknowledgments, a) how much is the transmission time for
a 10 MB file?
Transmission time=
Data Length/Data Rate=L/R=
(10*10^6*8) /(1*10^9)=0.08secb)
how much is the propagation delay?
Propagation delay=Distance/propagation=D/V
=(2*10^3)/(200/10^-6)=10μsec
4 .What is the throughput when retrieving a 1MB file across a
1Gbps network with a round trip time of 100msec, again
disregarding ACKs?
Throughput =TransferSize/Transfer time
Transfer time =RTT+(1/Bandwidth)*Transfer Size
= (100*10^3)+(1/10^9)*(8*10^6)=108msec
Throughput=(8*10^6)/(108*10^-3)=74.074 Mbps
5. Consider a point-to-point link 2 km in length. At what
bandwidth would propagation delay (at a speed of 2 × 108
m/s) equal transmit delay for 100-byte packets? What about
512-byte packets?
Solution: Since transmit = size/bandwidth and propagation =
distance/speedoflight = 2*103 m / (2*108 m/s)
= 1*10-5 s, we can conclude that bandwidth = size /
(distance/speed of light).
100 byte / 1*10-5 s = 10 MB/s = 80 Mbps
512 byte / 1*10-5s = 409.6 Mbp
6. How “wide” is a bit on a 1-Gbps link? How long is a bit in
copper wire, where the speed wide” of propagation is
2.3*108m/s?
Solution:
1Gbps = 109bps, and 1 bit on a 1-Gbps link is 1/109bps= 109s wide.
If the speed of propagation is 2.3*108 m/s, then the length of a
bit is (2.3*108 m/s) * (10-9 s) =0.23m
7. Suppose a 100-Mbps point-to-point link is being set up between Earth and a new lunar
colony. The distance from the moon to Earth is 385000 km and data travels over the link
at colony. the speed 3 x 108m/s.
(a) Calculate the minimum RTT for the link.
(b) Using the RTT as the delay, calculate the delay*bandwidth product for the link.
(c) What is the significance of the delay*bandwidth product computed in (b).
(d) A camera on the lunar base takes pictures of Earth and saves them in digital format to
disk. Suppose Mission Control on Earth wishes to download the most current image,
which is 25MB. What is the minimum amount of time that will elapse between when the
request for the data goes out and the transfer is finished?
Solution:
(a) RTT = 2*distance/speed = 2*3.85*108m / (3*108m/s) = 2.567s
(b) Delay*bandwidth = 2.567s*100Mbps = 256.7Mb = 32.09 MB
(c) The delay*bandwidth product corresponds to how many bits the sender must transmit
before the first bit arrives at the receiver.
(d) Total transfer time = RTT + transmit time = 2.567s + size/bandwidth = 2.567s +
(25*8/100) s =4.567s
8. Calculate the latency (from first bit sent to last bit received) for the following:
(a) A 10-Mbps Ethernet with a single store-and-forward switch in the path, and
a packet size of 5000 bits. Assume that each link introduces a propagation
delay of 10 μs, and that the switch begins retransmitting immediately after it
has finished receiving the packet.
(b) Same as (a) but with 3 switches.
(c) Same as (a) but assume the switch implements “cut-through” switching: it
is able to begin cut-through” retransmitting the packet after the first 200 bits
have been received.
Solution:
(a) Each line has a transmit delay = size / bandwidth = 5000 bits / 10 Mbps =
5*10-4 s. There are 2 links, each produces 2*10*10-6 s = 2*10-5s. So the
latency is 2*transmit delay + propagation delay = 1.02*10-3 s.
(b) 3 switches have 4 lines. The latency is 4*transmit delay + 4*propagation
delay = 2.04*10-3 s.
(c) With “cut-through” switching, the transmit delay of the first line is 200 bits /
10 Mbps = 2*10-5s. So the latency = 2*propagation delay + first transmit delay
+ second transmit delay = 2*10-5s +2*10-5 s + 5*10-4 s = 5.4*10-4 s.
9. Suppose that a certain communications protocol involves a per-packet
overhead of 100 bytes for headers and framing. We send 1 million bytes of
data using this protocol; however, one data type is corrupted and the entire
packet containing it is thus lost. Give the total lost. number of overhead + loss
bytes for packet data sizes of 1000, 5000, 10000, and 20000 bytes. Which size
is optimal?
Solution:
Packet number = 1*106 / Data Size.
Total overhead = 100 bytes * Packet number = 100*(1*106 / Data Size)
Lose = data size
Total size = total overhead + lose = 100*(1*106 / Data Size) + data size
1000 bytes: 1*108 / 1000 + 1000 =1.01*105
5000 bytes: 1*108 / 5000 + 5000 = 2.5*104
10000 bytes: 1*108 / 10000 + 10000 =2*104
20000 bytes: 1*108 / 20000 + 20000 =2.5*104
The 10000 bytes is optimal.
10. Assume you wish to transfer an n-byte file along a path composed of the
source, destination, seven point-to-point links, and five switches. Suppose
each link has a propagation delay of 2 ms, bandwidth of 4 Mbps, and that the
switches support both circuit and packet switching. Thus you can either break
the file up into 1-KB packets, or set up a circuit through the switches and send
the file as one contiguous bit stream. Suppose that packets have 24 bytes of
packet header information and 1000 bytes of payload, that store-and-forward
packet processing at each switch incurs a 1-ms delay after the packet has
been completely received, that packets may be sent continuously without
waiting for acknowledgments, and that circuit setup requires a 1-KB message
to make one round-trip on the path incurring a 1-ms delay at each switch after
the message has been completely received. Assume switches introduce no
delay to data traversing a circuit. You may also assume that file size is a
multiple of 1000 bytes.
(a) For what file size n bytes is the total number of bytes sent across the network
less for circuits than for packets?
(b) For what file size n bytes is the total latency incurred before the entire file
arrives at the destination less for circuits than for packets?
Solution:
A) Packet switching: packets pay an ongoing per packet cost
of 24 Bytes for a total count of 1024 x (number of packets) =
1024 * n / 1000 bytes
Circuit switching: circuits pay an up-front penalty of 1KB being
sent across the network twice for a total data count of 2000
+ n bytes
So the question really asks how many packet headers does it
take to exceed 2000 bytes, which is ~ 83.34.
Thus, for files longer than ~ 83,340 bytes, using packets
results in more total data sent.
B Packet switching:
Total transfer latency = (transmit delay for all pkts at the source)+
(per-pkt transmit delay introduced by each switch)+ (per-pkt
processing delay introduced by each switch)+ (propagation delay
for all links)
= (c*t) + (x*t) + (x*0.001) + ((x+1) * 0.001)
= 2.731*10-6n + 0.01992
c: number of packets = n/1000
t: transmission time of a packet = 1024*8bit/3Mbps
x: number of switches = 4
b: bandwidth = 3Mbps
Circuit switching:
Total transfer latency = (transmit delay for the whole file at
source)+(propagation delay for all links)+ (setup time for the
circuit)
= 8n/b + ((x+1)*0.001)+ 2*(1000*8/b +
(x*1000*8/b) + (x*0.001) + ((x+1)*0.001))
= 2.667*10-6n + 0.04967seconds
2.731*10-6n + 0.01992 > 2.667*10-6n + 0.04967
Solving
=> n > 465KB (rounding to the next 1000 bytes)
11. Consider a point-to-point link 2km in length. At what
bandwidth would propagation delay (at a speed of 2 x
108m/s) equal transmit time for 100-byte packets? What
about 512-byte packets?-Propagation delay=D/V=
(2*10^3)/(2*10^8)=10^-5 secTransmission time for 100
bytespackets=L/R=(100*8)/R=10^-5, so R= 80Mbps
Transmission time for 512
bytespackets=L/R=(512*8)/R=10^-5, R= 409.6Mbps