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The assignment content consists of three questions related to computer networks. The first question asks to calculate the bandwidth required for a LAN with a maximum distance of 2 km and packet sizes of 512 bytes and 2000 bytes. The second question calculates the minimum round-trip time (RTT) for a point-to-point link between the earth and a new lunar colony, and then uses this RTT to compute the delay x bandwidth product. Finally, the third question analyzes the transmission time required to send 10,000 bits from host A to host B through a switch via two 10-Mbps links.

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CIS 527 – COMPUTER NETWORKS

SOLUTION – HOMEWORK #1

1. (10 points) Consider a LAN with a maximum distance of 2 km. At what bandwidth

would propagation delay (at a speed of 2 x 108 m/s) equal transmit delay (insertion

delay) for 512 byte packets? What about 2000 byte packets?

Solution:

Case (a): Packet Size = 512 bytes

Maximum Distance = 2 km

Speed of Light = 2 x 108 m/s

Propagation = Distance / Speed of Light

= 2000 m / 2 x 108 m/s

Transmit = Size / Bandwidth

= 512 x 8 bits / Bandwidth

Therefore,

Bandwidth = Size x Speed of Light / Distance

= (512 x 8) bits x 2 x 108 m/s / 2000 m

= 4096 x 105 bits / sec

= 409.6 Mbits/sec

Case (b): Packet Size = 2000 bytes

Maximum Distance = 2 km

Speed of Light = 2 x 108 m/s

Propagation = Distance / Speed of Light

= 2000 m / 2 x 108 m/s

Transmit = Size / Bandwidth

= 2000 x 8 bits / Bandwidth

Therefore,

Bandwidth = Size x Speed of Light / Distance

= (2000 x 8) bits x 2 x 108 m/s / 2000 m

= 16000 x 105 bits / sec

= 1600 Mbits/sec

SOLUTION – HOMEWORK #1

1. (10 points) Consider a LAN with a maximum distance of 2 km. At what bandwidth

would propagation delay (at a speed of 2 x 108 m/s) equal transmit delay (insertion

delay) for 512 byte packets? What about 2000 byte packets?

Solution:

Case (a): Packet Size = 512 bytes

Maximum Distance = 2 km

Speed of Light = 2 x 108 m/s

Propagation = Distance / Speed of Light

= 2000 m / 2 x 108 m/s

Transmit = Size / Bandwidth

= 512 x 8 bits / Bandwidth

Therefore,

Bandwidth = Size x Speed of Light / Distance

= (512 x 8) bits x 2 x 108 m/s / 2000 m

= 4096 x 105 bits / sec

= 409.6 Mbits/sec

Case (b): Packet Size = 2000 bytes

Maximum Distance = 2 km

Speed of Light = 2 x 108 m/s

Propagation = Distance / Speed of Light

= 2000 m / 2 x 108 m/s

Transmit = Size / Bandwidth

= 2000 x 8 bits / Bandwidth

Therefore,

Bandwidth = Size x Speed of Light / Distance

= (2000 x 8) bits x 2 x 108 m/s / 2000 m

= 16000 x 105 bits / sec

= 1600 Mbits/sec

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2. (20 points) Suppose a 100-Mbps point-to-point link is being set up between the earth

a new lunar colony. The distant from the moon to the earth is approximately 385,000

km, and data travels over the link at the speed of light – 3 x 108 m/s.

Solution:

a) Calculate the minimum RTT for the link

Minimum RTT = 2 x Propagation

Propagation = Distance / Speed of Light

= 2 x 385000 km / 3 x 108 m/s

= 2 x 385000000 m / 3 x 108 m/s

= 2 x 385 / 300 sec

= 2.57 sec

b) Using the RTT as the delay, calculate the delay x bandwidth product for the

link.

Delay x Bandwidth = 2.57 sec x 100 Mbits/sec

= 257 Mbits

= 257/8 MB

= 32MB

c) What is the significance of the delay x bandwidth product computed in (b)?

This represents the amount of data the sender can send before it would be

possible to receive a response.

d) A camera on the lunar base takes pictures of the earth and saves them in

digital format to disk. Suppose Mission Control on earth wishes to

download the most current images, which is 30MB. What is the minimum

amount of time that will elapse between the request for the data goes out

and the transfer is finished?

We require at least one RTT before the picture could begin arriving at the ground

(TCP would take two RTTs). Assuming bandwidth delay only, it would then take

= 30MB/100Mbps

= (30 x 8 Mbits) / 100Mbps

= 240Mbits / 100 Mbits/sec

= 2.4 sec to finish sending,

for a total time of 2.4 + 2.57 = 4.97 sec until the last picture bit arrives on earth.

a new lunar colony. The distant from the moon to the earth is approximately 385,000

km, and data travels over the link at the speed of light – 3 x 108 m/s.

Solution:

a) Calculate the minimum RTT for the link

Minimum RTT = 2 x Propagation

Propagation = Distance / Speed of Light

= 2 x 385000 km / 3 x 108 m/s

= 2 x 385000000 m / 3 x 108 m/s

= 2 x 385 / 300 sec

= 2.57 sec

b) Using the RTT as the delay, calculate the delay x bandwidth product for the

link.

Delay x Bandwidth = 2.57 sec x 100 Mbits/sec

= 257 Mbits

= 257/8 MB

= 32MB

c) What is the significance of the delay x bandwidth product computed in (b)?

This represents the amount of data the sender can send before it would be

possible to receive a response.

d) A camera on the lunar base takes pictures of the earth and saves them in

digital format to disk. Suppose Mission Control on earth wishes to

download the most current images, which is 30MB. What is the minimum

amount of time that will elapse between the request for the data goes out

and the transfer is finished?

We require at least one RTT before the picture could begin arriving at the ground

(TCP would take two RTTs). Assuming bandwidth delay only, it would then take

= 30MB/100Mbps

= (30 x 8 Mbits) / 100Mbps

= 240Mbits / 100 Mbits/sec

= 2.4 sec to finish sending,

for a total time of 2.4 + 2.57 = 4.97 sec until the last picture bit arrives on earth.

3. (20 points) In the following figure, hosts A and B are each connected to a switch via

10-Mbps links. The propagation delay on each link is 40μs. S is a store-and-forward

device; it begins retransmitting a received packet 20μs after it has finished received

it. Calculate the total time required to transmit 10,000 bits from A to B

(a) as a single packet

(b) as two 5,000-bit packets sent one right after the other.

Solution:

Case (a): as a single packet

Per link Transmit Delay = Size / Bandwidth

=104 bits / 10 x 106 bits/sec

= 1000 s

Total Transmission Time = (2 x 1000 + 2 x 40 + 20) s

= (2000 + 80 + 20) s

= 2100 s

Case (b): as two 5,000-bit packets sent one right after the other

When sending as two packets, here is a table of times for various events:

Transmit delay = 5000 bits/ 10 x 106 bits/sec

= 500s

T=0 start

T=500 A finishes sending packet 1, starts packet 2

T=540 packet 1 finishes arriving at S

T=560 packet 1 departs for B

T=1000 A finishes sending packet 2

T=1060 packet 2 departs for B

T=1100 bit 1 of packet 2 arrives at B

T=1600 last bit of packet 2 arrives at B

Expressed algebraically,

Transmit delay = 5000 bits/ 10 x 106 bits/sec

= 500s

Total Transmit time = (3 x 500 + 2 x 40 + 1 x 20) s

= (1500 + 80 + 20) s

= 1600 s

Hence, Small packets are faster.

10-Mbps links. The propagation delay on each link is 40μs. S is a store-and-forward

device; it begins retransmitting a received packet 20μs after it has finished received

it. Calculate the total time required to transmit 10,000 bits from A to B

(a) as a single packet

(b) as two 5,000-bit packets sent one right after the other.

Solution:

Case (a): as a single packet

Per link Transmit Delay = Size / Bandwidth

=104 bits / 10 x 106 bits/sec

= 1000 s

Total Transmission Time = (2 x 1000 + 2 x 40 + 20) s

= (2000 + 80 + 20) s

= 2100 s

Case (b): as two 5,000-bit packets sent one right after the other

When sending as two packets, here is a table of times for various events:

Transmit delay = 5000 bits/ 10 x 106 bits/sec

= 500s

T=0 start

T=500 A finishes sending packet 1, starts packet 2

T=540 packet 1 finishes arriving at S

T=560 packet 1 departs for B

T=1000 A finishes sending packet 2

T=1060 packet 2 departs for B

T=1100 bit 1 of packet 2 arrives at B

T=1600 last bit of packet 2 arrives at B

Expressed algebraically,

Transmit delay = 5000 bits/ 10 x 106 bits/sec

= 500s

Total Transmit time = (3 x 500 + 2 x 40 + 1 x 20) s

= (1500 + 80 + 20) s

= 1600 s

Hence, Small packets are faster.

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