Assignment 1 Soln
ECS 152A Fall 2013: Assignment 1 - Solution
Date Assigned: 10/2/2013
Date Due: 10/9/2013
Problem 1
Consider two hosts A and B connected by a single link of rate R bps. Suppose the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Suppose host A begins to transmit a packet of size L bits at time t = 0.
1. Let dtrans denote the time to transmit the packet. At time t = dtrans where is the last bit of the packet?
2. Let dprop denote the propagation delay. Suppose dprop is greater than dtrans . At time t = dtrans where is the first bit of the packet?
3. Suppose s = 2.5 × 108 , L = 120 bits, and R = 56 kbps. Find the distance m such that dprop is equal to dtrans .
Answers
1. The bit has just left Host A.
2. The first bit is in the link and has not reached Host B.
3. dprop = m/s and dtrans = L/R. Setting dprop = dtrans , we have m = we get m = 536 Km.
L
R
s. Incorporating the given values,
Problem 2
Suppose two hosts A and B are separated by 20,000 kilometers and are connected by a direct link of rate R = 2
Mbps. Suppose the propagation speed over the link is 2.5 × 108 meters/sec.
1. Calculate the bandwidth delay product, R × dprop .
2. Consider sending a file of size 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?
3. Provide an interpretation of the bandwidth delay product.
4. What is the width (in meters) of a bit in the link? Is it longer than a football field?
1
Answers
1. BDP = R × dprop = 160, 000 bits.
2. 160,000 bits
3. The bandwidth-delay product of a link is the maximum number of bits that can be in the link.
4. The width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field
Problem 3
Consider sending a large file of F bits from Host A to Host B. There are three links (and
Date Assigned: 10/2/2013
Date Due: 10/9/2013
Problem 1
Consider two hosts A and B connected by a single link of rate R bps. Suppose the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Suppose host A begins to transmit a packet of size L bits at time t = 0.
1. Let dtrans denote the time to transmit the packet. At time t = dtrans where is the last bit of the packet?
2. Let dprop denote the propagation delay. Suppose dprop is greater than dtrans . At time t = dtrans where is the first bit of the packet?
3. Suppose s = 2.5 × 108 , L = 120 bits, and R = 56 kbps. Find the distance m such that dprop is equal to dtrans .
Answers
1. The bit has just left Host A.
2. The first bit is in the link and has not reached Host B.
3. dprop = m/s and dtrans = L/R. Setting dprop = dtrans , we have m = we get m = 536 Km.
L
R
s. Incorporating the given values,
Problem 2
Suppose two hosts A and B are separated by 20,000 kilometers and are connected by a direct link of rate R = 2
Mbps. Suppose the propagation speed over the link is 2.5 × 108 meters/sec.
1. Calculate the bandwidth delay product, R × dprop .
2. Consider sending a file of size 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?
3. Provide an interpretation of the bandwidth delay product.
4. What is the width (in meters) of a bit in the link? Is it longer than a football field?
1
Answers
1. BDP = R × dprop = 160, 000 bits.
2. 160,000 bits
3. The bandwidth-delay product of a link is the maximum number of bits that can be in the link.
4. The width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field
Problem 3
Consider sending a large file of F bits from Host A to Host B. There are three links (and