Do only #2. Problems Related to Propagation Delay and Transmission Time 1 Proble

Question

Do only #2.

Problems Related to Propagation Delay and Transmission Time 1 Problem 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 I. 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? 2 Problem The New Horizon spacecraft that recently flew past Pluto 1 has a high resolution camera LORRI (Long Range Reconnaissance Imager) which has a 1024 pixel by 1024 pixel resolution with each pixel being 12 bits. After applying some compression each image is approximately 2.5 Mb (Megabits). The distance to Pluto is 7.5 billion kilometers and the speed of light is 3 108 meters per second 1. Suppose the data rate of a direct link between New Horizon and an Earth receiver is 1200 bps. What is the time to receive one image? 2. Mars is 225 million kilometers from Earth. The Mars Atmosphere and Volatile EvolutioN (MAVEN)2 is a spacecraft that is currently orbiting Mars. We can achieve a data rate of 1 Mbps between New Horizon and MAVEN as well as between MAVEN and the Earth receiver. Suppose we deploy an intermediate router in MAVEN. If an entire image is put in a packet with an header that is 100 bytes what is the time to receive an image at the Earth receiver? 3. If the image is partitioned into 1000 parts. What is the time to receive an image? 4. If the the data rate between MAVEN and Earth receiver is doubled3 What will be the time to receive an image? https://www.nasa.gov/mission_pages/newhorizons/main/index.html http://www.nasa.gov/mission_pages/maven/main/index.htm.l This is achieved when Mars moves closer to sun so there is more power generated by the solar panels and hence more transmit power

Explanation / Answer

Hello there,

* As you mentioned to solve only 2nd question, i directly come to that question. In every subpart they asked about time to receive an image, which will be equal to total of transmission time and propagation time.

* So first come to transmission time, which is basically the time to load every bit of message(in our case an image) over physical channel.

Transmission time = packet size / bandwidth of link

* And,Propagation time is the time to travel every bit from sender to receiver through physical medium.

Propagation time = distance / propagation speed in medium

1) time to receive an image is = transmission time + propagation time

= (2.5 Mb/1200 bps) + (7.5 billion Km / 3*108 meterper second)

As per the both formulaes, I put the values to each required field,As you can see size of an image is 2.5 Mb and bandwidth is 1200 bps. Distance between spacecraft and receiver is 7.5 billion km and speed is 3*108 which is speed of light, now you can easily calculate the answer by calculating this above equation. and same i will make equation to othe part also.

2) in this case distance is added i.e. 225 million km because we setup an intermediate router between new horizon and earth receiver,and now data rate is 1 Mbps and now added one header of size 100 bytes, so packet size will be (2.5 Mb +100 Bytes) i.e. (2.5 Mb + 800 bits) as 1 bytes = 8 bits.

Now, time to receive an image = (3.3 Mb/1 Mbps) + (7.425 billion km / 3*108 meter per second)

3) in this image is divided into 1000 parts(if i consider original image size 2.5 Mb) so each image size is now, 2.5 kb so replace this with original and multiply by 1000 so this will not affect because we have same band width through out the channel.

4) if the data rate got doubled between maven and earth receiver then we have to make two equation for transmission time, one for new horizon to maven with 1 Mbps and from maven to earth 2 Mbps.

so, time to receive an image = [{(3.3 Mb/1 Mbps) + (7.2 billion km / 3*108 meter per second)} + {(3.3 Mb/2 Mbps) + (225 million km / 3*108 meter per second)} ]

Hope, You got your solution, feel free to ask any queries and to give feedback.

Thank you.

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