Assume that we are sending a 30 Mbit MP4 file from a source host to a destination host. All links in the path between source

Assume that we are sending a 30 Mbit MP4 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2* 10" meters/sec, and the distance between source and destination is 10,000 km.
a. Initially suppose there is only one link between source and destination. Also suppose that the entire MP4 file is sent as one packet. How much is the transmission delay?
b. Referring to the above question, how much is the propagation delay?
c. Now suppose there are two links between source and destination, with one router connecting the two links. Each link is 5,000 km long. Again suppose the MP4 file is sent as one packet. Suppose there is no congestion and ignore router processing delays, so that the packet is transmitted onto the second link as soon as the router receives the entire packet. How much is the end-to-end delay?

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  1. (a). The distance between started point and end point is 13.92 miles.

    (b), The direction is 68.9° north - west.

    Explanation:

    Given that,

    You  drive north for 10 miles, turn right and drive east for 5 miles, turn to the north and drive for 3  miles, drive west for 10 miles, and arrive at your destination.

    According to diagram

    (a).We need to calculate the distance

    Using pythagorean theorem

    [tex]EO=\sqrt{(ED)^2+(DO)^2}[/tex]

    Put the value into the formula

    [tex]EO=\sqrt{5^2+(13)^2}[/tex]

    [tex]EO=13.92\ miles[/tex]

    (b). We need to calculate the direction

    Using formula of direction

    [tex]\theta=\tan^{-1}(\dfrac{OD}{DE})[/tex]

    Put the value into the formula

    [tex]\theta=\tan^{-1}(\dfrac{13}{5})[/tex]

    [tex]\theta=68.9^{\circ}[/tex]

    The direction is 68.9° north - west.

    (c). We need to draw a vector diagram

    According to vector diagram,

    The distance between started point and end point will be same.

    Hence, (a). The distance between started point and end point is 13.92 miles.

    (b), The direction is 68.9° north - west.

    [tex]You are going for a drive, but a detour takes you out of the way of your destination. You drive no[/tex]

  2. The mean number of bags that arrive on time to its intended destination is 1260 with a standard deviation of 21.59.

    Step-by-step explanation:

    For each bag, there are only two possible outcomes. Either it arrives on time to it's intended destination, or it does not. The probability of a bag arriving on time is independent of other bags. So we use the binomial probability distribution to solve this question.

    Binomial probability distribution

    Probability of exactly x sucesses on n repeated trials, with p probability.

    The expected value of the binomial distribution is:

    [tex]E(X) = np[/tex]

    The standard deviation of the binomial distribution is:

    [tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

    Luggage checked-in at Airline A arrives on time to its intended destination with a probability of 0.63.

    This means that [tex]p = 0.63[/tex]

    In a random sample of 2000 bags

    This means that [tex]n = 2000[/tex]

    Mean and standard deviation of the number of bags that arrive on time to its intended destination:

    [tex]E(X) = np = 2000*0.63 = 1260[/tex]

    [tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.63*0.37} = 21.59[/tex]

    The mean number of bags that arrive on time to its intended destination is 1260 with a standard deviation of 21.59.

  3. 1)we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Initially suppose there is only one link between source and destination. Also suppose that the entire MP3 file is sent as one packet. The TRANSMISSION DELAY is:

    3 Seconds

    2)

    we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. The END TO END DELAY(transmission delay plus propagation delay) is

    3.05 seconds

    3)

    we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. how many bits will the source have transmitted when the first bit arrives at the destination.

    500,000 bits

    4)

    we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there are two links between source and destination, with one router connecting the two links. Each link is 5,000 km long. Again suppose the MP3 file is sent as one packet. Suppose there is no congestion, so that the packet is transmitted onto the second link as soon as the router receives the entire packet. The end-to-end delay is

    6.1 seconds

    5)

    we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose that the MP3 file is broken into 3 packets, each of 10 Mbits. Ignore headers that may be added to these packets. Also ignore router processing delays. Assuming store and forward packet switching at the router, the total delay is

    4.05 seconds

    6)

    we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 TDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

    30.05 seconds

    7)

    we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 FDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

    30.05 seconds

  4. The net sales revenue is $19,600

    Explanation:

    In order to calculate the net sales revenue we would have to make the following calculation:

    Net Sales Revenue=Sales Revenue-Sales Returns and Allowances-Sales Discounts

    Sales Revenue=$25,000

    Sales Returns and Allowances= $5,000

    Sales Discounts [($25,000-$5,000)*2%]= $400

    Net Sales Revenue=$25,000-$5,000 -$400

    Net Sales Revenue=$19,600

    The net sales revenue is $19,600

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