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Suppose you would like to urgently deliver 40 terabytes data from Boston to Los Angeles. You have available a 100 Mbps dedicated link for data transfer. Would you prefer to transmit the data via this link or instead use FedEx overnight delivery? Explain.

Short Answer

Expert verified
Using FedEx is faster, as it takes one day versus approximately 40.74 days over the internet.

Step by step solution

01

Convert Data to Bits

First, we need to convert the data size from terabytes to bits because our data transfer rate is in bits per second. We have 40 terabytes of data. Since 1 byte = 8 bits, 1 terabyte = 1,024 gigabytes = 1,024 * 1,024 megabytes = 1,024 * 1,024 * 1,024 kilobytes = 1,024 * 1,024 * 1,024 * 1,024 bytes. Therefore, 40 terabytes = 40 * 1,024 * 1,024 * 1,024 * 1,024 * 8 bits.
02

Calculate Total Data in Bits

Calculate the total number of bits: 40 terabytes = 40 * 1,024^4 * 8 bits. This equals 40 * 2^{40} * 8 bits, which is 3.52 * 10^{14} bits.
03

Calculate Transfer Time Over the Link

Using a 100 Mbps link, calculate the time it takes to transfer the data. The transfer rate is 100 megabits per second = 100 * 10^6 bits/second. To find out how long it will take, divide the total number of bits by the transfer rate: \[ \text{Time} = \frac{3.52 \times 10^{14} \text{ bits}}{100 \times 10^{6} \text{ bits/second}} = 3.52 \times 10^{6} \text{ seconds} \]
04

Convert Time to Hours and Days

Convert the time from seconds to hours and then to days. There are 3,600 seconds in an hour and 24 hours in a day: \[ \text{Hours} = \frac{3.52 \times 10^{6}}{3,600} \approx 977.78 \text{ hours} \] \[ \text{Days} = \frac{977.78}{24} \approx 40.74 \text{ days} \]
05

Compare with FedEx Overnight Delivery

FedEx overnight can deliver your data storage device the next day, assuming it takes one day for delivery. Compare this with the 40.74 days required for the data transfer over the internet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bandwidth Calculation
When discussing data transfer, bandwidth is a core concept. Bandwidth refers to the volume of data that can be transmitted over a network in a given amount of time. It is usually expressed in bits per second (bps), such as megabits per second (Mbps) or gigabits per second (Gbps). In our example, we have a link with a bandwidth of 100 Mbps.

To calculate the bandwidth needed for a data transfer, it's crucial to convert all data measurements to the same unit. For instance, to send 40 terabytes of data, we first convert this amount into bits. Here, 40 terabytes is equivalent to 3.52 x 1014 bits. Understanding this conversion is key in assessing how much bandwidth is required and how long a transfer will take.

Knowing the total number of bits and the link's bandwidth allows us to estimate the transfer time. For example, with a 100 Mbps link, we divide the total bits by our bandwidth to find the time in seconds. This calculation is vital for planning and understanding data transfers.
Data Transmission Speed
The speed at which data is transmitted over a network is distinct yet related to bandwidth. Data transmission speed measures how fast data travels from one point to another. While bandwidth gives the maximum data rate, actual data transmission speed can be influenced by several factors, including network congestion and signal quality.

In scenarios like ours, transmitting a large data set such as 40 terabytes over a 100 Mbps link takes considerable time. The calculation for transmission speed involves dividing the total data in bits (3.52 x 1014 bits) by the available bandwidth (100 Mbps), to get the time frame in seconds. This gives us a clue as to whether internet-based transfer is efficient compared to other methods like physical delivery.
  • Higher bandwidth generally equates to faster transmission speed.
  • Transmission speed might be lower than bandwidth due to network conditions.
Calculating transmission speed provides insights on the feasibility of different data transfer options.
Network Delay
Network delay, or latency, is the time it takes for data to travel from the source to the destination across a network. Network delay affects the overall time taken for a data transfer to complete. It is comprised of several components: propagation delay, transmission delay, queuing delay, and processing delay.

Propagation delay depends on the distance that data must travel, and transmission delay relates to the size of the data packet and the speed of the link. Queuing delay occurs when data packets wait in line to be sent due to traffic on the network, while processing delay refers to the time routers take to process packet headers.

In our scenario, although a 100 Mbps link can maintain a constant data rate, actual delays can occur due to these factors. A delay can influence the decision to opt for an alternative such as overnight delivery, which avoids network conditions entirely.
  • Latency is affected by distance—the farther the data travels, the more delay.
  • Delays can bottleneck even high-bandwidth connections.
  • Avoiding network delays would mean seeking non-digital alternatives for urgent transfers.
Network delay is an important factor to consider when planning large-scale data transfers.

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Most popular questions from this chapter

This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate \(R\) bps. Suppose that the two hosts are separated by \(m\) meters, and suppose the propagation speed along the link is \(s\) meters/sec. Host A is to send a packet of size \(L\) bits to Host B. a. Express the propagation delay, \(d_{\text {prop }}\), in terms of \(m\) and \(s\). b. Determine the transmission time of the packet, \(d_{\text {trans }}\), in terms of \(L\) and \(R\). c. Ignoring processing and queuing delays, obtain an expression for the endto- end delay. d. Suppose Host A begins to transmit the packet at time \(t=0\). At time \(t=d_{\text {trans }}\). where is the last bit of the packet? e. Suppose \(d_{\text {prop }}\) is greater than \(d_{\text {trans }} .\) At time \(t=d_{\text {trans }}\), where is the first bit of the packet? f. Suppose \(d_{\text {prop }}\) is less than \(d_{\text {trans }} .\) At time \(t=d_{\text {trans }}\), where is the first bit of the packet? g. Suppose \(s=2.5 \cdot 10^{8}, L=120\) bits, and \(R=56 \mathrm{kbps}\). Find the distance \(m\) so that \(d_{\text {prop }}\) equals \(d_{\text {trans }}^{-}\)

In this problem, we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP). Host A converts analog voice to a digital \(64 \mathrm{kbps}\) bit stream on the fly. Host A then groups the bits into 56 -byte packets. There is one link between Hosts A and B; its transmission rate is 2 Mbps and its propagation delay is \(10 \mathrm{msec}\). As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet's bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)?

(a) Suppose \(N\) packets arrive simultaneously to a link at which no packets are currently being transmitted or queued. Each packet is of length \(L\) and the link has transmission rate \(R\). What is the average queuing delay for the \(N\) packets? (b) Now suppose that \(N\) such packets arrive to the link every \(L N / R\) seconds. What is the average queuing delay of a packet?

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Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between the switch and the receiving host are \(R_{1}\) and \(R_{2}\), respectively. Assuming that the switch uses store-and-forward packet switching, what is the total end-to-end delay to send a packet of length \(L ?\) (Ignore queuing, propagation delay, and processing delay.)

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