Review Problems Lecture 4 Delay, Loss and Throughput

Solutions to Review Problems (Lecture 4: Delay, Loss and Throughput)

1.     How long was a bit in the original 802.3 standard in meters? Use a transmission speed of 10 Mbps and assume the propagation speed in coax is 2/3 the speed of light in vacuum.

Solution:

The speed of light in coax is about 200,000 km/sec, which is 200 meters/µsec.

At 10 Mbps, it takes 0.1µsec to transmit a bit.

Thus, the bit lasts 0.1 µsec in time, during which it propagates 20 meters.

Thus, a bit is 20 meters long here.

2.     An image is 1600 × 1200 pixels with 3 bytes/pixel. Assume the image is uncompressed. Determine How long does it take to transmit it over a 56-kbps modem channel?

Solution:

The image is 1600 × 1200 × 3 bytes or 5,760,000 bytes. This is 46,080,000 bits.

At 56,000 bits/sec, it takes about 822.857 sec.

3.     Five equal-size datagrams belonging to the same message leave for the destination one after another. However, they travel through different paths as shown in the below Table.

We assume that the delay for each switch (including waiting and processing) is 3, 10, 20, 7, and 20 ms respectively. Assuming that the propagation speed is 2 x 10⁸ m, find the order the datagrams arrive at the destination and the delay for each. Ignore any other delays in transmission.

Solution:

We assume that the transmission time is negligible in this case. This means that we suppose that all datagrams start at time 0. The arrival time is calculated as:

4.     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_prop, in terms of m and s.

b. Determine the transmission time of the packet, d_trans, in terms of L and R.

c. Ignoring processing and queuing delays, obtain an expression for the end-to-end delay.

d. Suppose Host A begins to transmit the packet at time t = 0. At time t = d_trans, where is the last bit of the packet?

e. Suppose d_prop is greater than d_trans. At time t = d_trans, where is the first bit of the packet?

f. Suppose d_prop is less than d_trans. At time t = d_trans, where is the first bit of the packet?

g. Suppose s = 2.5 × 10⁸, L = 120 bits, and R = 56 kbps. Find the distance m so that d_prop equals d_trans.

5.     How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5×10⁸ m/s, and transmission rate 2 Mbps? More generally, how long does it take a packet of length L to propagate over a link of distance d, propagation speed s, and transmission rate R bps? Does this delay depend on packet length? Does this delay depend on transmission rate?

Solution:

10msec;

d/s;

no;

no

6.     Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of rates R1 = 500 kbps, R2 = 2 Mbps, and R3 = 1 Mbps.

a. Assuming no other traffic in the network, what is the throughput for the file transfer?

b. Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how long will it take to transfer the file to Host B?

c. Repeat (a) and (b), but now with R2 reduced to 100 kbps.

Solution:

a) 500 kbps

b) 64 seconds

c) 100kbps; 320 seconds

7.     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 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 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)?

Solution:

Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires 

The time required to transmit the packet is

 

8.     Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These three links are connected by two packet switches. Let d_i, s_i, and R_i denote the length, propagation speed, and the transmission rate of link i, for i = 1, 2, 3. The packet switch delays each packet by d_proc. Assuming no queuing delays, in terms of d_i, s_i, R_i,(i = 1,2,3), and L, what is the total end-to-end delay for the packet? Suppose now the packet is 1,500 bytes, the propagation speed on all three links is 2.5×10⁸ m/s, the transmission rates of all three links are 2 Mbps, the packet switch processing delay is 3 msec, the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the end-to-end delay?

Solution:

The first end system requires L/R₁ to transmit the packet onto the first link; the packet propagates over the first link in d₁/s₁; the packet switch adds a processing delay of dproc; after receiving the entire packet, the packet switch connecting the first and the second link requires L/R₂ to transmit the packet onto the second link; the packet propagates over the second link in d₂/s₂. Similarly, we can find the delay caused by the second switch and the third link: L/R₃, d_proc, and d₃/s₃.

Adding these five delays gives

d_end-end = L/R₁ + L/R₂ + L/R₃ + d₁/s₁ + d₂/s₂ + d₃/s₃+ d_proc+ d_proc

To answer the second question, we simply plug the values into the equation to get:

6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec.

9.     Consider an image of dimensions 800 × 600 pixels with 8 bytes/pixel 1280 × 800 pixels with 4 bytes/pixel which begins at end server 0 and travels over three paths (3G, FTTH, and DSL) to three destinations (Tablet PC 0, PC 0, and Laptop 0) as shown in the following figure.

a.     Determine the throughput for the image transfer for each path: 3G, FTTH, and DSL.

b.     Calculate how long it will take to transfer the image to each destination: Tablet PC 0, PC 0, and Laptop 0.

Solution:

The image is 800 × 600 × 8 bytes = 3,840,000 bytes =  30,720,000 bits.

·       Throughput of 3G path = 42 Mbps

·       Throughput of FTTH path = 100 Mbps

·       Throughput of DSL path = 12 Mbps

·       The time it takes to transfer the image to Tablet PC0 = 30,720,000 bits/42*10⁶=  0.73 seconds

·       The time it takes to transfer the image to PC0 = 30,720,000 bits/100*10⁶= 0.3072 seconds

·       The time it takes to transfer the image to lap top 0 = 30,720,000 bits/12*10⁶=  2.56 seconds