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pogonyaev
2 years ago
9

Many real-time video applications run over UDP rather than TCP because they cannot tolerate retransmission delays. However, this

means video applications are not constrained by TCP’s congestion-control algorithm. What impact does this have on TCP traffic? Be specific about the consequences
Computers and Technology
1 answer:
alexdok [17]2 years ago
4 0

Answer:

Video applications typically run over UDP rather than TCP because they cannot tolerate  Re-transmission delays. However, this means video applications are not constrained by  TCP’s congestion control algorithms.

(a). What impact does this have on TCP traffic? Be specific about the consequences.

Fortunately, these video applications often use RTP, which results in RTCP “receiver  reports” being sent from the sink back to the source. These reports are sent periodically  (e.g. once a second) and include the percentage of packets successfully received in the  last reporting period.  

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Suppose that the data mining task is to cluster points (with (x, y) representing location) into three clusters, where the points
solong [7]

Answer:

Explanation:

K- is the working procedure:

It takes n no. of predefined cluster as input and data points.

It also randomly initiate n centers of the clusters.

In this case the initial centers are given.

Steps you can follow

Step 1. Find distance of each data points from each centers.

Step 2. Assign each data point to the cluster with whose center is nearest to this data point.

Step 3. After assigning all data points calculate center of the cluster by taking mean of data points in cluster.

repeat above steps until the center in previous iteration and next iteration become same.

A1(4,8), A2(2, 4), A3(1, 7), B1(5, 4), B2(5,7), B3(6, 6), C1(3, 7), C2(7,8)

Centers are X1=A1, X2=B1, X3=C1

A1 will be assigned to cluster1, B1 will be assigned to cluster2 ,C1 will be assigned to cluster3.

Go through the attachment for the solution.

5 0
2 years ago
In this problem we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP). Host A converts
Minchanka [31]

Answer:

<u>The total time elapsed 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 is </u><u>25.11 ms</u>

Explanation:

Host A first converts the analog signal to a digital 64kbps stream and then groups it into 56-byte packets. The time taken for this can be calculated as:

time taken 1= \frac{Packet Size in Bits}{Bit Rate}

                 = (56 x 8) bits / 64 x 10³ bits/s

                 = 7 x 10⁻³s

time taken 1= 7 ms

The transmission rate of the packet from Host A to Host B is 4 Mbps. The time taken to transfer the packets can be calculated as:

time taken 2= (56 x 8) bits / 4 x 10⁶ bits/s

                    = 1.12 x 10⁻⁴ s

time taken 2= 112 μs

The propagation delay is 18 ms.

To calculate the total time elapsed, we need to add up all the time taken at each individual stage.

<u />Time_{total}<u> = Time taken 1 + Time taken 2 + Propagation Delay</u>

                 = 7 ms + 112 μs + 18 ms

                 = 0.025112 s

Time_{total} = 25.11 ms

5 0
2 years ago
Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac
Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

  = 15 bits/sec

10 independent connections are established, during parallel downloads,  and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required  is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation  of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

               = 0.03 micro seconds

The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

4 0
2 years ago
Write a program that creates a dictionary containing the U.S. states as keys and their capitals as values. (Use the Internet to
Dima020 [189]

Answer:

  1. import random  
  2. states = {
  3.    "Alabama": "Montgomery",
  4.    "California": "Sacramento",
  5.    "Florida": "Tallahassee",
  6.    "Hawaii": "Honolulu",
  7.    "Indiana": "Indianapolis",
  8.    "Michigan": "Lansing",
  9.    "New York": "Albany",
  10.    "Texas" : "Austin",
  11.    "Utah" : "Salt Lake City",
  12.    "Wisconsin": "Madison"
  13. }
  14. correct = 0
  15. wrong = 0
  16. round = 1
  17. while(round <= 5):
  18.    current_state = random.choice(list(states))
  19.    answer = input("What is the capital of " + current_state + ": ")
  20.    
  21.    if(answer == states[current_state]):
  22.        correct += 1
  23.    else:
  24.        wrong += 1
  25.    
  26.    round += 1
  27. print("Correct answer: " + str(correct))
  28. print("Wrong answer: " + str(wrong))

Explanation:

The solution code is written in Python 3.

Line 3 -14

Create a dictionary of US States with capital as each of their corresponding value. Please note only ten sample states are chosen here.

Line 16 - 18

Create variables to track the number of correct and inaccurate response and also round counter.

Line 19 - 28

Set the while condition to enable user to play the quiz for five questions and use random.choice to randomly pick a state from the dictionary and prompt user to input the capital of selected stated.

If the answer matched with the capital value of the selected state, increment the correct counter by one. Otherwise the wrong counter will be incremented by one. Increment the round counter by one before proceed to next round.

Line 30 - 31

Print the number of correct responses and wrong responses.

7 0
2 years ago
A 1.17 g sample of an alkane hydrocarbon gas occupies a volume of 674 mL at 28°C and 741 mmHg. Alkanes are known to have the gen
Nina [5.8K]

Answer:

C3H8

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of alkane = 1.17g

Volume (V) = 674 mL

Temperature (T) = 28°C

Pressure (P) = 741 mmHg.

Gas constant (R) = 0.08206 atm.L/Kmol

Step 2:

Conversion to appropriate unit.

For Volume:

1000mL = 1L

Therefore, 674mL = 674/1000 = 0.674L

For Temperature:

Temperature (Kelvin) = Temperature (celsius) + 273

Temperature (celsius) = 28°C

Temperature (Kelvin) = 28°C + 273 = 301K

For Pressure:

760mmHg = 1atm

Therefore, 741 mmHg = 741/760 = 0.975atm

Step 3:

Determination of the number of mole of the alkane..

The number of mole of the alkane can be obtained by using the ideal gas equation. This is illustrated below:

Volume (V) = 0.674L

Temperature (T) = 301K

Pressure (P) = 0.975atm

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

PV = nRT

n = PV /RT

n = (0.975 x 0.674)/(0.08206x301)

n = 0.0266 mole

Step 4:

Determination of the molar mass of the alkane.

Mass of alkane = 1.17g

Number of mole of the alkane = 0.0266mole

Molar Mass of the alkane =?

Number of mole = Mass/Molar Mass

Molar Mass = Mass/number of mole

Molar Mass of the alkane = 1.17/0.0266 = 44g/mol

Step 5:

Determination of the molecular formula of the alkane.

This is illustrated below:

The general formula for the alkane is CnH2n+2

To obtain the molecular formula for the alkane we shall assume n = 1, 2, 3 or more till we arrive at molar Mass of 44.

When n = 1

CnH2n+2 = CH4 = 12 + (4x1) = 16g/mol

When n = 2

CnH2n+2 = C2H6 = (12x2) + (6x1) = 30g/mol

When n = 3

CnH2n+2 = C3H8 = (12x3) + (8x1) = 44g/mol

We can see that when n is 3, the molar mass is 44g/mol.

Therefore, the molecular formula for the alkane is C3H8.

7 0
2 years ago
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