BL40A2030 Wireless Communication Networks

Week 1: Introduction to Jupyter NB

Author: Pedro Nardelli

References

import numpy as np import matplotlib.pyplot as plt from matplotlib import style style.use('bmh') #Not to show warning messages (to keep the notebook clean) import warnings warnings.filterwarnings('ignore')

This first cell imports the libraries to be used in the tutorial. Everytime that you use a command related to numpy you need to use a np.. Likewise, whenever you call a command from matplotlib you use plt..

We call a plot style so every plot has a default template and another library is to cut the warning messages (usually annoying).

Basic commands

Define a vector with timesteps
Compute a sine and a cosine using this time steps
Plot it

#Example 1 t = np.linspace(0, 20, 2000) # t is defined from 0 to 20 with 2000 points plt.figure(figsize=(16,6)) plt.plot(t, np.sin(t)-2) plt.plot(t, np.cos(t)) plt.xlabel('$t$') plt.ylabel('$x(t)$') plt.title('Plot functions') plt.xlim([0, 20]) plt.grid(True) plt.show()

#Example 2 t = np.linspace(-5, 5, 100) # t is defined from -5 to 5 with 100 points plt.figure(figsize=(16,6)) plt.plot(t, t**2,label='$t^2$') #plt.plot(t, 1/3 * t**3, label='$t^3/3$') plt.xlabel('$t$') plt.ylabel('$x(t)$') plt.grid(True) plt.legend() plt.show()

Try yourself:

Plot: $e^{-t}$ and $\log_2 t$
Use $t$ between 0 and 4 with 100 steps

#Example 2 t = np.linspace(0, 5, 100) # t is defined from -5 to 5 with 100 points plt.figure(figsize=(16,6)) plt.plot(t, np.exp(-t)) #plt.plot(t, 1/3 * t**3, label='$t^3/3$') plt.xlabel('$t$') plt.ylabel('$x(t)$') plt.grid(True) plt.show()

Experiments using random variables

Let us make a computer experiment to simulate when someone throws a coin.

Defining the function of throwing the coin

def flip(p): return 'H' if np.random.random() < p else 'T'

In this case $p$ is the probability of being Head, so that Tail is $1-p$. If $p=0.5=50\%$, this is a fair coin, otherwise is a biased coin.

Experiment 1: 1 trial
Experiment 2: 10 trials
Experiment 3: 10000 trials
Get from experiment 3 the frequency that H and T appears
Compare this with the bias probability

### Experiment 1 p=0.65 flip(p)

### Experiment 2 - 10 trials a = [None] * 10 #create an empty list with 10 elements for i in range(10): a[i]=flip(p) print(a)

### Experiment 3 - Empirical probability based on frequency #Number o realizations x_max = 10000 count = 0 #Experiment for x in range(0, x_max): a = flip(p) #print a if a == 'H': count = count + 1 print('Number of heads:', count) print('Number of tails:', x_max - count)

#Probability from experiments: Number of Heads or Tails diveded by the number of realizations print('Empirical frequency') print('Probability (heads):', count / x_max) print('Probability (tails):', (x_max-count) / x_max) print('Bias probability') print('Probability (heads):', p) print('Probability (tails):', 1-p)

Try yourself:

set $p=0.85$ and $p=0.05$
run the experiments and check what happens

Generate Poisson random variables

Not explaining what is a Poisson random variable now
The idea is that during a fixed period of time, discrete events happens
The number of events in each period is independent
The only parameter needed is the expected value of how many packets will arrive in a given period
First experiment: We will generante an array with 10 numbers with mean of 4 packets per period.
Second experiment: Array with 10000 with means 2 and 4 and draw a histogram

#Simulating Poisson distribution with mean 4 mean=4 samples = np.random.poisson(mean, 10) print(samples)

### Drawing a histogram (empirical distribution) mean=2 samples_mean2 = np.random.poisson(mean, 10000) n_bins2 = np.max(samples_mean2) ## plt.figure(figsize=(14,5)) plt.hist(samples_mean2,n_bins2) plt.xlabel('$n$') plt.ylabel('Frequency of $N = n$') plt.title('Poisson distribution') plt.show()

mean=4 samples_mean4 = np.random.poisson(mean, 10000) n_bins4 = np.max(samples_mean4) ## plt.figure(figsize=(14,5)) plt.hist(samples_mean4,n_bins4,density=True) plt.xlabel('$n$') plt.ylabel('Frequency of $N = n$') plt.title('Poisson distribution') plt.show()

mean=6 samples_mean6 = np.random.poisson(mean, 10000) n_bins6 = np.max(samples_mean6) ## plt.figure(figsize=(14,5)) plt.hist(samples_mean6,n_bins6) plt.xlabel('$n$') plt.ylabel('Frequency of $N = n$') plt.title('Poisson distribution') plt.show()

n_bins = np.max([n_bins2, n_bins4,n_bins6]) plt.figure(figsize=(14,5)) plt.hist([samples_mean2, samples_mean4,samples_mean6],n_bins) plt.xlabel('$n$') plt.ylabel('Frequency of $N = n$') plt.title('Poisson distribution') plt.show()

Example of simple queue

Packets or persons arrive following a Poisson distribution with mean $\lambda$ in a given period of time
Packets or persons that have arrived service rate follows a Poisson distribution with mean $\mu$
Can you tell what is the worst case in relation to delay?

Case 1: Arrivals smaller than service rate

#time periods size=10000 #Arrival mean=1 arrival = np.random.poisson(mean, size) #Service mean=1.5 service = np.random.poisson(mean, size) # queue = np.zeros(size+1) for i in range(size): if ((arrival[i] - service[i]) + queue[i]) > 0: queue[i+1] = max(0,(arrival[i] - service[i]) + queue[i]) plt.figure(figsize=(14,5)) #plt.plot( arrival, 'o',label='Arrival') #plt.plot( service, 'x',label='Sevice') plt.plot( queue, 'ok') plt.xlabel('$n$') plt.ylabel('Size') plt.title('Queue') #plt.ylim([0, 1]) #plt.xlim([0, 10]) plt.grid(True) #grid plt.show() print('Worst delay:', np.max(queue))

Case 2: Arrivals greater than service rate

#time periods size=10000 #Arrival mean=1.5 arrival = np.random.poisson(mean, size) #Service mean=1 service = np.random.poisson(mean, size) # queue = np.zeros(size+1) for i in range(size): if ((arrival[i] - service[i]) + queue[i]) > 0: queue[i+1] = max(0,(arrival[i] - service[i]) + queue[i]) plt.figure(figsize=(14,5)) #plt.plot( arrival, '--o',label='Arrival') #plt.plot( service, '-x',label='Sevice') plt.plot( queue, 'ok') plt.xlabel('$n$') plt.ylabel('Size') plt.title('Queue') #plt.ylim([0, 1]) #plt.xlim([0, 10]) plt.grid(True) #grid plt.show() print('Worst delay:', np.max(queue))

Case 3: Arrivals equal to service rate

#time periods size=50000 #Arrival mean=1 arrival = np.random.poisson(mean, size) #Service mean=1 service = np.random.poisson(mean, size) # queue = np.zeros(size+1) for i in range(size): if ((arrival[i] - service[i]) + queue[i]) > 0: queue[i+1] = max(0,(arrival[i] - service[i]) + queue[i]) plt.figure(figsize=(14,5)) #plt.plot( arrival, '--o',label='Arrival') #plt.plot( service, '-x',label='Sevice') plt.plot( queue, 'ok') plt.xlabel('$n$') plt.ylabel('Size') plt.title('Queue') #plt.ylim([0, 1]) #plt.xlim([0, 10]) plt.grid(True) #grid plt.show() print('Worst delay:', np.max(queue))

Try yourself as follows:

$\lambda = 1$ and $\mu = 1.1$
$\lambda = 1.1$ and $\mu = 1$

To think about

What if:

The queue has a maximum value (overflow)?
If some packets are served in error and can be restransmitted?

BL40A2030 Wireless Communication Networks

Week 1: Introduction to Jupyter NB

Author: Pedro Nardelli

References

Markdown syntax for writing in the notebook

Library for numerical calculation in Python

Plot in pyhton

Basic commands

Try yourself:

Experiments using random variables

Defining the function of throwing the coin

Try yourself:

Generate Poisson random variables

Example of simple queue

Case 1: Arrivals smaller than service rate

Case 2: Arrivals greater than service rate

Case 3: Arrivals equal to service rate

Try yourself as follows:

To think about

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