1.1)
Nowadays, technology has been rapidly developing and companies are competing with each other to produce a
better product. It’s also same for the mobile wireless network. From generation to generation, companies
researched on technology to produce wireless network that can work faster and as a result, it changed from 1G,
2G, 3G, 4G and now 5G consecutively. Now, it is 5G era. When you buy a mobile phone, most of the handsets
are equipped with 5G technology. However, many people do not exactly acknowledge what 5G has over previous
generations. They just buy phone with 5G just because it is faster. Of course, it is true that 5G is a faster mobile
communication than the previous generations but there are other differences in other areas besides speed. Let’s
dig in to the other differences 5G have compared to 4G.
4G
5G
Carrier Technology
Filter Bank Multi carrier
Network Architecture
Ultra-broadband architecture capable of delivering a
theoretical peak of 10 Gbit/s and around 1-2Gbit/s for
individual users.
Mobility and Handover
Spectral efficiency
Downlink Spectral efficiency:15bits/second/Hz
Downlink Spectral efficiency:23bits/second/Hz
Uplink Spectral efficiency:3.75bits/second/Hz
Uplink Spectral efficiency:24bits/second/Hz
2000 devices/km2
1000000 devices/km2
Bearer-based: EPS bearer level
Flow based: QoS flow level
QoS Identifier: QCI (Quality Class Indicator)
QoS Identifier: 5QI (5G QoS Identifier)
Maximum 6,877W
Maximum 11,577W
Connection Density
Quality of Service
Power Consumption
6.
Now, lets calculate X’X
Then, lets get the inverse of X’X
Then, find X’Y
Now, we should calculate b=(X’X)-1(X’Y) to get coefficients b0, b1, b2, b3, b4
b0=-0.12315, b1=0.00009, b2=0.003942, b3=0.000283, b4=-0.03787
We can get equation of regression line as
Y=-0.12315+0.00009X1+0.00394X2+0.00028X3-0.03787X4
To estimate the NO density with data vector of {1436, 28.0, 68, 2.00}, plug in these values for Y
Y=-0.12315+0.00009(1436) +0.00394(28.0) +0.00028(68)-0.03787(2.00) = 0.06151
NO density with data vector of {1436, 28.0, 68, 2.00} is 0.06151
7.
To find parent node for the tree, find the entropy for class variables.
Total 14 Yes/No-> 9Yes and 5No
E(S)= -((9/14log2(9/14) + 5/14log2(5/14))
E(S)= 0.94
E(S, Outlook)= (5/14)[-(3/5)log(3/5) - (2/5)log(2/5)]+ (4/14)(0) + (5/14)[-(2/5)log(2/5) – (3/5)log(3/5)]
E(S, Outlook)= 0.693
Information Gain(S, Outlook)= 0.94-0.693=0.247
E(S, Humidity)= (7/14)[-(3/7)log(3/7)-(4/7)log(4/7)] + (7/14)[-(6/7)log(6/7) – (1/7)log(1/7)]
E(S, Humidity)= 0.788
Information Gain(S, Humidity)= 0.94-0.7880= 0.152
E(S, Windy)= (8/14)[-(6/8)log(6/8)-(2/8)log(2/8)] + (6/14)[-(3/6)log(3/6) – (3/6)log(3/6)]
E(S, Windy)=0.8932
Information Gain(S, Windy)= 0.94-0.8932=0.048
Since Overcast have only Yes, it is suitable to play
Now Decision Tree looks like this
Then, find high Information Gain
E(Sunny)=(-(3/5)log2(3/5)-(2/5)log2(2/5))=0.971
Information Gain(Sunny, Humidity)=0.971
Information Gain(Sunny, Windy)= 0.020
Humidity becomes next node because it is highest.
Finally the Decision Tree looks as follows
8.1)
P(Yes)=2/5, P(No)=3/5
P(Bad/Yes)=1/2
P(Good/Yes)=1/2
P(Bad/No)=2/3
P(Good/No)=1/3
P(Single/Yes)=1/2
P(Married/Yes)=1/2
P(Single/No)=2/3
P(Married/No)=1/3
P(High/Yes)=1/2
P(Low/Yes)=1/2
P(High/No)=1/3
P(Low/No)=2/3
2)
P(A1,A2,A3/Yes)=P(Bad/Yes)*P(Good/Yes)*P(Single/Yes)*P(Married/Yes)*P(High/Yes)*P(Low/Yes)
= (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)= 1/64
P(A1,A2,A3/No)= P(Bad/No)*P(Good/No)*P(Single/No)*P(Married/No)*P(High/No)*P(Low/No)
=(2/3)*(1/3)*(2/3)*(1/3)*(1/3)*(2/3)= 8/729
3)
P(Yes/A1,A2,A3)=P(Yes)*P(A1,A2,A3/Yes)=(2/5)*(1/64)=1/160
P(No/A1,A2,A3)= P(No)*P(A1,A2,A3/No)=(3/5)*(8/729)=8/1215
4)
P(Yes/A1,A2,A3) is 0.00625
P(No/A1,A2,A3) is approximately 0.0066
Since P(No/A1,A2,A3) is greater than . P(Yes/A1,A2,A3), loan should be denied.
9.1)
Zero Padding=1
P=1
Input size =32
Stride =1
Output size =28
According to formula of Output size,
Output size = [(Input size+2p)-Filter size)/Stride ]+1
28 = [(32+2(1))-Filter size)/1]+1
Filter size = 7
Number of input signals that each neuron in C1 receives is 7x7
2)
Number of neurons required in convolutional C1 layer = 6x28x28 = 4704neurons
3)
Filter size =2
Input size =28
Output size = 14
P=0
According to formula of Output size,
14 = [(28+2(0))-2)/Stride]+1
13=26/Stride
Stride=2
4)
Number of neurons required in S2 layer = 6x14x14 = 1176