Multiple Linear Regression

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We previously discussed the Simple Linear Regression Model. This time, we will look at Hypothesis, Cost Function, and Minimizing Cost of Multiple Linear Regression Model.

(Review) Simple Linear Regression Model

Hypothesis

\[H(x) = Wx + b\]

Cost Function

\[cost(W) = {1 \over 2m} {\sum_{i=1}^m} (Wx_i-y_i)^2\]

Gradient Descent Algorithm

\[W := W - \alpha{1 \over m} {\sum_{i=1}^m} (Wx_i-y_i) x_i\]

Multiple Linear Regression Model

Let’s take a look at the model with 3 features.

Hypothesis and Cost Function

\[H(x_1, x_2, x_3) = w_1x_1+w_2x_2+w_3x_3+b\] \[cost(W, b) = {1 \over m} {\sum_{i=1}^m} (H(x_{i1}, x_{i2}, x_{i3})-y_i)^2\]

Hypothesis Using Matrix

\[H(X)=XW= \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} \cdot \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} = \begin{pmatrix} x_1w_1 + x_2w_2 + x_3w_3 \end{pmatrix}\]

Predicting exam score

Suppose you have a quiz1, quiz2, or midterm score that predicts a final exam score. Five data are given as follows.

$x_1$(Quiz1) $x_2$(Quiz2) $x_3$(Midterm) $y$(Final)
80 75 90 85
65 55 60 70
35 55 45 60
78 85 80 78
95 90 94 98

Hypothesis in this case

\[H(X) = XW\] \[\begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \\ x_{41} & x_{42} & x_{43} \\ x_{51} & x_{52} & x_{53} \end{pmatrix} \cdot \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} = \begin{pmatrix} x_{11}w_{1} + x_{12}w_{2} + x_{13}w_{3} \\ x_{21}w_{1} + x_{22}w_{2} + x_{23}w_{3} \\ x_{31}w_{1} + x_{32}w_{2} + x_{33}w_{3} \\ x_{41}w_{1} + x_{42}w_{2} + x_{43}w_{3} \\ x_{51}w_{1} + x_{52}w_{2} + x_{53}w_{3} \\ \end{pmatrix}\]

Libraries

import numpy as np
import tensorflow as tf
print("TensorFlow Version: %s" % tf.__version__)

TensorFlow Version: 2.0.0

A. Computing without Matrix operations

Problems can be solved without matrix operations. However, it is troublesome to assign variables individually as follows.

# Data
x1 = [80., 65., 35., 78., 95.]
x2 = [75., 55., 55., 85., 90.]
x3 = [90., 60., 45., 80., 94.]
y  = [85., 70., 60., 78., 98.]

# Weights
tf.random.set_seed(2020)
w1 = tf.Variable(tf.random.normal([1], mean=0.0))
w2 = tf.Variable(tf.random.normal([1], mean=0.0))
w3 = tf.Variable(tf.random.normal([1], mean=0.0))
b  = tf.Variable(tf.random.normal([1], mean=0.0))

# Learning Rate
learning_rate = 0.0000001

# Training
for i in range(2000+1):

    with tf.GradientTape() as tape:
        hypothesis = w1*x1 + w2*x2 + w3*x3 + b
        cost = tf.reduce_mean(tf.square(hypothesis - y))
        w1_grad, w2_grad, w3_grad, b_grad = tape.gradient(cost, [w1, w2, w3, b])

    w1.assign_sub(learning_rate * w1_grad)
    w2.assign_sub(learning_rate * w2_grad)
    w3.assign_sub(learning_rate * w3_grad)
    b.assign_sub(learning_rate * b_grad)

    if i % 200 == 0:
        print("#%s \t cost: %s" % (i, cost.numpy()))

#0 	 cost: 19095.684
#200 	 cost: 5163.9473
#400 	 cost: 1452.9823
#600 	 cost: 464.43408
#800 	 cost: 201.04555
#1000 	 cost: 130.81319
#1200 	 cost: 112.03182
#1400 	 cost: 106.954956
#1600 	 cost: 105.52873
#1800 	 cost: 105.075134
#2000 	 cost: 104.88092

B. Computing with matrix operations

Unlike the method introduced above, using matrix computation can solve the problem very concisely.

Predicting exam score

$x_1$(Quiz1) $x_2$(Quiz2) $x_3$(Midterm) $y$(Final)
80 75 90 85
65 55 60 70
35 55 45 60
78 85 80 78
95 90 94 98 
# Data
data = np.array([

    [80., 75., 90., 85.],
    [65., 55., 60., 70.],
    [35., 55., 45., 60.],
    [78., 85., 80., 78.],
    [95., 90., 94., 98.]
], dtype = np.float32)

# Slice Data
X = data[:, :-1]
Y = data[:, [-1]]

X

array([[80., 75., 90.],
       [65., 55., 60.],
       [35., 55., 45.],
       [78., 85., 80.],
       [95., 90., 94.]], dtype=float32)

Y

array([[85.],
       [70.],
       [60.],
       [78.],
       [98.]], dtype=float32)

# Weights
tf.random.set_seed(2020)
W = tf.Variable(tf.random.normal([3, 1], mean=0.0))
b = tf.Variable(tf.random.normal([1], mean=0.0))

# Learning Rate
learning_rate = 0.0000001

# Hypothesis and Prediction Function
def predict(X):
    return tf.matmul(X, W) + b

# Training
for i in range(2000+1):

    with tf.GradientTape() as tape:
        cost = tf.reduce_mean(tf.square(predict(X) - Y))
        W_grad, b_grad = tape.gradient(cost, [W, b])

    W.assign_sub(learning_rate * W_grad)
    b.assign_sub(learning_rate * b_grad)

    if i % 200 == 0:
        print("#%s \t cost: %s" % (i, cost.numpy()))

#0 	 cost: 7769.522
#200 	 cost: 2113.2505
#400 	 cost: 606.6113
#600 	 cost: 205.28519
#800 	 cost: 98.37294
#1000 	 cost: 69.8826
#1200 	 cost: 62.28067
#1400 	 cost: 60.242634
#1600 	 cost: 59.686646
#1800 	 cost: 59.52544
#2000 	 cost: 59.46943


List of Tensorflow 2.0 Tutorials

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