## Workflows optimization in a warehouse

[This work is based on this course: Artificial Intelligence for Business.]

Our client has asked us to implement an algorithm. This robot has to run through a warehouse in the most efficient way possible. This is de warehouse plan:

A Q-learning algorithm must be applied. This is a model-free reinforcement learning algorithm.

Q-learning is a values-based learning algorithm. Value based algorithms updates the value function based on an equation(particularly Bellman equation). Whereas se sign column(bar) the other type, policy-based estimates the value function with a greedy policy obtained from the last policy improvement.

Q-learning uses Temporal Differences(TD) to estimate the value of $Q(s,a)$. Temporal difference is an agent learning from an environment through episodes with no prior knowledge of the environment:

$latex TD_t(s_t,a_t) = R(s_t,a_t)+ \gamma max_a(Q(s_{t+1},a)) – Q(s_t,a_t)$

The Q-function uses the Bellman equation:

$latex Q_t(s_t,a_t) = Q_{t-1}(s_t,a_t) + \alpha TD_t (s_t,a_t)$

## 1 Import Libraries

import numpy as np

## 2 Configuration of $latex \gamma$ coefficent*(Discount Rate)* and $latex \alpha$ coefficent*(Learning Rate)*

gamma = 0.75
alpha = 0.9

## 3 Definition of the environment

### 3.1 Definition of the states

The state is the location of our robot in the warehouse.

location_to_state = {'A': 0,
'B': 1,
'C': 2,
'D': 3,
'E': 4,
'F': 5,
'G': 6,
'H': 7,
'I': 8,
'J': 9,
'K': 10,
'L': 11}

## Definition of actions

The action is the next position where our robot can move.

actions = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

## Definition of rewards

We define the rewards, creating a reward matrix, where the rows correspond to current states $latex s_t$, the columns correspond to the actions $latex a_t$ that lead to the next state $latex s_{t} + 1$, and the cells contain the rewards $latex R(s_t, a_t)$. If a cell $latex (s_t, a_t)$ has a $latex 1$, that means that we can perform action a from the current state $latex s_t$ to get to the next state $latex s_t + 1$. If a cell $latex (s_t, a_t)$ has a $latex 0$, that means we cannot perform the action at from the current state $latex s_t$ to get to any next state $latex s_t + 1$.

– Columns are The action and the rows are The State:

R = np.array([[0,0,0,0,1,0,0,0,0,0,0,0], # A
[0,0,1,0,0,1,0,0,0,0,0,0], # B
[0,1,0,1,0,0,0,0,0,0,0,0], # C
[0,0,1,0,0,0,0,1,0,0,0,0], # D
[1,0,0,0,0,1,0,0,0,0,0,0], # E
[0,1,0,0,1,0,0,0,0,1,0,0], # F
[0,0,0,0,0,0,0,0,0,0,1,0], # G
[0,0,0,1,0,0,0,0,0,0,0,1], # H
[0,0,0,0,0,0,0,0,0,1,0,0], # I
[0,0,0,0,0,1,0,0,1,0,1,0], # J
[0,0,0,0,0,0,1,0,0,1,0,0], # K
[0,0,0,0,0,0,0,1,0,0,0,0]])# L 

## 4 Building a AI with Q-learning

– Inverse transformation from states to locations (1,2,3,…)—>(A,B,C,…):

state_to_location = {state : location for location, state in location_to_state.items()}

print(state_to_location)

– Create a final function that returns us the optimal route:

def route(starting_location, ending_location):
R_new = np.copy(R)
ending_state = location_to_state[ending_location]
R_new[ending_state, ending_state] = 1000 #to give 1000(points) to the final state

Q = np.array(np.zeros([12, 12])) #Q-value
for i in range(1000):
current_state = np.random.randint(0, 12)#random number between 0-12
playable_actions = []
for j in range(12): #thorugh the columns
if R_new[current_state, j] > 0:
playable_actions.append(j)
next_state = np.random.choice(playable_actions) #current_action
TD = R_new[current_state, next_state] + gamma*Q[next_state, np.argmax(Q[next_state,])] - Q[current_state, next_state] #Temporal Diferences(TD)
Q[current_state, next_state] = Q[current_state, next_state] + alpha*TD #Bellman equation

route = [starting_location]
next_location = starting_location
while(next_location != ending_location):
starting_state = location_to_state[starting_location]
next_state = np.argmax(Q[starting_state, ])
next_location = state_to_location[next_state]
route.append(next_location)
starting_location = next_location
return route


## 5 Final function

def best_route(starting_location, intermediary_location, ending_location):
return route(starting_location, intermediary_location) + route(intermediary_location, ending_location)[1:]

print("Chosen route:")
print(best_route('L', 'B', 'G'))
Chosen route:
['L', 'H', 'D', 'C', 'B', 'F', 'J', 'K', 'G']

We can check in our plan that this route is right.