Microsoft Word - Assignment 3.docx MATH7232 Operations Research & Mathematical Planning 2018 Assignment 3 – Dynamic Programming...

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Importing Packages¶
In [1]:

import numpy as np
import pandas as pd
Importing Demand data¶
In [2]:

df = pd.read_excel('demand.xlsx', index_col = 0)
demand = df.iloc[0].tolist()
df
Out[2]:
                1        2        3        4        5        6        7
        Day                                                        
        Demand        6        8        4        11        5        7        11
Variables¶
orders everyday should be between 0 to 15 bottles
In [3]:

orders = np.zeros(df.shape[1])
Defining the profit function
In [4]:

def profit(initial, demand, order):
'''
where:
initial is the amount of bottles available in the fridge from previous days: between 0 and 10
demand is the daily demand of bottles
order is the amount of bottles ordered a day before for selling: between 0 to 15
'''
if order == 0:
# No order, no fixed cost
return min([initial+order, demand])*(5 - 1.5)
else:
# As there is a fixed cost of $10 per delivery, therefore:
return min([initial+order, demand])*(5 - 1.5) - 10
Storing the profits for all possible values of variables: initial, demand and orde
In [5]:

# Creating an empty Dataframe
columns = ['Profit']
ows = (1+10)*(len(np.unique(demand)))*(1+15)
profitdb = pd.DataFrame(columns = columns, index = range(rows))
idx = 0
newIndex = list()
# Looping for every variable
for init in range(1+10):
for dmnd in np.unique(demand):
for ordr in range(1+15):
profitdb.iloc[idx] = [profit(init, dmnd, ordr)]
index = str(init) + '.' + str(dmnd) + '.' + str(ordr)
newIndex.append(index)
idx += 1
profitdb.index = newIndex
profitdb.to_csv('profit.csv')
Constraints¶
Anything left after Day 7 will be wasted. Therefore:
order.sum()
where, demand.sum() = 52
Objective¶
Each deliveries cost $10 extra so we have to keep the number of orders/deliveries as low as possible to increase the overall profit
In [6]:

# Creating a tree class
class Tree(object):
def __init__(self):
# name for the name of the node can be defined as day.initial.demand.orde
self.name = None
self.childrens = list()
self.parents = list()
self.path = list()

def setProfit(self):
temp = self.name.split('.')
init = int(temp[1])
demand = int(temp[2])
order = int(temp[3])
self.profit = profit(init, demand, order)
In [7]:

def find2numbers(sm):
'''
The function is used to find all the combinations of initial bottles and the order to be placed
to satisfy the orde
'''
result = []
for initial in range(min([10, sm])+1):
for order in range(min([sm+1, 15+1])):
if initial+order == sm:
result.append([initial, order])
return result
In [8]:

# Creating a tree for the problem
nodesAll = []
day7nodes = []
# Day 7
day = 7
for combination in find2numbers(demand[day-1]):
init = combination[0]
order = combination[1]
name = str(day) + '.' + str(init) + '.' + str(demand[day-1]) + '.' + str(order)
node = Tree()
node.name = name
node.setProfit()
day7nodes.append(node)
nodesAll.append(node)
In [9]:

# Day 6 to 1
nodes = day7nodes
for day in range(6, 0, -1):
nodesTemp = []
for nde in nodes:
temp = nde.name
temp = temp.split('.')
initTemp = int(temp[1])
for combination in find2numbers(initTemp+demand[day-1]):
init = combination[0]
order = combination[1]
name = str(day) + '.' + str(init) + '.' + str(demand[day-1]) + '.' + str(order)
node = Tree()
node.name = name
node.setProfit()
nde.parents.append(node)
node.childrens.append(nde)
nodesTemp.append(node)
nodesAll.append(node)
nodes = nodesTemp
In [10]:

maxNode = None
maxProfit = 0
for node in nodesAll:
order = node.name.split('.')[3]
node.path.append(order)
bestChild = None
best = 0
for child in node.childrens:
if child.profit > best:
best = child.profit
bestChild = child
if bestChild is not None:
node.profit += best
node.path += bestChild.path
if node.profit > maxProfit:
maxProfit = node.profit
maxNode = node
maxNode.path.reverse()
Results¶
In [11]:

df.loc['Order'] = maxNode.path[:7]
df
Out[11]:
                1        2        3        4        5        6        7
        Day                                                        
        Demand        6        8        4        11        5        7        11
        Order        11        7        0        15        0        13        0
In [12]:

print('The maximum weekly profit for the Juice shop is: ${}'.format(maxProfit))
The maximum weekly profit for the Juice shop is:...