Microsoft Word - Assignment 2.docx MATH7232 Operations Research & Mathematical Planning 2018 Assignment 2 – Integer Programming...

Optimization.html


Importing Packages¶
In [1]:

import pandas as pd
import numpy as np
from gurobipy import *
import warnings
warnings.filterwarnings("ignore")
Data¶
In [2]:

nrows = 8
ncols = 3
init = [3200, 4000, 3800]
fileHandle = 'Demand.xlsx'
df = pd.read_excel(fileHandle)
df
Out[2]:
                Quarter        Brisbane        Melbourne        Adelaide        Cost
        0        Q1        1700        1950        2750        832
        1        Q2        2550        3000        2850        965
        2        Q3        2900        2400        1200        968
        3        Q4        2800        1600        2250        874
        4        Q5        1850        2200        2900        966
        5        Q6        2050        3200        1950        1007
        6        Q7        3300        3150        900        827
        7        Q8        2650        2150        1700        914
Variables¶
In [3]:

m = Model("mip1")
var = {}
for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
var[name] = m.addVar(ub=10000, name=name)
var[name].vType = GRB.INTEGER

m.update()
In [4]:

idx = 0
for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
colname = df.columns[col+1]
if idx-3 >= 0:
init.append(init[idx-3] + var[name] - df.iloc[row][colname])
idx += 1
else:
init[idx] = init[idx] + var[name] - df.iloc[row][colname]
idx += 1
Objective¶
In [5]:

obj = 0
idx = 0
for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
colname = df.columns[col+1]
obj += var[name]*df.Cost[row] + init[idx]*35
idx += 1
m.setObjective(obj)
Constraints¶
In [6]:

contraint = 0
for row in range(nrows):
m.addConstr(var[str(row) + str(0)] + var[str(row) + str(1)] + var[str(row) + str(2)] <= 10000, "c"+str(contraint))
contraint += 1
idx = 0
for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
colname = df.columns[col+1]
m.addConstr(init[idx] >= 0, "c"+str(contraint))
contraint += 1
idx += 1
Optimization¶
In [7]:

m.optimize()
for v in m.getVars():
print('%s %g' % (v.varName, v.x))
print('Obj: %g' % obj.getValue())
Optimize a model with 32 rows, 24 columns and 132 nonzeros
Variable types: 0 continuous, 24 integer (0 binary)
Coefficient statistics:
Matrix range [1e+00, 1e+00]
Objective range [9e+02, 1e+03]
Bounds range [1e+04, 1e+04]
RHS range [1e+03, 2e+04]
Found heuristic solution: objective 4.279545e+07
Presolve removed 3 rows and 0 columns
Presolve time: 0.00s
Presolved: 29 rows, 24 columns, 129 nonzeros
Variable types: 0 continuous, 24 integer (0 binary)
Root relaxation: objective 4.076690e+07, 19 iterations, 0.00 seconds
Nodes | Cu
ent Node | Objective Bounds | Work
Expl Unexpl | Obj Depth IntInf | Incumbent BestBd Gap | It/Node Time
* 0 0 0 4.076690e+07 4.0767e+07 0.00% - 0s
Explored 0 nodes (19 simplex iterations) in 0.08 seconds
Thread count was 4 (of 4 available processors)
Solution count 2: 4.07669e+07 4.27955e+07
Optimal solution found (tolerance 1.00e-04)
Best objective 4.076690000000e+07, best bound 4.076690000000e+07, gap 0.0000%
00 3650
01 3350
02 3000
10 -0
11 -0
12 -0
20 300
21 -0
22 -0
30 5900
31 1850
32 2250
40 -0
41 5150
42 4850
50 800
51 -0
52 -0
60 3300
61 4100
62 2600
70 2650
71 1200
72 -0
Obj: 4.07669e+07
Result¶
In [8]:

db = df
for v in m.getVars():
row = int(list(str(v.varName))[0])
col = int(list(str(v.varName))[1])
colname = db.columns[col+1]
db[colname][row] = v.x

d
Out[8]:
                Quarter        Brisbane        Melbourne        Adelaide        Cost
        0        Q1        3650        3350        3000        832
        1        Q2        0        0        0        965
        2        Q3        300        0        0        968
        3        Q4        5900        1850        2250        874
        4        Q5        0        5150        4850        966
        5        Q6        800        0        0        1007
        6        Q7        3300        4100        2600        827
        7        Q8        2650        1200        0        914
Boss.docx
Importing Packages
import pandas as pd
import numpy as np
from gurobipy import *
import warnings
warnings.filterwarnings("ignore")
Data
nrows = 8
ncols = 3
init = [3200, 4000, 3800]
fileHandle = 'Demand.xlsx'
df = pd.read_excel(fileHandle)
df
        
        Quarte
        Brisbane
        Melbourne
        Adelaide
        Cost
        0
        Q1
        1700
        1950
        2750
        832
        1
        Q2
        2550
        3000
        2850
        965
        2
        Q3
        2900
        2400
        1200
        968
        3
        Q4
        2800
        1600
        2250
        874
        4
        Q5
        1850
        2200
        2900
        966
        5
        Q6
        2050
        3200
        1950
        1007
        6
        Q7
        3300
        3150
        900
        827
        7
        Q8
        2650
        2150
        1700
        914
Variables¶
Defining the variables. There are in total 24 (3 X 8) variables as there are 3 cities and 8 quarters.
m = Model("mip1")
var = {} for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
var[name] = m.addVar(ub=10000, name=name)
var[name].vType = GRB.INTEGER
m.update()
Defining the number of ba
els left at every city at the end of every quarter.
idx = 0
for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
colname = df.columns[col+1]
if idx-3 >= 0:
init.append(init[idx-3] + var[name] - df.iloc[row][colname])
idx += 1
else:
init[idx] = init[idx] + var[name] - df.iloc[row][colname]
idx += 1
Objective
Objective is to minimize the total cost of buying new ba
els and concentrating the existing ones with a summation over every city and every quarter.
obj = 0
idx = 0
for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
colname = df.columns[col+1]
obj += var[name]*df.Cost[row] + init[idx]*35
idx += 1
m.setObjective(obj)
Constraints
There are two types of constraints:
1. Total number of ba
els imported in any quarter (sum over all the cities) should be less than, equal to 10000.
2. The number of ba
els left after any quarter in any city should be greater than or equal to zero.
constraint = 0
for row in range(nrows):
m.addConstr(var[str(row) + str(0)] + var[str(row) + str(1)] + var[str(row) + str(2)] <= 10000, "c"+str(constraint))
constraint += 1
idx = 0
for row in range(nrows):
for col in range(ncols):
name = str(row) + str(col)
colname = df.columns[col+1]
m.addConstr(init[idx] >= 0, "c"+str(constraint))
constraint += 1
idx += 1
Optimization¶
m.optimize()
for v in m.getVars():
print('%s %g' % (v.varName, v.x)
print('Obj: %g' % obj.getValue())
Optimize a model with 32 rows, 24 columns and 132 nonzeros
Variable types: 0 continuous, 24 integer (0 binary)
Coefficient statistics:
Matrix range [1e+00, 1e+00]
Objective range [9e+02, 1e+03]
Bounds range [1e+04, 1e+04]
RHS range [1e+03, 2e+04]
Found heuristic solution: objective 4.279545e+07
Presolve removed 3 rows and 0 columns
Presolve time: 0.00s
Presolved: 29 rows, 24 columns, 129 nonzeros
Variable types: 0 continuous, 24 integer (0 binary)
Root relaxation: objective 4.076690e+07, 19 iterations, 0.00 seconds
Nodes | Cu
ent Node | Objective Bounds | ...