Disclaimer¶

This notebook is only working under the versions:

• JuMP 0.19 (unreleased, but currently in master)

• MathOptInterface 0.4.1

• GLPK 0.6.0

Description: This notebook describes how to make an optimization model more efficient by exploiting sparsity.

Author: Shuvomoy Das Gupta

Using Julia+JuMP for optimization - exploiting sparsity¶

Introduction¶

What is a sparse data structure?

A sparse data structure is one that has a lot of zeros in it. If a matrix has many more zeros than nonzeros, then it is a sparse matrix.

Why do we need to exploit sparsity?

• Sparsity in the input data increases with the dimension.

• Exploiting sparsity

• keeps the data size small
• saves memory
• reduces the running time
• improves the efficiency of the model

How to exploit sparsity in Julia?

• Define struct and create a dictionary or an array of it.

A test example: Transportation Problem¶

Consider a transportation problem which is going to be our test example:

• Problem setup: Some products have to transported from origin cities to destination cities

• Objective:

• Minimize the total cost of shipment over all relevant routes
• Decision Variables

• Find the optimum amount of every product to be shipped from one city to another
• Constraints

• How much of a product a city can supply to other cities is fixed.
• The amount of any product demanded by a city is also fixed.
• The total amount of products shipped between every pair of different cities can not exceed a given limit.

Suppose there are ten cities and three products in our problem.

In [1]:
cities =
[
:BANGKOK; :LONDON; :PARIS; :SINGAPORE;  :NEWYORK;  :ISTANBUL;  :DUBAI;  :KUALALUMPUR;  :HONGKONG;  :BARCELONA
]

products   =
[
:smartphone; :tablet; :laptop
]

Out[1]:
3-element Array{Symbol,1}:
:smartphone
:tablet
:laptop    

Defining structs¶

If we do not exploit sparsity, then number of ways we can ship the products from one city to other will be $3 \times (^{10}P_2+10)=300$.

Clearly many of them will be redundant, because of reasons like

• a product might not be needed by a city
• a product might not be produced by a city etc.

We just need to consider relevant routes, where a product can be shipped from one production city to the other demand city. So a relevant route can be defined by 3 features:

• a product
• a city that produces that product
• a city that demands that product

So, we define an struct Route as follows:

In [2]:
struct Route
p::Symbol # p stands for product
o::Symbol # o stands for origin
d::Symbol # d stands for destination
end


Here the datatype Symbol is a special type of immutable string. Then we create an array of only relevant routes.

In [3]:
routesExample =
[
Route(:smartphone,:BANGKOK,:SINGAPORE);
Route(:smartphone,:BANGKOK,:NEWYORK);
Route(:smartphone,:BANGKOK,:ISTANBUL);
Route(:smartphone,:BANGKOK,:DUBAI);
]

Out[3]:
4-element Array{Route,1}:
Route(:smartphone, :BANGKOK, :SINGAPORE)
Route(:smartphone, :BANGKOK, :NEWYORK)
Route(:smartphone, :BANGKOK, :ISTANBUL)
Route(:smartphone, :BANGKOK, :DUBAI)    

If we want to access $i$th element of the array by typing in routes[i]. When we want to access the product name associated with the $it$h element of the array, we can do so by typing routes[i].p.

In [4]:
routesExample[2] # Will give the second route

Out[4]:
Route(:smartphone, :BANGKOK, :NEWYORK)
In [5]:
routesExample[4].d # Will give the demand city of the 4th route

Out[5]:
:DUBAI

Creating new arrays efficiently from existing arrays¶

Often we need to create new arrays, where the elements of them are extracted from some already existing array conditionally.

Consider the immutable type Supply

In [6]:
struct Supply
p::Symbol # p stands for product name
o::Symbol # o stands for the origin city
end


We want to create an array suppliesExample, that contains all relevant product-city pairs, where the particular product is produced in that city. Clearly we can construct this array by plucking each product and corresponding city producing it from routesExample. This is how we do it efficiently:

• Create an empty array of type Supply
• Add elements to this array by

• selecting the product and origin from the elements of routes
• pushing them one by one in supplies
In [7]:
suppliesExample = Supply[] # Creates a 0 element array of immutable type Supply
for r in routesExample # For every element of the route route
push!(suppliesExample, Supply(r.p, r.o)) # pick the product and origin city and push it in  supplies
end

In [8]:
suppliesExample

Out[8]:
4-element Array{Supply,1}:
Supply(:smartphone, :BANGKOK)
Supply(:smartphone, :BANGKOK)
Supply(:smartphone, :BANGKOK)
Supply(:smartphone, :BANGKOK)

Dictionary¶

What is a dictionary? A dictionary is a data type which can be useful in exploiting sparsity.

Why is it needed? Often we might be interested to index a variable by a composite data type, rather than a number.

For example, for the transportation problem in consideration, it would be more convenient to index the decision variables in the routes that are present. Let

\begin{align} R=\{(p,o,d) \in P \times C \times C: \text{product } p \text{ has to be transported from city } o \text{ to city } d\} \end{align}

be the set of all the routes that are relevant for the problem. So, we can define our decision variables as below:

\begin{align} \forall (p,o,d) \in R \left(x_{(p,o,d)}= \text{the amount of a product } p \text{ that is transported from city } o \text{ city d} \right) \end{align}

From a data structure point of view, $\left(x_{(p,o,d)}\right)_{(p,o,d) \in R}$ is a dictionary which

• takes $(p,o,d) \in R$ as its key and
• has the value the optimum amount of the product $p$ to be shipped from city $o$ to city $d$.

Efficiently Constructing Dictionary of structs¶

Suppose we want to create a dictionary called costRoutes. Every element of the dictionary costRoutes contains the value of shipping cost along a particular route belonging to the array routesExample. So,

• the key to an element belonging to the dictionary is a specific route belonging to the array routesExample
• the value is the cost for that shipment.

Suppose the values of the costs are stored in an array named costCofExample.

In [9]:
costCofExample = [120; 205; 310; 45.0]

Out[9]:
4-element Array{Float64,1}:
120.0
205.0
310.0
45.0

We create the dictionary costRoutes similar to an array:

• we create an empty dictionary, and then
• use the command setindex!(name_of_dictionary, value, key) or name_of_dictionary[key]=value to add new elements in the dictionary one by one
In [10]:
costRoutesExample=Dict{Route, Float64}()# Create an empty dictionary
# where the key is Route and the value is Float64

Out[10]:
Dict{Route,Float64} with 0 entries
In [11]:
for i in 1:length(routesExample)
costRoutesExample[routesExample[i]]=costCofExample[i]
# routesExample[i] is the key, and costCofExample[i] is the value
end

In [12]:
costRoutesExample

Out[12]:
Dict{Route,Float64} with 4 entries:
Route(:smartphone, :BANGKOK, :NEWYORK)   => 205.0
Route(:smartphone, :BANGKOK, :DUBAI)     => 45.0
Route(:smartphone, :BANGKOK, :SINGAPORE) => 120.0
Route(:smartphone, :BANGKOK, :ISTANBUL)  => 310.0

After the dictionary is initialized, we can access the cost associated with some route routes[i] by typing in costRoutes[routes[i]]

In [13]:
costRoutesExample[routesExample[4]]

Out[13]:
45.0

Or we can input the description of the route itself:

In [14]:
routesExample[3]

Out[14]:
Route(:smartphone, :BANGKOK, :ISTANBUL)
In [15]:
costRoutesExample[Route(:smartphone,:BANGKOK,:ISTANBUL)]

Out[15]:
310.0

Mathematical representation of the transportation problem¶

The problem is a classic transportation problem. We will consider the sparse representation of the problem. Let

$C=\text{Set of all cities}$

$P=\text{Set of all products}$

$R=\{(p,o,d) \in P \times C \times C: \text{product } p \text{ has to be transported from city } o \text{ to city } d\}$

$\forall (p,o,d) \in R \quad \left(c_{(p,o,d)} = \text{cost of transporting some product } p \text{ from city } o \text{ to city } d \right)$

$\forall p \in P \quad \left(O_p = \text{Set of all origin cities where a product } p \text{ is produced}\right)$

$\forall p \in P \quad \left(D_p = \text{Set of all destination cities where a product } p \text{ has to be delivered}\right)$

$\forall p \in P \; \forall o \in O_p \quad \left(s_{(p,o)} = \text{the total amount of the product } p \text{ that can be supplied by city } o\right)$

$\forall p \in P \; \forall d \in D_p \quad \left(d_{(p,d)} = \text{the total amount of the product } p \text{ that is demanded by city } d\right)$

Total amount of shipped product between each pair of cities cannot exceed $\gamma$

The decision variable for this problem is

\begin{align} \forall (p,o,d) \in R \quad \left(x_{(p,o,d)}= \text{the amount of a product } p \text{ that is trasported from city } o \text{ city d} \right) \end{align}

The optimization problem can be described as below:

\begin{align} &\text{minimize} && \sum_{(p,o,d) \in R}{c_{(p,o,d)} x_{(p,o,d)}} \\ &\text{subject to} && \\ & && \forall p \in P \; \forall o \in O_p \quad \left(\sum_{d \in D_p} x_{(p,o,d)}=s_{(p,o)}\right) \\ & && \text{ : The amount of any product a city can supply to other cities is fixed} \\ & && \forall p \in P \; \forall d \in D_p \quad \left(\sum_{o \in O_p}{x_{(p,o,d)}}=d_{(p,d)}\right) \\ & && \text{ : The amount of any product demanded by a city is also fixed.} \\ & && \forall o \in C \; \forall d \in C \setminus \{o\} \quad \left(\sum_{(p,\bar{o},\bar{d}) \in R \; : \; o=\bar{o} \wedge d=\bar{d}}{x_{(p,\bar{o},\bar{d})}}\leq \gamma\right) \\ & && \text{ :The total amount of products shipped between every pair of different cities can not exceed a given limit.} \end{align}

Mapping of the mathematical symbols to JuMP¶

In the data file, the symbols in the model above are mapped as follows:

Mathematical Symbol In the code Comment
$C$ cities cities is an array of Symbols
$P$ products products is an array of Symbols
$R$ routes routes is an array of immutable type Route
$(O_p)_{p \in P}$ orig orig is a dictionary
$(D_p)_{p \in P}$ dest dest is a dictionary
$((s_{(p,o)})_{o \in O_p})_{p \in P}$ suppliedAmount suppliedAmount is a dictionary
$((d_{(p,d)})_{d \in D_p})_{p \in P}$ demandedAmount demandedAmount is a dictionary with key Customer and value Float64
$(x_{(p,o,d)})_{(p,o,d) \in R}$ trans trans is a dictionary and the variable in the problem
$(c_{(p,o,d)})_{(p,o,d) \in R}$ costRoutes costRoutes is a dictionary
$\gamma$ capacity It is of type Float64

Data file¶

In [16]:
#
cities =
[
:BANGKOK; :LONDON; :PARIS; :SINGAPORE;  :NEWYORK;  :ISTANBUL;  :DUBAI;  :KUALALUMPUR;  :HONGKONG;  :BARCELONA
]

products   =
[
:smartphone; :tablet; :laptop
]

capacity    = 700

struct Route
p::Symbol # p stands for product
o::Symbol # o stands for origin
d::Symbol # d stands for destination
end

routes =
[
Route(:smartphone,:BANGKOK,:SINGAPORE);
Route(:smartphone,:BANGKOK,:NEWYORK);
Route(:smartphone,:BANGKOK,:ISTANBUL);
Route(:smartphone,:BANGKOK,:DUBAI);
Route(:smartphone,:BANGKOK,:KUALALUMPUR);
Route(:smartphone,:BANGKOK,:HONGKONG);
Route(:smartphone,:BANGKOK,:BARCELONA);
Route(:smartphone,:LONDON,:SINGAPORE);
Route(:smartphone,:LONDON,:NEWYORK);
Route(:smartphone,:LONDON,:ISTANBUL);
Route(:smartphone,:LONDON,:DUBAI);
Route(:smartphone,:LONDON,:KUALALUMPUR);
Route(:smartphone,:LONDON,:HONGKONG);
Route(:smartphone,:LONDON,:BARCELONA);
Route(:smartphone,:PARIS,:SINGAPORE);
Route(:smartphone,:PARIS,:NEWYORK);
Route(:smartphone,:PARIS,:ISTANBUL);
Route(:smartphone,:PARIS,:DUBAI);
Route(:smartphone,:PARIS,:KUALALUMPUR);
Route(:smartphone,:PARIS,:HONGKONG);
Route(:smartphone,:PARIS,:BARCELONA);

Route(:tablet,:BANGKOK,:SINGAPORE);
Route(:tablet,:BANGKOK,:NEWYORK);
Route(:tablet,:BANGKOK,:ISTANBUL);
Route(:tablet,:BANGKOK,:DUBAI);
Route(:tablet,:BANGKOK,:KUALALUMPUR);
Route(:tablet,:BANGKOK,:HONGKONG);
Route(:tablet,:BANGKOK,:BARCELONA);

Route(:tablet,:LONDON,:SINGAPORE);
Route(:tablet,:LONDON,:NEWYORK);
Route(:tablet,:LONDON,:ISTANBUL);
Route(:tablet,:LONDON,:DUBAI);
Route(:tablet,:LONDON,:KUALALUMPUR);
Route(:tablet,:LONDON,:HONGKONG);
Route(:tablet,:LONDON,:BARCELONA);

Route(:tablet,:PARIS,:SINGAPORE);
Route(:tablet,:PARIS,:NEWYORK);
Route(:tablet,:PARIS,:ISTANBUL);
Route(:tablet,:PARIS,:DUBAI);
Route(:tablet,:PARIS,:KUALALUMPUR);
Route(:tablet,:PARIS,:HONGKONG);
Route(:tablet,:PARIS,:BARCELONA);

Route(:laptop,:BANGKOK,:SINGAPORE);
Route(:laptop,:BANGKOK,:NEWYORK);
Route(:laptop,:BANGKOK,:ISTANBUL);
Route(:laptop,:BANGKOK,:DUBAI);
Route(:laptop,:BANGKOK,:KUALALUMPUR);
Route(:laptop,:BANGKOK,:HONGKONG);
Route(:laptop,:BANGKOK,:BARCELONA);

Route(:laptop,:LONDON,:SINGAPORE);
Route(:laptop,:LONDON,:NEWYORK);
Route(:laptop,:LONDON,:ISTANBUL);
Route(:laptop,:LONDON,:DUBAI);
Route(:laptop,:LONDON,:KUALALUMPUR);
Route(:laptop,:LONDON,:HONGKONG);
Route(:laptop,:LONDON,:BARCELONA);

Route(:laptop,:PARIS,:SINGAPORE);
Route(:laptop,:PARIS,:NEWYORK);
Route(:laptop,:PARIS,:ISTANBUL);
Route(:laptop,:PARIS,:DUBAI);
Route(:laptop,:PARIS,:KUALALUMPUR);
Route(:laptop,:PARIS,:HONGKONG);
Route(:laptop,:PARIS,:BARCELONA);
]

struct Supply
p::Symbol
o::Symbol
end

#Creating the array supplies
#                   --------
supplies = Supply[] # Creates a 0 element array of immutable type Supply
for r in routes
push!(supplies, Supply(r.p, r.o))
end

# Creating suppliedAmount dictionary
#          --------------
#It might be better to create this as a dictionary, where the key is the
# element of the array supplies and the value is the corresponding supplied
#amount

suppliedAmount = Dict{Supply, Float64}()
for s in supplies
if s.p == :smartphone && s.o == :LONDON
suppliedAmount[s]=800
elseif s.p == :smartphone && s.o==:BANGKOK
suppliedAmount[s]=500
elseif s.p == :smartphone && s.o==:PARIS
suppliedAmount[s]=600
elseif s.p == :tablet && s.o==:BANGKOK
suppliedAmount[s]=1000
elseif s.p == :tablet && s.o==:LONDON
suppliedAmount[s]=1500
elseif s.p == :tablet && s.o == :PARIS
suppliedAmount[s]=1700
elseif s.p == :laptop && s.o == :BANGKOK
suppliedAmount[s]=150
elseif s.p == :laptop && s.o == :LONDON
suppliedAmount[s]=250
elseif s.p == :laptop && s.o == :PARIS
suppliedAmount[s]=400
end #if
end #for

struct Customer
p::Symbol
d::Symbol
end

# Creating customers array, which is an array of custom immutable Customer
#          ---------
customers = Customer[]
for r in routes
push!(customers, Customer(r.p, r.d))
end

demandedAmount = Dict{Customer, Float64}()
for c in customers
#1
if c.p==:smartphone && c.d==:SINGAPORE
demandedAmount[c]=400
#2
elseif c.p==:tablet && c.d==:SINGAPORE
demandedAmount[c]=600
#3
elseif c.p==:laptop && c.d==:SINGAPORE
demandedAmount[c]=90
#4
elseif c.p==:smartphone && c.d==:NEWYORK
demandedAmount[c]=200
#5
elseif c.p==:tablet && c.d==:NEWYORK
demandedAmount[c]=650
#6
elseif c.p==:laptop && c.d==:NEWYORK
demandedAmount[c]=110
#7
elseif c.p==:smartphone && c.d==:ISTANBUL
demandedAmount[c]=100
#8
elseif c.p==:tablet && c.d==:ISTANBUL
demandedAmount[c]=300
#9
elseif c.p==:laptop && c.d==:ISTANBUL
demandedAmount[c]=0
#10
elseif c.p==:smartphone && c.d==:DUBAI
demandedAmount[c]=175
#11
elseif c.p==:tablet && c.d==:DUBAI
demandedAmount[c]=350
#12
elseif c.p==:laptop && c.d==:DUBAI
demandedAmount[c]=65
#13
elseif c.p==:smartphone && c.d==:KUALALUMPUR
demandedAmount[c]=550
#14
elseif c.p==:tablet && c.d==:KUALALUMPUR
demandedAmount[c]=950
#15
elseif c.p==:laptop && c.d==:KUALALUMPUR
demandedAmount[c]=185
#16
elseif c.p==:smartphone && c.d==:HONGKONG
demandedAmount[c]=200
#17
elseif c.p==:tablet && c.d==:HONGKONG
demandedAmount[c]=750
#18
elseif c.p==:laptop && c.d==:HONGKONG
demandedAmount[c]=150
#19
elseif c.p==:smartphone && c.d==:BARCELONA
demandedAmount[c]=275
#20
elseif c.p==:tablet && c.d==:BARCELONA
demandedAmount[c]=600
#21
elseif c.p==:laptop && c.d==:BARCELONA
demandedAmount[c]=200
end
end

costCof =
[34; 7; 8; 10; 11; 74; 9; 18; 5; 15; 6; 23; 81; 18; 20; 10; 9;
13; 25; 85; 13; 40; 17; 7; 16; 20; 80; 9; 24; 5; 15; 11; 23;
90; 22; 19; 15; 16; 15; 24; 100; 21; 37; 12; 9; 16; 14;
88; 9; 28; 13; 17; 8; 32; 100; 18; 28; 15; 18; 16; 30; 102; 15]

# Creating costRoutes dictionary which contains the costs of the relevant routes
#          ----------
costRoutes=Dict{Route, Float64}()
for i in 1:length(routes)
costRoutes[routes[i]]=costCof[i]
end

# Creating orig, which takes the product as the input and gives the set of origins of that product
#          ----
orig = Dict{Symbol, Array}()
for i in 1:length(products)
dummy_array = Symbol[]
for j in 1:length(routes)
#println(i, j, products[i] == routes[j].p)
if products[i] == routes[j].p
push!(dummy_array, routes[j].o)
#println(orig[products[i]])
else
#println("Oops, something is not right")
end #if
end #for
orig[products[i]]=unique(dummy_array)
end #for

# Creating dest, which takes the product as the input and gives the set of destinations of that product
#          ----
dest = Dict{Symbol, Array}()
for i in 1:length(products)
dummy_array = Symbol[]
for j in 1:length(routes)
#println(i, j, products[i] == routes[j].p)
if products[i] == routes[j].p
push!(dummy_array, routes[j].d)
#println(orig[products[i]])
else
#println("Oops, something is not right")
end #if
end #for
dest[products[i]]=unique(dummy_array)
end #for


Model File¶

In [17]:
using JuMP  # Need to say it whenever we use JuMP
using MathOptInterface
# shortcuts
const MOI = MathOptInterface
const MOIU = MathOptInterface.Utilities

#MODEL CONSTRUCTION
#------------------
transpModel = Model(optimizer = GLPK.GLPKOptimizerLP())

#VARIABLES
#---------
@variable(transpModel, trans[routes] >= 0)

#OBJECTIVE
#---------

@objective(transpModel, Min, sum(costRoutes[l]*trans[l] for l in routes))

#Constraints
#-----------

# First Constraint
# ----------------

for pr in products
for org in orig[pr]
@constraint(transpModel, sum(trans[Route(pr, org, de)] for de in dest[pr])
==
suppliedAmount[Supply(pr,org)])
end
end

#Second Constraint
#-----------------

for pr in products
for de in dest[pr]
@constraint(transpModel, sum(trans[Route(pr, org, de)] for org in orig[pr])
==
demandedAmount[Customer(pr,de)])
end
end

# Final constraint:
# ----------------
for org in cities
for de in cities
if org!=de
@constraint(transpModel,
sum(
trans[r] for r in routes
if r.o == org && r.d==de # This will be used as an filtering condition
)
<=
capacity)
else
continue
end
end
end

statusMipModel = JuMP.optimize(transpModel) # solves the model

println("The optimal objective value is: ", JuMP.objectivevalue(transpModel))

println("The optimal solution is, trans= \n", JuMP.resultvalue.(trans))

The optimal objective value is: 178330.0
The optimal solution is, trans=
1-dimensional JuMPArray{Float64,1,...} with index sets:
Dimension 1, Route[Route(:smartphone, :BANGKOK, :SINGAPORE), Route(:smartphone, :BANGKOK, :NEWYORK), Route(:smartphone, :BANGKOK, :ISTANBUL), Route(:smartphone, :BANGKOK, :DUBAI), Route(:smartphone, :BANGKOK, :KUALALUMPUR), Route(:smartphone, :BANGKOK, :HONGKONG), Route(:smartphone, :BANGKOK, :BARCELONA), Route(:smartphone, :LONDON, :SINGAPORE), Route(:smartphone, :LONDON, :NEWYORK), Route(:smartphone, :LONDON, :ISTANBUL)  …  Route(:laptop, :LONDON, :KUALALUMPUR), Route(:laptop, :LONDON, :HONGKONG), Route(:laptop, :LONDON, :BARCELONA), Route(:laptop, :PARIS, :SINGAPORE), Route(:laptop, :PARIS, :NEWYORK), Route(:laptop, :PARIS, :ISTANBUL), Route(:laptop, :PARIS, :DUBAI), Route(:laptop, :PARIS, :KUALALUMPUR), Route(:laptop, :PARIS, :HONGKONG), Route(:laptop, :PARIS, :BARCELONA)]
And data, a 63-element Array{Float64,1}:
0.0
0.0
0.0
0.0
500.0
0.0
0.0
355.0
50.0
0.0
175.0
20.0
200.0
0.0
45.0
150.0
100.0
0.0
30.0
0.0
275.0
0.0
0.0
0.0
0.0
0.0
565.0
435.0
0.0
650.0
0.0
350.0
315.0
185.0
0.0
600.0
0.0
300.0
0.0
635.0
0.0
165.0
0.0
0.0
0.0
0.0
150.0
0.0
0.0
35.0
0.0
-0.0
65.0
0.0
150.0
0.0
55.0
110.0
0.0
0.0
35.0
0.0
200.0

In [ ]: