Changing password for the Postgres user

I recently updated my Ubuntu from 13.04 to the next Long Term Support (LTS) version which is 14.04. Unfortunately my postgreSQL database did not allow the OS user to login, saying it was a bad password.

I recalled my earlier post from some time back, where I had addressed this issue. However even switching to postgres user did not seem to work, as the system complained that I was entering the wrong password.

I needed to reset my password for the OS level postgres user. For this, I entered the following command on my terminal.

$ sudo passwd postgres

This prompted me to enter the sudo password followed by the new password for the postgres user.

Now I followed the steps from my previous post to reinstate the passwords for my OS level user.

Installing Node.js on Ubuntu

In this post, we are going to see how to install Node.js (a server side Javscript engine) on Ubutntu. Installing Node.js on Ubuntu is straightforward.

Open a terminal window and enter the following command.

$ sudo apt-get install nodejs

The system will show a prompt to download and install the requisite binaries as follows.

After the installation is complete, open an editor and enter the code to create a small test programme. Our test programme will create an http server.

var http = require('http');

http.createServer(function (req, res) {

res.writeHead(200, {'Content-Type': 'text/plain'});

res.end('Hello World\n');

}).listen(1337, '127.0.0.1');

console.log('Server running at http://127.0.0.1:1337/');

Save the program (I saved mine as test_http.js), and open a terminal window and navigate to the place you have saved your test program. Now enter the following command.

$ nodejs test_http.js

That's it. Our test http server is running.

Now lets open a browser and enter the URL for the server.

http://127.0.0.1:1337

There you have it folks. Our first test program with server side javascript with Node.js on Ubuntu.

Building apps on Facebook - Registering as a Developer

To build apps on Facebook, you need to first register as a Developer. To register as a developer, access lhttp://developers.facebook.com and login to your account. You will see a top level menu item called “Apps” with an option titled “Register as a Developer” as shown in the screenshot below. Go ahead and select that option.

You will be shown an option to review the Facebook policy and accept it. Please read the policy as it contains some important things you should know before signing up.

If you have read the policy and agree with it, go ahead and say, yes.

Next FB will ask you to provide a contact number for a code that may be sent as text or through an automated calling system.

Enter the code in the box provided, and go ahead and register. You will be shown a message that you have been accepted.

That is it. That was simple enough.

Binary Linear Programming using R

In the following series we will perform Linear optimizations using R. The first thing to do is to install the necessary packages in R. We will be using a package called lpSolveAPI for this exercise. We will start by downloading and installing the necessary packages followed by selecting a problem for project selection, and solving it within R

Installing the necessary R packages

First start R as an administrator. In Ubuntu, it will be sudo R.

Type command to install lpSolveAPI package as shown below.

install.packages(“lpSolveAPI”)

R will prompt us to choose an appropriate mirror. Choose the mirror closest to you.

Next install ggplot2

Next install reshape

Next install gridExtra

We can quit running R as an administrator, and start it again as a normal user, as shown below.

Now we are ready to formulate the linear programming problem and solve it within R.

Project Portfolio Selection Using R

The next thing we want to do is a pick up a relevant problem. For purpose of illustration, we are going to use an example of project portfolio selection example from a relevant paper “Strategic Selection of the Most Feasible Projects Using Linear Programming Models” by Yara Hamdan. The purpose is to pick up projects that are going to have the maximum benefit, and minimize costs along multiple dimensions.

In the paper, the author has cited 20 hypothetical projects that compete with each other for cost, labour and other similar constraints. The objective is to select as many projects as possible within the given constraints.

The overall data for all the projects is given in the following table, that is identical to the data given in the paper.

	Project	Cost	Labor	Priority	Quality	Risk	Historical Info	New Skills	Commercial Complex	Tech. Advance
1	A	100000	300000	0	70	80	1200	720	1400	230
2	B	2600000	550000	0	130	130	2300	830	1950	410
3	C	500000	90000	0	420	40	550	560	670	1230
4	D	3200000	700000	0	360	140	3500	1000	3200	1200
5	E	2700000	500000	0	400	100	2500	870	2300	1000
6	F	4500000	850000	0	200	175	3800	1200	3600	750
7	G	700000	120000	0	150	65	850	660	800	560
8	H	250000	75000	0	90	30	400	400	470	320
9	I	750000	200000	0	320	70	900	680	890	900
10	J	2300000	450000	0	120	120	2000	880	2200	290
11	K	3500000	560000	0	300	155	3700	1100	3400	1000
12	L	1700000	350000	0	90	100	1500	750	1800	320
13	M	900000	270000	1	360	70	1000	700	1300	980
14	N	3600000	750000	0	400	160	2100	950	2400	1500
15	O	1200000	320000	0	250	110	1300	730	1000	900
16	P	2200000	480000	0	220	125	2100	860	2400	720
17	Q	600000	100000	0	160	60	780	700	800	520
18	R	800000	160000	0	100	75	950	790	900	250
19	S	3000000	500000	0	340	135	2900	920	3200	900
20	T	542663	70000	0	210	60	620	560	590	640

The paper provides the formulation in LINGO language. We shall solve the same problem using R programming language.

The R script for the above is as follows

# Load the lpSolveAPI library

library(lpSolveAPI)
maxrow <-9
maxcolumn <-20

# We create a new empty model with 9 rows and 20 columns

lpmodel <-make.lp(maxrow,maxcolumn)

# Setting the appropriate data for each project, that translates to a column in the lpSolve equation

set.column(lpmodel, 1, c(100000,300000,0,70,80,1200,720,1400,230))

set.column(lpmodel, 2, c(2600000,550000,0,130,130,2300,830,1950,410))

set.column(lpmodel, 3, c(500000,90000,0,420,40,550,560,670,1230))

set.column(lpmodel, 4, c(3200000,700000,0,360,140,3500,1000,3200,1200))

set.column(lpmodel, 5, c(2700000,500000,0,400,100,2500,870,2300,1000))

set.column(lpmodel, 6, c(4500000,850000,0,200,175,3800,1200,3600,750))

set.column(lpmodel, 7, c(700000,120000,0,150,65,850,660,800,560))

set.column(lpmodel, 8, c(250000,75000,0,90,30,400,400,470,320))

set.column(lpmodel, 9, c(750000,200000,0,320,70,900,680,890,900))

set.column(lpmodel, 10, c(2300000,450000,0,120,120,2000,880,2200,290))

set.column(lpmodel, 11, c(3500000,560000,0,300,155,3700,1100,3400,1000))

set.column(lpmodel, 12, c(1700000,350000,0,90,100,1500,750,1800,320))

set.column(lpmodel, 13, c(900000,270000,1,360,70,1000,700,1300,980))

set.column(lpmodel, 14, c(3600000,750000,0,400,160,2100,950,2400,1500))

set.column(lpmodel, 15, c(1200000,320000,0,250,110,1300,730,1000,900))

set.column(lpmodel, 16, c(2200000,480000,0,220,125,2100,860,2400,720))

set.column(lpmodel, 17, c(600000,100000,0,160,60,780,700,800,520))

set.column(lpmodel, 18, c(800000,160000,0,100,75,950,790,900,250))

set.column(lpmodel, 19, c(3000000,500000,0,340,135,2900,920,3200,900))

set.column(lpmodel, 20, c(542663,70000,0,210,60,620,560,590,640))

# Declaring the rhs values

cost <-10000000
hours <-2000000
priority  <- 1
quality <-1000
risk <-600
histinfo <-10000
newexp <-5000
comm <-10000
tech <-3000

# To start, each project has uniform weightage

set.objfn(lpmodel,rep(1,maxcolumn))

# Each constraint provides the upper bound of rhs

set.constr.type(lpmodel,rep("<=", maxrow))

# Set the RHS values

set.rhs(lpmodel, c(cost,hours,priority,quality,risk,histinfo,newexp,comm,tech))

# Setting all variable types to be binary

set.type(lpmodel,1,"binary")

set.type(lpmodel,2,"binary")

set.type(lpmodel,3,"binary")

set.type(lpmodel,4,"binary")

set.type(lpmodel,5,"binary")

set.type(lpmodel,6,"binary")

set.type(lpmodel,7,"binary")

set.type(lpmodel,8,"binary")

set.type(lpmodel,9,"binary")

set.type(lpmodel,10,"binary")

set.type(lpmodel,11,"binary")

set.type(lpmodel,12,"binary")

set.type(lpmodel,13,"binary")

set.type(lpmodel,14,"binary")

set.type(lpmodel,15,"binary")

set.type(lpmodel,16,"binary")

set.type(lpmodel,17,"binary")

set.type(lpmodel,18,"binary")

set.type(lpmodel,19,"binary")

set.type(lpmodel,20,"binary")

#Setting the objective function direction

lp.control(lpmodel,sense='max')

#Finally lets name the model variables

Rownames <- c("Cost","Labor","Priority","Quality","Risk","Historical Info","New Skills","Commercial Complex","Tech. Advance")

Colnames <- c("Project A","Project B","Project C","Project D","Project E","Project F","Project G","Project H","Project I","Project J","Project K","Project L","Project M","Project N","Project O","Project P","Project Q","Project R","Project S","Project T")
dimnames(lpmodel) <- list(Rownames, Colnames)

# Solving the model

solve(lpmodel)

get.objective(lpmodel)

get.variables(lpmodel)

get.constraints(lpmodel)

Solving the model, we can see that the same 7 projects got selected using R lpSolveAPI

> solve(lpmodel)
[1] 0
> get.objective(lpmodel)
[1] 7
> get.variables(lpmodel)
 [1] 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 1 0 0
> get.constraints(lpmodel)
[1] 6750000 1655000       0     790     540    7980    4850    8120    2610

Here is the screenshot

That's it folks.... We solved a Binary LP problem using R.