Wednesday, 13 February 2013

IT LAB: Assignment 6

The class on 12-February focussed on how to create historical volatility of a data and the Auto Correlated plot (ACF)  of the data.

Assignment 1:  create log of returns data (from 01.01.2012 to 01.01.2013) and calculate historical volatility

Syntax:

> stockprice<-read.csv(file.choose(),header=T)
> head(stockprice)
> closingprice<-stockprice[,5]
> closingprice.ts<-ts(closingprice,frequency=252)
> returns<-(closingprice.ts-lag(closingprice.ts,k=-1))/lag(closingprice.ts,k=-1)
> z<-scale(returns)+10
> logreturns<-log(z)
> logreturns
> acf(logreturns)






From the above graph, we can see that the measurements lie with in the 95% confidence interval. Therefore, the time series is stationary.

Assignment 2: Create ACF plot for the log returns data ,perform adf test and interpret.

Syntax:

T=252^0.5
> historicalvolatility<-sd(logreturns)*T
> historicalvolatility
> adf.test(logreturns)




From the test results, we can see that p-value=0.01 (<0.05).
 Therefore, we reject the null hypothesis and accept the alternate hypothesis which states that the time series is stationary.











Thursday, 7 February 2013

IT LAB : ASSIGNMENT 5


Assignment 1:


 The Data set is downloaded from  CNX Mid-cap Index downloaded from NSE from August 2012-January 2013. The data should be read in such a manner that start data is the 10th reading and the end data is the 95th reading. 

Syntax:

> z<-read.csv(file.choose(),header=T)
> close.ts<-z$Close
> close.ts
> close.time<-ts(close.ts)
> close.time<-ts(Close.ts[10:95],deltat= 1/252)
> z.diff<- diff(close.time) 
> returns<- cbind(close.time,z.diff,lag(close.time,k= -1))
> returns<- cbind(close.time,z.diff,lag(close.time,k= -1), z.diff/lag(close.time,k= -1))
> returns
> plot(returns)

>returns<- z.diff/lag(close.time,k= -1)
>plot(returns)




Assignment 2:


Predicting the new data by doing an estimation(regression) with already available data from 1-700.

Syntax:


> z<-read.csv(file.choose(),header=T)
> z1<-z[1:700,1:9]
> head(z1)
> z1$ed<-factor(z1$ed)
> z1.est<-glm(default ~ age + ed + employ + address + income, data=z1, family ="binomial")
> summary(z1.est)
> forecast<-z[701:850,1:8]
> forecast$ed<-factor(forecast$ed)
> forecast$probability<-predict(z1.est,newdata=forecast,type="response")
> head(forecast)






Wednesday, 23 January 2013

IT Labs:: Assignment 3



ASSIGNMENT 1a: Mileage-groove

Fit ‘lm’ and comment on the applicability of ‘lm’
Plot1: Residual vs Independent curve
Plot2: Standard Residual vs independent curve

> file<-read.csv(file.choose(),header=T)
> file
  mileage groove
1       0 394.33
2       4 329.50
3       8 291.00
4      12 255.17
5      16 229.33
6      20 204.83
7      24 179.00
8      28 163.83
9      32 150.33
> x<-file$groove
> x
[1] 394.33 329.50 291.00 255.17 229.33 204.83 179.00 163.83 150.33
> y<-file$mileage
> y
[1]  0  4  8 12 16 20 24 28 32
> reg1<-lm(y~x)
> res<-resid(reg1)
> res
         1          2          3          4          5          6          7          8          9
 3.6502499 -0.8322206 -1.8696280 -2.5576878 -1.9386386 -1.1442614 -0.5239038  1.4912269  3.7248633
> plot(x,res)

Output:



 As the plot is parabolic, so we will not be able to do regression.


Assignment 1 (b) -Alpha-Pluto Data

Fit ‘lm’ and comment on the applicability of ‘lm’
Plot1: Residual vs Independent curve
Plot2: Standard Residual vs independent curve
Also plot the following:
Qq plot
Qqline

> file<-read.csv(file.choose(),header=T)
> file
   alpha pluto
1  0.150    20
2  0.004     0
3  0.069    10
4  0.030     5
5  0.011     0
6  0.004     0
7  0.041     5
8  0.109    20
9  0.068    10
10 0.009     0
11 0.009     0
12 0.048    10
13 0.006     0
14 0.083    20
15 0.037     5
16 0.039     5
17 0.132    20
18 0.004     0
19 0.006     0
20 0.059    10
21 0.051    10
22 0.002     0
23 0.049     5
> x<-file$alpha
> y<-file$pluto
> x
 [1] 0.150 0.004 0.069 0.030 0.011 0.004 0.041 0.109 0.068 0.009 0.009 0.048
[13] 0.006 0.083 0.037 0.039 0.132 0.004 0.006 0.059 0.051 0.002 0.049
> y
 [1] 20  0 10  5  0  0  5 20 10  0  0 10  0 20  5  5 20  0  0 10 10  0  5
> reg1<-lm(y~x)
> res<-resid(reg1)
> res
         1          2          3          4          5          6          7
-4.2173758 -0.0643108 -0.8173877  0.6344584 -1.2223345 -0.0643108 -1.1852930
         8          9         10         11         12         13         14
 2.5653342 -0.6519557 -0.8914706 -0.8914706  2.6566833 -0.3951747  6.8665650
        15         16         17         18         19         20         21
-0.5235652 -0.8544291 -1.2396007 -0.0643108 -0.3951747  0.8369318  2.1603874
        22         23
 0.2665531 -2.5087486
> plot(x,res)

Output: 



> qqnorm(res)

Output:



 > qqline(res)

Output:





Assignment 2: Justify Null Hypothesis using ANOVA

> file<-read.csv(file.choose(),header=T)
> file

   Chair Comfort.Level Chair1
1      I             2      a
2      I             3      a
3      I             5      a
4      I             3      a
5      I             2      a
6      I             3      a
7     II             5      b
8     II             4      b
9     II             5      b
10    II             4      b
11    II             1      b
12    II             3      b
13   III             3      c
14   III             4      c
15   III             4      c
16   III             5      c
17   III             1      c
18   III             2      c
> file.anova<-aov(file$Comfort.Level~file$Chair1)
> summary(file.anova)

            Df Sum Sq Mean Sq F value Pr(>F)
file$Chair1  2  1.444  0.7222   0.385  0.687

Since p-value is greater than 5%, we cannot reject the null hypothesis

Wednesday, 16 January 2013

IT Lbs: Assignment Set 2

Today we have learnt about creation,inverse,transpose and multiplication of matrices.Then we moved on to regression and residual analysis by taking NSE historical data for NIFTY index for a certain period.Finally we had an introductory idea about how to plot normally distributed curve.

Question 1:

Create two matrices of say size 3 X 3 and select the column 1 from one matrix and column 3 from second matrix. After selecting the columns in objects say x and y merge these two columns using cbind to create a new matrix .

Syntax & Output: 



Question 2:

Multiplication of two matrices. The two matrices which were created in the previous question were multiplied and results were seen.

Syntax : z1%*%z2

Output:





Question 3:

Read historical data of NIFTY indices from NSE for the period 1st Dec 2012 to 31st Dec 2012. Find regression and residuals.

Syntax:




Output:

Question 4: Generate a normal distribution data and plot it.

Syntax:




Output:





Thursday, 29 November 2012

Tata Steel Ideation

# The post is created as requirement for MIS Course,Autumn Semester,VGSoM Batch of 2014.

#This assignment was submitted jointly by Muralidharan U (12BM60070) and Siddharth Himakar (12BM60062)

Problem Statement:

Tata Steel is looking for an original idea that will use any aspect of digital technology (cloud, mobile, Android, IoS etc ) to create a application or platform that can be integrated with the strategic management requirements of the company. Since this is an ideation contest, no software need to be built but paper must contain the following sections.


  1. Identify a need or a gap in the current management architecture of Tata Steel. Information available in the public domain can be used for this purpose.
  2. Propose a solution that will show how the current gap can be closed or narrowed significantly.
  3. If possible give examples where a similar solution has or is being implemented.
  4. List benefits -- economic or otherwise -- that will accrue to the company if it is implemented. Also identify potential problems or difficulties that need to be guarded against.
  5. Provide rough estimate of costs including people costs that the company would have to incur as a one time expense and as a regular expense.

The Solution for the above statement is submitted in the below link:

Solution In Slide Share:

Wednesday, 17 October 2012

The Search ends for an application in a web hosting platform (X10)


The application which I aim to build for my non-profit organisation (Parivartan) is a Mangers portal with a CRM, where I can handle my needs in a customised manner.


Parivartan:  The organization which aims to bring up the people in the rural sections by providing training in hand craft making. The organizations sell the various hand crafts made by these people and generates revenue to support for various awareness programs and one such awareness is rural sanitation.

Requirement: As the parivartan is mushrooming in its operations, there required a system to handle its Sales details, Client details, Lead details, Vendor details, Product details in an automated manner, also the events and happenings as to be noted in a single platform. The customers feedback can also be accommodated  using Trouble tickets and this trouble tickets can be assigned to various division/persons within the organisation.

Necessity is the mother of invention. So the parivartan necessitated to search for some online application which can be customised and used at almost free of cost. Finally found out the X10(web hosting) and its softaculous.

Softaculous has various application modules and it offers different applications in each module. I choose Vtiger application in the CRM module category. The reason being,it is user friendly and a simple application that encompasses all my needs.

The application is configured to update our products, Contacts, Invoices, Leads, Campaigning events, our calendar and other CRM related stuffs like Trouble ticket creation etc.

The username/password for this application is admin/pass
.
A user named user (credentials are user/password) is created to access this application. But then since the organization is carried by only few, most of the roles and the responsbilites are tagged to the user "admin".

The application can be accessed at http://murali.x10.bz/vtiger

Parivartan: http://www.facebook.com/Parivartan.VGSoM?fref=ts

Thursday, 20 September 2012

Requirement Specifications for an Order Management System: Sports Store

Requirement: Sports Order Management System
Version:0.1.Date:-19/09/2012.Author:- Muralidharan U.RollNo:- 12BM60070.

Require an order management system for a sports store for ordering sports equipment. The application should be developed in such a manner that there should be two users accessing the application for an order. One would be the customer for the store and other would be the admin/owner of the store managing inventory and orders.


Customer: This is the user with which our business system is associated. Initially the customer should register with the store to do any transactions. The customer must be assigned a unique customer ID and this customer ID should be quoted for all future correspondence. The customer should be intimated via email for all the transactions. 

Admin: Admin is the owner of the store who should manage all the transactions. He /she should manage the customer details and the customer related issues and concerns. He should also manage the inventory like adding/updating/deleting inventory details. He should collect the money from the customer to keep up with the business
 
Login page: The customer should login in this page. The customer credentials should be validated here. If the customer is a new user then it should be directed to registration page.


The registration page should contain the following mandatory fields.
1.       Name
2.       Address
3.       Email ID (Should be a valid email Address)
4.       Mobile number (Should be  a number with 10 digits)



Once the customer done with the registration/ login page he should be directed to order page where he/she can choose available products.

Order Page:The Customer ID should be mapped with the customer id for the order placed. The customer should choose a product from the available list of product category. For example there should be a product category called cricket where the customer can choose different bat types, Pad, gloves, ball etc..


Once the customer chooses the product, the customer should be able to view the quantity available in the inventory and unit price of the product. Once the customer selects the number of units, the total amount of the order including any tax should be displaced, and then the customer should proceed for payment.


Payment:Each order should be mapped with the unique order ID and customer ID. The customer can pay for the order, once the order is accepted by the admin. Different modes payment, like online payment and cash on delivery types should be available for the customer to make the payment. Once the customer makes the payment he/she should be intimated with the order id and transaction details and the cut-off date with which the goods would be delivered.


Sequence diagram for the System: