Can anyone help me with a method that calculates the IRR of a series of stock trades?
Let's say the scenario is:
$10,000 of stock #1 purchased 1/1 and sold 1/7 for $11,000 (+10%)
$20,000 of stock #2 purchased 1/1 and sold 1/20 for $21,000 (+5%)
$15,000 of stock #3 purchased on 1/5 and sold 1/18 for $14,000 (-6.7%)
This should be helpful: http://www.rubyquiz.com/quiz156.html
But I couldn't figure out how to adapt any of the solutions since they assume the period of each return is over a consistent period (1 year).
I finally found exactly what I was looking for: http://rubydoc.info/gems/finance/1.1.0/Finance/Cashflow
gem install finance
To solve the scenario I posted originally:
include Finance
trans = []
trans << Transaction.new( -10000, date: Time.new(2012,1,1) )
trans << Transaction.new( 11000, date: Time.new(2012,1,7) )
trans << Transaction.new( -20000, date: Time.new(2012,1,1) )
trans << Transaction.new( 21000, date: Time.new(2012,1,20) )
trans << Transaction.new( -15000, date: Time.new(2012,1,5) )
trans << Transaction.new( 14000, date: Time.new(2012,1,18) )
trans.xirr.apr.to_f.round(2)
I also found this simple method: https://gist.github.com/1364990
However, it gave me some trouble. I tried a half dozen different test cases and one of them would raise an exception that I was never able to debug. But the xirr() method in this Finance gem worked for every test case I could throw at it.
For an investment that has an initial value and final value, as is the case with your example data that includes purchase price, sell price and a holding period, you only need to find holding period yield.
Holding period yield is calculated by subtracting 1 from holding period return
HPY = HPR - 1
HPR = final value/initial value
HPY = 11,000/10,000 - 1 = 1.1 - 1 = 0.10 = 10%
HPY = 21,000/20,000 - 1 = 1.05 - 1 = 0.05 = 5%
HPY = 14,000/15,000 - 1 = 0.9333 - 1 = -0.0667 = -6.7%
This article explains holding period return and yield
You can also annualize the holding period return and holding period yield using following formula
AHPR = HPR^(1/n)
AHPY = AHPR - 1
The above formulas only apply if you have a single period return as is the case with your example stock purchase and sale.
Yet if you had multiple returns, for example, you purchased a stock A on 1/1 for 100 and it's closing price over the next week climbed and fell to 98, 103, 101, 100, 99, 104
Then you will have to look beyond what HPR and HPY for multiple returns. In this case you can calculate ARR and GRR. Try out these online calculators for arithmetic rate of return and geometric rate of return.
But then if you had a date schedule for your investments then none of these would apply. You would then have to resort to finding IRR for irregular cash flows. IRR is the internal rate of return for periodic cash flows. For irregular cash flows such as for stock trade, the term XIRR is used. XIRR is an Excel function that calculates internal rate of return for irregular cash flows. To find XIRR you would need a series of cash flows and a date schedule for the cash flows.
Finance.ThinkAndDone.com explains IRR in much more detail than the articles you cited on RubyQuiz and Wiki. The IRR article on Think & Done explains IRR calculation with Newton Raphson method and Secant method using either the NPV equation set to 0 or the profitability index equation set to 1. The site also provides online IRR and XIRR calculators
I don't know anything about finance, but it makes sense to me that if you want to know the rate of return over 6 months, it should be the rate which equals the yearly rate when compounded twice. If you want to know the rate for 3 months, it should be the rate which equals the yearly rate when compounded 4 times, etc. This implies that converting from a yearly return rate to a rate for an arbitrary period is closely related to calculating roots. If you express the yearly return rate as a proportion of the original amount (i.e. express 20% return as 1.2, 100% return as 2.0, etc), then you can get the 6-month return rate by taking the square root of that number.
Ruby has a very handy way to calculate all kinds of complex roots: the exponentiation operator, **.
n ** 0.5 # square root
n ** (1.0/3.0) # 3rd root
...and so on.
So I think you should be able to convert a yearly rate of return to one for an arbitrary period by:
yearly_return ** (days.to_f / 365)
Likewise to convert a daily, weekly, or monthly rate or return to a yearly rate:
yearly_return = daily_return ** 365
yearly_return = weekly_return ** 52
yearly_return = monthly_return ** 12
...and so on.
As far as I can see (from reading the Wikipedia article), the IRR calculation is not actually dependent on the time period used. If you give a series of yearly cash flows as input, you get a yearly rate. If you give a series of daily cash flows as input, you get a daily rate, and so on.
I suggest you use one of the solutions you linked to to calculate IRR for daily or weekly cash flows (whatever is convenient), and convert that to a yearly rate using exponentiation. You will have to add 1 to the output of the irr() method (so that 10% return will be 1.1 rather than 0.1, etc).
Using the daily cash flows for the example you gave, you could do this to get daily IRR:
irr([-30000,0,0,0,-15000,0,11000,0,0,0,0,0,0,0,0,0,0,14000,0,21000])
You can use the Exonio library:
https://github.com/Noverde/exonio
and use it like this:
Exonio.irr([-100, 39, 59, 55, 20]) # ==> 0.28095
I believe that the main problem in order to be able to understand your scenario is the lack of a cash flow for each of the stocks, which is an essential ingredient for computing any type of IRR, without these, none of the formulas can be used. If you clarify this I can help you solve your problem
Heberto del Rio
There is new gem 'finance_math' that solves this problem very easy
https://github.com/kolosek/finance_math
Related
Im using pycharm
Write a program that will calculate tax on the user's annual salary. It must :
1. ask the user to enter their name,
2. ask the user to enter their annual salary
3. print their tax bill on screen
However, Australian tax laws are complicated.
They follow these rules:
•0 – $18,200 Nil ($0 tax paid)
•$18,201 – $45,000 19 cents for each $1 over $18,200
•$45,001 – $120,000 $5,092 plus 32.5 cents for each $1 over $45,000
•$120,001 – $180,000 $29,467 plus 37 cents for each $1 over $120,000
•$180,001 and over, $51,667 plus 45 cents for each $1 over $180,000
This function works and does not require any dependencies to work.
def taxesDue(x:float):
'''Function that takes in a person's yearly salary (unit: AUD) and returns the taxes due (unit: AUD)'''
if(x <= 18200):
return 0 # lucky person
elif(x <= 45000):
return round(0.19*(x-18200), 2)
elif(x<= 120000):
return round(5092+0.325*(x-45000), 2)
elif(x <= 180000):
return round(29467+0.37*(x-120000),2)
else:
return round(51667+0.45*(x-180000)*0.45, 2)
The sample output is
taxesDue(16500)
>0
taxesDue(18201)
>0.19
taxesDue(1e6) # scientific notation for 1 million (float)
>217717.0
Since all of us were new to coding at one point. Some explanation on things you will likely encounter on your journey deeper into Python.
The function's input is the salary in AUD (can be an integer like 20000 or a float such as 20000.95 where the decimals represent cents. Therefore, I rounded the taxes due to two digits through round(y, 2). In case the input salary is always of type int you can leave the rounding out as the output will naturally only have two decimals.
Speaking of float and int. Types in Python are dynamic so the float:x in the function's argument list is syntactic sugar (nice to look at for the developer/user but no impact on the rest of the code) to emphasize that a floating point number (the salary) goes in rather than a string str like x=Hello IRS. Note that int is a subset of float so float is more general.
The if/elif/else iterates through the conditions (e.g. x <= 45000). elif and the final else is only checked if none of the previous conditions was met. Note that this naturally reflects your task at hand.
Any function is exited as soon as any of the return's is reached.
Comments such as #lucky or the the comment right underneath the function's head '''Function... will go into the docstring. In turn, the developer can retrieve it when running
?taxesDue
If you need to print the result run
x = 475000 # or whatever salary you can think of
print(taxesDue(x))
In a fairly balanced binomial classification response problem, I am observing unusual level of error in h2o.gbm classification for determining class 0, on train set itself. It is from a competition which is over, so interest is only towards understanding what is going wrong.
Confusion Matrix (vertical: actual; across: predicted) for F1-optimal threshold:
0 1 Error Rate
0 147857 234035 0.612830 =234035/381892
1 44782 271661 0.141517 =44782/316443
Totals 192639 505696 0.399260 =278817/698335
Any expert suggestions to treat the data and reduce the error is welcome.
Following approaches are tried and error is not found decreasing.
Approach 1: Selecting top 5 important variables via h2o.varimp(gbm)
Approach 2: Converting the negative normalized variable as zero and possitive as 1.
#Data Definition
# Variable Definition
#Independent Variables
# ID Unique ID for each observation
# Timestamp Unique value representing one day
# Stock_ID Unique ID representing one stock
# Volume Normalized values of volume traded of given stock ID on that timestamp
# Three_Day_Moving_Average Normalized values of three days moving average of Closing price for given stock ID (Including Current day)
# Five_Day_Moving_Average Normalized values of five days moving average of Closing price for given stock ID (Including Current day)
# Ten_Day_Moving_Average Normalized values of ten days moving average of Closing price for given stock ID (Including Current day)
# Twenty_Day_Moving_Average Normalized values of twenty days moving average of Closing price for given stock ID (Including Current day)
# True_Range Normalized values of true range for given stock ID
# Average_True_Range Normalized values of average true range for given stock ID
# Positive_Directional_Movement Normalized values of positive directional movement for given stock ID
# Negative_Directional_Movement Normalized values of negative directional movement for given stock ID
#Dependent Response Variable
# Outcome Binary outcome variable representing whether price for one particular stock at the tomorrow’s market close is higher(1) or lower(0) compared to the price at today’s market close
temp <- tempfile()
download.file('https://github.com/meethariprasad/trikaal/raw/master/Competetions/AnalyticsVidhya/Stock_Closure/test_6lvBXoI.zip',temp)
test <- read.csv(unz(temp, "test.csv"))
unlink(temp)
temp <- tempfile()
download.file('https://github.com/meethariprasad/trikaal/raw/master/Competetions/AnalyticsVidhya/Stock_Closure/train_xup5Mf8.zip',temp)
#Please wait for 60 Mb file to load.
train <- read.csv(unz(temp, "train.csv"))
unlink(temp)
summary(train)
#We don't want the ID
train<-train[,2:ncol(train)]
# Preserving Test ID if needed
ID<-test$ID
#Remove ID from test
test<-test[,2:ncol(test)]
#Create Empty Response SalePrice
test$Outcome<-NA
#Original
combi.imp<-rbind(train,test)
rm(train,test)
summary(combi.imp)
#Creating Factor Variable
combi.imp$Outcome<-as.factor(combi.imp$Outcome)
combi.imp$Stock_ID<-as.factor(combi.imp$Stock_ID)
combi.imp$timestamp<-as.factor(combi.imp$timestamp)
summary(combi.imp)
#Brute Force NA treatment by taking only complete cases without NA.
train.complete<-combi.imp[1:702739,]
train.complete<-train.complete[complete.cases(train.complete),]
test.complete<-combi.imp[702740:804685,]
library(h2o)
y<-c("Outcome")
features=names(train.complete)[!names(train.complete) %in% c("Outcome")]
h2o.shutdown(prompt=F)
#Adjust memory size based on your system.
h2o.init(nthreads = -1,max_mem_size = "5g")
train.hex<-as.h2o(train.complete)
test.hex<-as.h2o(test.complete[,features])
#Models
gbmF_model_1 = h2o.gbm( x=features,
y = y,
training_frame =train.hex,
seed=1234
)
h2o.performance(gbmF_model_1)
You've only trained a single GBM with the default parameters, so it doesn't look like you've put enough effort into tuning your model. I'd recommend a random grid search on GBM using the h2o.grid() function. Here is an H2O R code example you can follow.
I am a beginner in prolog and was wondering if there was an easy way to convert numbers to time, for comparison.
For example:
The below two lists show bus name, capacity, time it arrives at city, time it departs city.
bus_info(bus1,150, 12:30, 14:30).
bus_info(bus2, 200, 16:00, 18:00).
passenger_info(mike, 21, 17:30). -shows name, age, and time available
I want to check which bus Mike can catch. The answer is bus 2, but how do I calculate this in prolog?
You're just comparing times for a given day so you don't need to convert the numbers to any kind of system time encoding. You only need, say "minutes past midnight" or something like that. For example, 12:30 would be (12*60)+30 minutes past midnight. And you can use that as your comparison units for a daily schedule.
To capture your hours and minutes to do this calculation, if you were to "ask" in Prolog:
bus_info(Bus, Num, StartHH:StartMM, EndHH:EndMM).
You would get two results:
Bus = bus1
Num = 150
StartHH = 12
StartMM = 30
EndHH = 14
EndHH = 30
And
Bus = bus2
Num = 200
StartHH = 16
StartMM = 0
EndHH = 18
EndMM = 0
To assign a numeric value of an expression in Prolog, you need the is predicate. For example:
StartTime is (StartHH * 60) + StartMM.
That basic information should get you started if you've learned how Prolog predicates basically work.
I'm trying to program a line of Ruby code that calculates the probable success rate of a simple skill test in a text adventure game. The test is "if x + random(12) => y then". How do I calculate the probable rate of success of this statement being true in Ruby?
In the game the player has certain skills and will occasionally have to test those skills plus a random number to get greater or equal to a given difficulty number. I want to calculate the success rate percentage of being able to win that skill test.
As an example in the adventure game your trying to track some animal through the jungle. To do this you must test your skill at tracking. If for example you have a tracking skill of 3 and you add a random number between 1-12 to that, you need to score at least a 9 or greater to succeed. Basically: Skill + random(12) => Difficulty_Number. I want to show a Success Rate percentage before they play to see what their chance of succeeding will be.
So in Ruby, what would be the algorithm to figure out the chance of success with my current Skill score? Thanks!
You could do this.
def success_rate(skill, success_level, random_range=12)
delta = success_level - skill
return [100 - (delta.to_f / random_range * 100), 100].min.round(2)
end
I have a question,
In Matlab, I have a vector of 20 years of daily data (X) and a vector of the relevant dates (DATES). In order to find the mean value of the daily data per year, I use the following script:
A = fints(DATES,X); %convert to financial time series
B = toannual(A,'CalcMethod', 'SimpAvg'); %calculate average value per year
C = fts2mat(B); %Convert fts object to vector
C is a vector of 20 values. showing the average value of the daily data for each of the 20 years. So far, so good.. Now I am trying to do the same thing but instead of calculating mean values annually, i need to calculate std annually but it seems there is not such an option with function "toannual".
Any ideas on how to do this?
THANK YOU IN ADVANCE
I'm assuming that X is the financial information and it is an even distribution across each year. You'll have to modify this if that isn't the case. Just to clarify, by even distribution, I mean that if there are 20 years and X has 200 values, each year has 10 values to it.
You should be able to do something like this:
num_years = length(C);
span_size = length(X)/num_years;
for n = 0:num_years-1
std_dev(n+1,1) = std(X(1+(n*span_size):(n+1)*span_size));
end
The idea is that you simply pass the date for the given year (the day to day values) into matlab's standard deviation function. That will return the std-dev for that year. std_dev should be a column vector that correlates 1:1 with your C vector of yearly averages.
unique_Dates = unique(DATES) %This should return a vector of 20 elements since you have 20 years.
std_dev = zeros(size(unique_Dates)); %Just pre allocating the standard deviation vector.
for n = 1:length(unique_Dates)
std_dev(n) = std(X(DATES==unique_Dates(n)));
end
Now this is assuming that your DATES matrix is passable to the unique function and that it will return the expected list of dates. If you have the dates in a numeric form I know this will work, I'm just concerned about the dates being in a string form.
In the event they are in a string form you can look at using regexp to parse the information and replace matching dates with a numeric identifier and use the above code. Or you can take the basic theory behind this and adapt it to what works best for you!