i have this problem in college for this assignment and no one seems to be able to help me the question is as follows:
1.2 Write a pseudocode to represent the logic of a program that allows a user to enter an hourly pay rate and hours worked. The program outputs the user’s gross pay
[25] 1.3 Modify the program that computes gross pay to allow the user to enter the income tax rate. The program outputs the net pay after taxes have been deducted.
I am not experienced in the IT field that well as this is my first year.
I have tried doing pseudocode but everyone whom i have taken it to review says it is wrong, but no one shows the right thing in which way i should do it.
start
initialize the known variables: max hours worked with no overtime (MAxNoOvertime). bonus rate for overtime hours (BonusRate), non taxable payroll amount (MAxNoDue), Tax rate (Due);
enter hours worked overtime (HoursWorked) and hourly rate (Hourrate0;
if(HOurWorked -MaxNoOvertime)<=0 Then grosspay=HoursWOrked*HourRate;
Else
GrossPay=HourRate*(MaxNoOvertime + BonusRate* (Hoursworked - MaxNoOverTime));
End;
Many people are telling me that question 1.3 is suppose to be linked up with 1.2, and some are saying it vice versa. please help me
Question 1.2 and 1.3 are indeed linked. Maybe I am oversimplifying but isn't it as easy as just doing the following?:
// 1.2
variable hoursWorked = <user entered>
variable payRate = <user entered>
variable grossPay = (hoursWorked * payRate)
output grossPay
// 1.3
variable hoursWorked = <user entered>
variable payRate = <user entered>
variable taxRate = <user entered>
variable grossPay = (hoursWorked * payRate)
variable netPay = grossPay - ((grossPay/100) * taxRate)
output netPay
In your assignments I don't see anything about overtime which you did include in your examples. Why did you decide to include it?
Furthermore, psuedocode is meant to convey an idea. It is not meant to be compiled and ran.
Related
I am struggling to compile my large dataset and am assuming syntax commands are the answer, however, I am not skilled at all with syntax. My questions are specific to what syntax commands (or other methods) I should use to create hundreds/thousands of new variable names so I do not need to do it manually.
I am working with a dataset involving intimate partner homicides and domestic violence services utilization among victims from 2012-2021 (10 years), examined monthly (120 months). Across that timeframe, I have a three variable name set (REC [number of clients who received services], CALL [number of calls for services], HOUR [number of hours advocates/employees spent providing services]) that needs to be repeated monthly Jan-Dec across 10 years 2012-2021 and 39 separate services. See below:
MonthYear_REC_ServiceName
MonthYear_CALL_ServiceName
MonthYear_HOUR_ServiceName
"Month" in the above is Jan-Dec (01-12), "Year" is 2012-2021 (12-21), and "ServiceName" would be replaced with 39 different services. As an example for the year 2017 and "Shelter" services:
0117_REC_Shelter
0117_CALL_Shelter
0117_HOUR_Shelter
0217_REC_Shelter
0217_CALL_Shelter
0217_HOUR_Shelter
0317_REC_Shelter
0317_CALL_Shelter
0317_HOUR_Shelter
.....so on and so forth until December of 2017.
To further explain: This sequential monthly order would need to be repeated for each year in the 2012-2021 timeframe for each of 39 services for which I have data. "Shelter" services are shown as an example above, but I also need the same set of variable names across 38 other service types such as group counseling, legal advocacy, economic assistance, etc.
My overall question is (sorry for the repetition)- What syntax commands would I need to input to create this MASSIVE amount of variable names/variables? I hope this makes sense to everyone in the same way it makes sense to me! Sorry for the length and thank you in advance.
Best,
Shannon H.
Assuming what you want is to create an empty dataset with all the variable names you described, this will do it:
INPUT PROGRAM.
LOOP ind = 1 to (12*10*3*39).
END CASE.
END LOOP.
END FILE.
END INPUT PROGRAM.
EXECUTE.
do repeat vr=month year set service/vl=12 10 3 39.
compute vr=mod(ind,vl).
recode vr (0=vl).
compute ind=trunc((ind-1)/vl)+1.
end repeat.
compute year=year+11.
formats all (f2).
alter type month year (a2) set (a4).
compute month = char.lpad(ltrim(month), 2, "0").
recode set (" 1"="REC")(" 2"="CALL")(" 3"="HOUR").
* I suggest at this point you use "match files" to match the service numbers here with a list of service names.
* The following code creates fictitious service names instead to demonstrate how to use them.
string serviceName (a20) vrnm (a50).
compute serviceName=concat("service", char.lpad(ltrim(string(service, f2)), 2, "0") ).
* now to create the final variable names.
compute vrnm=concat(month, year, "_", set, "_", serviceNAme).
flip NEWNAMES = vrnm.
select if CASE_LBL="".
exe.
How would I get the age of a file in days in Ruby?
NOTE that I need a way to accurately get the age of a given file; this means that leap years need to be taken into account.
I need this for a program that removes files after they reach a certain age in days, such as files that are 20 days or older.
And by age, I mean the last access time of a given file, so if a file hasn't been accessed in the past 20 days or more, it gets deleted.
In Perl, I know that you can use date::calc to calculate a date in terms of days since 1 AD, and I used to have a Common-Lisp program that used the Common-Lisp implementation of date::calc, but I don't have that anymore, so I've been looking for an alternative and Ruby seems to have the required capability.
path = '/path/to/file'
(Time.now - File.stat(path).mtime).to_i / 86400.0
#=> 1.001232
Here is the implementation of my above comment, it returns a floating point number expressing the number of days passed.
I know it is an old question, but I needed the same and came up with this solution that might be helpful for others.
As the difference is in days, there is not need to directly deal with seconds.
require 'date'
age_in_days = (Date.today - File.mtime(path).to_date).to_i
if age_in_days > 20
# delete logs
end
If using Rails, you can take advantage of ActiveSupport:
if File.mtime(path) < 20.days.ago
# delete logs
end
If you aren't using Rails, Eduardo's solution above would be my pick.
This seems simple enough in theory but I haven't found anything on it. I need it for a client. Please see this page as an example: http://www.customsportsteamuniforms.com/index.php/test-shirt.html
On that page, you will find the first option that says "What kind of Screen Printing do you want?" If you select 1 color and you also happen to want more than 1 quantity (let's say 5), you will end up with this formula for the product cost:
$5 (cost) x $25 (option) x 5 quantity = total.
I DO NOT want it to do that. The option should be a one time fee in this case. The formula should read:
$5 (cost) x 5 quantity = sub-total + $25 (option) = total
How do I do this?
Magento's additional fees are all calculated on a per-product basis. If you want to add a fee to the entire order, you'll need to add some custom code to add this as a sort of handling fee for the order. The semantics for this are up to you (for instance, what about 2 different shirts both with 1-color screen printing options).
I found a plugin that does this:
http://www.absolutepricing.com/
Although, in my opinion, this should have been part of the system to begin with, but that's neither here nor there...
I am new to Ruby and Shoes, I think I have everything. the program appears to work correctly except when I get to the last step. I, enter the loan amount, interest rate, in to edit_lines, when I press the calculate button, it performs the calculations, stores the calculated numbers to a variable. The last step is dividing the total loan (loan and interest) by the length of the loan in months to ge the monthly payment, so I can make a payment table for the entire loan, but I either get in-corredt results or I get no reeults.
I think I converted the integers to floats, etc. , but... not sure. It appears to add, multiply, subtrct, except it will not divide 2 qbjects. If I enter numbers it works ok.
What am I doing wrong. It does seem like it is that difficult. Example code of dividng the values in a varible by the value of another varible?
It looks like you're using eval(), which you almost never, ever want to use. You can do the exact same thing in normal ruby. I'm just guessing right now since the code I can see in your comment is lacking newlines, but I think this code would work:
#numberbox3.text = #totalinterest + #loadamount
#numberbox5.text = #totalloan / #lengthyears
Hope this helps!
Is there a way to represent dates like 12/25 without year information? I'm thinking of just using an array of [month, year] unless there is a better way.
You could use the Date class and hard set the year to a leap year (so that you could represent 2/29 if you wanted). This would be convenient if you needed to perform 'distance' calculations between two dates (assuming that you didn't need to wrap across year boundaries and that you didn't care about the off-by-one day answers you'd get when crossing 2/29 incorrectly for some years).
It might also be convenient because you could use #strftime to display the date as (for example) "Mar-3" if you wanted.
Depending on the usage, though, I think I would probably represent them explicitly, either in a paired array or something like YearlessDate = Struct.new(:month,:day). That way you're not tempted to make mistakes like those mentioned above.
However, I've never had a date that wasn't actually associated with a year. Assuming this is the case for you, then #SeanHill's answer is best: keep the year info but don't display it to the user when it's not appropriate.
You would use the strftime function from the Time class.
time = Time.now
time.strftime("%m/%d")
While #Phrogz answer makes perfect sense, it has a downside:
YearlessDate = Struct.new(:month,:day)
yearless_date = YearlessDate.new(5, 8)
This interface is prone to MM, DD versus DD, MM confusion.
You might want to use Date instead and consider the year 0 as "yearless date" (provided you're not a historian dealing with real dates around bc/ad of course).
The year 0 is a leap year and therefore accommodates every possible day/month duple:
Date.parse("0000-02-29").leap? #=> true
If you want to make this convention air tight, just define your own class around it, here's a minimalistic example:
class YearlessDate < Date
private :year
end
The most "correct" way to represent a date without a year is as a Fixnum between 001 and 365. You can do comparisons on them without having to turn it into a date, and can easily create a date for a given year as needed using Date.ordinal