14 Replies  666 Views  Last Post: 23 August 2012  11:47 PM
#1
calling all 5th graders! easy math question...into vb.net
Posted 22 August 2012  03:07 AM
If i have 10 minutes to complete a project but complete it in 11 minutes, i would be running at 90 percent efficiencey
HOWEVER...
If i have 10 minutes to complete a project. Yet i complete it in 9 minutes. I would be running at 110 percent efficiency.
the problem is that when i attempt to use the same formula for both i end up getting 90.9090 % for the first one. and that isnt correct!!! ahhh it is 90 percent.
i would think the solution would be
the timetaken / allowabletime *100 but that isnt working
im doing the program in vb.net but i cant get the logic. would i use a direct proportion? i must show everything in the form of... 90 percent of benchmark or 120 percent of benchmark
Replies To: calling all 5th graders! easy math question...into vb.net
#2
Re: calling all 5th graders! easy math question...into vb.net
Posted 22 August 2012  06:16 AM
#3
Re: calling all 5th graders! easy math question...into vb.net
Posted 22 August 2012  10:47 AM
rgfirefly24, on 22 August 2012  06:16 AM, said:
Can you take me back to Mrs Monahans 6th grade math class for a moment? Why is the answer 91 percent? 11 is 1 Minute more than 10 minute, so in everyone's head that would be 10 percent yielding an answer of 90 percent. Getbwhat I'm saying?
#4
Re: calling all 5th graders! easy math question...into vb.net
Posted 22 August 2012  11:07 AM
10 minutes per project is 100%, right?
If I solve it in 9 minutes, you say I am at 110%? Compared to what?
At what percentage would I be, if I solve it in 8 minutes? What about 5 minutes? And in 1 minute: 190%?
Sounds interesting?
Rethink what rgfirefly24 wrote back there.
This post has been edited by lucky3: 22 August 2012  11:10 AM
#5
Re: calling all 5th graders! easy math question...into vb.net
Posted 22 August 2012  11:31 AM
lucasjohnson00, on 22 August 2012  12:47 PM, said:
rgfirefly24, on 22 August 2012  06:16 AM, said:
Can you take me back to Mrs Monahans 6th grade math class for a moment? Why is the answer 91 percent? 11 is 1 Minute more than 10 minute, so in everyone's head that would be 10 percent yielding an answer of 90 percent. Getbwhat I'm saying?
Ok, so think of it this way. You can never really do more then 100% effort right? so if Boss A tells Employee B, "It should only take you 10 minutes to finish this project". That means that 100% of effort is needed for 10 minutes. In reality it took Employee B 11 minutes to complete the project, so now we say that 11 minutes is 100% effort.
(Projected Time/Actual Time) * effort = percentage of projected
so in your case we would take 10 (the amount of time it was projected to take) / 11 (Actual time it took) which is equal to .909 repeating. We then multiply it by 100 (percent of effort used) which gives us 90.9 rounded up to 91%.
So at 100% effort you are 91% as efficient as someone that can do it in 10 minutes.
Lets then reverse this and say you did it in 9 minutes
so 10/9 = 1.111 repeating. Multiply it by the effort of 100% 1.111 * 100 = 111% So if it took you 9 minutes you are said to be 111% more efficient then someone who does it in 10 minutes.
This post has been edited by rgfirefly24: 22 August 2012  11:41 AM
#6
Re: calling all 5th graders! easy math question...into vb.net
Posted 22 August 2012  07:10 PM
Planned_Completion_TimeActual_Completion_TimeEfficiency
10 min11 min90%
10 min10 min100%
10 min9 min110%
Analyzing the table, the formula to get the Efficiency in Percent is:
(
Planned_Completion_Time/Planned_Completion_Time (In Fraction)
Minus
(Actual_Completion_Time  Planned Completion Time)/Planned_Completion_Time (In Fraction)
) * 100
Ex:
(10/10  (1110)/10) * 100 = (10/10  1/10) * 100 = 9/10 * 100 = 90%
(10/10  (1010)/10) * 100 = (10/10  0/10) * 100 = 10/10 * 100 = 100%
(10/10  (910)/10) * 100 = (10/10  1/10) * 100 = 11/10 * 100 = 110%
This post has been edited by November06: 22 August 2012  07:14 PM
#7
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  12:09 AM
November06, on 22 August 2012  07:10 PM, said:
Planned_Completion_TimeActual_Completion_TimeEfficiency
10 min11 min90%
10 min10 min100%
10 min9 min110%
Analyzing the table, the formula to get the Efficiency in Percent is:
(
Planned_Completion_Time/Planned_Completion_Time (In Fraction)
Minus
(Actual_Completion_Time  Planned Completion Time)/Planned_Completion_Time (In Fraction)
) * 100
Ex:
(10/10  (1110)/10) * 100 = (10/10  1/10) * 100 = 9/10 * 100 = 90%
(10/10  (1010)/10) * 100 = (10/10  0/10) * 100 = 10/10 * 100 = 100%
(10/10  (910)/10) * 100 = (10/10  1/10) * 100 = 11/10 * 100 = 110%
Thank you so much! I knew that the original formula wasn't practical for this application. To make sure I have the logic correct, can u attach a paint document or something to show the traditional format wIth the fractions.
#8
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  12:23 AM
I only wrote in fraction so it can be understood easily while computing manually.
You see the 10/10 or the (1110)/10? Those are the fractions used.
Since you are going to do a program anyway, you just have to code the formula and the computer should perform it. The fractions will be converted to decimal number but the end result should be the same as what you need.
Ex.
Using Fractions:
(10/10  (1110)/10) * 100 = (10/10  1/10) * 100 = 9/10 * 100 = 9 * 10 = 90%
Using Decimals:
First, just code the formula, then it should be evaluated as this way...
(10/10  (1110)/10) * 100
= (1  1/10) * 100
= (1  0.1) * 100
= 0.9 * 100
= 90%
Doesn't matter whether it's fraction or decimal... just code the formula and the application should be able to compute it properly.
#9
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  12:30 AM
%Efficiency = (1  (ACTPCT)/PCT) * 100
where ACT = ActualCompletion Time and PCT = Planned Completion Time
Just be careful about dividing by zero. You should validate that the planned completion Time should not be equal to 0 or there will be an exception thrown.
This post has been edited by November06: 23 August 2012  12:35 AM
#10
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  12:41 AM
November06, on 22 August 2012  07:10 PM, said:
Planned_Completion_TimeActual_Completion_TimeEfficiency
10 min11 min90%
10 min10 min100%
10 min9 min110%
Analyzing the table, the formula to get the Efficiency in Percent is:
(
Planned_Completion_Time/Planned_Completion_Time (In Fraction)
Minus
(Actual_Completion_Time  Planned Completion Time)/Planned_Completion_Time (In Fraction)
) * 100
Ex:
(10/10  (1110)/10) * 100 = (10/10  1/10) * 100 = 9/10 * 100 = 90%
(10/10  (1010)/10) * 100 = (10/10  0/10) * 100 = 10/10 * 100 = 100%
(10/10  (910)/10) * 100 = (10/10  1/10) * 100 = 11/10 * 100 = 110%
I just wrote it out and it works perfectly! When using this in vb.net, do I use this format the way you did on the bottom? Also, why doesn't the formula everyone else says work? I knew It didn't work... more curious as to why... is it because 100 percent isn't counted in the calculation? When you think of it.... isn't 100 percent the neutral where neither a positive nor negative exists? a representatIon of nothIng. Sort of like 0.. you don't count 0 when counting... does the computer treat the 100 percent the same?
#11
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  12:59 AM
Why does the formula everyone says doesn't work? Maybe it is intended to answer another sort of problem. I just read your post and based on the examples you've given, I formulated my solution. Hopefully, it's what you need.
I am not sure about what you mean by 100% isn't counted in the calculation and the next lines you wrote. Can you clarify it more?
Anyway, if my posts help you out, I would really appreciate it if you vote my posts up.
This post has been edited by November06: 23 August 2012  01:01 AM
#12
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  01:37 AM
If an average individual (let's call him/her person A (as average)) needs 10 minutes for a task, than if the other one (person S as super guy) solves it in 1 minute, person S is 10 times better than person A, meaning S is 1000% as good as an ordinary guy would be, not just 190%. It is because in 10 minutes, person S could do the same task 10 times, if he/she only needs 1 minute to do it, not only 1,9 times.
On the other hand, if someone needs 20 minutes for the same task (let's call this one person U as underperformer), U is 50% as efficient as an average person is. Because person A would do the same task twice in 20 minutes, not only once. Meaning U is half as good as A is, and that's why U is at 50% not at 0% when he/she needs 20 minutes for the same task. Person U is half way done working the task, when he/she reaches 10 minutes time frame, while A is done at the same time.
#13
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  05:01 AM
November06, on 22 August 2012  09:10 PM, said:
Planned_Completion_TimeActual_Completion_TimeEfficiency
10 min11 min90%
10 min10 min100%
10 min9 min110%
Analyzing the table, the formula to get the Efficiency in Percent is:
(
Planned_Completion_Time/Planned_Completion_Time (In Fraction)
Minus
(Actual_Completion_Time  Planned Completion Time)/Planned_Completion_Time (In Fraction)
) * 100
Ex:
(10/10  (1110)/10) * 100 = (10/10  1/10) * 100 = 9/10 * 100 = 90%
(10/10  (1010)/10) * 100 = (10/10  0/10) * 100 = 10/10 * 100 = 100%
(10/10  (910)/10) * 100 = (10/10  1/10) * 100 = 11/10 * 100 = 110%
I'm curious, where did you get the formula from? I'm not trying to say it's wrong and mine is right, but everything I've seen for calculating efficiency is W_{out}/W_{in} * 100 = efficiency
#14
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  05:59 PM
It is he who gave the problem so it is by the details he gave that we are going to base our answers not by the general ideas.
If we consider him as our client, then we should consider the fact that he has a different requirement than the usual. Say we compute efficiency as W_{out}/W_{in} * 100 just as you said but maybe they compute it differently in their part. We also have to consider that different companies have different ways on how they carry out processes and sometimes, the way they compute things are also different.
#15
Re: calling all 5th graders! easy math question...into vb.net
Posted 23 August 2012  11:47 PM
Many things get lost in translation, so I wrote in one of my previous posts, that OP was wrong. My intention was/is to help not to offend.
So, back to the problem:
lucasjohnson00, on 22 August 2012  03:07 AM, said:
...
If i have 10 minutes to complete a project but complete it in 11 minutes, i would be running at 90 percent efficiencey
...
If i have 10 minutes to complete a project. Yet i complete it in 9 minutes. I would be running at 110 percent efficiency.
...
the problem is that when i attempt to use the same formula for both i end up getting 90.9090 % for the first one. and that isnt correct!!! ahhh it is 90 percent.
I understand lucasjohnson00's post as call for helping him to understand why the formula gives him back 90.9090%, and that is what I'm trying to do once more:
If to complete the task (to get it done 100%) should take one on average 10 minutes, than if I could do it in 1 minute, I'll do 100% of the task in that minute. In 10 minutes I could do 10 times 100% of the task, and that is why I am (10 x 100%) 1000% more efficient as an average person. On the other hand, if I need 20 minutes to get 100% of the task done, I'd be at 50% of the task, when an average person would complete it, when we both hit 10 minute mark. That is the reason I'm 50% as efficient, as an average person is.
In my opinion, if I consider lucasjohnson00 as my client, as November06 proposes, I fill duty to think of him as my potential long term client. I'd not want to give him "wrong" solution, because when later he finds out that he needs to pay for the same solution once again to get it right the 2nd time, I bet he'd not hire me again. That is the reason why I'd do my best to explain him why his way of thinking might cause him trouble later.
But as I said before, I think of the OP's question as question, and call for help to understanding, not as search for the solution to his way of thinking, and that is another reason why I personally try to explain to him how I think is better way of looking at the problem he gave.
If he would be business owner, he could actually gain from paying his workers by efficiency formula November06 offered as the solution to original way of OP's thinking. But what if he would be on the other side of the table? If he is a worker and gets paid based on that kind of formula, wouldn't be fair to help him understand how he should be paid?
BTW: I'm happy to "reward" your effort November06 for giving such great response why you formulated an equation out of OP's example, and I give you + for that
This post has been edited by lucky3: 23 August 2012  11:56 PM
