I am fascinated by board games. They provide immense fun, anyone can enjoy them, they are unpredictable and best of all they are great value for money. That is why whenever I get sometime I experiment with simulating games to know them better [read Why Monopoly board game is not as random as it appears]. So, out of curiosity I have created an excel sheet that can generate bingo / housie (housey) tickets – 24 of them at a time. To get new set of tickets you would hit F9 (recalculate).
Click here to download the bingo / housie ticket generator.
Note that these are Bingo UK / India / Australia variant I am talking about, not the US 5*5 type of bingo tickets.
Read on if you want to know how this is done:
According to Wikipedia:
A typical housie/bingo ticket .., contains fifteen numbers, arranged in nine columns by three rows. Each row contains five numbers and four blank spaces. Each column contains either one, two, or very rarely three, numbers:
* The first column contains numbers from 1 to 9,
* The second column numbers from 10 to 19,
* The third 20 to 29 and so on up until the last column, which contains numbers from 80 to 90.
I have removed number “90” from the list in order to reduce some complexity in generating the tickets.
The problem is now to “generate 15 random number between 1 to 89 and fill them in 15 random spots in a grid of 3 rows by 9 columns such that each row has exactly 5 numbers”
Now I could write a function in VBA to do this, but I wanted to do this only using formulas. So I started breaking the problem.
The first challenge is to select any random 5 cells in a 9 cell row
Once we select any random 5 cells in a 9 cell row, we will fill them with bingo numbers. Now, excel has a function to generate random numbers between 1 to 9 (=round(rand()*9,0)), but this is not good for us since each time we call this function we will get a random number between 1 to 9, where as we need a 5 random numbers without repetition between 1 to 9. The function is memoryless and could repeat numbers when called 5 times.
Instead we can list all the possible “5 cells with numbers and others are empty” combinations of a 9 cells region and select a random combination every time. There are essentially 9C5 i.e. 126 ways in which you can select any 5 cells out of 9 cells (without repetition of course).
So I listed all these combinations in a table and then randomly selected one of the combinations. You can see the first five such combinations in the image below:
Selecting any five cells out of nine cells
Now I created a 3*9 region and filled the cells with 1s or 0s, “1” when the cell in bingo ticket is supposed to have a value and “0” if the cell is empty as shown below:
Next challenge is to show random values in each cell
The trick here is that first column in our 3*9 bingo ticket has any number(s) from 1 to 9, second column has any number(s) from 10 to 19 …
Again, the challenge is the numbers should not repeat, otherwise we could simply use rand()*10, rand()*10+10, rand()*10+20 ... to generate the numbers.
This time it gets even more trickier because each column can have either no values, or 1 value or 2 values or 3 values.
The ticket generation logic now looks like:
- If the column has no values in it, then we will leave all the cells in that column of bingo ticket empty
- If the column has 1 value, we will generate any random number from that column’s range of possible values (1-9, 10-19,20-29,…80-89) and place it in the cell that is supposed to have a value and leave other cells empty.
- If the column has 2 values, we will generate 2 random numbers without repetition from that column’s range of possible values and place them in cells that are supposed to have them
- If the column has 3 values, we will generate 3 random numbers without repetition from that column’s range of possible values and place them in cells that are supposed to have them
As you can see, it is easy when the column has no or 1 value in it. But when the column has 2 or 3 I used the combinations trick described earlier.
First I created all 2 number combinations and 3 number combinations. Since the numbers on Bingo ticket are always sorted from top to bottom in a column, I just had to list down 45 combinations (10C2) for 2 numbers and 120 combinations (10C3) for 3 numbers.
The rest of the details are small enough that I can leave them to your imagination. So when the ticket is generated, it looks like this:
Remember to download housie / bingo ticket generator excel sheet and print your tickets at home. Just F9 to generate new set of tickets. Un-hide the rows from 43 if you want to see how this is done.
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15 Responses to “Highlight Employees by Performance Rating – Conditional Formatting Challenge”
While this might solve the question Shelly asked, there is another option that might be more useful - a pivot table could make a list of people who fall into the various categories, so, if you needed to simply see who got in the top bracket to give them a bonus, you would have that list
Simply sorting by the rankings would work too, but you would knock them out of alphabetical order.
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The solution I chose makes use of the percentile formula.
The percentile formula returns the value representing the K-th percentile of a range of values. The range of values is the first criteria, and K is the second criteria in the formula.
I applied Conditional Formatting according to the formulas in the order below:
5% =$C6>=PERCENTILE($C$6:$C$33,0.95) Dark Blue
15% =$C6>=PERCENTILE($C$6:$C$33,0.85) Light Blue
65% =$C6>=PERCENTILE($C$6:$C$33,0.1) Green
10% =$C6>=PERCENTILE($C$6:$C$33,0.05) Light Red
5% =$C6<PERCENTILE($C$6:$C$33,0.05) Dark Red
The issue I noted with this approach is that Zambi was not highlighted in my solution as it is in the solution provided. Unless I am mistaken, and I very well may be, the 10th percentile for this data set is at 2.21, so Zambi would fall above the 10th percentile with a PR of 2.3.
The first step to this was figuring out the 'buckets'; what scores should fall into each range. In attempting to match the formatting of the spreadsheet, I determined the buckets below.
5% = 95% to 100%
10% = 90% up to but not including 95%
65% = 10% up to but not including 90%
10% = 5% up to but not including 10%
5% = under 5%
After that, it is a relatively simple matter to plug the necessary values into the conditional formatting formulas as shown above.
One final consideration is that while the buckets above match the color banding on the spreadsheet, I believe that the original request suggests a different color banding with 6 buckets shown below.
Top 5% = 95 to 100% Dark blue
Top 10% = 85 up to but not including 95% Light blue
Top 65% = 35 up to but not including 85% Green
Bottom 10% = 10% down to but not including 5% Light Red
Bottom 5% = 5% or under Dark Red
This leaves one final bucket of 10 to 35% (exclusive of both values) that is not highlighted and so would remain white.
Thank you Chandoo and Shelly for an interesting and useful exercise. This is certainly a valuable technique to have in my reporting bag of tricks.
Use of PERCENTILE is a smarter way of doing it. Below is my solution.
First 5 % = Apply conditional formatting (Dark Blue) as highlight ">=" =PERCENTILE(C:C,0.95)
Next 15% = Apply conditional formatting (Lighter Blue) as highlight between =PERCENTILE(C:C,0.95)-0.01 and =PERCENTILE(C:C,0.8)
Next 65% = Apply conditional formatting as highlight (Olive Green) between =PERCENTILE(C:C,0.8)-0.01 and =PERCENTILE(C:C,0.15)
Next 10% = Apply conditional formatting as highlight (Lighter Red) between =PERCENTILE(C:C,0.15)-0.01 and =PERCENTILE(C:C,0.05)
Bottom 5% = Apply conditional formatting (Red) as less than =PERCENTILE(C:C,0.05)
I agree, this is a challenge faced by HR managers every year and use of percentile formulae is the most popular solution which permits further processing like making bell curve, applying increments based on segmentation etc.
Hi Chandoo,
I came at the same solution as yours (not looking at yours first) but I have hard coded the conditions in the conditional formatting. For example:
=AND($C6>=$D$10,$C6<$D$9)
I have done the same thing 5 times for each condition. This makes the formatting independent of the order of specification. I think it will work better across versions of excel.
To copy the same thing in all sheets, Shelly can copy these formatted cells with format painter and apply it to the relevant cells in next sheet and so on! I know 700 sheets will be difficult but I dont know of any other way to apply conditional formating rules to the whole sheet.
First i have used percentile formula in the next column of "percentile Threshold" where E5, E6.. is input to colour code.
The idea behind doing this is to replicate the formula for any range and any threshold
=PERCENTILE($C$3:$C$30,1-E5)
=PERCENTILE($C$3:$C$30,1-E6)
=PERCENTILE($C$3:$C$30,1-E7)
=PERCENTILE($C$3:$C$30,1-E8)
=PERCENTILE($C$3:$C$30,1-E9)
Now i have given logic to different employee by applying "if Formula"
=+IF(J3>=$G$5,1,IF(J3>=$G$6,2,IF(J3>=$G$7,3,IF(J3>=$G$8,4,5))))
where 'J" referes to PR and "G" refers to percentile derived from above mentioned formula.
once again it is replicable (just change reference points)
Now comes the major part of Conditional Formatting, i have used "use a formula to determine which cells to be formatted"
Formula =$j=5, format "required colour" Applies to "$I$3:$J$30"
plus put tick on stop if true
This solves the query, important point that this is repeatable and can be done for n number of departments
Thanks !
I had done some reading on it and in Excel 2010 a new function has been introduced, percentile.exc. Attaching a video which also talks why the old percentile function shouldn't be used as it acts erroneous at times. Might be worth a watch Chandoo,
http://www.itechtalk.com/thread10579.html
@Deepa
Quit correct.
Where ever you use statistical spreadsheet functions and are using excel 2010 you should use the new versions of the functions as MS did a lot of work to speed up and fix errors in the old functions.
Warning: If you use the new Excel 2010 statistical functions in Named Formulas most of them will crash excel so do keep that in mind.
Hello Chandoo,
When i first read the challenge file, i thought, the color that need to be applied for a given rule, also need to be picked dynamically as given in rule set. But in the solution file, i found that color is hard Coded. So in case, someone has same data, but wants different colors, he/she needs to goto manage rules and change colors.
Let me know if my understanding is correct, and if yes, can we also make the color to be applied dynamic?
Thanks
Kishore
HI I ALSO USED THE PERCENTILE FUNCTION. HOWEVER, I WENT A STEP FURTHER AND USING THE SMALL() FUNCTION I SORTED THE DATA BY PERCENTILE SO THE COLOSCHEME WOULD BE GROUPED BASED ON THE VALUE. THIS WAY IT IS BETTER AND EASIER TO VIEW.
[...] recently posted a challenge to help a reader with a [...]
Hi, i have got doubt regarding to the percentages that has been put in chandoo's spreadsheet, i cant understadn how he put directly. can some one please explain how chandoo put the percetages straight way that i stated below..
5%
15%
60%
10%
5%
I have stumbled on this post as the solution has been already given so I have taken the liberty to record a video where I show the implementation of it as well as adding a filtering feature which I hope can prove to be useful.
Thank you
http://www.xlninja.com/2012/06/28/how-to-use-excel-to-highlight-employee-performance-rating/
[...] scriu nici macar un cuvant din urmatorul articol. Astazi mi-am citit mailul si hopa challenge de la Chandoo. Cum puteam sa refuz asa ceva si m-am apucat de citit, iar dupa 5 min i-am spus sotului ca pe asta [...]
Question for Chandoo:
I came to your site late but am totally loving these challenges 🙂
I guess it all boils down to how the bins are set up.
I agree with the PERCENTILE.INC function.
pls help me understand where I am wrong.
I have determined following the bins:
bottom 5% <=2.00 (F6:F33 <=PERCENTILE(range,.05))
lower 15% (5+10) <= 2.40 (F6:F33 <=PERCENTILE(range,.15))
lower 80% (5+10+65) <=3.46 (F6:F33 <=PERCENTILE(range,.80))
lower 95% (5+10+65+15) <=4.00 (F6:F33 =PERCENTILE(range,.95))
top 5% <=4.20 (F6:F33 <=PERCENTILE(range,1.00))
I find that only Tom is highest scorer and unique top 5% achiever.
I notice that Chandoo has included Christy and Daniel in top 5% achievers. How can there be 3 people in top 5% out of a population of 28 (5% of 28 = 1.4, i.e. only one person can achieve that status)?
I tried different ways but cannot get to that distribution.
Rest of the work is simply organizing the conditional formatting rules with Stop If True box checked.
Thanks for your insights