Van Gysel asked in a recent post at Chandoo.org for a way to calculate the costs of running a plantation. The twist is that the costs vary by year, and based on the age of the trees.
The following is a slightly simplified version of the solution I offered:
=SUM(IFERROR(LOOKUP(“Year”&MMULT(N($B$3:B$7>0),TRANSPOSE(COLUMN($B$3:B$7)^0)), $B$11:$I$11, $B12:$I12),0)*B$3:B$7) Ctrl+Shift+Enter
Today I am going to try and explain how the formula works.
As always at Formula Forensics, you can follow along with a sample file: Download Here
The Problem
In a plantation, the costs for planting and maintaining trees vary based on the age of the trees and by year. The table below shows the acres of trees planted per year and the yield and costs per year that vary based on the age of the trees.
Let us look at the calculations needed for each year.
Year 2013
- 300 acres of trees were planted in 2013. Calculations for 2013 are as follows. (Only Yield calculation is shown, but the process is similar for Nursery costs, Fertilizers, etc.)
- The trees do not yield any fruits in the first year. As such, Yield for year1=300*0=0
That was easy!
Year 2014
- 700 additional acres of trees will be planted in 2014. Calculations for 2014 are as follows. (Again, only Yield calculation is shown, but others are calculated similarly.)
- 300 acres of trees are 2 years old. 700 acres are 1 year old.
- The 300 acres from 2013 now yield fruit since it is year2. However, the new trees (700 acres) do not yield any fruits yet. So total yield for 2014=300*Year2Yield+700*Year1Yield=300*5+700*0
Year 2015
- 1000 additional acres are to be planted in 2015. Calculations for 2015 are as follows:
- 300 acres are from 2013 (3 years old); 700 acres are from 2014 (2 years old); 1000 acres are from 2015 (1 year old).
- Yield for 2015=300*Year3Yield + 700*Year2Yield + 1000*Year1Yield = 300*10 + 700*5 + 1000*0
Year 2016
- 1000 additional acres are to be planted in 2016. Calculations for 2016 are as follows:
- Yield for 2016=300*15+700*10+1000*5+1000*0
How do we simulate the above calculation in an Excel formula?
A Solution
Let us first look at how we performed the calculations above manually, using the 2016 Yield as an example.
- We took each acreage value in 2016, and determined its age by counting how many years it has been since that acreage was planted. You might have observed that the age can be counted by the number of times a value has been repeated up to that point. (In other words, if I planted 300 acres in 2013, I should see that same amount in 2014, 2015 and 2016.) As such, 300 acres is repeated 4 times. 700 acres is repeated 3 times. 1000 acres is repeated 2 times. And the latest planting of 1000 acres exists only once.
- Once we determine the age for a given acreage, we looked up the yield for that age in the second table
- We then multiplied the acreage with the corresponding yield value.
Calculation #1 can be expressed as follows:
- Age for acreage 1 (first planted in 2013)=count of B3:E3 where value is greater than zero. i.e. COUNTIF(B3:E3,”>0”)
- Age for acreage 2 (planted in 2014)=count of B4:E4 where value is greater than zero. i.e. COUNTIF(B4:E4,”>0”)
- Age for acreage 3 (planted in 2015)=count of B5:E5 where value is greater than zero. i.e. COUNTIF(B5:E5,”>0”)
- Age for acreage 4 (planted in 2016)=count of B6:E6 where value is greater than zero. i.e. COUNTIF(B6:E6,”>0”)
- Age for acreage 5 is zero since nothing has been planted for 2017 yet in 2016
The above approach would work if we were calculating the age one row at a time. However, that can become tedious really fast. We need to perform the calculation for the full range (B3:E7) together, but return the counts for each row individually.
Excel’s MMULT function comes to the rescue!
MMULT (which stands for Matrix Multiply) multiplies two matrices and returns a third matrix based on rules for matrix multiplication. I am planning to devote a whole article to explain the MMULT function. As such, for this article, we will summarize the utility of the function as “take a 2-dimensional array, add each column’s value for each row, and return a 1-column array”.
MMULT requires that its arguments be numeric.
So to obtain the counts for the year 2016, we can use the following:
MMULT(N($B$3:E$7>0),TRANSPOSE(COLUMN($B$3:E$7)^0))
As you can see from the picture below, MMULT’s results are the addition of each column for each row.
In the above formula, you may have noticed that the range uses absolute and relative referencing (signified by the $ sign or lack thereof). This is to ensure that the range grows or shrinks as needed. The upper left address is held constant ($B$3). However, the lower right address for the range has columns that vary but row that is fixed on row #7. This ensures that the formula would work if we copy to the left, right, etc. in the final results.
Now that we have the age for each acreage value, we can look up the corresponding yield value using (what else?) LOOKUP function.
But before we can use LOOKUP, we will need to convert the numeric values returned from MMULT into the strings Year1, Year2, etc. found in the Costs table. Of course, you know how to do that… concatenate the string “Year” to the result from MMULT
“Year”&MMULT(N($B$3:E$7>0),TRANSPOSE(COLUMN($B$3:E$7)^0))
For the 2016 example, we get {“Year4″;”Year3″;”Year2″;”Year1″;”Year0”}
We can now use LOOKUP as follows:
LOOKUP(“Year”&MMULT(N($B$3:E$7>0),TRANSPOSE(COLUMN($B$3:E$7)^0)), $B$11:$I$11, $B12:$I12)
You may recall that LOOKUP looks up a value in the array indicated by the second argument, and returns the corresponding value from the third array argument. In this case, instead of looking up a single value, we look up an array of values (supplied in the first argument) to the function.
The above formula translates to the following:
LOOKUP({“Year4″;”Year3″;”Year2″;”Year1″;”Year0”}, {“Year1″,”Year2″,”Year3″,”Year4″,”Year5″,”Year6″,”Year7″,”Year8”}, {0,5,10,15,20,25,30,35})
The result from LOOKUP is {15;10;5;0;#N/A}
(The last value is #N/A because there is no acreage value for 2017 yet (as of 2016 column). The concatenation resulted in Year0 which does not exist in the “Age of The Trees” range (B11:I11) above.)
By using IFERROR(LOOKUP(…),0) we get {15;10;5;0;0}
We can now multiply the above result with the acreage values for 2016 to get {4500;7000;5000;0;0}
Finally, we SUM the values to get 16500
Putting it all together, we get the following formula (shown for Production for year 2016)
=SUM(IFERROR(LOOKUP(“Year”&MMULT(N($B$3:E$7>0),TRANSPOSE(COLUMN($B$3:E$7)^0)), $B$11:$I$11, $B12:$I12),0)*E$3:E$7)
One of the benefits of the above formula is that you can copy the same formula to calculate values for additional years, as well as other plantation costs.
Download
You can download a copy of the above file and follow along: Download sample file.
Let me know (using the comments below) what you think of the above approach and solution, as well as any other approaches you have utilized to solve a similar problem. In the meantime, I wish you continued Excellence!
-Sajan.
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26 Responses to “Get busy this weekend, with OR XOR AND [Excel Homework]”
first solution for AND
The two numbers are in A1 and B1
= SUBSTITUTE (SUBSTITUTE (A1+B1*9*9, 9, 1), 8, 0)
regards
Stef@n
next solution for OR
=1*SUBSTITUTE (A1+A2;2;1)
regards
Stef@n
last solution for XOR
=1*SUBSTITUTE (A1+A2;2;0)
regards
Stef@n
Or you could make use of the VBA logical operators!
Define the following as custom functions
Public Function BITXOR(x As Long, y As Long)
BITXOR = x Xor y
End Function
Public Function BITAND(x As Long, y As Long)
BITAND = x And y
End Function
Public Function BITOR(x As Long, y As Long)
BITOR = x Or y
End Function
and then use them such:
A B =BITOR(A,B) =BITAND(A,B) =BITXOR(A,B)
0101 0100 0101 0100 0001
an another solution for AND
=1*SUBSTITUTE (SUBSTITUTE (A1+A2;1;0);2;1)
note:
the binary numbers are in A1 and A2 !
regards
Stef@n
I was obviously playing hooky at the beach during the bit-wise math lesson – you lost me at “Understanding bit-wise operations” 🙂
After looking at the above solutions, I find my solution silly, but still:
For the following formulae,
Row 1: headers,
Row 2: OR
Row 3: AND
Row 4: XOR
Column 1: Input 1
Column 2: Input 2
Column 3: Result
OR
{=SUM(IF(MID(A2,ROW(OFFSET($A$1,0,0,LEN(A2),1)),1)+MID(B2,ROW(OFFSET($A$1,0,0,LEN(B2),1)),1)>0,1,0)*10^(LEN(A2)-ROW(OFFSET($A$1,0,0,LEN(B2),1))))}
AND
{=SUM(IF(MID(A3,ROW(OFFSET($A$1,0,0,LEN(A3),1)),1)+MID(B3,ROW(OFFSET($A$1,0,0,LEN(B3),1)),1)=2,1,0)*10^(LEN(A3)-ROW(OFFSET($A$1,0,0,LEN(B3),1))))}
XOR
{=SUM(IF(MID(A4,ROW(OFFSET($A$1,0,0,LEN(A4),1)),1)+MID(B4,ROW(OFFSET($A$1,0,0,LEN(B4),1)),1)=1,1,0)*10^(LEN(A4)-ROW(OFFSET($A$1,0,0,LEN(B4),1))))}
@Anup
Please don't consider your solution silly
Firstly, You are the 3rd person to submit an answer
Secondly, The best formula/function is the one that you know and understand.
I think I have a very tedious solution, which people won't have the patience to do except in small numbers.
I used the same problem setup as "Anup Agarwal"
AND =IF(AND(MID(B2,1,1)="1",MID(C2,1,1)="1"),1,0)&IF(AND(MID(B2,2,1)="1",MID(C2,2,1)="1"),1,0)&IF(AND(MID(B2,3,1)="1",MID(C2,3,1)="1"),1,0)&IF(AND(MID(B2,4,1)="1",MID(C2,4,1)="1"),1,0)
OR =IF(OR(MID(B3,1,1)="1",MID(C3,1,1)="1"),1,0)&IF(OR(MID(B3,2,1)="1",MID(C3,2,1)="1"),1,0)&IF(OR(MID(B3,3,1)="1",MID(C3,3,1)="1"),1,0)&IF(OR(MID(B3,4,1)="1",MID(C3,4,1)="1"),1,0)
=IF(OR(AND(MID(B4,1,1)="1",MID(C4,1,1)="0"),AND(MID(B4,1,1)="0",MID(C4,1,1)="1")),1,0)&IF(OR(AND(MID(B4,2,1)="1",MID(C4,2,1)="0"),AND(MID(B4,2,1)="0",MID(C4,2,1)="1")),1,0)&IF(OR(AND(MID(B4,3,1)="1",MID(C4,3,1)="0"),AND(MID(B4,3,1)="0",MID(C4,3,1)="1")),1,0)&IF(OR(AND(MID(B4,4,1)="1",MID(C4,4,1)="0"),AND(MID(B4,4,1)="0",MID(C4,4,1)="1")),1,0)
Sorry my last post was totally messed up
AND
=IF(AND(MID(B2,1,1)="1",MID(C2,1,1)="1"),1,0)&IF(AND(MID(B2,2,1)="1",MID(C2,2,1)="1"),1,0)&IF(AND(MID(B2,3,1)="1",MID(C2,3,1)="1"),1,0)&IF(AND(MID(B2,4,1)="1",MID(C2,4,1)="1"),1,0)
OR
=IF(OR(MID(B3,1,1)="1",MID(C3,1,1)="1"),1,0)&IF(OR(MID(B3,2,1)="1",MID(C3,2,1)="1"),1,0)&IF(OR(MID(B3,3,1)="1",MID(C3,3,1)="1"),1,0)&IF(OR(MID(B3,4,1)="1",MID(C3,4,1)="1"),1,0)
XOR
=IF(OR(AND(MID(B4,1,1)="1",MID(C4,1,1)="0"),AND(MID(B4,1,1)="0",MID(C4,1,1)="1")),1,0)&IF(OR(AND(MID(B4,2,1)="1",MID(C4,2,1)="0"),AND(MID(B4,2,1)="0",MID(C4,2,1)="1")),1,0)&IF(OR(AND(MID(B4,3,1)="1",MID(C4,3,1)="0"),AND(MID(B4,3,1)="0",MID(C4,3,1)="1")),1,0)&IF(OR(AND(MID(B4,4,1)="1",MID(C4,4,1)="0"),AND(MID(B4,4,1)="0",MID(C4,4,1)="1")),1,0)
@stefan,
I just couldn't get your solutions to work.
01010101010 + 01010101110 = 02020210120
what am i doing wrong?
@anup
...I got yours to work!
@Stephen - I get the same, but Stef@an's second solution for AND does work (at least for the test cases I used)
@ Stephen / Rich
yes , you are right ! - only this works:
OR
=1*SUBSTITUTE (A1+A2;2;1)
XOR
=1*SUBSTITUTE (A1+A2;2;0)
AND
=1*SUBSTITUTE (SUBSTITUTE (A1+A2;1;0);2;1)
@Stef@n - You're answer is really smart, I never knew about the substitute function before. Great Work!
Thx Michael 🙂
yes - it is simply easy 😉
if you add 1 and 1 - excel calculate 2
and then you have to substitute the 2 - new = 0 respectively 1
Here is a good resource for people wanting to learn binary and hexadecimal.
http://justwebware.com/bitwise/bitwise.html
Three that weren't asked for:
NOT
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1+A2,0,3),1,0),3,1)
EQV
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1+A2,0,3),2,3),1,0),3,1)
IMP
=SUBSTITUTE(SUBSTITUTE(A1+SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A2,0,3),1,0),3,1),0,1),2,0)
(was using Daniel Ferry's bitwise file to verify against)
@ Kyle
Not only takes one parameter and inverts 0 -1 and 1-0
Took out the +A2
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,0,3),1,0),3,1)
Great solutions!
I'll add two:
NAND =1*SUBSTITUTE (A1+A2,2,0)
NOR=1*SUBSTITUTE(SUBSTITUTE (SUBSTITUTE(A1+A2,0,2),1,0),2,1)
This will work for binary numbers of any size (although the text format mask will have to have as many zeroes as there are digits in the longest addend)
Assume binary #s are in C35 & C36, then add and format as text in C37:
=TEXT(C36+C35,"000000000000")
-sum- = 101112211112
AND - SUBSTITUTE 0s for 1s in -sum-, then sub 1s for 2s
=SUBSTITUTE(SUBSTITUTE(C37,"1","0"),"2","1")
OR - sub 1s for 2s in -sum-
=SUBSTITUTE(C37,"2","1")
XOR - sub 0s for 2s in -sum-
=SUBSTITUTE(C37,"2","0")
Just wandered by:
AND:
=SUBSTITUTE(A1+A2,1,0)/2
Clever, Shane. I like that.
[…] post http://www.excelhero.com/blog/2010/01/5-and-3-is-1.html for examples using Sumproduct, and http://chandoo.org/wp/2011/07/29/bitwise-operations-in-excel/ for examples using Text […]
Hi Chandoo,
I am not (yet) really into bitwise calculation, but I am looking for a way to speed up my vba calculation with very big numbers. Would is ben convenient to use bitwise notation for this?
Best regards,
Ronald (the Netherlands)
p.s. love your country!
@Ronald
I'd suggest asking this in the Chandoo.org Forums
https://chandoo.org/forum/
Attach a sample file with an example of some data and describe what you want to achieve