Comparing Two Values to get the percentage increase?
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Hey guys, How do you compare two values and say that value 1 is X% larger than value 2 or value 1 is x fold larger than value 2? Thanks! :)
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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Hey guys, How do you compare two values and say that value 1 is X% larger than value 2 or value 1 is x fold larger than value 2? Thanks! :)
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
Ian Uy wrote:
How do you compare two values and say that value 1 is X% larger than value 2
double value1 = 150.0
double value2 = 100.0double difference = value1 - value2;
double percentageIncrease = (difference / value2) * 100;It would be said that value1 is 50% larger than value2.
Ian Uy wrote:
value 1 is x fold larger than value 2
Technically, "x fold larger" is a suffix you add to a number to say "multipled by."
double value1 = 200.0;
double value2 = 50.0;double foldIncrease = value1 / value2;
It would be said that value1 is four-fold larger than value2. Enjoy, Robert C. Cartaino
modified on Tuesday, August 19, 2008 4:48 PM
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Ian Uy wrote:
How do you compare two values and say that value 1 is X% larger than value 2
double value1 = 150.0
double value2 = 100.0double difference = value1 - value2;
double percentageIncrease = (difference / value2) * 100;It would be said that value1 is 50% larger than value2.
Ian Uy wrote:
value 1 is x fold larger than value 2
Technically, "x fold larger" is a suffix you add to a number to say "multipled by."
double value1 = 200.0;
double value2 = 50.0;double foldIncrease = value1 / value2;
It would be said that value1 is four-fold larger than value2. Enjoy, Robert C. Cartaino
modified on Tuesday, August 19, 2008 4:48 PM
Uhmm. I have a problem. The two values are MILES APART and I can't find a calculator powerful enough to compute the percentage increase. Value 1 = 1.36 x 10^58 Value 2 = 2.36 x 10^19,709 Is there any mathematical technique or approach to simplify this?
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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Uhmm. I have a problem. The two values are MILES APART and I can't find a calculator powerful enough to compute the percentage increase. Value 1 = 1.36 x 10^58 Value 2 = 2.36 x 10^19,709 Is there any mathematical technique or approach to simplify this?
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
Ian Uy wrote:
Value 2 = 2.36 x 10^19,709 Is there any mathematical technique or approach to simplify this?
The rules for addition/subtraction/multiplication/division of scientific notation are not difficult. Here is an article on Scientific Notation[^] which covers how to multiply and divide powers of ten. Here is another one[^] which also covers addition and subtraction.
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Ian Uy wrote:
Value 2 = 2.36 x 10^19,709 Is there any mathematical technique or approach to simplify this?
The rules for addition/subtraction/multiplication/division of scientific notation are not difficult. Here is an article on Scientific Notation[^] which covers how to multiply and divide powers of ten. Here is another one[^] which also covers addition and subtraction.
Yeah, I already did that before asking. However, I am not sure if its correct. To subtract value 1 from value 2, I need to move value 1's decimal point (19709-58)=19651 times to the right? this will produce a relatively small number. And if I subtract that small number to value 2, the result will not be different. It is still 2.36 x 10^19,709. Right?
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
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Yeah, I already did that before asking. However, I am not sure if its correct. To subtract value 1 from value 2, I need to move value 1's decimal point (19709-58)=19651 times to the right? this will produce a relatively small number. And if I subtract that small number to value 2, the result will not be different. It is still 2.36 x 10^19,709. Right?
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
Ian Uy wrote:
this will produce a relatively small number. And if I subtract that small number to value 2, the result will not be different. It is still 2.36 x 10^19,709. Right?
Pretty much. Consider your numbers.
Value 2 = 2.36 x 10^19,709
There are about 1.0 x 10^80 atoms in the known universe. If we discovered that there were actually a billion other universes out there, there would still only be 1.0 x 10^89 atoms in all of them combined. So your number (2.36 x 10^19,709) is what would be known in the mathematical community as "big... really big." Relatively speaking (compared to your first number)...
Value 1 = 1.36 x 10^58
...is really, really small. Pretty much, for all intents and purposes, practically zero. So your question boils down to something like, "If I take all the matter in the in the known universe -- every tree, every person, every planet, every sun, every black hole contained in all 100-billion known galaxies -- and manage to sneeze into a parallel dimension, what percentage did I decrease the weight of the universe? And you would still be way off from your example. Enjoy, Robert C. Cartaino
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Ian Uy wrote:
this will produce a relatively small number. And if I subtract that small number to value 2, the result will not be different. It is still 2.36 x 10^19,709. Right?
Pretty much. Consider your numbers.
Value 2 = 2.36 x 10^19,709
There are about 1.0 x 10^80 atoms in the known universe. If we discovered that there were actually a billion other universes out there, there would still only be 1.0 x 10^89 atoms in all of them combined. So your number (2.36 x 10^19,709) is what would be known in the mathematical community as "big... really big." Relatively speaking (compared to your first number)...
Value 1 = 1.36 x 10^58
...is really, really small. Pretty much, for all intents and purposes, practically zero. So your question boils down to something like, "If I take all the matter in the in the known universe -- every tree, every person, every planet, every sun, every black hole contained in all 100-billion known galaxies -- and manage to sneeze into a parallel dimension, what percentage did I decrease the weight of the universe? And you would still be way off from your example. Enjoy, Robert C. Cartaino
Comparing the larger number to the smaller, it is (2.36/1.36)*10^19651 larger. 10^19728 is just less than 2^64K. This can be expressed exactly by an extended arithmetic binary number 64K BITS long, or 8K BYTES long, or 2048 DWORDS long.
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Uhmm. I have a problem. The two values are MILES APART and I can't find a calculator powerful enough to compute the percentage increase. Value 1 = 1.36 x 10^58 Value 2 = 2.36 x 10^19,709 Is there any mathematical technique or approach to simplify this?
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
Ian Uy wrote:
Value 1 = 1.36 x 10^58 Value 2 = 2.36 x 10^19,709 Is there any mathematical technique or approach to simplify this?
In that case Value2 - Value1 = Value2 accurate to 19,651 decimal places. Of course, one would have to question whether percentage increase is a meaningful method of displaying the difference.
Peter "Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Ian Uy wrote:
this will produce a relatively small number. And if I subtract that small number to value 2, the result will not be different. It is still 2.36 x 10^19,709. Right?
Pretty much. Consider your numbers.
Value 2 = 2.36 x 10^19,709
There are about 1.0 x 10^80 atoms in the known universe. If we discovered that there were actually a billion other universes out there, there would still only be 1.0 x 10^89 atoms in all of them combined. So your number (2.36 x 10^19,709) is what would be known in the mathematical community as "big... really big." Relatively speaking (compared to your first number)...
Value 1 = 1.36 x 10^58
...is really, really small. Pretty much, for all intents and purposes, practically zero. So your question boils down to something like, "If I take all the matter in the in the known universe -- every tree, every person, every planet, every sun, every black hole contained in all 100-billion known galaxies -- and manage to sneeze into a parallel dimension, what percentage did I decrease the weight of the universe? And you would still be way off from your example. Enjoy, Robert C. Cartaino
Thank you for taking the time to answer my question. I understand it now. :)
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
-
Uhmm. I have a problem. The two values are MILES APART and I can't find a calculator powerful enough to compute the percentage increase. Value 1 = 1.36 x 10^58 Value 2 = 2.36 x 10^19,709 Is there any mathematical technique or approach to simplify this?
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.
Ian Uy wrote:
Is there any mathematical technique or approach to simplify this?
Sure! Given V1 = 1.36 x 10^58 and V2 = 2.36 x 10^19709, and %Increase = [(V2 - V1)/V1] x 100: 1. (V2 - V1)/V1 = V2/V1 - V1/V1 = V2/V1 - 1 2. V2/V1 = 10^(log V2 - log V1) 3. Subtract 1, then multiply by 100 to get % so, 1. (2.36 x 10^19709 - 1.36 x 10^58)/1.36 x 10^58 = (2.36 x 10^19709/1.36 x 10^58) - 1 2. V2/V1 = 10^(log( 2.36 x 10^19709 ) - log( 1.36 x 10^58)) = 10^ (19709.00864 - 58.13354) = 10^ 19650.87510 3. % = (10^19650.87510 -1) x 100. You're still going to have a hell of a time getting a computer to calculate a number that large, but the use of logarithms reduces the size of the intermediate results to a manageable magnitude. :)
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
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Ian Uy wrote:
Is there any mathematical technique or approach to simplify this?
Sure! Given V1 = 1.36 x 10^58 and V2 = 2.36 x 10^19709, and %Increase = [(V2 - V1)/V1] x 100: 1. (V2 - V1)/V1 = V2/V1 - V1/V1 = V2/V1 - 1 2. V2/V1 = 10^(log V2 - log V1) 3. Subtract 1, then multiply by 100 to get % so, 1. (2.36 x 10^19709 - 1.36 x 10^58)/1.36 x 10^58 = (2.36 x 10^19709/1.36 x 10^58) - 1 2. V2/V1 = 10^(log( 2.36 x 10^19709 ) - log( 1.36 x 10^58)) = 10^ (19709.00864 - 58.13354) = 10^ 19650.87510 3. % = (10^19650.87510 -1) x 100. You're still going to have a hell of a time getting a computer to calculate a number that large, but the use of logarithms reduces the size of the intermediate results to a manageable magnitude. :)
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
Roger Wright wrote:
You're still going to have a hell of a time getting a computer to calculate a number that larg
I beg to differ, it does not take that long. I made the calculations (I am only reporting the digit count of the various values, but I can report the actual values if desired): 14.984 msec to calculate (14,984 seconds for 1000 loops). Chars in 10^56 = 57 Chars in 136*10^56 (=V1) = 63 Chars in 10^19707 = 19708 Chars in 236*10^19707 (=V2) = 19714 Chars in (V2-V1) = 19714 Chars in quotient of (V2-V1)/V1 = 19652 Chars in remainder of (V2-V1)/V1 = 63 Chars in quotient of V2/V1 = 19652 Chars in remainder of V2/V1 = 63 Dave Augustine.