I would like to create a random, square table of 0's and 1's like so
0 1 0
0 0 1
0 1 0
but only bigger (around 14 x 14). The diagonal should be all 0's and the table should be symmetric across the diagonal. For example, this would not be good:
0 1 1
0 0 0
0 1 0
I would also like to have control over the number of 1 's that appear, or at least the probability that 1's appear.
I only need to make a few such tables, so this does not have to be fully automated by any means. I do not mind at all doing a lot of the work by hand.
If possible, I would greatly prefer doing this without coding in VBA (small code in cells is okay of course) since I do not know it at all.
Edit: amended so as to return a symmetrical array, as requested by the OP.
=LET(λ,RANDARRAY(ξ,ξ),IF(1-MUNIT(ξ),GESTEP(MOD(MMULT(λ,TRANSPOSE(λ)),1),1-ζ),0))
where ξ is the side length of the returned square array and ζ is an approximation as to the probability of a non-diagonal entry within that array being unity.
As ξ increases, so does the accuracy of ζ.
Can somebody please help me understand this question:
"(c) A wants to send a 500 byte packet to D through B. B is supposed to follow the store-andforward model, that is, B will receive the whole packet from A and then start transmitting the
packet to D. What is the end-to-end delay seen by the packet?"
A --> B (4 Mbps & 3000 km) and B --> D (10 Mbps & 900 km)
This is also assuming all data is sending at the speed of light (3 * 10pow(5) km/s
Im just really stuck on this question like I get the calculations for the most part, however I have no idea how to determine any of this.
It sounds like the questions aims at helping you understand different transit times of data of different speed/length links.
For A->B you should calculate how long the packet takes to transmit on a 4 megabit link. You then need to add the physical transit time, using the distance and the speed of light.
I.e. first you need to know how long it takes until the last bit is put on the link and then how long it takes for that bit to travel to the receiver.
When B has received that last bit it will forward the packet to D. You therefor need to repeat the calculation for the B->D part.
The sum of the two parts should be your answer.
I won't to the calculations for you, though.
Edit:
Ok, I get the feeling you really tried yourself, so here goes.
Transmission time
A->B:
4 Mbps = 4 000 000 bits/s
500 bytes = 500*8 bits = 4000 bits
Transmission time = Packet size / Bit rate => 4 000 / 4 000 000 => 0.001 s
Distance = 3 000 km
Propagation speed = 300 000 km/s
Propagation time = Distance / propagation speed => 3 000 / 300 000 = 0.01 s
Total time = 0.001 + 0.01 = 0.011 s
Now you do B->D and add the two parts.
Thanks for the help, I was actually able to figure it out. So I had to calculate the delay from A -> B, which after calculations turned out to be 11ms. Then I had to add that to the delay of B -> D which was 3.4ms which makes the delay from A -> D 14.4ms. Thanks for the help.
x-.y includes all items of x except for those that are cells of y
But what if I want to get all items that are cells of x and of y?
I can achieve this by
x -.^:2 y
But it require running expensive operation twice.
Is there a better solution?
e. is often useful when working with sets.
x e. y
gives a list of matches:
for each item of x return 1 if it exists in the "set" y, 0 otherwise.
1 2 3 4 e. 5 9 2
0 1 0 0
Then,
x (e. # [) y
selects those elements that do exist in both lists.
1 2 3 4 (e. # [) 5 9 2
2
5 8 (e. # [) i.12
5 8
Doing -. twice is the classic way of implementing intersection in J.
The inefficiency is minor (a constant factor - and, in general, you should not concern yourself with efficiency issues in J unless they exceed a factor of 2 - when you have resource problems you're generally going to want to focus on the factor of 1000 or greater issues).
Put differently, if ([-.-.) or -.^:2 is too slow for you then -. would also be too slow for you. (This can happen on extremely large data sets where the underlying implementation has been inefficient. Recent versions of J have had some work done, to correct this issue.)
Disappointing, perhaps, but practical.
Imagine you're generating the Fibonacci numbers using the obvious, brute-force algorithm. If I know the number of Fibonaccis I want to generate in advance, I can do something like this using the power conjunction ^::
(, [: +/ _2&{.)^:20 i.2
How can I instead stop when the Fibonaccis reach some limit, say 1e6? (I know how to do this inside a function using while., but that's no fun.)
I want to stress that this is a general question about J, not a specific question about Fibonacci. Don't get too distracted by Fibonacci numbers. The heart of the question is how to keep appending to a list until some condition is met.
Power has also a verb form u^:v^:n where the second verb can be used as a check. Eg: double (+:) while (n is _) less than 100 (100&>):
+:^:(100&>)^:_ ] 1
128
+:^:(100&>)^:_ ] 3
192
As usual, to append to the result of power, you box the noun:
+:^:(100&>)^:(<_) ] 3
3 6 12 24 48 96 192
I think the best answer to this is in Henry Rich's book J for C programmers. Specifically, it using the Power Conjunction ^: . You can also use it to converge until there is no change, so that the limit would not need to be defined. Henry uses the example that:
2 *^:(100&>#:])^:_"0 (1 3 5 7 9 11)
128 192 160 112 144 176
The ^:_ Power Conjunction repeats until there is no change and the ^:(100&>#:]) tests for the result being less than 100. If it is then ^: is applied to 1 and the loop 2* is done again, if it is not less than 100 then ^: would be applied to 0 and that results in it doing nothing and nothing changes and the loop exits. The fact that it use "0 as the rank means that it can apply the doubling function 2* to each of 1 3 5 7 9 11 individually.
Henry really does explain the process better than I, so here is the reference for further reading.
http://www.jsoftware.com/help/jforc/loopless_code_iv_irregular_o.htm#_Toc191734389
I am trying to calculate percentage growth in excel with a positive and negative number.
This Year's value: 2434
Last Year's value: -2
formula I'm using is:
(This_Year - Last_Year) / Last_Year
=(2434 - -2) / -2
The problem is I get a negative result. Can an approximate growth number be calculated and if so how?
You could try shifting the number space upward so they both become positive.
To calculate a gain between any two positive or negative numbers, you're going to have to keep one foot in the magnitude-growth world and the other foot in the volume-growth world. You can lean to one side or the other depending on how you want the result gains to appear, and there are consequences to each choice.
Strategy
Create a shift equation that generates a positive number relative to the old and new numbers.
Add the custom shift to the old and new numbers to get new_shifted and old_shifted.
Take the (new_shifted - old_shifted) / old_shifted) calculation to get the gain.
For example:
old -> new
-50 -> 30 //Calculate a shift like (2*(50 + 30)) = 160
shifted_old -> shifted_new
110 -> 190
= (new-old)/old
= (190-110)/110 = 72.73%
How to choose a shift function
If your shift function shifts the numbers too far upward, like for example adding 10000 to each number, you always get a tiny growth/decline. But if the shift is just big enough to get both numbers into positive territory, you'll get wild swings in the growth/decline on edge cases. You'll need to dial in the shift function so it makes sense for your particular application. There is no totally correct solution to this problem, you must take the bitter with the sweet.
Add this to your excel to see how the numbers and gains move about:
shift function
old new abs_old abs_new 2*abs(old)+abs(new) shiftedold shiftednew gain
-50 30 50 30 160 110 190 72.73%
-50 40 50 40 180 130 220 69.23%
10 20 10 20 60 70 80 14.29%
10 30 10 30 80 90 110 22.22%
1 10 1 10 22 23 32 39.13%
1 20 1 20 42 43 62 44.19%
-10 10 10 10 40 30 50 66.67%
-10 20 10 20 60 50 80 60.00%
1 100 1 100 202 203 302 48.77%
1 1000 1 1000 2002 2003 3002 49.88%
The gain percentage is affected by the magnitude of the numbers. The numbers above are a bad example and result from a primitive shift function.
You have to ask yourself which critter has the most productive gain:
Evaluate the growth of critters A, B, C, and D:
A used to consume 0.01 units of energy and now consumes 10 units.
B used to consume 500 units and now consumes 700 units.
C used to consume -50 units (Producing units!) and now consumes 30 units.
D used to consume -0.01 units (Producing) and now consumes -30 units (producing).
In some ways arguments can be made that each critter is the biggest grower in their own way. Some people say B is best grower, others will say D is a bigger gain. You have to decide for yourself which is better.
The question becomes, can we map this intuitive feel of what we label as growth into a continuous function that tells us what humans tend to regard as "awesome growth" vs "mediocre growth".
Growth a mysterious thing
You then have to take into account that Critter B may have had a far more difficult time than critter D. Critter D may have far more prospects for it in the future than the others. It had an advantage! How do you measure the opportunity, difficulty, velocity and acceleration of growth? To be able to predict the future, you need to have an intuitive feel for what constitutes a "major home run" and a "lame advance in productivity".
The first and second derivatives of a function will give you the "velocity of growth" and "acceleration of growth". Learn about those in calculus, they are super important.
Which is growing more? A critter that is accelerating its growth minute by minute, or a critter that is decelerating its growth? What about high and low velocity and high/low rate of change? What about the notion of exhausting opportunities for growth. Cost benefit analysis and ability/inability to capitalize on opportunity. What about adversarial systems (where your success comes from another person's failure) and zero sum games?
There is exponential growth, liner growth. And unsustainable growth. Cost benefit analysis and fitting a curve to the data. The world is far queerer than we can suppose. Plotting a perfect line to the data does not tell you which data point comes next because of the black swan effect. I suggest all humans listen to this lecture on growth, the University of Colorado At Boulder gave a fantastic talk on growth, what it is, what it isn't, and how humans completely misunderstand it. http://www.youtube.com/watch?v=u5iFESMAU58
Fit a line to the temperature of heated water, once you think you've fit a curve, a black swan happens, and the water boils. This effect happens all throughout our universe, and your primitive function (new-old)/old is not going to help you.
Here is Java code that accomplishes most of the above notions in a neat package that suits my needs:
Critter growth - (a critter can be "radio waves", "beetles", "oil temprature", "stock options", anything).
public double evaluate_critter_growth_return_a_gain_percentage(
double old_value, double new_value) throws Exception{
double abs_old = Math.abs(old_value);
double abs_new = Math.abs(new_value);
//This is your shift function, fool around with it and see how
//It changes. Have a full battery of unit tests though before you fiddle.
double biggest_absolute_value = (Math.max(abs_old, abs_new)+1)*2;
if (new_value <= 0 || old_value <= 0){
new_value = new_value + (biggest_absolute_value+1);
old_value = old_value + (biggest_absolute_value+1);
}
if (old_value == 0 || new_value == 0){
old_value+=1;
new_value+=1;
}
if (old_value <= 0)
throw new Exception("This should never happen.");
if (new_value <= 0)
throw new Exception("This should never happen.");
return (new_value - old_value) / old_value;
}
Result
It behaves kind-of sort-of like humans have an instinctual feel for critter growth. When our bank account goes from -9000 to -3000, we say that is better growth than when the account goes from 1000 to 2000.
1->2 (1.0) should be bigger than 1->1 (0.0)
1->2 (1.0) should be smaller than 1->4 (3.0)
0->1 (0.2) should be smaller than 1->3 (2.0)
-5-> -3 (0.25) should be smaller than -5->-1 (0.5)
-5->1 (0.75) should be smaller than -5->5 (1.25)
100->200 (1.0) should be the same as 10->20 (1.0)
-10->1 (0.84) should be smaller than -20->1 (0.91)
-10->10 (1.53) should be smaller than -20->20 (1.73)
-200->200 should not be in outer space (say more than 500%):(1.97)
handle edge case 1-> -4: (-0.41)
1-> -4: (-0.42) should be bigger than 1-> -9:(-0.45)
Simplest solution is the following:
=(NEW/OLD-1)*SIGN(OLD)
The SIGN() function will result in -1 if the value is negative and 1 if the value is positive. So multiplying by that will conditionally invert the result if the previous value is negative.
Percentage growth is not a meaningful measure when the base is less than 0 and the current figure is greater than 0:
Yr 1 Yr 2 % Change (abs val base)
-1 10 %1100
-10 10 %200
The above calc reveals the weakness in this measure- if the base year is negative and current is positive, result is N/A
It is true that this calculation does not make sense in a strict mathematical perspective, however if we are checking financial data it is still a useful metric. The formula could be the following:
if(lastyear>0,(thisyear/lastyear-1),((thisyear+abs(lastyear)/abs(lastyear))
let's verify the formula empirically with simple numbers:
thisyear=50 lastyear=25 growth=100% makes sense
thisyear=25 lastyear=50 growth=-50% makes sense
thisyear=-25 lastyear=25 growth=-200% makes sense
thisyear=50 lastyear=-25 growth=300% makes sense
thisyear=-50 lastyear=-25 growth=-100% makes sense
thisyear=-25 lastyear=-50 growth=50% makes sense
again, it might not be mathematically correct, but if you need meaningful numbers (maybe to plug them in graphs or other formulas) it's a good alternative to N/A, especially when using N/A could screw all subsequent calculations.
You should be getting a negative result - you are dividing by a negative number. If last year was negative, then you had negative growth. You can avoid this anomaly by dividing by Abs(Last Year)
Let me draw the scenario.
From: -303 To 183, what is the percentage change?
-303, -100% 0 183, 60.396% 303, 100%
|_________________ ||||||||||||||||||||||||________|
(183 - -303) / |-303| * 100 = 160.396%
Total Percent Change is approximately 160%
Note: No matter how negative the value is, it is treated as -100%.
The best way to solve this issue is using the formula to calculate a slope:
(y1-y2/x1-x2)
*define x1 as the first moment, so value will be "C4=1"
define x2 as the first moment, so value will be "C5=2"
In order to get the correct percentage growth we can follow this order:
=(((B4-B5)/(C4-C5))/ABS(B4))*100
Perfectly Works!
Simplest method is the one I would use.
=(ThisYear - LastYear)/(ABS(LastYear))
However it only works in certain situations. With certain values the results will be inverted.
It really does not make sense to shift both into the positive, if you want a growth value that is comparable with the normal growth as result of both positive numbers. If I want to see the growth of 2 positive numbers, I don't want the shifting.
It makes however sense to invert the growth for 2 negative numbers. -1 to -2 is mathematically a growth of 100%, but that feels as something positive, and in fact, the result is a decline.
So, I have following function, allowing to invert the growth for 2 negative numbers:
setGrowth(Quantity q1, Quantity q2, boolean fromPositiveBase) {
if (q1.getValue().equals(q2.getValue()))
setValue(0.0F);
else if (q1.getValue() <= 0 ^ q2.getValue() <= 0) // growth makes no sense
setNaN();
else if (q1.getValue() < 0 && q2.getValue() < 0) // both negative, option to invert
setValue((q2.getValue() - q1.getValue()) / ((fromPositiveBase? -1: 1) * q1.getValue()));
else // both positive
setValue((q2.getValue() - q1.getValue()) / q1.getValue());
}
These questions are answering the question of "how should I?" without considering the question "should I?" A change in the value of a variable that takes positive and negative values is fairly meaning less, statistically speaking. The suggestion to "shift" might work well for some variables (e.g. temperature which can be shifted to a kelvin scale or something to take care of the problem) but very poorly for others, where negativity has a precise implication for direction. For example net income or losses. Operating at a loss (negative income) has a precise meaning in this context, and moving from -50 to 30 is not in any way the same for this context as moving from 110 to 190, as a previous post suggests. These percentage changes should most likely be reported as "NA".
Just change the divider to an absolute number.i.e.
A B C D
1 25,000 50,000 75,000 200%
2 (25,000) 50,000 25,000 200%
The formula in D2 is: =(C2-A2)/ABS(A2) compare with the all positive row the result is the same (when the absolute base number is the same). Without the ABS in the formula the result will be -200%.
Franco
Use this code:
=IFERROR((This Year/Last Year)-1,IF(AND(D2=0,E2=0),0,1))
The first part of this code iferror gets rid of the N/A issues when there is a negative or a 0 value. It does this by looking at the values in e2 and d2 and makes sure they are not both 0. If they are both 0 then it will place a 0%. If only one of the cells are a 0 then it will place 100% or -100% depending on where the 0 value falls. The second part of this code (e2/d2)-1 is the same code as (this year - lastyear)/Last year
Please click here for example picture
I was fumbling for answers today, and think this would work...
=IF(C5=0, B5/1, IF(C5<0, (B5+ABS(C5)/1), IF(C5>0, (B5/C5)-1)))
C5 = Last Year, B5 = This Year
We have 3 IF statements in the cell.
IF Last Year is 0, then This Year divided by 1
IF Last Year is less than 0, then This Year + ABSolute value of Last Year divided by 1
IF Last Year is greater than 0, then This Year divided by Last Year minus 1
Use this formula:
=100% + (Year 2/Year 1)
The logic is that you recover 100% of the negative in year 1 (hence the initial 100%) plus any excess will be a ratio against year 1.
Short one:
=IF(D2>C2, ABS((D2-C2)/C2), -1*ABS((D2-C2)/C2))
or confusing one (my first attempt):
=IF(D2>C2, IF(C2>0, (D2-C2)/C2, (D2-C2)/ABS(C2)), IF(OR(D2>0,C2>0), (D2-C2)/C2, IF(AND(D2<0, C2<0), (D2-C2)/ABS(C2), 0)))
D2 is this year, C2 is last year.
Formula should be this one:
=(thisYear+IF(LastYear<0,ABS(LastYear),0))/ABS(LastYear)-100%
The IF value if < 0 is added to your Thisyear value to generate the real difference.
If > 0, the LastYear value is 0
Seems to work in different scenarios checked
This article offers a detailed explanation for why the (b - a)/ABS(a) formula makes sense. It is counter-intuitive at first, but once you play with the underlying arithmetic, it starts to make sense. As you get used to it eventually, it changes the way you look at percentages.
Aim is to get increase rate.
Idea is following:
At first calculate value of absolute increase.
Then value of absolute increase add to both, this and last year values. And then calculate increase rate, based on the new values.
For example:
LastYear | ThisYear | AbsoluteIncrease | LastYear01 | ThisYear01 | Rate
-10 | 20 | 30 = (10+20) | 20=(-10+30)| 50=(20+30) | 2.5=50/20
-20 | 20 | 40 = (20+20) | 20=(-20+40)| 60=(20+40) | 3=60/2
=(This Year - Last Year) / (ABS(Last Year))
This only works reliably if this year and last year are always positive numbers.
For example last_year=-50 this_year = -1. You get -100% growth when in fact the numbers have improved a great deal.