[2 days left] Whatâ€™s wrong with your cloud strategy? Learn why multicloud solutions matter with Nimble Storage.Register Now

x
Solved

# Random Number + Triangular Distribution

Posted on 2000-03-24
Medium Priority
848 Views
I need to generate a series of random numbers based on a TRIANGULAR DISTRIBUTION. Preference would be not to utilise the inbuilt Rnd function
0
Question by:jsteve
[X]
###### Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

• Help others & share knowledge
• Earn cash & points

LVL 28

Expert Comment

ID: 2656166
what do you mean by a triangular distribution?
0

LVL 14

Expert Comment

ID: 2656167
Why not the Rnd function? Its statistically appropriate and built in. The only statistical fault it has is that first you have to seed it.. but.. ALL random number genrators have to do that. As such.. there is no difference between the VB Rnd and anything else on the market.. <smile>.

As to the Triangular Distribution.. if you think I'm gonna blow the dust off of my old Statistic books.. you got another think coming.. <lol>. Let's see some empirical formulae.. <smile>.
0

LVL 3

Expert Comment

ID: 2656676
Why not

Dim N As Single
Randomize(0, N)
0

LVL 1

Expert Comment

ID: 2659142
What is the shape of the triangular distribution.? A right triangle with a positive slope? A symetrical triangle with a peak at the midpoint? or?

The way to generate your distribution is to write an equation for Y vs. X. A triangle will have to be in segments. Integerate this and get a funtion of the integral of Y vs X. Normalize the Y values so they sum to 1. Then using the rnd function, generate a number from 0 to 1. Use this as a Y value and determine the related value for X from the equation of the normalized integral of Y vs. X.
0

LVL 1

Expert Comment

ID: 2659454
A symterical triangular distribution that goes from X1 to X2 with a peak at (X1+X2)/2 (i.e. the midpoint) can be formed by the sum of two uniform distributions that go from X1/2 to X2/2. Each of these can be easily generated from the VB rnd function that gives a uniform distribution from 0 to 1 by multiplying and adding a constant to a variable formed by the rnd function.

Try the following code:

Z1 = 1/2(X1 + rnd(X2-X1))
Z2 = 1/2(X1 + rnd(X2-X1))
Z3 = Z1 + Z2

Z3 should be a random variable with a symetrical triangular distribution from X1 to X2 with a peak at (X1 + X2)/2.

If you want a different triangular distribution, define its shape and I will try to provide some code.
0

Author Comment

ID: 2659749
Adjusted points from 150 to 200
0

Author Comment

ID: 2659750
Almost there. The distribution that I will be using is not symetrical. Could you provide the code. I will increase the points to 200.

Thanks
0

LVL 1

Expert Comment

ID: 2661814
I need the values describing the triangle. What are the 3 X values for
the begining of the triangle, the peak, and the end point. I don't need the Y values, since I will assume the triangle area corresponding to the probability must sum to one.

If you provide these I think I can give code to do this. But it will use the VB rnd function somewhere in the code.
0

LVL 1

Accepted Solution

MinnEE earned 800 total points
ID: 2662657
After some trial and error, I derived general equations for this. I checked the code by running 10000 variables for various shaoes and did a histogram on them.

Triangular distribution

Note: pdf = probability density function
cdf = cumulative probability

Assume the pdf goes from X1 to X2 with a peak at XP. Then the cdf =
the area under the curve of the pdf.

Let A1 = the area from X1 to Xp
A2 = the area from Xp to X2
A1 + A2 = 1   (sum of all possible probs = 1)

Writing equations for Ax as a function of X and solving for X gives
the following code. z will be a random variable with the desired
triangular distribution.

A1 = (Xp-X1)/(X2-X1) ' area up to peak
A2 = 1-A1          ' area after peak
r = Rnd
If r < = A1 Then
z = X1+Sqr(r*(X2-X1)*(Xp-X1))
Else
z = X2-Sqr((1-r)*(X2-Xp)*(X2-X1))
End If
0

LVL 1

Expert Comment

ID: 2714649
Calculate mean and standard deviation of triangular pdf distribution.

Let a = lowest value in distribution
b = peak of distribution
c = largest value in distribution

Then
Mean = (a + b + c)/3

Let
V = (c^3-a^3+3*b*(c^2-a^2))/(9*(c-a))
W = ((b-a)^3+(c-b)^3)/(18*(c-a))

Then
Std Dev = Sqrt(V + W - Mean^2)

Derived by RJP
RJPetsch @aol.com
0

## Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Here are a few simple, working, games that you can use as-is or as the basis for your own games. Tic-Tac-Toe This is one of the simplest of all games. Â  The game allows for a choice of who goes first and keeps track of the number of wins forâ€¦
This article describes some techniques which will make your VBA or Visual Basic Classic code easier to understand and maintain, whether by you, your replacement, or another Experts-Exchange expert.
Get people started with the utilization of class modules. Class modules can be a powerful tool in Microsoft Access. They allow you to create self-contained objects that encapsulate functionality. They can easily hide the complexity of a process fromâ€¦
Show developers how to use a criteria form to limit the data that appears on an Access report. It is a common requirement that users can specify the criteria for a report at runtime. The easiest way to accomplish this is using a criteria form that aâ€¦
###### Suggested Courses
Course of the Month14 days, 9 hours left to enroll