November/December 2006
Palisade User Conference Americas Big Success
From Tacoma, Washington to Bucaramanga, Colombia, scores of risk management
professionals from all corners of the globe descended on Miami, Florida for the 2006
Palisade User Conference: Americas on
November 13-14.

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Seminar Schedule
:: Regional Seminars
Risk and Decision Assessment using @RISK and the DecisionTools Suite
December 4-5, Chicago
December 11-12, Houston
January 8-9, San Diego
January 15-16, Atlanta
January 29-30, San Juan

Risk Assessment:
Oil & Gas Focus

December 14-15, Houston

Project Risk Assessment
Using @RISK for Project

January 17-18, Atlanta

:: Live Web Training
Risk and Decision Assessment using @RISK, Part I
December 14-15
January 22-23

Risk and Decision Assessment using @RISK, Part II
December 18-19
January 25-26

full schedule and registration

Free Live Webcasts
Overview of @RISK for Project
December 14

Selecting the Right Distribution
December 20

Data Collection with
@RISK for Project

January 9

full schedule and registration

Ask Amy
Expert Answers to
Technical Questions

Dear Amy,

Can I determine how many iterations I need to run in my simulation so that the estimate of the mean is calculated within a specific confidence interval?

— J.H.

Dear J.H.,

Yes you can. In the example here, let’s suppose that we want to use simulation to estimate the mean of the output in cell B11 and be accurate within 10 units 95% of the time. The number of iterations needed to meet these requirements can be calculated using the following formula:


In this formula,

n is the number of iterations needed.
S is the estimated standard deviation of the output.
D is the desired width of the confidence interval (in this case, 10 units).

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Ask Amy continued from above

za/2 is the number that satisfies P(Z>za/2) = a/2, where Z follows a normal distribution with mean 0 and standard deviation 1. In words, za/2 is the z value such that the area of the right-hand tail is a/2. a/2 can be found by setting the desired confidence level equal to 100(1−a) and solving for a.

In this example (a 95% confidence level):

95 = 100(1−a)

Then a is 0.05 and a/2 is 0.025. To compute za/2 in Excel, use the NORMSINV function and enter =NORMSINV(1−a/2, 0, 1). Cell E13 of the attached example shows a Z value of approximately 1.96 for a 95% confidence interval.

To obtain an estimate for the standard deviation of the output, the @RISK statistics function RiskStdDev was placed in cell B14 and a simulation was run with just 100 iterations. This gave us a standard deviation of approximately 58.7. If we plug the above information into our formula we get

n = [ 1.96 × 58.7 / 10 ] ² = 133

Thus, at least 133 iterations should be run to be 95% sure that our estimate of the mean of the output in cell B11 is accurate within 10 units.

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