This document discusses Monte Carlo simulation techniques for low power VLSI circuit design. It explains that Monte Carlo simulation involves running simulations with different input vectors multiple times to estimate power consumption statistics. It describes how the mean and variance of power consumption values are calculated from multiple samples and how the central limit theorem can be applied. It presents an equation to determine the minimum number of samples needed to estimate the mean power within a given error tolerance with a certain confidence level.

1. By ANURAG JAISWAL
10SETECE447
SHARDA UNIVERSITY

2.  SIMULATION VECTORS :- the primary inputs to the circuit to trigger the circuit activities.
 SIMULATION PROCESS :- designing the whole circuitry using simulation vectors and
verifying the output of the circuit.
 STOPPING CRITERIA :- when do we stop the simulation process so that the result is
accurate enough for the purpose.
 PROBABILITY DISTRIBUTION :- probability of occurrence of a random event in an
experiment.
 MEAN AND VARIANCE :- mean is the average. Variance is the deviation from the mean.
 CENTRAL LIMIT THEOREM :- the mean of a sufficiently large outcome of an experiment
will be approximately normally distributed.
 CONFIDENCE LEVEL :- given by 100(1-α)%. α is the variation from the exact mean µ.

3.  It is a technique of designing a circuit with low power consumption and by
repeatedly using some primary factors like the circuit block, simulation vectors like
power dissipation and clock time.
 The name “Monte Carlo” is given from the name of capital of Monaco where
gambling is major. The idea of Monte Carlo simulation itself came from the probability
of success while betting over a particular outcome in a casino game.
 In Low Power VLSI, several simulation vectors are kept repeatedly using in a Monte
Carlo simulation.

4.  We define a basic sample period T in which a single power dissipation value is
observed.
 T may be several vectors or several clock cycles.
 After a particular simulation period Ti, the power dissipation of the circuit „i.e. Pi‟ is
computed.
 Hence a series of power sample is obtained : P0,P1,P2,……PN.
 Estimated power dissipation is given by mean value of all Ps. i.e.
P = (P0+P1+P2+…….+PN) / N ….(i)

5. 1. If N is too small, P will not be nearby the actual mean µ.
2. If N is too large, unnecessary computation will be there without meaningful
accuracy.
Hence, there is a trade-off between sample size and accuracy. We nee d to find the
optimum sample size N.

6. ASSUMPTIONS:
I. Let us assume Pi has normal distribution pattern.
then according to the Central Limit Theorem, the sample mean P approaches the
normal distribution of Pi.
The basic statistical theory states that the average of normally distributed random
variables also has normal distribution. The mean of P is exactly µ and its variance is
given by:
(σp)^2 = (σ^2)/N ….(ii)
With increase in the sample size N, (σp)^2 decreases. So that we obtain more accurate
measures of true mean µ.

7.  Let us assume µ is positive. To quantify the accuracy of the sample mean P, we
describe a maximum error tolerance ε, typically with the values less than 10%.
 we consider the probability of occurrence of the same i.e. what is the probability that P
is within the ε error range of true mean µ.
OR
the probability of occurrence of the condition:
0 <= (|P- µ|)/µ <= ε … (iii)
if the probability is high then P is acceptable. Otherwise we’ll have to increase the
sample size N.

8.  The required probability is obtained by the normal distribution graph:

9.  The probability is more conveniently expressed by a confident level variable α.
 The confidence level is defined as 100(1-α)%.
 A confidence level of 100% means α =0. which means P is absolutely within the error
tolerance ε.
 Typically the confidence within the error tolerance of ε.
NOW, we have to form a relation between ε, α, and N. Also we have a variable Z α/2.
we know the area between
µ - (Z α/2)(σp) and µ - (Z α/2)(σp) is (1-α).

10.  From figure, the confidence level of the condition is:
|P-µ| <= (Z α/2)(σp) …. (v)
To ensure the condition;
(|P-µ|)/µ <= ε …. (vi)
We have;
(|P-µ|)/µ <= {(Z α/2)(σp)}/µ <= ε …. (vii)
Hence:
{(Z α/2)(σ)}/µ√N <= ε …. (viii) (using eq
i)

11. The above equation tells us that the minimum number of samples N to be taken in order
to be at least (1-α) confident that the error of P is within the tolerance of ε.