HO 4
CSC 354S
SYSTEMS MODELING AND DISCRETE SIMULATION

	SPRING 1997
	UNIVERSITY OF TORONTO

MIDTERM TEST

DURATION - 50 minutes

Test aids allowed:  One 8.5 X 11" fact sheet, possibly double-sided

Notes to students: 
A.  There are a total of 50 marks and 50 minutes to obtain them.  
B.  The test paper has 6 questions.  Write your answers on the paper or booklets provided.
C.  In general, state your assumptions and show your intermediate work leading to an answer.  

1.  [5 marks]
Explain in sufficient detail how the Utilization Law is really just Little's Law in disguise.
	Little's Law, applied to some system, says that the mean number in that system is equal  to 
	the product of the arrival rate to the system and the mean time spent in the system.  If the 
	system represents the server only, then the mean number there is the utilization (the 
	probability of finding someone there) and the mean time spent there is the mean service 
	time.  If the mean arrival rate to the server is equal to the mean rate of work that flows 
	through the server, then the Utilization Law (the utilization of a server is equal to the 
	product of the throughput through the server and the mean service time) is really just 
	Little's Law.

2.  [6 marks] 
State 2 reasons why we want the generation of a random number stream to be reproducible.
	I. It makes the debugging of a computer program for a simulation easier, because you can 
	re-test execution flow with the same inputs after doing a bug fix.
	II.  It makes the comparison of 2 or more systems easier when doing simulations, because 
	the systems are subjected to the same workload.

3.  [5 marks]
In a multiplicative LCG with modulus m , why is the maximum period  m-1 ?
	The absolute maximum period for any LCG is m . For a multiplicative LCG, 0 cannot be a 
	valid value, as that would cause immediate cycling in a multiplicative LCG.  Therefore, the 
	maximum period for such an LCG is m - 1 .

4.  [10 marks]
Let X  be a random variable that is normally distributed with a mean m of 10 and a variance of  
4. Using reasonable interpolations in the absence of a calculator, find the values of   a  and  b  
such that
	i)  P ( a < X < b )  =  0.90 , and
	ii)  |  m - a |  =  |  m - b |
	P ( X < b )  =  P ( X > a )  =  0. 95 , by symmetry
	F ( [ ( b - 10 ) / 2 ]  =  .95 , so by table lookup,  b = 13.3
	Immediately by symmetry,  a  =  6.7

5.  [10 marks]
Suppose that  X and  Y  are random variables such that  Y  =  aX + b , where  a  and  b  are 
constants. Show that
	correlation ( X , Y )  =  + 1  if  a  >  0
	                                         - 1  if  a  <  0
	cov ( X , Y )  =  cov ( X , aX + b )  =  a Var ( X )
	Var ( Y )  =  Var ( aX + b )  =  a2 Var ( X )
	corr ( X , Y )  =  cov ( X , Y ) / ™ [ Var ( X ) Var ( Y ) ]
	                       =  a Var ( X ) / ™ [ Var (X ) a2 Var ( X ) ]
	                       =  a / ™ (a2)
	                       =  + 1 if a > 0 , and -1 if a < 0

6.  [14 marks]
Show how the composition method can be used to generate a variate with probability density 
function
	                 x	0  œ  x  œ  1
	 f ( x )  =    2  -  x	1  œ  x  œ  2
	                 0	elsewhere
	f  is a triangular distribution.  Rewrite  f  as  f(x)  =  0.5 f1(x) + 0.5 f2(x), 
	where
		f1 ( x )  =  2 x ,          0  œ  x  œ  1
		f2 ( x )  =  4  -  2 x  ,  1  œ  x  œ 2
then F1( x )  =  x2  on the appropriate interval, and  F2( x )  =  4 x - x2 - 3  
on the appropriate interval.  F1-1( u )  =  ™  ( u ) , and  F2-1( u )  =  2  -  ™ (1  - u) . 
Generate a u1 ~  U (0,1) and if u1 œ 0.5, then sample from f1;  otherwise, sample from f2 
Generate u2  ~ U(0,1) to sample from f1 or f2.