Institute Of Technology. - 131 Questions and Answers Already
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Subject Area Simulation and Modeling
Description This exam covers foundational topics in discrete-event simulation, including
random number and variate generation, input modeling, output analysis, variance
reduction, and model validation. Designed for Georgia Tech's ISYE 6644 course
at the graduate level.
Expected Grade A+
Total Questions 131
Duration 3 hours
Learning Outcomes 1. Apply inverse transform and acceptance-rejection methods to generate random
variates
2. Construct and validate random number generators using theoretical conditions
3. Design and analyze simulation experiments with proper output analysis
techniques
4. Implement variance reduction techniques to improve estimator precision
Accreditation ABET-accredited program; adheres to rigorous US university standards for
industrial engineering graduate education.
Page 1
,Question 1 of 131
A linear congruential generator (LCG) with parameters m = 2^31 - 1, a = 7^5, c = 0, and seed X0 = 1
is used. Which condition ensures that the LCG has full period (m-1)?
A. m is prime and a is a primitive root modulo m
B. c 0 and gcd(c, m) = 1
C. a 1 mod p for every prime p dividing m, and a 1 mod 4 if m is a multiple of 4
D. m is a power of 2 and a 5 mod 8
The correct answer is:
Correct Action: m is prime and a is a primitive root modulo m
Rationales
• m is prime and a is a primitive root modulo m (Correct):
This is the correct action. For a multiplicative LCG (c=0), full period (m-1) is achieved if m is prime and a is a primitive
root modulo m. Conditions in B are for mixed LCGs with c0. C gives conditions for full period in mixed LCGs when m is a
product of distinct primes. D is for power-of-two moduli with c0.
• c 0 and gcd(c, m) = 1 (Incorrect):
This option is not appropriate. Conditions in B are for mixed LCGs with c0. C gives conditions for full period in mixed
LCGs when m is a product of distinct primes
• a 1 mod p for every prime p dividing m, and a 1 mod 4 if m is a multiple of 4 (Incorrect):
This option is not appropriate. Conditions in B are for mixed LCGs with c0. C gives conditions for full period in mixed
LCGs when m is a product of distinct primes
• m is a power of 2 and a 5 mod 8 (Incorrect):
This option is not appropriate. Conditions in B are for mixed LCGs with c0. C gives conditions for full period in mixed
LCGs when m is a product of distinct primes
Page 2
,Question 2 of 131
Using the inverse transform method to generate a random variate from a distribution with CDF F(x)
= 1 - exp(-x) for x 0, which of the following transformations of a U(0,1) random variable U yields the
correct variate?
A. X = -ln(1-U) /
B. X = -ln(U) /
C. X = -ln(1-U) *
D. X = -ln(U) *
The correct answer is:
Correct Action: X = -ln(U) /
Rationales
• X = -ln(U) / (Correct):
This is the correct action. Inverse transform: set U = F(X) => U = 1 - exp(-X) => 1-U = exp(-X) => X = -ln(1-U)/. Since
1-U is also U(0,1), often simplified to X = -ln(U)/. Options A is correct but B is equivalent. However, B is the common
form. All others are incorrect due to factor placement.
• X = -ln(1-U) / (Incorrect):
This option is not appropriate. Since 1-U is also U(0,1), often simplified to X = -ln(U)/. Options A is correct but B is
equivalent
• X = -ln(1-U) * (Incorrect):
This option is not appropriate. Since 1-U is also U(0,1), often simplified to X = -ln(U)/. Options A is correct but B is
equivalent
• X = -ln(U) * (Incorrect):
This option is not appropriate. Since 1-U is also U(0,1), often simplified to X = -ln(U)/. Options A is correct but B is
equivalent
Page 3
, Question 3 of 131
Consider using acceptance-rejection to generate random variates from a target density f(x) = 2x on
[0,1] using a uniform proposal density g(x)=1 on [0,1]. What is the expected number of iterations
until acceptance?
A. 0.5
B. 1
C. 2
D. 4
The correct answer is:
Correct Action: 2
Rationales
• 2 (Correct):
This is the correct action. Acceptance probability is 1/c where c = sup f(x)/g(x) = max 2x / 1 = 2. So expected iterations = c
= 2. Thus the efficiency is 1/2.
• 0.5 (Incorrect):
This option is not appropriate. So expected iterations = c = 2. Thus the efficiency is 1/2
• 1 (Incorrect):
This option is not appropriate. So expected iterations = c = 2. Thus the efficiency is 1/2
• 4 (Incorrect):
This option is not appropriate. So expected iterations = c = 2. Thus the efficiency is 1/2
Page 4