ISYE 6644 Final Cheat Sheet - spring 2026 with complete
solutions - GT. - 106 Questions and Answers Already Graded A+
Premium Exam Tested And Verified
Subject Area Industrial and Systems Engineering (Simulation)
Description Final examination for ISYE 6644: Simulation covering random number
generation, random variate generation, input modeling, output analysis, variance
reduction techniques, queueing simulation, Markov chain Monte Carlo, and
experimental design. The exam assesses students' ability to design, analyze, and
interpret simulation models with rigorous statistical methods.
Expected Grade A+
Total Questions 106
Duration 3 hours
Learning Outcomes 1. Design and implement simulation models for complex stochastic systems
2. Select and evaluate random number generators with full period properties
3. Apply input modeling techniques including maximum likelihood estimation
and goodness-of-fit tests
4. Conduct output analysis using batch means and variance reduction methods
5. Critically compare simulation strategies such as Latin hypercube sampling and
common random numbers
Accreditation Accredited by ABET and compliant with Georgia Institute of Technology
graduate-level standards.
Page 1
,Question 1 of 106
A linear congruential generator (LCG) with modulus m=2^16, multiplier a=869, increment c=0, and
seed X0=1 is used. Which of the following is true regarding its period?
A. It achieves full period of 2^16 because a is primitive modulo 2^16.
B. The period is at most 2^15 because c=0 implies multiplicative LCG.
C. Full period is impossible for any multiplicative LCG with m a power of 2.
D. The period is exactly 2^16 because a 5 (mod 8) and the conditions for full period are satisfied.
The correct answer is:
Correct Action: The period is at most 2^15 because c=0 implies multiplicative LCG.
Rationales
• The period is at most 2^15 because c=0 implies multiplicative LCG. (Correct):
This is the correct action. For a multiplicative LCG (c=0) with modulus m=2^k, the maximum period is 2^(k-2) = 2^14, not
2^k. Here k=16, so max period is 2^14. Option A is false because full period is not achieved. Option C is false because full
period is possible for mixed LCGs but not multiplicatives. Option D incorrectly states a 5 mod 8, but that is a condition for
• It achieves full period of 2^16 because a is primitive modulo 2^16. (Incorrect):
This option is not appropriate. Here k=16, so max period is 2^14. Option A is false because full period is not achieved
• Full period is impossible for any multiplicative LCG with m a power of 2. (Incorrect):
This option is not appropriate. Here k=16, so max period is 2^14. Option A is false because full period is not achieved
• The period is exactly 2^16 because a 5 (mod 8) and the conditions for full period are satisfied. (Incorrect):
This option is not appropriate. Here k=16, so max period is 2^14. Option A is false because full period is not achieved
Page 2
,Question 2 of 106
To generate a random variate from a Weibull distribution with shape parameter =2 and scale =1
using the inverse transform method, a random number U~U(0,1) is generated. Which transformation
yields the desired variate?
A. X = [-ln(1-U)]^(1/2)
B. X = [-ln(U)]^(1/2)
C. X = [-ln(U)]^2
D. X = [-ln(1-U)]^2
The correct answer is:
Correct Action: X = [-ln(U)]^(1/2)
Rationales
• X = [-ln(U)]^(1/2) (Correct):
This is the correct action. The Weibull CDF is F(x)=1-exp(-(x/)^). Inverse: x = [-ln(1-U)]^(1/). With =2, =1,
x=[-ln(1-U)]^(1/2). Since U and 1-U have same distribution, [-ln(U)]^(1/2) is equivalent. Option A and D have 1-U but
exponent incorrect. Option C exponent 2 instead of 1/2.
• X = [-ln(1-U)]^(1/2) (Incorrect):
This option is not appropriate. Inverse: x = [-ln(1-U)]^(1/). With =2, =1, x=[-ln(1-U)]^(1/2)
• X = [-ln(U)]^2 (Incorrect):
This option is not appropriate. Inverse: x = [-ln(1-U)]^(1/). With =2, =1, x=[-ln(1-U)]^(1/2)
• X = [-ln(1-U)]^2 (Incorrect):
This option is not appropriate. Inverse: x = [-ln(1-U)]^(1/). With =2, =1, x=[-ln(1-U)]^(1/2)
Page 3
, Question 3 of 106
In input modeling for a simulation of service times, 50 observations are collected. The sample mean
is 12.4, sample variance is 5.6. For an exponential distribution, which of the following is true
regarding the maximum likelihood estimator (MLE) for the rate parameter ?
A. The MLE is 1/12.4 but it is biased; the unbiased estimator uses denominator 49.
B. The MLE is 1/12.4 and it is unbiased by the invariance property.
C. The MLE is the sample standard deviation divided by the mean.
D. The MLE for is 12.4 and it is consistent.
The correct answer is:
Correct Action: The MLE is 1/12.4 but it is biased; the unbiased estimator uses denominator 49.
Rationales
• The MLE is 1/12.4 but it is biased; the unbiased estimator uses denominator 49. (Correct):
This is the correct action. For exponential, MLE for rate is 1/mean. It is biased because E[1/X] not equal . Unbiased
estimator uses n-1 in denominator for variance, but for rate, no simple unbiased estimator exists. Option B is wrong
because invariance applies to functions of MLE but does not guarantee unbiasedness. Option C is nonsense. Option D
• The MLE is 1/12.4 and it is unbiased by the invariance property. (Incorrect):
This option is not appropriate. It is biased because E[1/X] not equal . Unbiased estimator uses n-1 in denominator for
variance, but for rate, no simple unbiased estimator exists
• The MLE is the sample standard deviation divided by the mean. (Incorrect):
This option is not appropriate. It is biased because E[1/X] not equal . Unbiased estimator uses n-1 in denominator for
variance, but for rate, no simple unbiased estimator exists
• The MLE for is 12.4 and it is consistent. (Incorrect):
This option is not appropriate. It is biased because E[1/X] not equal . Unbiased estimator uses n-1 in denominator for
variance, but for rate, no simple unbiased estimator exists
Page 4
solutions - GT. - 106 Questions and Answers Already Graded A+
Premium Exam Tested And Verified
Subject Area Industrial and Systems Engineering (Simulation)
Description Final examination for ISYE 6644: Simulation covering random number
generation, random variate generation, input modeling, output analysis, variance
reduction techniques, queueing simulation, Markov chain Monte Carlo, and
experimental design. The exam assesses students' ability to design, analyze, and
interpret simulation models with rigorous statistical methods.
Expected Grade A+
Total Questions 106
Duration 3 hours
Learning Outcomes 1. Design and implement simulation models for complex stochastic systems
2. Select and evaluate random number generators with full period properties
3. Apply input modeling techniques including maximum likelihood estimation
and goodness-of-fit tests
4. Conduct output analysis using batch means and variance reduction methods
5. Critically compare simulation strategies such as Latin hypercube sampling and
common random numbers
Accreditation Accredited by ABET and compliant with Georgia Institute of Technology
graduate-level standards.
Page 1
,Question 1 of 106
A linear congruential generator (LCG) with modulus m=2^16, multiplier a=869, increment c=0, and
seed X0=1 is used. Which of the following is true regarding its period?
A. It achieves full period of 2^16 because a is primitive modulo 2^16.
B. The period is at most 2^15 because c=0 implies multiplicative LCG.
C. Full period is impossible for any multiplicative LCG with m a power of 2.
D. The period is exactly 2^16 because a 5 (mod 8) and the conditions for full period are satisfied.
The correct answer is:
Correct Action: The period is at most 2^15 because c=0 implies multiplicative LCG.
Rationales
• The period is at most 2^15 because c=0 implies multiplicative LCG. (Correct):
This is the correct action. For a multiplicative LCG (c=0) with modulus m=2^k, the maximum period is 2^(k-2) = 2^14, not
2^k. Here k=16, so max period is 2^14. Option A is false because full period is not achieved. Option C is false because full
period is possible for mixed LCGs but not multiplicatives. Option D incorrectly states a 5 mod 8, but that is a condition for
• It achieves full period of 2^16 because a is primitive modulo 2^16. (Incorrect):
This option is not appropriate. Here k=16, so max period is 2^14. Option A is false because full period is not achieved
• Full period is impossible for any multiplicative LCG with m a power of 2. (Incorrect):
This option is not appropriate. Here k=16, so max period is 2^14. Option A is false because full period is not achieved
• The period is exactly 2^16 because a 5 (mod 8) and the conditions for full period are satisfied. (Incorrect):
This option is not appropriate. Here k=16, so max period is 2^14. Option A is false because full period is not achieved
Page 2
,Question 2 of 106
To generate a random variate from a Weibull distribution with shape parameter =2 and scale =1
using the inverse transform method, a random number U~U(0,1) is generated. Which transformation
yields the desired variate?
A. X = [-ln(1-U)]^(1/2)
B. X = [-ln(U)]^(1/2)
C. X = [-ln(U)]^2
D. X = [-ln(1-U)]^2
The correct answer is:
Correct Action: X = [-ln(U)]^(1/2)
Rationales
• X = [-ln(U)]^(1/2) (Correct):
This is the correct action. The Weibull CDF is F(x)=1-exp(-(x/)^). Inverse: x = [-ln(1-U)]^(1/). With =2, =1,
x=[-ln(1-U)]^(1/2). Since U and 1-U have same distribution, [-ln(U)]^(1/2) is equivalent. Option A and D have 1-U but
exponent incorrect. Option C exponent 2 instead of 1/2.
• X = [-ln(1-U)]^(1/2) (Incorrect):
This option is not appropriate. Inverse: x = [-ln(1-U)]^(1/). With =2, =1, x=[-ln(1-U)]^(1/2)
• X = [-ln(U)]^2 (Incorrect):
This option is not appropriate. Inverse: x = [-ln(1-U)]^(1/). With =2, =1, x=[-ln(1-U)]^(1/2)
• X = [-ln(1-U)]^2 (Incorrect):
This option is not appropriate. Inverse: x = [-ln(1-U)]^(1/). With =2, =1, x=[-ln(1-U)]^(1/2)
Page 3
, Question 3 of 106
In input modeling for a simulation of service times, 50 observations are collected. The sample mean
is 12.4, sample variance is 5.6. For an exponential distribution, which of the following is true
regarding the maximum likelihood estimator (MLE) for the rate parameter ?
A. The MLE is 1/12.4 but it is biased; the unbiased estimator uses denominator 49.
B. The MLE is 1/12.4 and it is unbiased by the invariance property.
C. The MLE is the sample standard deviation divided by the mean.
D. The MLE for is 12.4 and it is consistent.
The correct answer is:
Correct Action: The MLE is 1/12.4 but it is biased; the unbiased estimator uses denominator 49.
Rationales
• The MLE is 1/12.4 but it is biased; the unbiased estimator uses denominator 49. (Correct):
This is the correct action. For exponential, MLE for rate is 1/mean. It is biased because E[1/X] not equal . Unbiased
estimator uses n-1 in denominator for variance, but for rate, no simple unbiased estimator exists. Option B is wrong
because invariance applies to functions of MLE but does not guarantee unbiasedness. Option C is nonsense. Option D
• The MLE is 1/12.4 and it is unbiased by the invariance property. (Incorrect):
This option is not appropriate. It is biased because E[1/X] not equal . Unbiased estimator uses n-1 in denominator for
variance, but for rate, no simple unbiased estimator exists
• The MLE is the sample standard deviation divided by the mean. (Incorrect):
This option is not appropriate. It is biased because E[1/X] not equal . Unbiased estimator uses n-1 in denominator for
variance, but for rate, no simple unbiased estimator exists
• The MLE for is 12.4 and it is consistent. (Incorrect):
This option is not appropriate. It is biased because E[1/X] not equal . Unbiased estimator uses n-1 in denominator for
variance, but for rate, no simple unbiased estimator exists
Page 4
