ISYE 6644 Exam 2 Actual Exam 2026/2027: Complete
Exam-Style Questions with Detailed Rationales | 100%
Verified | Pass Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Random Number Generation and Testing | Q1 – Q10
Section 2 | Input Modeling and Distribution Fitting | Q11 – Q20
Section 3 | Output Analysis and Confidence Intervals | Q21 – Q30
Section 4 | Variance Reduction Techniques | Q31 – Q40
Section 5 | Comparing Alternative System Designs | Q41 – Q50
Instructions: Choose the single best answer. Pass: 80% in 90 minutes.
══════════════════════════════════════
SECTION 1: RANDOM NUMBER GENERATION AND TESTING Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 34-year-old simulation analyst at a semiconductor fab in Austin is validating a new
multiplicative congruential generator with modulus m = 2³¹ − 1 = 2,147,483,647 and
multiplier a = 16,807. She generates the first five values starting from seed X₀ = 12,345
and obtains 2,074,941,799, 1,220,136,621, 1,527,485,461, 1,710,229,131, and
1,873,047,241. She suspects the generator may be cycling prematurely. What is the
maximum possible period of this generator?
A. 2,147,483,646
B. 2,147,483,647
C. 2,147,483,646 ✓ CORRECT
D. 2,147,483,648
Correct Answer: C
,Rationale: For a multiplicative congruential generator with prime modulus m, the
maximum possible period is m − 1, which occurs when the multiplier a is a primitive
root modulo m. Option B incorrectly treats the modulus itself as the period, which is
impossible since zero can never be reached from a non-zero seed in a multiplicative
generator. In practice, analysts at fabs verify period length before running
million-replication experiments, because a short period causes severe correlation in
output statistics.
Question 2 of 50
A 41-year-old operations research consultant in Chicago is auditing a legacy simulation
code written in Fortran for a logistics client. He discovers the code uses a mixed
congruential generator with m = 2¹⁶ = 65,536, a = 1,291, and c = 12,345. The client has
been using this generator for 30 years to simulate warehouse throughput. What is the
most serious statistical flaw in this generator for modern simulation work?
A. The modulus is too small for 64-bit architecture
B. The period of at most 65,536 is far too short for any nontrivial simulation ✓
CORRECT
C. The increment c is not relatively prime to the modulus
D. The multiplier fails the spectral test in three dimensions
Correct Answer: B
Rationale: A period of 65,536 means the generator will cycle after producing only 65,536
unique values, which is completely inadequate for simulations requiring millions or
billions of random numbers. Option A is incorrect because modulus size has nothing to
do with processor architecture compatibility. Consultants who audit legacy code
frequently encounter this exact problem—generators that were marginally acceptable in
the 1990s are now hopelessly inadequate for the scale of modern discrete-event
simulation.
,Question 3 of 50
A 29-year-old PhD student in the ISYE department at Georgia Tech is implementing the
Kolmogorov-Smirnov test to evaluate a random number generator. She generates n =
100 uniform(0,1) variates and computes the empirical CDF F (x). The test statistic D is
the supremum of |F (x) − x| over all x in [0,1]. She obtains D₁₀₀ = 0.092. At α = 0.05, what
is the critical value she should compare against?
A. 0.086
B. 0.122
C. 0.134 ✓ CORRECT
D. 0.162
Correct Answer: C
Rationale: For the Kolmogorov-Smirnov test with n = 100 and α = 0.05, the critical value
is approximately 1.358/√n = 0.1358, which rounds to 0.134. Option B is the critical value
for n = 50 at the same significance level, a common mistake when students look up the
wrong row in standard tables. Graduate students in simulation courses learn that the KS
test is sensitive in the tails, so they often supplement it with the chi-square test for a
more complete assessment of uniformity.
Question 4 of 50
A 47-year-old reliability engineer at an automotive plant in Detroit is testing a new linear
congruential generator for Monte Carlo analysis of warranty claims. She runs the
runs-up test on a sequence of 10,000 generated values and counts the number of
runs-up of length 1, 2, 3, 4, and 5 or more. The observed counts are 4,175, 1,875, 850,
395, and 205 respectively. For a truly random sequence, the expected number of runs-up
of length 1 is approximately 4,167. What is the approximate expected number of
runs-up of length 2?
A. 1,833
, B. 1,875
C. 1,833 ✓ CORRECT
D. 2,083
Correct Answer: C
Rationale: For the runs-up test on a sequence of n values, the expected number of
runs-up of length k is given by n/(k+1)! − 1/(k+1)! for large n, so for length 2 the
expectation is approximately 10,000/6 = 1,666.7, but more precisely using the exact
formula it is (10,000+1)/6 − 1/2 = 1,833.5. Option B matches the observed count rather
than the expected value, which is a common trap for students who confuse observed
with expected frequencies. Automotive reliability engineers use runs tests extensively
because they are particularly good at detecting the kind of subtle trends that indicate
generator defects.
Question 5 of 50
A 38-year-old data scientist at a fintech startup in San Francisco is building a random
number generator for pricing exotic options. He considers using the Mersenne Twister
MT19937, which has a period of 2¹⁹⁹³⁷ − 1. He needs to generate 10¹⁰ random numbers
per trading day across 250 trading days per year. How many years could he run before
exhausting one percent of the generator's period?
A. Approximately 10²⁰ years
B. Approximately 10⁵⁹⁸⁰ years
C. Approximately 10⁵⁹⁸⁰ years ✓ CORRECT
D. Approximately 10¹⁹⁹³⁷ years
Correct Answer: C
Rationale: One percent of 2¹⁹⁹³⁷ − 1 is approximately 2¹⁹⁹³⁷/100, and at 2.5 × 10¹²
numbers per year, the time to exhaust one percent is roughly 2¹⁹⁹³⁷/(100 × 2.5 × 10¹²) ≈
10⁵⁹⁸⁰ years, since log₁₀(2¹⁹⁹³⁷) ≈ 19937 × 0.301 ≈ 6003. Option D confuses the exponent
of the period with the actual time calculation. Fintech practitioners rarely worry about
Exam-Style Questions with Detailed Rationales | 100%
Verified | Pass Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Random Number Generation and Testing | Q1 – Q10
Section 2 | Input Modeling and Distribution Fitting | Q11 – Q20
Section 3 | Output Analysis and Confidence Intervals | Q21 – Q30
Section 4 | Variance Reduction Techniques | Q31 – Q40
Section 5 | Comparing Alternative System Designs | Q41 – Q50
Instructions: Choose the single best answer. Pass: 80% in 90 minutes.
══════════════════════════════════════
SECTION 1: RANDOM NUMBER GENERATION AND TESTING Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A 34-year-old simulation analyst at a semiconductor fab in Austin is validating a new
multiplicative congruential generator with modulus m = 2³¹ − 1 = 2,147,483,647 and
multiplier a = 16,807. She generates the first five values starting from seed X₀ = 12,345
and obtains 2,074,941,799, 1,220,136,621, 1,527,485,461, 1,710,229,131, and
1,873,047,241. She suspects the generator may be cycling prematurely. What is the
maximum possible period of this generator?
A. 2,147,483,646
B. 2,147,483,647
C. 2,147,483,646 ✓ CORRECT
D. 2,147,483,648
Correct Answer: C
,Rationale: For a multiplicative congruential generator with prime modulus m, the
maximum possible period is m − 1, which occurs when the multiplier a is a primitive
root modulo m. Option B incorrectly treats the modulus itself as the period, which is
impossible since zero can never be reached from a non-zero seed in a multiplicative
generator. In practice, analysts at fabs verify period length before running
million-replication experiments, because a short period causes severe correlation in
output statistics.
Question 2 of 50
A 41-year-old operations research consultant in Chicago is auditing a legacy simulation
code written in Fortran for a logistics client. He discovers the code uses a mixed
congruential generator with m = 2¹⁶ = 65,536, a = 1,291, and c = 12,345. The client has
been using this generator for 30 years to simulate warehouse throughput. What is the
most serious statistical flaw in this generator for modern simulation work?
A. The modulus is too small for 64-bit architecture
B. The period of at most 65,536 is far too short for any nontrivial simulation ✓
CORRECT
C. The increment c is not relatively prime to the modulus
D. The multiplier fails the spectral test in three dimensions
Correct Answer: B
Rationale: A period of 65,536 means the generator will cycle after producing only 65,536
unique values, which is completely inadequate for simulations requiring millions or
billions of random numbers. Option A is incorrect because modulus size has nothing to
do with processor architecture compatibility. Consultants who audit legacy code
frequently encounter this exact problem—generators that were marginally acceptable in
the 1990s are now hopelessly inadequate for the scale of modern discrete-event
simulation.
,Question 3 of 50
A 29-year-old PhD student in the ISYE department at Georgia Tech is implementing the
Kolmogorov-Smirnov test to evaluate a random number generator. She generates n =
100 uniform(0,1) variates and computes the empirical CDF F (x). The test statistic D is
the supremum of |F (x) − x| over all x in [0,1]. She obtains D₁₀₀ = 0.092. At α = 0.05, what
is the critical value she should compare against?
A. 0.086
B. 0.122
C. 0.134 ✓ CORRECT
D. 0.162
Correct Answer: C
Rationale: For the Kolmogorov-Smirnov test with n = 100 and α = 0.05, the critical value
is approximately 1.358/√n = 0.1358, which rounds to 0.134. Option B is the critical value
for n = 50 at the same significance level, a common mistake when students look up the
wrong row in standard tables. Graduate students in simulation courses learn that the KS
test is sensitive in the tails, so they often supplement it with the chi-square test for a
more complete assessment of uniformity.
Question 4 of 50
A 47-year-old reliability engineer at an automotive plant in Detroit is testing a new linear
congruential generator for Monte Carlo analysis of warranty claims. She runs the
runs-up test on a sequence of 10,000 generated values and counts the number of
runs-up of length 1, 2, 3, 4, and 5 or more. The observed counts are 4,175, 1,875, 850,
395, and 205 respectively. For a truly random sequence, the expected number of runs-up
of length 1 is approximately 4,167. What is the approximate expected number of
runs-up of length 2?
A. 1,833
, B. 1,875
C. 1,833 ✓ CORRECT
D. 2,083
Correct Answer: C
Rationale: For the runs-up test on a sequence of n values, the expected number of
runs-up of length k is given by n/(k+1)! − 1/(k+1)! for large n, so for length 2 the
expectation is approximately 10,000/6 = 1,666.7, but more precisely using the exact
formula it is (10,000+1)/6 − 1/2 = 1,833.5. Option B matches the observed count rather
than the expected value, which is a common trap for students who confuse observed
with expected frequencies. Automotive reliability engineers use runs tests extensively
because they are particularly good at detecting the kind of subtle trends that indicate
generator defects.
Question 5 of 50
A 38-year-old data scientist at a fintech startup in San Francisco is building a random
number generator for pricing exotic options. He considers using the Mersenne Twister
MT19937, which has a period of 2¹⁹⁹³⁷ − 1. He needs to generate 10¹⁰ random numbers
per trading day across 250 trading days per year. How many years could he run before
exhausting one percent of the generator's period?
A. Approximately 10²⁰ years
B. Approximately 10⁵⁹⁸⁰ years
C. Approximately 10⁵⁹⁸⁰ years ✓ CORRECT
D. Approximately 10¹⁹⁹³⁷ years
Correct Answer: C
Rationale: One percent of 2¹⁹⁹³⁷ − 1 is approximately 2¹⁹⁹³⁷/100, and at 2.5 × 10¹²
numbers per year, the time to exhaust one percent is roughly 2¹⁹⁹³⁷/(100 × 2.5 × 10¹²) ≈
10⁵⁹⁸⁰ years, since log₁₀(2¹⁹⁹³⁷) ≈ 19937 × 0.301 ≈ 6003. Option D confuses the exponent
of the period with the actual time calculation. Fintech practitioners rarely worry about
