ISYE 6644 Final Exam Actual Exam 2026/2027: Complete
Exam-Style Questions with Detailed Rationales | 100%
Verified | Pass Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Simulation Basics and Random Number Generation | Q1 – Q10
Section 2 | Input Modeling and Probability Distributions | Q11 – Q20
Section 3 | Output Analysis and Confidence Intervals | Q21 – Q30
Section 4 | Variance Reduction Techniques | Q31 – Q40
Section 5 | Discrete-Event Simulation and Model Validation | Q41 – Q50
Instructions: Choose the single best answer. Pass: 80% in 90 minutes.
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SECTION 1: SIMULATION BASICS AND RANDOM NUMBER GENERATION Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A graduate student is coding a multiplicative congruential generator for a course
project. He chooses modulus m = 2^31 - 1, multiplier a = 16807, and increment c = 0. He
runs a small test and notices the sequence eventually repeats.
A. The maximum possible period is m because a prime modulus always yields full
period regardless of c.
B. The maximum possible period is m - 1 because m is prime and 16807 is a primitive
root modulo m. ✓ CORRECT
C. The maximum possible period is m/2 because c = 0 forces the sequence to skip
every other integer.
D. The maximum possible period is 2^31 because the modulus determines the number
of bits in the output.
Correct Answer: B
,Rationale: For a multiplicative congruential generator with prime modulus m and
multiplier a that is a primitive root modulo m, the maximum period is m - 1, which is the
longest possible before repeating. The most tempting wrong answer is A because a
prime modulus does not guarantee full period m unless the increment c is carefully
chosen relative to a and m, and with c = 0 the period can never reach m. In practice, the
Park-Miller generator with these exact parameters is the classic minimal standard.
Question 2 of 50
A simulation modeler needs to generate exponential random variates with rate λ = 2 for
an interarrival process. She remembers the inverse transform method but is unsure
which formula to implement in her code.
A. She should set X = -ln(U) * 2 where U is uniform on (0,1), multiplying the logarithm by
the rate parameter.
B. She should set X = -ln(1-U) * 2 where U is uniform on (0,1), which incorrectly
multiplies by the rate instead of dividing.
C. She should set X = -ln(U) / 2 where U is uniform on (0,1), which correctly divides the
logarithm by the rate parameter. ✓ CORRECT
D. She should set X = ln(U) / 2 where U is uniform on (0,1), which omits the negative
sign and yields invalid negative interarrival times.
Correct Answer: C
Rationale: The inverse transform for an exponential distribution with rate λ yields X =
-ln(U)/λ, so dividing by 2 is correct. The most tempting wrong answer is A because
students often confuse whether to multiply or divide by the rate parameter, and
multiplying by 2 is a common algebraic error. In practice, most simulation libraries
implement this as -log(U)/rate, and forgetting the reciprocal produces wildly wrong
arrival processes.
Question 3 of 50
,A simulation analyst is running 100 independent replications of a queueing model. He
wants each replication to use a non-overlapping sequence of random numbers to
ensure statistical independence across runs.
A. He should partition the generator's period into 100 disjoint streams by using different
seeds spaced far apart along the cycle. ✓ CORRECT
B. He should use the same seed for every replication and discard the first million
random numbers as a burn-in period.
C. He should use a different random number generator algorithm for each replication to
guarantee variety.
D. He should generate all 100 replications simultaneously using parallel threads sharing
a single seed.
Correct Answer: A
Rationale: Partitioning a long-period generator into disjoint streams by carefully
selecting seeds that are far apart along the cycle ensures that sequences do not overlap
and that replications remain statistically independent. The most tempting wrong answer
is B because using the same seed produces identical random number sequences
across replications, which destroys independence regardless of any burn-in period. In
practice, modern generators like MRG32k3a provide built-in stream partitioning that is
more reliable than manual seed selection.
Question 4 of 50
A modeler wants to generate variates from a beta distribution with parameters that
make the PDF complex to invert directly. She decides to use the acceptance-rejection
method with a uniform envelope over the support [0,1].
A. The method will be highly efficient because the beta PDF is always bounded by 1 on
[0,1].
B. The method will fail because the acceptance-rejection method cannot be used for
bounded distributions.
, C. The efficiency depends on the ratio of the area under the envelope to the area under
the target PDF, and a uniform envelope may lead to many rejections if the beta PDF has
a sharp peak. ✓ CORRECT
D. The method requires inverting the beta CDF to generate candidate variates from the
envelope.
Correct Answer: C
Rationale: Acceptance-rejection efficiency is determined by the constant c such that f(x)
≤ c*g(x) everywhere, and if the target PDF has a high peak relative to a flat uniform
envelope, c becomes large and the acceptance probability drops. The most tempting
wrong answer is A because while the beta PDF on [0,1] is bounded, the maximum value
can be much greater than 1 depending on parameters, making a uniform envelope with
height 1 impossible or inefficient. In practice, choosing a better envelope shape like a
triangular or another beta distribution dramatically improves efficiency.
Question 5 of 50
A researcher is evaluating two LCG candidates for a high-dimensional simulation.
Generator A has a spectral test score indicating large gaps in its 5-dimensional lattice
structure, while Generator B passes the spectral test in dimensions up to 6.
A. Generator A is superior because large gaps mean the points are well-separated and
less correlated.
B. Generator B is superior because a good spectral test indicates that the lattice
structure is dense and hyperplanes are closely spaced in high dimensions. ✓ CORRECT
C. Generator A is superior because large gaps in the lattice reduce the chance of
sequence overlap.
D. Generator B is dangerous because closely spaced hyperplanes increase the risk of
numerical overflow.
Correct Answer: B
Rationale: The spectral test measures the maximal distance between adjacent
hyperplanes in the lattice structure of an LCG, and smaller distances indicate better
Exam-Style Questions with Detailed Rationales | 100%
Verified | Pass Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Simulation Basics and Random Number Generation | Q1 – Q10
Section 2 | Input Modeling and Probability Distributions | Q11 – Q20
Section 3 | Output Analysis and Confidence Intervals | Q21 – Q30
Section 4 | Variance Reduction Techniques | Q31 – Q40
Section 5 | Discrete-Event Simulation and Model Validation | Q41 – Q50
Instructions: Choose the single best answer. Pass: 80% in 90 minutes.
══════════════════════════════════════
SECTION 1: SIMULATION BASICS AND RANDOM NUMBER GENERATION Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A graduate student is coding a multiplicative congruential generator for a course
project. He chooses modulus m = 2^31 - 1, multiplier a = 16807, and increment c = 0. He
runs a small test and notices the sequence eventually repeats.
A. The maximum possible period is m because a prime modulus always yields full
period regardless of c.
B. The maximum possible period is m - 1 because m is prime and 16807 is a primitive
root modulo m. ✓ CORRECT
C. The maximum possible period is m/2 because c = 0 forces the sequence to skip
every other integer.
D. The maximum possible period is 2^31 because the modulus determines the number
of bits in the output.
Correct Answer: B
,Rationale: For a multiplicative congruential generator with prime modulus m and
multiplier a that is a primitive root modulo m, the maximum period is m - 1, which is the
longest possible before repeating. The most tempting wrong answer is A because a
prime modulus does not guarantee full period m unless the increment c is carefully
chosen relative to a and m, and with c = 0 the period can never reach m. In practice, the
Park-Miller generator with these exact parameters is the classic minimal standard.
Question 2 of 50
A simulation modeler needs to generate exponential random variates with rate λ = 2 for
an interarrival process. She remembers the inverse transform method but is unsure
which formula to implement in her code.
A. She should set X = -ln(U) * 2 where U is uniform on (0,1), multiplying the logarithm by
the rate parameter.
B. She should set X = -ln(1-U) * 2 where U is uniform on (0,1), which incorrectly
multiplies by the rate instead of dividing.
C. She should set X = -ln(U) / 2 where U is uniform on (0,1), which correctly divides the
logarithm by the rate parameter. ✓ CORRECT
D. She should set X = ln(U) / 2 where U is uniform on (0,1), which omits the negative
sign and yields invalid negative interarrival times.
Correct Answer: C
Rationale: The inverse transform for an exponential distribution with rate λ yields X =
-ln(U)/λ, so dividing by 2 is correct. The most tempting wrong answer is A because
students often confuse whether to multiply or divide by the rate parameter, and
multiplying by 2 is a common algebraic error. In practice, most simulation libraries
implement this as -log(U)/rate, and forgetting the reciprocal produces wildly wrong
arrival processes.
Question 3 of 50
,A simulation analyst is running 100 independent replications of a queueing model. He
wants each replication to use a non-overlapping sequence of random numbers to
ensure statistical independence across runs.
A. He should partition the generator's period into 100 disjoint streams by using different
seeds spaced far apart along the cycle. ✓ CORRECT
B. He should use the same seed for every replication and discard the first million
random numbers as a burn-in period.
C. He should use a different random number generator algorithm for each replication to
guarantee variety.
D. He should generate all 100 replications simultaneously using parallel threads sharing
a single seed.
Correct Answer: A
Rationale: Partitioning a long-period generator into disjoint streams by carefully
selecting seeds that are far apart along the cycle ensures that sequences do not overlap
and that replications remain statistically independent. The most tempting wrong answer
is B because using the same seed produces identical random number sequences
across replications, which destroys independence regardless of any burn-in period. In
practice, modern generators like MRG32k3a provide built-in stream partitioning that is
more reliable than manual seed selection.
Question 4 of 50
A modeler wants to generate variates from a beta distribution with parameters that
make the PDF complex to invert directly. She decides to use the acceptance-rejection
method with a uniform envelope over the support [0,1].
A. The method will be highly efficient because the beta PDF is always bounded by 1 on
[0,1].
B. The method will fail because the acceptance-rejection method cannot be used for
bounded distributions.
, C. The efficiency depends on the ratio of the area under the envelope to the area under
the target PDF, and a uniform envelope may lead to many rejections if the beta PDF has
a sharp peak. ✓ CORRECT
D. The method requires inverting the beta CDF to generate candidate variates from the
envelope.
Correct Answer: C
Rationale: Acceptance-rejection efficiency is determined by the constant c such that f(x)
≤ c*g(x) everywhere, and if the target PDF has a high peak relative to a flat uniform
envelope, c becomes large and the acceptance probability drops. The most tempting
wrong answer is A because while the beta PDF on [0,1] is bounded, the maximum value
can be much greater than 1 depending on parameters, making a uniform envelope with
height 1 impossible or inefficient. In practice, choosing a better envelope shape like a
triangular or another beta distribution dramatically improves efficiency.
Question 5 of 50
A researcher is evaluating two LCG candidates for a high-dimensional simulation.
Generator A has a spectral test score indicating large gaps in its 5-dimensional lattice
structure, while Generator B passes the spectral test in dimensions up to 6.
A. Generator A is superior because large gaps mean the points are well-separated and
less correlated.
B. Generator B is superior because a good spectral test indicates that the lattice
structure is dense and hyperplanes are closely spaced in high dimensions. ✓ CORRECT
C. Generator A is superior because large gaps in the lattice reduce the chance of
sequence overlap.
D. Generator B is dangerous because closely spaced hyperplanes increase the risk of
numerical overflow.
Correct Answer: B
Rationale: The spectral test measures the maximal distance between adjacent
hyperplanes in the lattice structure of an LCG, and smaller distances indicate better
