ISYE 6644 SIMULATION MODELING FINAL EXAM
SUMMER 2026 | Stochastic Processes | Graded A |
Georgia Tech OMS Analytics | Pass Guaranteed - A+
Graded
Section 1: Random Number Generation & Probability
Distributions (Q1-14)
Q1. A linear congruential generator (LCG) is defined by the recurrence relation
X_{n+1} = (aX_n + c) mod m. For the LCG with parameters a = 13, c = 0, m = 31, and
seed X_0 = 1, what is the maximum possible period length?
A. 15
B. 30
C. 31
D. 62
Correct Answer: B. 30 [CORRECT]
Rationale: For a multiplicative LCG (c = 0), the maximum period is m - 1 = 30 when
m is prime and a is a primitive root modulo m. Since 31 is prime and 13 is a primitive
root mod 31 (its powers generate all non-zero residues), the period is 30. Option A is
incorrect—13 does not produce half period. Option C is only achievable with mixed
LCG (c ≠ 0) and Hull-Dobell conditions. Option D exceeds the modulus.
Q2. Which condition is NECESSARY for a mixed LCG (c ≠ 0) to achieve full period
m?
A. m must be prime, and a must be a primitive root modulo m
B. c and m must be relatively prime; a - 1 must be divisible by all prime factors of m;
a - 1 must be divisible by 4 if m is divisible by 4
,C. c must be zero, and m must be a power of 2
D. a must equal 1, and c must be odd
Correct Answer: B. c and m must be relatively prime; a - 1 must be divisible by all
prime factors of m; a - 1 must be divisible by 4 if m is divisible by 4 [CORRECT]
Rationale: These are the Hull-Dobell theorem conditions for full period of a mixed
LCG. Option A describes conditions for multiplicative LCG maximum period. Option C
describes a multiplicative LCG with power-of-2 modulus (which cannot achieve full
period). Option D would produce a trivial generator with period dependent only on c.
Q3. To generate an exponential random variate with mean β = 5 using the inverse
transform method, given a uniform random number U ~ U(0,1), which
transformation is CORRECT?
A. X = -5 * ln(U)
B. X = -ln(1 - U) / 5
C. X = -5 * ln(1 - U)
D. X = 5 * ln(1/(1-U))
Correct Answer: C. X = -5 * ln(1 - U) [CORRECT]
Rationale: The inverse transform for exponential with mean β: X = -β * ln(1 - U).
Since 1-U is also uniform, -β * ln(U) is equivalent, but the standard form uses 1-U for
numerical stability near U=0. Option A is missing the 1-U term (though equivalent in
distribution, it's less stable). Option B incorrectly divides by 5 instead of multiplying.
Option D is algebraically equivalent to C but not the standard form.
Q4. A simulation requires generating standard normal random variates using the
acceptance-rejection method. Which statement about this approach is TRUE?
A. The acceptance-rejection method for normal distribution uses a uniform
majorizing function over the entire real line
B. The Box-Muller transform is a more efficient alternative that avoids rejection
,C. The acceptance-rejection method requires finding a majorizing density g(x) such
that f(x)/g(x) ≤ c for all x, where c is minimized
D. The inverse transform method is preferred for normal distribution because its CDF
has a closed-form inverse
Correct Answer: C. The acceptance-rejection method requires finding a
majorizing density g(x) such that f(x)/g(x) ≤ c for all x, where c is minimized
[CORRECT]
Rationale: Acceptance-rejection requires a majorizing density g(x) and constant c
where f(x) ≤ c·g(x) for all x. Minimizing c maximizes efficiency. Option A is incorrect—
a uniform function cannot majorize the normal density over infinite support. Option
B is true about Box-Muller but doesn't describe acceptance-rejection. Option D is
false—the normal CDF has no closed-form inverse.
Q5. Using the inverse transform method, how would you generate a random
variate from a triangular distribution with parameters min = 2, mode = 5, and
max = 10?
A. Generate U ~ U(0,1), then if U ≤ (5-2)/(10-2), return 2 + sqrt(U * 8 * 3), else return
10 - sqrt((1-U) * 8 * 5)
B. Generate U ~ U(0,1), then if U ≤ 0.5, return 2 + U * 3, else return 5 + (U - 0.5) * 5
C. Generate U ~ U(0,1), then if U ≤ (5-2)/(10-2), return 2 + sqrt(U * (10-2) * (5-2)), else
return 10 - sqrt((1-U) * (10-2) * (10-5))
D. Generate two uniform random numbers and return their average scaled to [2, 10]
Correct Answer: C. Generate U ~ U(0,1), then if U ≤ (5-2)/(10-2), return 2 + sqrt(U
* (10-2) * (5-2)), else return 10 - sqrt((1-U) * (10-2) * (10-5)) [CORRECT]
Rationale: For triangular distribution, the CDF has two pieces. The breakpoint is at
F(mode) = (mode-min)/(max-min) = 3/8. The inverse for the ascending portion: X =
min + sqrt(U * (max-min) * (mode-min)). For the descending portion: X = max -
sqrt((1-U) * (max-min) * (max-mode)). Option A uses incorrect constants. Option B
uses a simplified incorrect piecewise linear approach. Option D generates a different
distribution.
, Q6. Which random number generator would be MOST appropriate for a large-
scale simulation requiring billions of random numbers with minimal correlation?
A. A simple LCG with m = 2^31 - 1
B. A combined multiple recursive generator (CMRG) such as MRG32k3a
C. A multiplicative LCG with m = 2^16
D. A manual dice-rolling approach
Correct Answer: B. A combined multiple recursive generator (CMRG) such as
MRG32k3a [CORRECT]
Rationale: CMRGs like MRG32k3a (L'Ecuyer) have extremely long periods (~2^191),
excellent statistical properties, and are designed for large-scale simulations. Option A
has period ~2 billion, which may be exhausted. Option C has far too short a period
(65,536). Option D is impractical and non-reproducible.
Q7. To generate a Poisson random variate with λ = 3 using the inverse transform
method, which approach is CORRECT?
A. Generate U ~ U(0,1), find the smallest k such that F(k) ≥ U where F(k) = Σ(i=0 to k)
e^(-3) * 3^i / i!
B. Generate exponential variates with mean 3 until their sum exceeds 1, then count
how many were generated
C. Generate U ~ U(0,1), return 3 * U
D. Generate three uniform random numbers and return their sum
Correct Answer: A. Generate U ~ U(0,1), find the smallest k such that F(k) ≥ U
where F(k) = Σ(i=0 to k) e^(-3) * 3^i / i! [CORRECT]
Rationale: The inverse transform method for discrete distributions searches the CDF
for the first k where F(k) ≥ U. Option B describes an alternative algorithm
(exponential interarrival counting) but with wrong mean (should be mean 1, not 3).
Option C generates uniform, not Poisson. Option D generates Irwin-Hall distribution,
not Poisson.
SUMMER 2026 | Stochastic Processes | Graded A |
Georgia Tech OMS Analytics | Pass Guaranteed - A+
Graded
Section 1: Random Number Generation & Probability
Distributions (Q1-14)
Q1. A linear congruential generator (LCG) is defined by the recurrence relation
X_{n+1} = (aX_n + c) mod m. For the LCG with parameters a = 13, c = 0, m = 31, and
seed X_0 = 1, what is the maximum possible period length?
A. 15
B. 30
C. 31
D. 62
Correct Answer: B. 30 [CORRECT]
Rationale: For a multiplicative LCG (c = 0), the maximum period is m - 1 = 30 when
m is prime and a is a primitive root modulo m. Since 31 is prime and 13 is a primitive
root mod 31 (its powers generate all non-zero residues), the period is 30. Option A is
incorrect—13 does not produce half period. Option C is only achievable with mixed
LCG (c ≠ 0) and Hull-Dobell conditions. Option D exceeds the modulus.
Q2. Which condition is NECESSARY for a mixed LCG (c ≠ 0) to achieve full period
m?
A. m must be prime, and a must be a primitive root modulo m
B. c and m must be relatively prime; a - 1 must be divisible by all prime factors of m;
a - 1 must be divisible by 4 if m is divisible by 4
,C. c must be zero, and m must be a power of 2
D. a must equal 1, and c must be odd
Correct Answer: B. c and m must be relatively prime; a - 1 must be divisible by all
prime factors of m; a - 1 must be divisible by 4 if m is divisible by 4 [CORRECT]
Rationale: These are the Hull-Dobell theorem conditions for full period of a mixed
LCG. Option A describes conditions for multiplicative LCG maximum period. Option C
describes a multiplicative LCG with power-of-2 modulus (which cannot achieve full
period). Option D would produce a trivial generator with period dependent only on c.
Q3. To generate an exponential random variate with mean β = 5 using the inverse
transform method, given a uniform random number U ~ U(0,1), which
transformation is CORRECT?
A. X = -5 * ln(U)
B. X = -ln(1 - U) / 5
C. X = -5 * ln(1 - U)
D. X = 5 * ln(1/(1-U))
Correct Answer: C. X = -5 * ln(1 - U) [CORRECT]
Rationale: The inverse transform for exponential with mean β: X = -β * ln(1 - U).
Since 1-U is also uniform, -β * ln(U) is equivalent, but the standard form uses 1-U for
numerical stability near U=0. Option A is missing the 1-U term (though equivalent in
distribution, it's less stable). Option B incorrectly divides by 5 instead of multiplying.
Option D is algebraically equivalent to C but not the standard form.
Q4. A simulation requires generating standard normal random variates using the
acceptance-rejection method. Which statement about this approach is TRUE?
A. The acceptance-rejection method for normal distribution uses a uniform
majorizing function over the entire real line
B. The Box-Muller transform is a more efficient alternative that avoids rejection
,C. The acceptance-rejection method requires finding a majorizing density g(x) such
that f(x)/g(x) ≤ c for all x, where c is minimized
D. The inverse transform method is preferred for normal distribution because its CDF
has a closed-form inverse
Correct Answer: C. The acceptance-rejection method requires finding a
majorizing density g(x) such that f(x)/g(x) ≤ c for all x, where c is minimized
[CORRECT]
Rationale: Acceptance-rejection requires a majorizing density g(x) and constant c
where f(x) ≤ c·g(x) for all x. Minimizing c maximizes efficiency. Option A is incorrect—
a uniform function cannot majorize the normal density over infinite support. Option
B is true about Box-Muller but doesn't describe acceptance-rejection. Option D is
false—the normal CDF has no closed-form inverse.
Q5. Using the inverse transform method, how would you generate a random
variate from a triangular distribution with parameters min = 2, mode = 5, and
max = 10?
A. Generate U ~ U(0,1), then if U ≤ (5-2)/(10-2), return 2 + sqrt(U * 8 * 3), else return
10 - sqrt((1-U) * 8 * 5)
B. Generate U ~ U(0,1), then if U ≤ 0.5, return 2 + U * 3, else return 5 + (U - 0.5) * 5
C. Generate U ~ U(0,1), then if U ≤ (5-2)/(10-2), return 2 + sqrt(U * (10-2) * (5-2)), else
return 10 - sqrt((1-U) * (10-2) * (10-5))
D. Generate two uniform random numbers and return their average scaled to [2, 10]
Correct Answer: C. Generate U ~ U(0,1), then if U ≤ (5-2)/(10-2), return 2 + sqrt(U
* (10-2) * (5-2)), else return 10 - sqrt((1-U) * (10-2) * (10-5)) [CORRECT]
Rationale: For triangular distribution, the CDF has two pieces. The breakpoint is at
F(mode) = (mode-min)/(max-min) = 3/8. The inverse for the ascending portion: X =
min + sqrt(U * (max-min) * (mode-min)). For the descending portion: X = max -
sqrt((1-U) * (max-min) * (max-mode)). Option A uses incorrect constants. Option B
uses a simplified incorrect piecewise linear approach. Option D generates a different
distribution.
, Q6. Which random number generator would be MOST appropriate for a large-
scale simulation requiring billions of random numbers with minimal correlation?
A. A simple LCG with m = 2^31 - 1
B. A combined multiple recursive generator (CMRG) such as MRG32k3a
C. A multiplicative LCG with m = 2^16
D. A manual dice-rolling approach
Correct Answer: B. A combined multiple recursive generator (CMRG) such as
MRG32k3a [CORRECT]
Rationale: CMRGs like MRG32k3a (L'Ecuyer) have extremely long periods (~2^191),
excellent statistical properties, and are designed for large-scale simulations. Option A
has period ~2 billion, which may be exhausted. Option C has far too short a period
(65,536). Option D is impractical and non-reproducible.
Q7. To generate a Poisson random variate with λ = 3 using the inverse transform
method, which approach is CORRECT?
A. Generate U ~ U(0,1), find the smallest k such that F(k) ≥ U where F(k) = Σ(i=0 to k)
e^(-3) * 3^i / i!
B. Generate exponential variates with mean 3 until their sum exceeds 1, then count
how many were generated
C. Generate U ~ U(0,1), return 3 * U
D. Generate three uniform random numbers and return their sum
Correct Answer: A. Generate U ~ U(0,1), find the smallest k such that F(k) ≥ U
where F(k) = Σ(i=0 to k) e^(-3) * 3^i / i! [CORRECT]
Rationale: The inverse transform method for discrete distributions searches the CDF
for the first k where F(k) ≥ U. Option B describes an alternative algorithm
(exponential interarrival counting) but with wrong mean (should be mean 1, not 3).
Option C generates uniform, not Poisson. Option D generates Irwin-Hall distribution,
not Poisson.
