ISYE 6644 / ISYE6644 – Spring 2025 Exam 1 Solutions Slides and Solutions from ISYE6644 Course materi

Exam 1 Solutions ISYE6644 – Spring 2025 TA: Brian Kearney Slides and Solutions from ISYE6644 Course materials and Solution sets

Question 1 TRUE or FALSE? In a single-server queue simulation, if the arrival rate of customers showing up is less than the service rate of the single-server, then the queue will not explode to infinity.

Question 1 TRUE or FALSE? In a single-server queue simulation, if the arrival rate of customers showing up is less than the service rate of the single-server, then the queue will not explode to infinity.

Solution: If the arrival rate is less than the service rate the queue will get cleaned out eventually.

So, the answer is TRUE.

□

Question 2 Consider a family that has 2 children. Assume that P(kid is a boy) = 1Τ3 and P(kid is a girl) = 2Τ3, and that the genders of the 2 kids are independent. What is the probability that both kids are girls given that we know at least 1 of them is a girl? (a) 1/16 (b) 1/8 (c) 1/5 (d) 1/4 (e) 1/2

Question 2 Consider a family that has 2 children. Assume that P(kid is a boy) = 1Τ3 and P(kid is a girl) = 2Τ3, and that the genders of the 2 kids are independent. What is the probability that both kids are girls given that we know at least 1 of them is a girl? (a) 1/16 (b) 1/8 (c) 1/5 (d) 1/4 (e) 1/2

What is this question asking?

Sample Space 4Τ9

2Τ9

2Τ9

1Τ9

Pr = 2Τ3 ∗ 2Τ3 Pr = 1Τ3 ∗ 2Τ3

Pr = 2Τ3 ∗ 1Τ3 Pr = 1Τ3 ∗ 1Τ3

Question 2 Consider a family that has 2 children. Assume that P(kid is a boy) = 1Τ3 and P(kid is a girl) = 2Τ3, and that the genders of the 2 kids are independent. What is the probability that both kids are girls given that we know at least 1 of them is a girl?

What is this question asking?

(a) 1/16 (b) 1/8 (c) 1/5 (d) 1/4 (e) 1/2

Sample Space 2 =1 2

4Τ9

2Τ9

2Τ9

1Τ9

2 =2 1

2 =1 0

Question 2 Consider a family that has 2 children. Assume that P(kid is a boy) = 1Τ3 and P(kid is a girl) = 2Τ3, and that the genders of the 2 kids are independent. What is the probability that both kids are girls given that we know at least 1 of them is a girl? (a) 1/16 (b) 1/8 (c) 1/5 (d) 1/4 (e) 1/2

What is this question asking?

Sample Space 4Τ9

2Τ9

2Τ9

1Τ9

This doesn’t fit.

Question 2 Consider a family that has 2 children. Assume that P(kid is a boy) = 1Τ3 and P(kid is a girl) = 2Τ3, and that the genders of the 2 kids are independent. What is the probability that both kids are girls given that we know at least 1 of them is a girl? (a) 1/16 (b) 1/8 (c) 1/5 (d) 1/4 (e) 1/2

What is this question asking?

What we are looking for.

Sample Space 4Τ9

2Τ9

2Τ9

1Τ9

Question 2 Consider a family that has 2 children. Assume that P(kid is a boy) = 1Τ3 and P(kid is a girl) = 2Τ3, and that the genders of the 2 kids are independent. What is the probability that both kids are girls given that we know at least 1 of them is a girl? (a) 1/16 (b) 1/8 (c) 1/5 (d) 1/4 (e) 1/2

Solution: The sample space is 𝑆 = 𝐺𝐺, 𝐺𝐵, 𝐵𝐺, 𝐵𝐵 where each entry in an ordered double. Let 𝐴 be the event that both kids are girls, i.e., 𝐴 = 𝐺𝐺 ; and let 𝐵 be the event that at least 1 of the children is a girl, i.e., 𝐵 = 𝐺𝐺, 𝐺𝐵, 𝐵𝐺 . Then, using the probabilities given: 4ൗ 𝑃 𝐴∩𝐵 1 𝑃 𝐴|𝐵 = = 9= 8ൗ 𝑃 𝐵 2 9 So, the answer is (e).

□

Question 2 (practical application) Punnett Squares | Ask A Biologist (asu.edu) What is the chance of a Homozygous dominant (AA) within the *yellow* pea population of the offspring (at least one A).

Question 3 TRUE or FALSE? Discrete-event simulations are particularly suitable for analyzing continuous-flow phenomena such as weather patterns propagating across the country.

Question 3 TRUE or FALSE? Discrete-event simulations are particularly suitable for analyzing continuous-flow phenomena such as weather patterns propagating across the country.

Solution: This is more suited for continuous time models.

So, the answer is FALSE.

□

Question 4 The planet Tranya has 240-day years. Suppose there are three Tranyanians in the room.

What is the probability that at least two of them share the same birthday? (a) 0.0125 (b) 0.025 (c) 0.035 (d) 0.149 (e) 0.9975

Question 4 The planet Tranya has 240-day years. Suppose there are three Tranyanians in the room.

What is the probability that at least two of them share the same birthday? (a) 0.0125 (b) 0.025 (c) 0.035 (d) 0.149 (e) 0.9975 Solution: 240 239 238 P none share the same birthday = 2403 = 0.9875 P none share the same birthday =1 − P none share the same birthday = 1 − 0.9875 = 0.0125

So, the answer is (a) .

□

Number of available days Number of total days

Question 4 The planet Tranya has 240-day years. Suppose there are three Tranyanians in the room.

What is the probability that at least two of them share the same birthday? (a) 0.0125 (b) 0.025 (c) 0.035 (d) 0.149 (e) 0.9975 Solution: 240 239 238 P none share the same birthday = 2403 = 0.9875 P none share the same birthday =1 − P none share the same birthday = 1 − 0.9875 = 0.0125

So, the answer is (a) .

□

Question 5 TRUE or FALSE? The function 𝑓 𝑥

𝑥Τ − = 5𝑒 5 for 𝑥 ≥ 0 is a valid p.d.f.

Question 5

Incorrect 1Τ𝜆

TRUE or FALSE? The function 𝑓 𝑥

𝑥Τ − = 5𝑒 5 for 𝑥 ≥ 0 is a valid p.d.f.

𝑥

𝑓 𝑥 = 1Τ5 𝑒 − Τ5

Correct PDF

Question 5 TRUE or FALSE? The function 𝑓 𝑥

𝑥Τ − = 5𝑒 5 for 𝑥 ≥ 0 is a valid p.d.f.

∞

𝑎 𝑥 𝑥 −𝑥ൗ5 න 𝑓 𝑥 𝑑𝑥 = න 5𝑒 𝑑𝑥 = lim න 5𝑒 − ൗ5 𝑑𝑥 = lim −25𝑒 − ൗ5 |𝑎0 𝑎→∞ 0 𝑎→∞ ℝ 0 𝑎 = lim −25𝑒 − ൗ5 + 25 = 0 + 25 = 25 𝑎→∞

Solution: Note that‫׬‬ℝ 𝑓 𝑥 𝑑𝑥 = 25, so FALSE.

Question 5 TRUE or FALSE? The function 𝑓 𝑥

𝑥Τ − = 5𝑒 5 for 𝑥 ≥ 0 is a valid p.d.f.

∞

𝑎 𝑥 𝑥 −𝑥ൗ5 න 𝑓 𝑥 𝑑𝑥 = න 5𝑒 𝑑𝑥 = lim න 5𝑒 − ൗ5 𝑑𝑥 = lim −25𝑒 − ൗ5 |𝑎0 𝑎→∞ 0 𝑎→∞ ℝ 0 𝑎 = lim −25𝑒 − ൗ5 + 25 = 0 + 25 = 25 𝑎→∞

Solution: Note that‫׬‬ℝ 𝑓 𝑥 𝑑𝑥 = 25, so FALSE.

Question 6 𝑥

Suppose that 𝑋 is a continuous random variable with p.d.f. 𝑓 𝑥 = for 0 < 𝑥 < 2. Find 2 Pr 𝑋 > 1 𝑋 > 1Τ2 . (a) 0 (b) 0.2 (c) 0.25 (d) 0.75 (e) 0.8

Question 6 𝑥

Suppose that 𝑋 is a continuous random variable with p.d.f. 𝑓 𝑥 = for 0 < 𝑥 < 2. Find 2 Pr 𝑋 > 1 𝑋 > 1Τ2 . (a) 0 (b) 0.2 (c) 0.25 (d) 0.75 (e) 0.8

2𝑥

Solution: Pr 𝑋 > 1 𝑋 > 1Τ2 =

‫׬‬1 2𝑑𝑥 2 𝑥

‫׬‬1ൗ 2𝑑𝑥 2

So, the answer is (e).

□

=

𝑥2 2 | 4 1 𝑥2 2 | 4 1ൗ2

1

=

1−4

1 =

1−16

3Τ 4 15Τ 16

=

12Τ 16 15Τ 16

4 5

= = 0.8

Question 7 Suppose that 𝑋 ~ Bern(0.5). What is E 𝑒 4𝑋 ? (a) 1 (b) 4.38 (c) 𝑒 4 (d) 1 + 𝑒 4 (e) 1 + 𝑒 4 Τ2 (f) None of the above.

Question 7 Suppose that 𝑋 ~ Bern(0.5). What is E 𝑒 4𝑋 ? (a) 1 (b) 4.38 (c) 𝑒 4 (d) 1 + 𝑒 4 (e) 1 + 𝑒 4 Τ2 (f) None of the above.

Solution: By LOTUS, E 𝑒 4𝑋 = ෍ 𝑒 4𝑋 P 𝑋 = 𝑥 = 𝑒 0 1 − 0.5 + 𝑒 4 0.5 = 0.5 + 0.5𝑒 4 = 1 + 𝑒 4 Τ2 𝑥

So, the answer is (e).

□

Question 8 Suppose 𝑋 and 𝑌 are continuous random variables with joint p.d.f. 𝑓 𝑥, 𝑦 = 𝑐𝑥𝑦 2

for 0 ≤ 𝑥 ≤ 2 and 0 ≤ 𝑦 ≤ 1,

for some appropriate constant 𝑐. Find E 2𝑋 − 1 .

(a) 1Τ6 (b) 5Τ3 (c) 5Τ2 (d) 11Τ3 (e) 12

Question 8 Suppose 𝑋 and 𝑌 are continuous random variables with joint p.d.f. 𝑓 𝑥, 𝑦 = 𝑐𝑥𝑦 2

for 0 ≤ 𝑥 ≤ 2 and 0 ≤ 𝑦 ≤ 1,

for some appropriate constant 𝑐. Find E 2𝑋 − 1 .

(a) 1Τ6 (b) 5Τ3 (c) 5Τ2 (d) 11Τ3 (e) 12

Solution: 1

2

1

2 3 1 = න න 𝑐𝑥𝑦 𝑑𝑥𝑑𝑦 = න 2𝑐 𝑦 𝑑𝑦 = 𝑐 ∴ 𝑐 = 3 2 0 0 0 2

1

2

1 3 2 38 2 4 E 𝑋 = න න 𝑥 𝑥𝑦 𝑑𝑥𝑑𝑦 = න 𝑦 𝑑𝑦 = 2 3 0 0 0 23

So, the answer is (b).

2

4 8 3 5 E 2𝑋 − 1 = 2E 𝑋 − 1 = 2 − 1 = − = 3 3 3 3 □

Question 9 Suppose 𝑋 and 𝑌 are discrete random variables with joint p.m.f.

Find Var 𝑋 + 2𝑌 . (a) 0.16 (b) 0.21 (c) 0.49 (d) 1.24 (e) 3.14

Question 9 Suppose 𝑋 and 𝑌 are discrete random variables with joint p.m.f.

Find Var 𝑋 + 2𝑌 .

Solution: E 𝑋 + 2𝑌 = 0.2 2 + 2 0 + 0.5 3 + 2 0 + 0.0 2 + 2 1 + 0.3 3 + 2 1 (a) 0.16 = 0.4 + 1.5 + 0 + 1.5 = 3.4 (b) 0.21 2 2 2 2 E 𝑋 + 2𝑌 = 0.2 2 + 2 0 + 0.5 3 + 2 0 + 0.0 2 + 2 1 + 0.3 3 + 2 1 (c) 0.49 = 0.8 + 4.5 + 0 + 7.5 = 12.8 (d) 1.24 ∴ (e) 3.14 Var 𝑋 + 2𝑌 = 𝐸 𝑋 + 2𝑌 2 − E 𝑋 + 2𝑌 2

So, the answer is (d).

= 12.8 − 3.42 = 12.8 − 11.56 = 1.24 □

2

Question 10 YES or NO? Suppose 𝑋 and 𝑌 have joint p.d.f. 𝑓 𝑥, 𝑦 = 𝑐𝑥 𝑒 𝑦 Τ 1 + 𝑦 3

for 0 ≤ 𝑥 ≤ 1 and 1 ≤ 𝑦 ≤ 3,

where 𝑐 is whatever constant that makes this integrate to 1. Are 𝑋 and 𝑌 independent?

Question 10 YES or NO? Suppose 𝑋 and 𝑌 have joint p.d.f. 𝑓 𝑥, 𝑦 = 𝑐𝑥 𝑒 𝑦 Τ 1 + 𝑦 3

for 0 ≤ 𝑥 ≤ 1 and 1 ≤ 𝑦 ≤ 3,

where 𝑐 is whatever constant that makes this integrate to 1. Are 𝑋 and 𝑌 independent? 𝑎 𝑥 = 𝑐𝑥 𝑏 𝑦 = 𝑒𝑦Τ 1 + 𝑦3

Solution: Since 𝑓(𝑥, 𝑦) *can* be written as 𝑎 𝑥 𝑏(𝑦) and there are no “funny limits”, the random variables of 𝑋 and 𝑌 *are* independent. So, the answer is YES. □

Question 11 TRUE or FALSE? If the covariance 𝑋 and 𝑌 is zero, then this means that

𝑋 and 𝑌 must be independent.

Question 11 TRUE or FALSE? If the covariance 𝑋 and 𝑌 is zero, then this means that

𝑋 and 𝑌 must be independent.

Solution: Independent variables have COV 𝑋, 𝑌 = 0, but that isn’t a guarantee! FALSE

Question 12 Suppose 𝑋 and 𝑌 are identically distributed with a mean of 5, a variance of 20, and Cov 𝑋, 𝑌 = 10. Find Corr 2𝑋, 𝑌 . (a) 0 (b) 1Τ20 (c) 1Τ10 (d) 1Τ2 (e) 9Τ10

Question 12 Suppose 𝑋 and 𝑌 are identically distributed with a mean of 5, a variance of 20, and Cov 𝑋, 𝑌 = 10. Find Corr 2𝑋, 𝑌 . (a) 0 (b) 1Τ20 (c) 1Τ10 (d) 1Τ2 (e) 9Τ10 Solution: Corr 𝑋, 𝑌 =

So, the answer is (d).

□

Cov 2𝑋, 𝑌 Var 2𝑋 Var 𝑌

=

2 10

20 1 = = 40 2 80 20

Question 12 Suppose 𝑋 and 𝑌 are identically distributed with a mean of 5, a variance of 20, and Cov 𝑋, 𝑌 = 10. Find Corr 2𝑋, 𝑌 . (a) 0 (b) 1Τ20 (c) 1Τ10 (d) 1Τ2 (e) 9Τ10 Solution: Corr 𝑋, 𝑌 =

So, the answer is (d).

□

Cov 2𝑋, 𝑌 Var 2𝑋 Var 𝑌

=

2 10

20 1 = = 40 2 80 20

Question 13 TRUE or FALSE? Suppose the arrival of clients to a bank can be modeled as a Poisson process with a given hourly rate 𝜆. Then the expected number of clients arriving between 12:00-2:00 pm is greater than the expected number of clients arriving between 10:00-11:00 am.

Question 13 TRUE or FALSE? Suppose the arrival of clients to a bank can be modeled as a Poisson process with a given hourly rate 𝜆. Then the expected number of clients arriving between 12:00-2:00 pm is greater than the expected number of clients arriving between 10:00-11:00 am.

Solution: The expected number of arrivals depends only on the elapsed time for homogeneous Poisson since it is memoryless.

Therefore, the answer is TRUE because the first interval is longer than the second.

Question 14 Joey's Lightbulb Company sells bulbs that have Exp 𝜆 = 1Τyear lifetimes. Suppose I just bought 100 lightbulbs from Joey. What is the approximate probability that the average of the 100 bulb lifetimes will be at most 1.1 years? (a) 0.14 (b) 0.16 (c) 0.50 (d) 0.84 (e) 1.00

Question 14 Joey's Lightbulb Company sells bulbs that have Exp 𝜆 = 1Τyear lifetimes. Suppose I just bought 100 lightbulbs from Joey. What is the approximate probability that the average of the 100 bulb lifetimes will be at most 1.1 years?

2 2 (a) 0.14 Solution: Since 𝑋𝑖 is Exp 𝜆 = 1 , it has mean 𝜇 = 1Τ𝜆 = 1 and variance 𝜎 = 1Τ𝜆 = 1. (b) 0.16 (c) 0.50 Then the CLT implies 100 (d) 0.84 2 ~ Nor 100,100 ෍ 𝑋 ≈ Nor 𝑛𝜇, 𝑛𝜎 𝑖 (e) 1.00 𝑖=1

This implies 100

Pr ෍ 𝑋𝑖 < 110 ≐ Pr 𝑍100 < 𝑖=1

So, the answer is (d).

□

110 − 100 100

= Φ 1 = 0.84

Φ 1

0.8413

Question 14 Joey's Lightbulb Company sells bulbs that have Exp 𝜆 = 1Τyear lifetimes. Suppose I just bought 100 lightbulbs from Joey. What is the approximate probability that the average of the 100 bulb lifetimes will be at least 1.1 years?

2 2 (a) 0.14 Solution: Since 𝑋𝑖 is Exp 𝜆 = 1 , it has mean 𝜇 = 1Τ𝜆 = 1 and variance 𝜎 = 1Τ𝜆 = 1. (b) 0.16 (c) 0.50 Then the CLT implies 100 (d) 0.84 2 ~ Nor 100,100 ෍ 𝑋 ≈ Nor 𝑛𝜇, 𝑛𝜎 𝑖 (e) 1.00 𝑖=1

This implies 100

Pr ෍ 𝑋𝑖 < 110 ≐ Pr 𝑍100 < 𝑖=1

So, the answer is (d).

□

110 − 100 100

= Φ 1 = 0.84

Question 15 Consider the following integral: 3

𝐼=න

1 −2 𝑥−2 2 𝑒 𝑑𝑥

2

Use the following four 𝑈𝑛𝑖𝑓(0,1) random numbers to compute a Monte Carlo estimate of 𝐼.

0.87 0.35 0.64 0.13 Here’s the formula from class that ought to be used to calculate the estimate: 𝑛 𝑏−𝑎 𝐼෡𝑛 = ෍ 𝑔 𝑎 + 𝑏 − 𝑎 𝑈𝑖 𝑛 𝑖=1

(a) 0.135 (b) 0.342 (c) 0.456 (d) 0.857 (e) 1.206

Question 15 Solution: Plug n’ Chug! 𝑛 𝑏−𝑎 3−2 𝑛 𝐼෡𝑛 = ෍ 𝑔 𝑎 + 𝑏 − 𝑎 𝑈𝑖 = ෍ 𝑔 2 + 3 − 2 𝑈𝑖 𝑛 𝑛 𝑖=1 𝑖=1 𝑛 𝑛 1 1 1 −2 2+𝑈𝑖 −2 2 = ෍ 𝑔 2 + 𝑈𝑖 = ෍ 𝑒 𝑛 𝑛 𝑖=1 𝑖=1

1 1 1 1 −1 0.87 2 − 0.35 2 − 0.64 2 − 0.13 2 2 2 2 = 𝑒 +𝑒 +𝑒 +𝑒 2 = 0.857 4

So, the answer is (d).

□

Question 15 Consider the following integral: 3

𝐼=න

1 −2 𝑥−2 2 𝑒 𝑑𝑥

2

Use the following four 𝑈𝑛𝑖𝑓(0,1) random numbers to compute a Monte Carlo estimate of 𝐼.

0.87 0.35 0.64 0.13 Here’s the formula from class that ought to be used to calculate the estimate: 𝑛 𝑏−𝑎 𝐼෡𝑛 = ෍ 𝑔 𝑎 + 𝑏 − 𝑎 𝑈𝑖 𝑛 𝑖=1

(a) 0.135 (b) 0.342 (c) 0.456 (d) 0.857 (e) 1.206

Question 16 Consider a circle inscribed in a unit square (so that the circle has area 𝜋Τ4). We toss 100 darts randomly into the square, and it turns out that 77 of those darts land in the circle.

Using the method described in class, estimate 𝜋.

(a) 3.04 (b) 3.08 (c) 3.146 (d) 3.16 (e) 3.20

Question 16 Consider a circle inscribed in a unit square (so that the circle has area 𝜋Τ4). We toss 100 darts randomly into the square, and it turns out that 77 of those darts land in the circle.

Using the method described in class, estimate 𝜋.

(a) 3.04 (b) 3.08 (c) 3.146 (d) 3.16 (e) 3.20

Question 16 Consider a circle inscribed in a unit square (so that the circle has area 𝜋Τ4). We toss 100 darts randomly into the square, and it turns out that 77 of those darts land in the circle.

Using the method described in class, estimate 𝜋.

(a) 3.04 (b) 3.08 (c) 3.146 (d) 3.16 (e) 3.20

𝑝Ƹ𝑛 =

77 = 0.77 100

Solution: 𝜋ො = 4 × proportion landing in circle = 4 0.77 = 3.08, i.e., (b). □

Question 16 Consider a circle inscribed in a unit square (so that the circle has area 𝜋Τ4). We toss 100 darts randomly into the square, and it turns out that 77 of those darts land in the circle.

Using the method described in class, estimate 𝜋.

(a) 3.04 (b) 3.08 (c) 3.146 (d) 3.16 (e) 3.20

Solution: 𝜋ො = 4 × proportion landing in circle = 4 0.77 = 3.08, i.e., (b). □

Question 17 Suppose that 𝑈 and 𝑉 are i.i.d. Unif(0,1) random variables. Select from the options below an appropriate expression to generate one realization of a 20-sided die toss.

(a) 10𝑈 + 10𝑉 (b) 0.5 × 20𝑈 + 0.5 × 20𝑉 (c) Either 20𝑈 or 20𝑉 (d) All of the above. (e) Only (a) and (c)

Question 17 Suppose that 𝑈 and 𝑉 are i.i.d. Unif(0,1) random variables. Select from the options below an appropriate expression to generate one realization of a 20-sided die toss.

Not always *INTEGER*

(a) 10𝑈 + 10𝑉 (b) 0.5 × 20𝑈 + 0.5 × 20𝑉 (c) Either 20𝑈 or 20𝑉 (d) All of the above. (e) Only (a) and (c)

Solution: Option B doesn’t always give integers.

Option A simulates a triangular p.m.f.

Option C is correct, they both have a uniform p.m.f with values from 1-20.

□

Question 17 Suppose that 𝑈 and 𝑉 are i.i.d. Unif(0,1) random variables. Select from the options below an appropriate expression to generate one realization of a 20-sided die toss.

(a) 10𝑈 + 10𝑉 (b) 0.5 × 20𝑈 + 0.5 × 20𝑉 (c) Either 20𝑈 or 20𝑉 (d) All of the above. (e) Only (a) and (c)

Solution: Option B doesn’t always give integers.

Option A simulates a triangular p.m.f.

Option C is correct, they both have a uniform p.m.f with values from 1-20.

https://math.stackexchange.com/questions/1204 396/why-is-the-sum-of-the-rolls-of-two-dices-abinomial-distribution-what-is-define

□

Question 17 Suppose that 𝑈 and 𝑉 are i.i.d. Unif(0,1) random variables. Select from the options below an appropriate expression to generate one realization of a 20-sided die toss.

(a) 10𝑈 + 10𝑉 (b) 0.5 × 20𝑈 + 0.5 × 20𝑉 (c) Either 20𝑈 or 20𝑉 (d) All of the above. (e) Only (a) and (c)

Solution: Option B doesn’t always give integers.

Option A simulates a triangular p.m.f.

Option C is correct, they both have a uniform p.m.f with values from 1-20.

□

Question 17 Suppose that 𝑈 and 𝑉 are i.i.d. Unif(0,1) random variables. Select from the options below an appropriate expression to generate one realization of a 20-sided die toss.

(a) 10𝑈 + 10𝑉 (b) 0.5 × 20𝑈 + 0.5 × 20𝑉 (c) Either 20𝑈 or 20𝑉 (d) All of the above. (e) Only (a) and (c)

Solution: Option B doesn’t always give integers.

Option A simulates a triangular p.m.f.

Option C is correct, they both have a uniform p.m.f with values from 1-20.

□

Question 18 Consider the discrete random variable 𝑋:

𝑃 𝑋=𝑥 =

0.3, 0.2, 0.5, 0,

𝑖𝑓 𝑥 = −3.5 𝑖𝑓 𝑥 = 2 𝑖𝑓 𝑥 = 6 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Use the discrete version of the Inverse Transform method from class and the Unif(0,1) pseudo-random number 𝑈 = 0.73 to generate one observation 𝑋 coming from this p.m.f.

(a) 𝑋 = −3.5 (b) 𝑋 = 0.73 (c) 𝑋 = 1 (d) 𝑋 = 2 (e) 𝑋 = 6

Question 18 Solution: Using the discrete version of the Inverse Transform Theorem, we can draw the following table: 𝒙

𝒇(𝒙) 𝑭(𝒙)

Unif(𝟎, 𝟏)

-3.5

0.3

0.3

[0.0,0.3]

2

0.2

0.5

(0.3,0.5]

6

0.5

1.0

(0.5,1.0]

By the Inverse Transform Theorem, 𝑋 = 𝐹 −1 𝑈 . For 𝑈 = 0.73, the corresponding xvalue is 𝑋 = 6, thus the answer is (e). □

Question 18 Consider the discrete random variable 𝑋:

𝑃 𝑋=𝑥 =

0.3, 0.2, 0.5, 0,

𝑖𝑓 𝑥 = −3.5 𝑖𝑓 𝑥 = 2 𝑖𝑓 𝑥 = 6 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Use the discrete version of the Inverse Transform method from class and the Unif(0,1) pseudo-random number 𝑈 = 0.73 to generate one observation 𝑋 coming from this p.m.f.

(a) 𝑋 = −3.5 (b) 𝑋 = 0.73 (c) 𝑋 = 1 (d) 𝑋 = 2 (e) 𝑋 = 6

Question 19 Suppose that we generate a Unif 0,1 PRN, 𝑈 = 0.5.

Now use the Inverse Transform Method to generate an Exponential random variate with 𝜆 = 1Τ2. (a) −1.386 (b) −0.387 (c) 0.347 (d) 0.693 (e) 1.386

Question 19 Suppose that we generate a Unif 0,1 PRN, 𝑈 = 0.5.

Now use the Inverse Transform Method to generate an Exponential random variate with 𝜆 = 1Τ2. (a) −1.386 (b) −0.387 (c) 0.347 (d) 0.693 (e) 1.386 Solution: Plug n’ chug

𝑋=

−1 −1 ln 1 − 𝑈 = ln 1 − 0.5 = −2 ln 0.5 = ln 4 = 1.386 𝜆 1Τ2

So, the answer is (e).

□

Question 20 Consider the linear congruential generator 𝑋𝑖+1 = 3𝑋𝑖 + 1 mod 4 . If 𝑋0 = 1, calculate the pseudo-random number 𝑈3 . (a) 0 (b) 0.125 (c) 0.25 (d) 0.5 (e) 1

Question 20 Consider the linear congruential generator 𝑋𝑖+1 = 3𝑋𝑖 + 1 mod 4 . If 𝑋0 = 1, calculate the pseudo-random number 𝑈3 . (a) 0 (b) 0.125 (c) 0.25 (d) 0.5 (e) 1 Solution: This generator has a short cycle length of 2 when you start with 𝑋0 = 1. 𝑖

0

1

2

3

4

𝑋𝑖

1

0

1

0

1

Thus 𝑋3 = 𝑋2 = 𝑋1 = 𝟎.

Therefore, 𝑈3 = 02 = 0, the answer is (a). □

Question 21 Consider a single-server system where the first customer arrives at time 2 and starts getting served immediately. The interarrival times for the next 3 customers that arrive are all equal to 2 --- they are so well-behaved!! This means that customer 2 shows up at time 4, customer 3 arrives at time 6, and customer 4 at time 8. The line rule is FIFO, and suppose service times decrease as more customers arrive. The service times for customers 1,2,3,4 are 12, 8, 4, 2, respectively.

When does the third customer (i.e., the customer who arrives at time 6) leave the system? (a) 14 (b) 16 (c) 20 (d) 22 (e) 26

Question 21

Time service lasted

Solution: If we can first determine the departure time 𝐷𝑖 = 𝑇𝑖 + 𝑆𝑖 for each customer, we can compare them to their arrival times: Time service started 𝐢

𝑨𝒊

𝑫𝒊

1

2

14

𝐢

𝑨𝒊

𝑻𝒊

𝑺𝒊

2

4

22

1

2

2

12

3

6

26

2

4

14

8

4

8

28

3

6

22

4

4

8

26

2

Arrival and Departure Times

We end up with a Departure time of 26 for customer 3. Answer (e). □

Event list for FIFO

Question 21 Solution: If we can first determine the departure time 𝐷𝑖 = 𝑇𝑖 + 𝑆𝑖 for each customer, we can compare them to their arrival times: 𝐢

𝑨𝒊

𝑫𝒊

1

2

14

𝐢

𝑨𝒊

𝑻𝒊

𝑺𝒊

2

4

22

1

2

2

12

3

6

26

2

4

14

8

4

8

28

3

6

22

4

4

8

26

2

Arrival and Departure Times

We end up with a Departure time of 26 for customer 3. Answer (e). □

Event list for FIFO

Question 21 Consider a single-server system where the first customer arrives at time 2 and starts getting served immediately. The interarrival times for the next 3 customers that arrive are all equal to 2 --- they are so well-behaved!! This means that customer 2 shows up at time 4, customer 3 arrives at time 6, and customer 4 at time 8. The line rule is FIFO, and suppose service times decrease as more customers arrive. The service times for customers 1,2,3,4 are 12, 8, 4, 2, respectively.

When does the third customer (i.e., the customer who arrives at time 6) leave the system? (a) 14 (b) 16 (c) 20 (d) 22 (e) 26

Question 22 Consider a single-server system where the first customer arrives at time 2 and starts getting served immediately. The interarrival times for the next 3 customers that arrive are all equal to 2 --- they are so well-behaved!! This means that customer 2 shows up at time 4, customer 3 arrives at time 6, and customer 4 at time 8. The line rule is LIFO, and suppose service times decrease as more customers arrive. The service times for customers 1,2,3,4 are 12, 8, 4, 2, respectively.

When does the third customer (i.e., the customer who arrives at time 6) leave the system? (a) 14 (b) 16 (c) 20 (d) 22 (e) 26

Question 22

Time service lasted

Solution: If we can first determine the departure time 𝐷𝑖 = 𝑇𝑖 + 𝑆𝑖 for each customer, we can compare them to their arrival times: Time service started 𝐢

𝑨𝒊

𝑫𝒊

1

2

14

𝐢

𝑨𝒊

𝑻𝒊

𝑺𝒊

2

4

22

1

2

2

12

3

6

26

2

4

20

8

4

8

28

3

6

16

4

4

8

14

2

Arrival and Departure Times

We end up with a Departure time of 20 for customer 3. Answer (c). □

Event list for LIFO

Question 22 Solution: If we can first determine the departure time 𝐷𝑖 = 𝑇𝑖 + 𝑆𝑖 for each customer, we can compare them to their arrival times: 𝐢

𝑨𝒊

𝑫𝒊

1

2

14

𝐢

𝑨𝒊

𝑻𝒊

𝑺𝒊

2

4

22

1

2

2

12

3

6

26

2

4

20

8

4

8

28

3

6

16

4

4

8

14

2

Arrival and Departure Times

We end up with a Departure time of 20 for customer 3. Answer (c). □

Event list for LIFO

Question 22 Consider a single-server system where the first customer arrives at time 2 and starts getting served immediately. The interarrival times for the next 3 customers that arrive are all equal to 2 --- they are so well-behaved!! This means that customer 2 shows up at time 4, customer 3 arrives at time 6, and customer 4 at time 8. The line rule is LIFO, and suppose service times decrease as more customers arrive. The service times for customers 1,2,3,4 are 12, 8, 4, 2, respectively.

When does the third customer (i.e., the customer who arrives at time 6) leave the system? (a) 14 (b) 16 (c) 20 (d) 22 (e) 26

Question 23 In the course of a simulation study at Disneyworld, we provide our client with confidence intervals for the mean customer waiting time in front of the Space Mountain attraction. What step of the simulation study is this? (a) Problem Formulation (b) Data Collection (c) Verification (d) Validation (e) Output Analysis

Question 23 In the course of a simulation study at Disneyworld, we provide our client with confidence intervals for the mean customer waiting time in front of the Space Mountain attraction. What step of the simulation study is this? (a) Problem Formulation (b) Data Collection (c) Verification (d) Validation (e) Output Analysis Solution: Confidence intervals on an output? That’s clearly Output Analysis!

Answer (e). □

Question 24 TRUE or FALSE? An event is a time, often associated with a system state change.

Question 24 TRUE or FALSE? An event is a time, often associated with a system state change.

Solution: TRUE!

□

Question 25 TRUE or FALSE? Entities can have various properties or attributes (e.g. the priority of a customer).

Question 25 TRUE or FALSE? Entities can have various properties or attributes (e.g. the priority of a customer).

Solution: TRUE!

□

Question 26 TRUE or FALSE? In a discrete-event simulation, it is possible for two events (e.g., an arrival and a departure) to happen simultaneously with respect to the simulation clock.

Question 26 TRUE or FALSE? In a discrete-event simulation, it is possible for two events (e.g., an arrival and a departure) to happen simultaneously with respect to the simulation clock.

Solution: TRUE!

□

Question 27 TRUE or FALSE? The simulation clock is a variable that always represents simulated time.

Question 27 TRUE or FALSE? The simulation clock is a variable that always represents simulated time.

Solution: TRUE, the simulation clock is *not* real time, but it *is* simulated time! □

Question 28 Which of the following is FALSE about the Next-Event time-advance mechanism? (a) Times of all known future events are determined and placed in the future events list (FEL). (b) The simulation clock is initialized at 0. (c) This approach is used in continuous-time models. (d) The clock advances to the most-imminent event. (e) At each event, the system state is updated, and the FEL is updated.

Question 28 Which of the following is FALSE about the Next-Event time-advance mechanism? (a) Times of all known future events are determined and placed in the future events list (FEL). (b) The simulation clock is initialized at 0. (c) This approach is used in continuous-time models. (d) The clock advances to the most-imminent event. (e) At each event, the system state is updated, and the FEL is updated.

Solution: Fixed increment is primarily for continuous time models or ones where data comes in … at the fixed increment. Since the clock *always* moves the same amount, there is nothing saying the state has to change! However, time still has to move *forward* in a fixed increment!

Next-Event time advance is, indeed, initialized at 0 time and all future events are in the FEL. The clock advances to the next most-imminent event (nothing can happen inbetween anyways!). The events then update the system state and FEL. Answer (c)

□

Question 29 Which modeling approach does Arena use? (a) Event-Scheduling. (b) Process-Interaction. (c) Next-Event-Advancement. (d) All of the above. (e) (a) and (c)

Question 29 Which modeling approach does Arena use? (a) Event-Scheduling. (b) Process-Interaction. (c) Next-Event-Advancement. (d) All of the above. (e) (a) and (c)

Solution: (b). ARENA is based on the Process-Interaction “world view”.

Question 30 TRUE or FALSE? Most of the programming effort involved in the Event-Scheduling modeling approach comes from keeping track of the event bookkeeping.

Question 30 TRUE or FALSE? Most of the programming effort involved in the Event-Scheduling modeling approach comes from keeping track of the event bookkeeping.

Solution: TRUE! That’s why we are studying Process-Interaction!

□

Question 31 In Arena, what modules would I need to generate customer arrivals, perform some action on them, and remove them from the system? (a) GENERATE, DELAY, PROCESS (b) CREATE, ASSIGN, DISPOSE (c) ARRIVE, PROCESS, DISPOSE (d) GENERATE, ASSIGN, DISPOSE (e) CREATE, PROCESS, DISPOSE

Question 31

SEE: Module05-09_-_AssignPlusFunnyQueuePriority

Question 31 In Arena, what modules would I need to generate customer arrivals, perform some action on them, and remove them from the system? (a) GENERATE, DELAY, PROCESS (b) CREATE, ASSIGN, DISPOSE (c) ARRIVE, PROCESS, DISPOSE (d) GENERATE, ASSIGN, DISPOSE (e) CREATE, PROCESS, DISPOSE

Solution: CREATE customers, PROCESS them to act on them, then DISPOSE of them to remove them from the system. Answer (e). □

Question 32 TRUE or FALSE? In Arena, you can use a single DECIDE module to route customers to that one of four servers having the smallest queue.

Question 32 TRUE or FALSE? In Arena, you can use a single DECIDE module to route customers to that one of four servers having the smallest queue.

Solution: TRUE!

□

Question 33 Which of the following modules can not be found in the Arena Basic Process template? (a) DECIDE (b) CREATE (c) GENERATE (d) ASSIGN (e) DISPOSE

Question 33 Which of the following modules can not be found in the Arena Basic Process template? (a) DECIDE (b) CREATE (c) GENERATE (d) ASSIGN (e) DISPOSE

Solution: (c).

□