Georgia Institute Of Technology ISYE 6644 EXTRA OFFICE HOUR – MIDTERM 1 SOLUTIONS
ISYE 6644
EXTRA OFFICE HOUR – MIDTERM 1 SOLUTIONS
(SORRY IN
... [Show More] ADVANCE FOR MY LACK OF POWERPOINT SKILLS!)
With only one server, customer arrivals will overwhelm the server and a queue WILL form.
10AM 11AM 1PM 2PM
This leaves us with 0 - 0(𝐸 𝑋2] = 0
*Extra example here; not needed to derive/calculate
this to answer this question.
The question was supposed to state “sum is less than or equal to 2” (or alternatively “sum is less than 3”). This would have made
the correct answer to be FALSE because
P(sum is less than or equal to 2)= 1
P(sum is greater than 11)= 1
P(sum is less than or equal to 2 AND sum is greater than 11)= 0 and then 0 ≠ 1
which means they are dependent.
However, the way the question is worded, the P(sum is less than 2)= 0, so we would have 0 = 0 1
Only possibility of being square is when 𝑋 = 4.7 and 𝑌 = 4.7, so we want to find 𝑃(𝑋 = 4.7, 𝑌 = 4.7)
Time between events (interarrival times) in a Poisson process follow an exponential distribution with 𝜆 = 1
Let 𝑋 = time between the 5th and 6th arrivals; 𝑋~𝐸𝑥𝑝(𝜆 = 1)
Note: The cumulative distribution function for 𝐸𝑥𝑝(𝜆) is 𝐹 𝑥 = 𝑃 𝑋 ≤ 𝑥 = 1 - 𝑒-𝜆𝑥
Could also try integrating the probability density function
et 𝑌 = number of arrivals in 1 hour; 𝑌~𝑃𝑜𝑖𝑠(𝜆 = 1)
f 𝑋𝑖 has mean 𝜇 = 𝜆 and variance 𝜎2 = 𝜆, then the Central Limit Theorem implies that
Would generate sum of two dice
Would include 1.5, 2.5, etc.
𝑛𝑈 would be creating 𝑈𝑛𝑖𝑓(0, 𝑛). Using the ceiling function would then create a discrete distribution with the
possible outcomes of 1, 2, 3, 4, … , 𝑛 which is a realization of an 𝑛-sided die toss.
Using properties of logarithms, -2 ln 𝑈4 = -8ln(𝑈)
This has the same form as - 1
𝜆 ln(𝑈) which is ~𝐸𝑥𝑝(𝜆), so -8 ln 𝑈 ~𝐸𝑥𝑝 1 8
Note, this is not Erlang because we only have one 𝑈 here, so we are not adding iid
exponentials. We would be adding up the same exponential four times.
Erlang would look like
-2 ln 𝑈1𝑈2𝑈3𝑈4 = -2 ln 𝑈1 - 2ln(𝑈2) - 2ln(𝑈3) - 2ln(𝑈4)
This is an application of inverse transform since Φ(𝑥) is the cumulative distribution function for 𝑁(0,1) distribution.
Integral of 𝑁 0, 1 distribution, so this is the same as 𝑃 0 ≤ 𝑍 ≤ 1 = Φ 1 - Φ 0 = 0.8413 - 0.5000 = 0.3413
Has a cycle length of 8, so one way is to take 100𝑚𝑜𝑑 8 = 4, so we want 𝑋4 = 7 [Show Less]