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CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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Problem 1 a) Given: n = 0 to 15; # of trucks s = 3; # of pavers M= total # of entities to be served P0 = Probability of having 0 trucks Pn = other probability 0<=n<=s S<=n<=M n>=M Multi-server Stochastic Queueing Model - Finite Population Matlab Script available on the course website Input parameters
Limited Source Queueing Model arrival rate (trucks/hr) = 2.2222 service rate (trucks/hr) = 14.9925 Po = 0.11403 1) Fraction of idle time (FTI) = 0.37072 ---------- ------------- --------- # of customers Prob. of n customers in the Q-S in the Q-S (n) (Pn) (n*Pn)
CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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---------- ------------- --------- 0 0.11403 0 1 0.25352 0.25352 2 0.26304 0.52608 3 0.16895 0.50685 4 0.10017 0.40068 5 0.05444 0.2722 6 0.026897 0.16138 7 0.01196 0.083722 8 0.0047274 0.03782 9 0.001635 0.014715 10 0.00048469 0.0048469 11 0.00011974 0.0013171 12 2.3663e-05 0.00028396 13 3.5074e-06 4.5596e-05 14 3.4658e-07 4.8522e-06 15 1.7124e-08 2.5686e-07 ---------- ------------- --------- SUM 1 2.2635 (=L) ---------- ------------- --------- 2) Expected # of working trucks(K=M-L) = 12.7365 3) Operating efficiency of truck (K/M) = 0.8491 4) Expected # of pavers that are working (R) = 1.8878 Operating efficiency of a paver (R/S) = 0.62928 Average arrival rate = 28.3034 <Sample Hand Calculation> lamda/mu = 2.22/15 = 0.148 Calculator P0 n M (M-
n) M! (M-n)! (
𝑙𝑎𝑚𝑑𝑎𝑚𝑢
)( 𝑀!
𝑀 − 𝑛 ! 𝑛!(𝑙𝑎𝑚𝑑𝑎𝑚𝑢
)(
s-3
0 15 15 15!= 1.30767E+12
15!= 1.30767E+12
(0.148)^0 1
s-2
1 15 14 1.30767E+12 14!= 87178291200
(0.148)^1 2.22
s-1
2 15 13 1.30767E+12 13!= 6227020800
(0.148)^2 2.30
CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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∑ 5.52 n M! (M-n)! (lamda/mu)^n s! s^(n-s) 𝑀!
𝑀 − 𝑛 ! 𝑠! 𝑠(./(𝑙𝑎𝑚𝑑𝑎𝑚𝑢
)(
3 1.30767E+12
12! (0.148)^3 3! =6 1 1.475015364 11! (0.148)^4 3 0.8732090935
10! (0.148)^5 9 0.473861468
6 9! (0.148)^6 27 0.2337716577 8! (0.148)^7 81 0.1037946168 7! (0.148)^8 243 0.0409642759 6! (0.148)^9 729 0.0141463310 5! (0.148)^10 2187 0.00418731411 4! (0.148)^11 6561 0.00103287112 3! (0.148)^12 19683 0.0002038213 2! (0.148)^13 59049 3.01653E-0514 1! (0.148)^14 177147 2.97631E-0615 0! (0.148)^15 531441 1.46831E-07 0
(1/
3.22
P0 = 1/(5.52+3.22) = 0.114 P1 P1 𝑀!
𝑀 − 𝑛 ! 𝑛!(𝑙𝑎𝑚𝑑𝑎𝑚𝑢
)( 𝑃0 =
15!14 ! 1!
0.148 : 0.114
0.253
P5 𝑀!𝑀 − 𝑛 ! 𝑠! 𝑠(./
𝑙𝑎𝑚𝑑𝑎𝑚𝑢
(
𝑃𝑜 =
15!10 ! 3! 3=
0.148 > 0.114
0.054
b)
CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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Problem 2 Multi-server Stochastic Queueing Model - Infinite Population (Matlab script) available on the course website
lamda = 2.8 ships per day mu = 1/1.4 day = 0.714 ships per day lamda/mu = 3.92 lamda/(s*mu) = 0.784 <Sample Hand Calculation> n n! (lamda/mu)^n (lamda/mu)^s (𝑙𝑎𝑚𝑑𝑎𝑚𝑢 )(
𝑛!
/.:
(1?
+(𝑙𝑎𝑚𝑑𝑎𝑚𝑢 )/
𝑠!1
1 − (𝑙𝑎𝑚𝑑𝑎𝑠 ∗ 𝑚𝑢)
s-5 0 1 (3.92)^0 = 1 (3.92)^5 = 925.61
CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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s-4 1 1 (3.92)^1 =3.92
(1+3.92+7.6832+10.04+9.84)+35.71=68.19P0=1/68.19=0.0147
s-3 2 4 (3.92)^2 =15.36
s-2 3 6 (3.92)^3 =60.24
s-1 4 24 (3.92)^4 = 234.13
P3 3! = 6 (𝑙𝑎𝑚𝑑𝑎𝑚𝑢
()
𝑛!𝑃0 =
(3.92)D
6∗ 0.0147 = 0.147
P8 8! = 40320 (𝑙𝑎𝑚𝑑𝑎𝑚𝑢()
𝑠! 𝑠(./𝑃0 =
(3.92)G
5! ∗ 5D∗ 0.0147 = 0.0546
b) 1-P8 = Probability to have more than 3 ships waiting for service. This means that 5 ships are at the servers (berths) and 3 are waiting for service at port. The sum of the probabilities is 1. The probability of more than 3 ships wait for service is then: P(x>8) = 1- P(x<=8) where x is the number of ships in the system. 1-(P0+P1+P2+P3+P4+P5+P6+P7+P8) = 1- 0.8022 = 0.1978
There is 19.78 % chance of having more than 3 ships in queue. c)
CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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d) $32,000 per hour *24 hour/day = $768,000/day
𝑊IJ/K = 𝑊L ∗ 𝑙𝑎𝑚𝑑𝑎 ∗ 𝑁 ∗ 𝐶OPQ
𝑊IJ/K = 0.67884𝑑𝑎𝑦𝑠 ∗ 2.8𝑠ℎ𝑖𝑝𝑠𝑑𝑎𝑦
∗ 365𝑑𝑎𝑦𝑠 ∗$768,000day
= $532,818,800
Problem 3
Deterministic Queueing Model - Main File (Matlab script)
Deterministic Queueing Model - Function File (Matlab script) available on the course website
CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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Average delay per vehicle = Total Delay/Max Queue Length = 1263.4/1344.7 = 0.939 hr <Sample Manual Calculation>
CEE 3604 – Introduction to Transportation Engineering Assignment 8 Solution Prepared by Amy Seo
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