1. CE 4640: Transportation Design Prof. Tapan Datta, Ph.D., P.E. Fall 2002

2. Example on Rate Quality Control Method Intersection Right Angle Crashes Rear End Crashes Side Swipe Crashes Left-Turn Head-On Crashes OtherTotal Crashes Entering ADT X 5 63141915,000 Y684232330,000 Z253111220,000 Three similar intersection locations with 4-lane divided lane configuration have the following annual crash data and total approach volumes. Using Rate Quality Control method, determine the hazardous location(s) based on total crashes.

3. Let us do the analysis considering total crashes. Crash Rate at a location, R sp = Calculation of Crash Rates Crashes per million vehicles where T = Period of study (years) V = Average Daily Traffic (Sum of all approach volumes) Freq. of Crashes*10 6 (365)(T)(V)

4. Crash Rate at a location X, R X = Calculation of Crash Rate Crashes per million vehicles 19*10 6 (365)(1)(15,000) = 3.47 crashes per million vehicles

5. Crash Rate at a location X, R Y = Calculation of Crash Rate Crashes per million vehicles 23*10 6 (365)(1)(30,000) = 2.10 crashes per million vehicles

6. Crash Rate at a location X, R Z = Calculation of Crash Rate Crashes per million vehicles 12*10 6 (365)(1)(20,000) = 1.64 crashes per million vehicles

7. Average crash rate for 3 locations = (3.47+2.10+1.64)/3 = 2.40 crashes per million vehicles Average ADT = (15,000+30,000+20,000)/3 = 21,667 Average Crash Rate & ADT

8. Calculation of Critical Crash Rate Critical crash rate is given by: R c = R a + K (R a / M) 1/2 + 1/(2M) where, R c = Critical rate for spot or section R a = Average crash rate for all spots of similar characteristics or on similar road types M = Millions of vehicles passing over a spot or millions of vehicles miles of travel on a section K = A probability factor (as shown on next slide)

9. P (Probability) 0.005 0.0075 0.05 0.075 0.10 K-value 2.576 1.960 1.645 1.440 1.282 The most commonly used K values are 2.576 (P =.005) and 1.645 (P = 0.05). Selection of Probability Factor (K)

10. Calculation of Critical Crash Rate In this problem, R a = 2.40 crashes per million entering vehicles M = (Average ADT*365)/10 6 = (21,667*365)/10 6 = 7.90 million vehicles per year K = 1.645 for a probability of 0.05  Critical crash rate, R c = 2.40 + 1.645 (2.40/7.90) 1/2 + 1/(2*7.90) = 3.37 crashes per million vehicles

11. Results and Conclusion Critical crash rate, R c = 3.37 crashes per million vehicles. Crash rates at locations X, Y and Z are 3.47, 2.10 and 1.64 million vehicles respectively.  The hazardous location based on Rate Quality Control method is X, where the crash rate is higher than the critical crash rate.

12. Queueing Theory Queueing is a common phenomena in Fast food store Bank Toll plaza A signalized intersection Elevator Other service centers

13. Queueing Process Assumptions in basic queueing process: Units requiring service are generated over time by an ‘input source’. These units enter the queueing system to join a ‘queue’. At a certain point of time, a member is selected for service by a rule known as ‘service discipline’.

14. Queueing Process INPUT SOURCE QUEUE SERVICE MECHANISM QUEUEING SYSTEM calling units served units

15. Few Terms Explained Queue Length – Number of units waiting for service Line Length – Number of units in the line including one or more units being served Service Discipline – rule for providing service “First In, First Out” … Fast food store “First In, Last Out” … Elevator

16. Few Terms Explained Service Mechanism – One or more service facilities each of which contain one or more “parallel service channels” (servers). If there is more than one service facility, calling unit may receive service from a sequence, called “service channels in series”. Pressure Coefficient – Situation with too much pressure, too many people in queue.

17. Example of Banking Process TELLERS BANK’S TELLER SERVICE QUEUE MORE TELLER COUNTERS WILL INCREASE SERVICE RATE & DECREASE QUEUE LENGTH QUEUE

18. Single Server Model Poisson Input & Exponential Service Times L = /(  - ) = P/(1-P) L q = 2 /(  - ) where L = expected line length Lq = expected queue length = arrival time  = service time P = / 

19. Single Server Model  = 1/(  - )  q =  - 1/  = /[  (  -1)] where  = expected waiting time in the system  q = expected waiting time in the queue = arrival time  = service time

20. Example Problem Customers arrive as per Poisson distribution with mean rate of arrival of 30/hr. Required time to serve a customer has an Exponential distribution and is 90 sec. Determine queue characteristics: L, L q, ,  q

21. Example Problem = 30 cust/hr = 30 cust/60 min = ½ cust/min 1/  = 90 sec/cust * (1 min/60 sec) = 3/2 min/cust P = /  = (1/2)/(2/3) = 3/4 L = /(  - ) = P/(1-P) = 3/4 /(1-3/4) = 3 L q = 2 /(  - ) = (1/2) 2 /[2/3(2/3 – 1/2)] = 2.25  = 1/(  - ) = 1/(2/3 – 1/2) = 6  q =  - 1/  = 6 – 1/(2/3) = 4.5

22. Poisson Input & Constant Service Time L q = P 2 /2(1-P) where Lq = expected queue length = arrival time  = service time P = / 

23. Poisson Input & Erlang’s Service Time Probability density function, f(t) = (  k) k t k-1 e -k  t (k-1)! for t  0 where  = service time k = parameter determining the dispersion of the distribution.

24. Poisson Input & Erlang’s Service Time L q = [(1+k)/(2k)]*[ 2 /  (  - 1)] L =  where Lq = expected queue length  = expected waiting time in the system  q = expected waiting time in the queue = arrival time  = service time k = parameter determining the dispersion of the distribution.  q = (1+k) 2k  (  - )

25. Poisson Input & Arbitrary Service Time L q = 2  2 + P 2 L = P + L q  =  q + 1/   q = L q / where Lq = expected queue length L = expected line length  = expected waiting time in the system  q = expected waiting time in the queue = arrival time  = service time P = /   2 = variance 2 (1- P)