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CE 4640: Transportation Design Prof. Tapan Datta, Ph.D., P.E. Fall 2002
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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.
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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)
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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
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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
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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
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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
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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)
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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)
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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
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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.
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Queueing Theory Queueing is a common phenomena in Fast food store Bank Toll plaza A signalized intersection Elevator Other service centers
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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’.
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Queueing Process INPUT SOURCE QUEUE SERVICE MECHANISM QUEUEING SYSTEM calling units served units
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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
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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.
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Example of Banking Process TELLERS BANK’S TELLER SERVICE QUEUE MORE TELLER COUNTERS WILL INCREASE SERVICE RATE & DECREASE QUEUE LENGTH QUEUE
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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 = /
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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
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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
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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
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Poisson Input & Constant Service Time L q = P 2 /2(1-P) where Lq = expected queue length = arrival time = service time P = /
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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.
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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 ( - )
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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)
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