Microscopic Flow Characteristics Time Headway - Distribution
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1 CE57: Traffic Flow Thory Spring 20 Wk 2 Modling Hadway Disribuion Microscopic Flow Characrisics Tim Hadway - Disribuion Tim Hadway Dfiniion Tim Hadway vrsus Gap Ahmd Abdl-Rahim Civil Enginring Dparmn, Univrsiy of Idaho Hadway Characrisics Som Applicaions Uninrrupd Traffic: Drivr Bhavior Sudis (minimum hadway Sauraion Flow Sudis Frway Simulaion modls Frway Mrging Characrisics Hadway Characrisics Som Applicaions Inrrupd Traffic: Gap Accpanc (Unsignalizd inrscion capaciy Sauraion Flow Sudis Traffic Signal Conrol
2 Hadway Frquancy Microscopic Flow Characrisics Tim Hadway - Disribuion 20 Hadway Obsrvaion NB = 250 vph EB= 320 vph Microscopic Flow Characrisics Tim Hadway - Disribuion Hadway Disribuion NB Traffic NB 4 EB Obsrvaion Hadway (sc Hadway Disribuion Norhbound Traffic Hadway Disribuion Easbound Traffic Microscopic Flow Characrisics Tim Hadway - Classificaion Random Hadway Sa (Ngaiv Exponnial-Poisson coun Disribuion Consan Hadway Sa (Normal Disribuion (Parson yp II, Gamma, Enlarg, Ngaiv Exponnial, shifd Ngaiv Exponnial Microscopic Flow Characrisics Random Hadway Sa-Poisson Disribuion mxm P(x P(x = Probabiliy ha xacly x numbr of vns occur during im inrval ( m = Avrag numbr of vns during im inrval ( = Napirian bas of logarihms (= Populaion man = populaion varianc 2
3 Exampls of using Poisson Disribuion On an inrscion approach wih a lf urn volum of 20 vph. (Poisson, wha is h probabiliy of skipping h grn phas for h lf urn raffic? Th inrscion is conrolld by am acuad signal wih an avrag cycl lngh of 90 sconds. m = Avrag numbr of lf urn vhicls pr Cycl numbr of cycls pr hour = 300/90 = 40 m = 20 vph / 40 = 3 vhicls/cycl Probabiliy of x = 0 P(0 = = 4.9 % (.9 cycls/hours mxm P(x Exampls of using Poisson Disribuion A parking sudy was conducd for a parking lo ha has 0 parking spacs. Th sudy usd 5 minus inrvals ovr a 2 hour priod for 5 days. Th numbr of mpy parking spacs obsrvd during h oal im priod is 200. Assuming a Poisson disribuion, wha is h probabiliy ha a parking spac will b availabl a any im? m = Avrag numbr of mpy spacs pr im priod oal numbr of im priods = 24 x 5 = 20 m = 200 vph / 20 =.7 spac/im priod Probabiliy of x > 0 = -P(0 P(0 = 0.8 mxm P(x P(x > 0 =- 0.8 = % Probabiliy of finding an mpy parking spac Exampls of using Poisson Disribuion An inrscion is conrolld by a fixd im signal having a cycl lngh of 55 sconds. From h norhbound, hr is a prmid lf urn movmn of 75 vph. If wo vhicls can urn ach cycl wihou causing dlay, on wha prcn of h cycls will dlay occur? m = Avrag numbr of lf urn vhicls pr Cycl numbr of cycls pr hour = 300/55 = 5.4 m = 75 vph / 5.4 = 2.7 LT vhicls/cycl mxm P(x Probabiliy of x > 2 P (x>2 = [P(0 + P( + P (2] = [ ] = 49.9% [ 32.7 Cycls/hour] Exampls of using Poisson Disribuion Using h following daa o consruc a disribuion curv, drmin h probabiliy ha a vhicl approaching on a sid sr will hav o wai, 2, or 3 gaps bfor nring h raffic sram. (h minimum gap accpanc is 4.0 sconds Gap siz Obsrvd frquncy P (h=x < > Toal of 70 Obsrvaions P (h x
4 Exampls of using Poisson Disribuion Using h following daa o consruc a disribuion curv, drmin h probabiliy ha a vhicl approaching on a sid sr will hav o wai, 2, or 3 gaps bfor nring h raffic sram. (h minimum gap accpanc is 4.0 sconds Probabiliy ha h gab in h main sr < 4.0 sconds P (h < 4.0 = 0.34 Probabiliy ha h gab in h main sr 4.0 sconds P (h 4.0 = =0.4 Probabiliy ha a vhicl on h sid sr will no wai (urn in h firs gap = Probabiliy of firs gap 4.0 = 0.4 (4% Probabiliy ha a vhicl on h sid sr will wai ONE gap = Probabiliy of firs gap < 4.0 AND scond gab 4 = 0.34 x 0.4 =0.27 Probabiliy ha a vhicl on h sid sr will wai TWO gaps = Probabiliy of firs gap < 4.0 AND scond gab <4 AND hird gab 4 = 0.34 x 0.34x0.4 =0.074 (7.4% Random Hadway Sa-Poisson Disribuion mxm P(x P (0 = m If no vhicls arriv in im inrval (, hn h im hadway mus b qual o or grar han ( Thrfor, P (h = P(0 = m If V = hourly flow ra, hn m = (V/300 (vh/im inrval And, P (h = m = v/300 Random Hadway Sa-Poisson Disribuion Man Tim Hadway _ = 300/v And, P (h = v/300 Random Hadway Sa-Poisson Disribuion Exampl: raffic volum = 720 vph Man Tim Hadway _ = 300/720 = sconds Thrfor, P (h = (/ _ = (/ 5 = 0.2 Thrfor, P (h = (/ _ P ( h < + = P (h - P (h + To obain h frquncy of hadways: F ( h < + = N [P ( h < +] Whr N is oal numbr of obsrvd hadway P (h P ( h < F ( h <
5 Consan Hadway Sa-Normal Disribuion Man im hadway (h = 300/V Normal disribuion wih man ( and sandard dviaion s 95% confidnc inrval in h rang ( 2s Minimum hadway ( = - 2s or = _ 2s Thn s=(h - /2 Consan Hadway Sa-Normal Disribuion Normal Disribuion: To g h probabiliy P (A X < B Z= (B A From h abl using h valu (Z/S W ar inrsd in h probabiliy: P ( h < _ Consan Hadway Sa-Normal Disribuion Exampl: V= 2000 vph Man im hadway= 300/2000 =.8 sconds Assuming a minimum hadway of 0. sconds s= (.8 0./2 = 0. To obain h probabiliy: P (.0 h < _ Z=.8.0 = 0.8 or Z/s= 0.8 / 0. =.333 P (.0 h <.8 = % of h hadway bwn h man hadway and.0 scond Composi Modl Approach Ohr Approachs 5
6 . Parson Typ III 2. Gamma 3. Erlang 4. Ngaiv Exponnial 5. Shifd Ngaiv Exponnial ( ( Parson Typ III f ( ( K f (: Probabiliy dnsiy funcion K and : Usr slc paramrs ha affc h shap and h shif of h disribuion : Paramr ha is a funcion of h man im hadway and h wo usr spcifid paramrs (K and : Tim hadway bing invsigad (: Gamma funcion quivaln o (K-! ( ( Parson Typ III f ( ( K ( ( Parson Typ III f ( ( K = 0 and K > 0 = 0 and K = Gamma Disribuion f ( ( K Ngaiv Exponnial Disribuion f (
7 ( ( Parson Typ III f ( ( K > 0 and K = For any Modl P ( h f ( d Shifd Ngaiv Exponnial Disribuion f ( ( Composi Modl Approach P P P NP P NP P P = Proporion of vhicls in Plaoon (Shifd Ngaiv Exponnial Disribuion P NP = Proporion of vhicls NOT in Plaoon Normal Disribuion ( ( Parson Typ III f ( ( K > 0 and K = = 0 and K = f ( f ( f ( = 0 and K > 0 Shifd Ngaiv Exponnial Disribuion Ngaiv Exponnial Disribuion ( K ( Gamma Disribuion ( k! K f ( Erlang Disribuion = 0 and K =, 2, 3 For any Modl P ( h f ( d 7
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