Vehicle Arrival Models : Headway

Chaper 12 Vehicle Arrival Models : Headway 12.1 Inroducion Modelling arrival of vehicle a secion of road is an imporan sep in raffic flow modelling. I has imporan applicaion in raffic flow simulaion where vehicles are o be generaed how vehicles arrive a a secion. The vehicle arrival is obviously a random process. This is eviden if one observe how vehicles are arriving a a cross secion. Some imes several vehicles come ogeher, while a oher imes, hey come sparsely. Hence, vehicle arrival a a secion need o be characerized saisically. Vehicle arrivals can be modelled in wo iner-relaed ways; namely modelling wha is he ime inerval beween he successive arrival of vehicles or modelling how many vehicle arrive in a given inerval of ime. In he former approach, he random variables he ime denoing inerval beween successive arrival of vehicle can be any posiive real values and hence some suiable coninuous disribuion can be used o model he vehicle arrival. In he laer approach, he random variables represen he number of vehicles arrived in a given inerval of ime and hence akes some ineger values. Here in his approach, a discree disribuion can be used o model he process. This chaper presens how some coninuous disribuions can be used o model he vehicle arrival process. 12.2 Headway modelling An imporan parameer o characerize he raffic is o model he iner-arrival ime of vehicle a a secion on he road. The iner-arrival ime or he ime headway is no consan due o he sochasic naure of vehicle arrival. A common way of modeling o rea he iner-arrival ime or he ime headway as a random variable and use some mahemaical disribuions o model hem. The behavior of vehicle arrival is differen a differen flow condiion. I may be possible ha differen disribuions may work beer a differen flow condiions. Suppose he vehicle arrive a a poin a ime 1, 2,.... Then he ime difference beween wo consecuive Dr. Tom V. Mahew, IIT Bombay 12.1 February 19, 2014

Disance Occupancy Time 00 11 01 00 11 01 00 11 01 Time gap 00 11 01 00 11 Observaion 01 poin 1 110011 0011 0011 2 3 4 00 11 Time Figure 12:1: Illusraion of headways arrivals is defined as headway. This is shown as a ime-disance diagram in figure 12:1. In fac he headway consis of wo componens, he occupancy ime which is he duraion required for he vehicle o pass he observaion poin and he ime gap beween he rear of he lead vehicle and fron of he following vehicle. Hence, he headways h 1 = 2 1, h 2 = 3 2,... I may be noed ha he headways h 1, h 2,... will no be consan, bu follows some random disribuion. Furher, under various raffic saes, differen disribuion may bes explain he arrival paern. A brief discussion of he various raffic saes and suiable disribuions are discussed nex. 12.2.1 Classificaion of raffic sae Generally, raffic sae can be divided ino hree; namely low, medium and high flow condiions. Salien feaures of each of he flow sae is presened below afer a brief discussion of he probabiliy disribuion. 1. Low volume flow (a) Headway follow a random process as here is no ineracion beween he arrival of wo vehicles. (b) The arrival of one vehicle is independen of he arrival of oher vehicle. (c) The minimum headway is governed by he safey crieria. Dr. Tom V. Mahew, IIT Bombay 12.2 February 19, 2014

(d) A negaive exponenial disribuion can be used o model such flow. 2. High volume flow (a) The flow is very high and is near o he capaciy. (b) There is very high ineracion beween he vehicle. (c) This is characerized by near consan headway. (d) The mean and variance of he headway is very low. (e) A normal disribuion can used o model such flow. 3. Inermediae flow (a) Some vehicle ravel independenly and some vehicle has ineracion wih oher vehicles. (b) More difficul o analyze, however, has more applicaion in he field. (c) Pearson Type III Disribuion can be used which is a very general case of negaive exponenial disribuion. 12.3 Negaive exponenial disribuion The low flow raffic can be modeled using he negaive exponenial disribuion. Firs, some basics of negaive exponenial disribuion is presened. The probabiliy densiy funcion of any disribuion has he following wo imporan properies: Firs, p[ < < + ] = + d = 1 (12.1) where is he random variable. This means ha he oal probabiliy defined by he probabiliy densiy funcion is one. Second: p[a b] = b a d (12.2) This gives an expression for he probabiliy ha he random variable akes a value wih in an inerval, which is essenially he area under he probabiliy densiy funcion curve. The probabiliy densiy funcion of negaive exponenial disribuion is given as: = λe λ, 0 (12.3) Dr. Tom V. Mahew, IIT Bombay 12.3 February 19, 2014

λ = 1.5 λ = 1 λ = 0.5 Figure 12:2: Shape of he Negaive exponenial disribuion for various values of λ where λ is a parameer ha deermines he shape of he disribuion ofen called as he shape parameer. The shape of he negaive exponenial disribuion for various values of λ (0.5, 1, 1.5) is shown in figure 12:2. The probabiliy ha he random variable is greaer han or equal o zero can be derived as follow, p( 0) = = λ = λ 0 0 λ e λ d (12.4) e λ λ e λ d = e λ 0 = e λ + e λ0 0 = 0 + 1 = 1 The probabiliy ha he random variable is greaer han a specific value h is given as p( h) = 1 p( < h) (12.5) = 1 h 0 [ e λ = 1 λ λ λ.e λ d ] h = 1 + e λ h 0 = 1 + [ e λh e λ0] = 1 + e λh 1 = e λh Dr. Tom V. Mahew, IIT Bombay 12.4 February 19, 2014 0

p( h) h h + δ h p( h + δh) p(h h + δh) h h + δ h Figure 12:3: Evaluaion of negaive exponenial disribuion for an inerval Unlike many oher disribuions, one of he key advanages of he negaive exponenial disribuion is he exisence of a closed form soluion o he probabiliy densiy funcion as seen above. The probabiliy ha he random variable lies beween an inerval is given as: p[h (h + δh)] = p[ h] p[ (h + δh)] (12.6) = e λ h e λ (h+δh) This is illusraed in figure 12:3. The negaive exponenial disribuion is closely relaed o he Poisson disribuion which is a discree disribuion. The probabiliy densiy funcion of Poisson disribuion is given as: p(x) = λx e λ (12.7) x! where, p(x) is he probabiliy of x evens (vehicle arrivals) in some ime inerval (), and λ is he expeced (mean) arrival rae in ha inerval. If he mean flow rae is q vehicles per hour, hen λ = q vehicles per second. Now, he probabiliy ha zero vehicle arrive in an inerval 3600, denoed as p(0), will be same as he probabiliy ha he headway (iner arrival ime) greaer han or equal o. Therefore, p(x = 0) = λ0 e λ 0! = e λ = p(h ) = e λ Dr. Tom V. Mahew, IIT Bombay 12.5 February 19, 2014

Here, λ is defined as average number of vehicles arriving in ime. If he flow rae is q vehicles per hour, hen, λ = q 3600 = µ (12.8) Since mean flow rae is inverse of mean headway, an alernae way of represening he probabiliy densiy funcion of negaive exponenial disribuion is given as = 1 µ e µ (12.9) where µ = 1 or λ = 1. Here, µ is he mean headway in seconds which is again he inverse λ µ of flow rae. Using equaion 12.6 and equaion 12.5 he probabiliy ha headway beween any inerval and flow rae can be compued. The nex example illusraes how a negaive exponenial disribuion can be fied o an observed headway frequency disribuion. Numerical Example An observaion from 2434 samples is given able below. Mean headway and he sandard deviaion observed is 3.5 and 2.6 seconds respecively. Fi a negaive exponenial disribuion. Table 12:1: Observed headway disribuion h h + dh p o i 0.0 1.0 0.012 1.0 2.0 0.178 2.0 3.0 0.316 3.0 4.0 0.218 4.0 5.0 0.108 5.0 6.0 0.055 6.0 7.0 0.033 7.0 8.0 0.022 8.0 9.0 0.013 9.0 > 0.045 Toal 1.00 Soluion: The soluion is shown in Table 12:2. The headway range and he observed probabiliy (or proporion) is given in column (2), (3) and (4). The observed frequency for he firs Dr. Tom V. Mahew, IIT Bombay 12.6 February 19, 2014

inerval (0 o 1) can be compued as he produc of observed frequency p i and he number of observaion (N). Tha is, fi o = p i N = 0.012 2434 = 29.21 and is shown in column (5). The probabiliy ha he headway greaer han = 0 is compued as p( 0) = e 0 = 1 (refer equaion 12.5) and is given in column (6). These seps are repeaed for he second inerval, ha is fi o = 0.178 2434 = 433.25, and p( 1) = e 1 = 0.751. Now, he probabiliy of headway lies beween 0 and 1 for he firs inerval is given by he probabiliy ha headway greaer han zero from he firs inerval minus probabiliy ha headway greaer han one from second inerval. Tha is p i (0 1) = p i ( > 0) p i ( > 1) = 1.00 0.751 = 0.249 and is given in column (7). Now he compued frequency fi c is p i N = 0.249 2434 = 604.904 and is given in column (8). This procedure is repeaed for all he subsequen iems. I may be noed ha probabiliy of headway > 9.0 is compued by 1-probabiliy of headway less han 9.0 = 1 (0.249 + 0.187 +...) = 0.076. Table 12:2: Illusraion of fiing a negaive exponenial disribuion No h h + dh p o i fi o p( >= h) p c i fi c (1) (2) (3) (4) (5) (6) (7) (8) 1 0.0 1.0 0.012 29.21 1.000 0.249 604.904 2 1.0 2.0 0.178 433.25 0.751 0.187 454.572 3 2.0 3.0 0.316 769.14 0.565 0.140 341.600 4 3.0 4.0 0.218 530.61 0.424 0.105 256.705 5 4.0 5.0 0.108 262.87 0.319 0.079 192.908 6 5.0 6.0 0.055 133.87 0.240 0.060 144.966 7 6.0 7.0 0.033 80.32 0.180 0.045 108.939 8 7.0 8.0 0.022 53.55 0.135 0.034 81.865 9 8.0 9.0 0.013 31.64 0.102 0.025 61.520 10 9.0 > 0.045 109.53 0.076 0.076 186.022 Toal 2434 1.000 2434 12.4 Normal disribuion The probabiliy densiy funcion of he normal disribuion is given by: = 1 σ 2π e ( µ) 2 2σ 2 ; < <, < µ <, σ > 0 (12.10) Dr. Tom V. Mahew, IIT Bombay 12.7 February 19, 2014

α 0 α Figure 12:4: Shape of normal disribuion curve where µ is he mean of he headway and σ is he sandard deviaion of he headways. The shape of he probabiliy densiy funcion is shown in figure 12:4. The probabiliy ha he ime headway () less han a given ime headway (h) is given by p( h) = h d (12.11) and he value of his is shown as he area under he curve in figure 12:5 (a) and he probabiliy of ime headway () less han a given ime headway (h + δh) is given by p( h + δh) = h+δh d (12.12) This is shown as he area under he curve in figure 12:5 (b). Hence, he probabiliy ha he ime headway lies in an inerval, say h and h + δh is given by p(h h + δh) = p( h + δh) p( h) (12.13) = h+δh d h d This is illusraed as he area under he curve in figure 12:5 (c). Alhough he probabiliy for headway for an inerval can be compued easily using equaion 12.13, here is no closed form soluion o he equaion 12.11. Evenhough i is possible o solve he above equaion by numerical inegraion, he compuaions are ime consuming for regular applicaions. One way o overcome his difficuly is o use he sandard normal disribuion able which gives he soluion o he equaion 12.11 for a sandard normal disribuion. A sandard normal disribuion is normal disribuion of a random variable whose mean is zero and sandard deviaion is one. The probabiliy for any random variable, having a mean (µ) and sandard deviaion (σ) can be compued by normalizing ha random variable wih respec o is mean Dr. Tom V. Mahew, IIT Bombay 12.8 February 19, 2014

p( h) p( h + δh) h h + δ h p(h h + δh) h h + δ h Figure 12:5: Illusraion of he expression for probabiliy ha he random variable lies in an inerval for normal disribuion and sandard deviaion and hen use he sandard normal disribuion able. This is based on he concep of normalizing any normal disribuion based on he assumpion ha if follows normal disribuion wih mean µ and sandard deviaion σ, hen ( µ)/σ follows a sandard normal disribuion having zero mean and uni sandard deviaion. The normalizaion seps shown below. [ h µ p[h (h + δh)] = p σ = p [ µ ] (h + δh) µ σ σ ] [ (h + δh) µ p h µ σ σ ] (12.14) The firs and second erm in his equaion be obained from sandard normal disribuion able. The following example illusraes his procedure. Numerical Example If he mean and sandard deviaion of cerain observed se of headways is 2.25 and 0.875 respecively, hen compue he probabiliy ha he headway lies in an inerval of 1.5 o 2.0 seconds. Dr. Tom V. Mahew, IIT Bombay 12.9 February 19, 2014

3σ 2σ σ µ σ 2σ Figure 12:6: Normal Disribuion Soluion: The probabiliy ha headway lies beween 1.5 and 2.0 can be obained using equaion 12.14, given ha µ = 2.25 and σ = 0.85 as: p[1.5 2.0] = p[ 2.0] p[ 1.5] [ ] [ ] 2.0 2.25 1.5 2.25 = p p 0.875 0.875 = p [ 0.29] p [ 0.86] = 0.3859 0.1949 (from ables) = 0.191. Noe ha he p( 0.29) and p( 0.80) are obained from he sandard normal disribuion ables. Since he normal disribuion is defined from α o +α unlike an exponenial disribuion which is defined only for posiive number, i is possible ha normal disribuion may generae negaive headways. A pracical way of avoiding his is o shif he disribuion by some value so ha i will mosly generae realisic headways. The concep is illusraed in figure 12:6. Suppose α is he minimum possible headway and if we se α = µ σ han abou 60% of headway will be greaer han α. Alernaively, if we se α = µ 2σ, han abou 90% of he headway will be greaer han α. Furher, if we se α = µ 3σ, han abou 99% of he headway will be greaer han α. To generalize, α = µ nσ where n is 1, 2, 3, ec and higher he value of n, hen is beer he precision. From his equaion, we can compue he value of σ o be used in normal disribuion calculaion when he random variable canno be negaive as: σ = µ α n (12.15) Dr. Tom V. Mahew, IIT Bombay 12.10 February 19, 2014

Numerical Example Given ha observed mean headway is 3.5 seconds and sandard disribuion is 2.6 seconds, hen compue he probabiliy ha he headway lies beween 0 and 0.5. Assume ha he minimum expeced headway is 0.5 seconds. Soluion: Firs, compue he sandard deviaion o be used in calculaion using equaion 12.15, given ha µ = 3.5, σ = 2.6, and α = 0.5. Then: σ = µ α 3.5 0.5 = = 1.5 (12.16) 2 2 Second, compue he probabiliy ha headway less han zero. ( p( < 0) p 0 3.5 ) 1.5 = p( 2.33) = 0.01 The value 0.01 is obained from sandard normal disribuion able. Similarly, compue he probabiliy ha headway less han 0.5 as ( ) 0.5 3.5 p( 0.5) p 1.5 = p( < 2) = 0.023 The value 0.23 is obained from he sandard normal disribuion able. Hence, he probabiliy ha headway lies beween 0 and 0.5 is obained using equaion 12.14 as p(0 0.5)=0.023 0.010 = 0.023. Numerical Example An observaion from 2434 samples is given able below. Mean headway observed was 3.5 seconds and he sandard deviaion observed was 2.6 seconds. Fi a normal disribuion, if we assume minimum expeced headway is 0.5. Soluions The given headway range and he observed probabiliy is given in column (2), (3) and (4). The observed frequency for he firs inerval (0 o 1) can be compued as he produc of observed frequency p i and he number of observaion (N) i.e. p o i = p i N = 0.012 2434 = 29.21 as shown in column (5). Compue he sandard deviaion o be used in calculaion, given ha µ = 3.5, σ = 2.6, and α = 0.5 as: σ = µ α 2 = 3.5 0.5 2 = 1.5 Dr. Tom V. Mahew, IIT Bombay 12.11 February 19, 2014

Table 12:3: Observed headway disribuion h h + dh p o i 0.0 1.0 0.012 1.0 2.0 0.178 2.0 3.0 0.316 3.0 4.0 0.218 4.0 5.0 0.108 5.0 6.0 0.055 6.0 7.0 0.033 7.0 8.0 0.022 8.0 9.0 0.013 9.0 > 0.045 Toal 1.00 Second, compue he probabiliy ha headway less han zero. ( p( < 0) p 0 3.5 ) 1.5 = p( 2.33) = 0.010 The value 0.01 is obained for sandard normal disribuion able is shown in column (6). Similarly, compue he probabiliy ha headway less han 1.0 as: ( ) 1.0 3.5 p( 1) p 1.5 = p( < 2) = 0.048 The value 0.048 is obained from he sandard normal disribuion able is shown in column (6). Hence, he probabiliy ha headway beween 0 and 1 is obained using equaion 12.14 as p(0 1)=0.048 0.010 = 0.038 and is shown in column (7). Now he compued frequency F c i is p( < h < +1) N = 0.038 2434 = 92.431 and is given in column (8). This procedure is repeaed for all he subsequen iems. I may be noed ha probabiliy of headway > 9.0 is compued by one minus probabiliy of headway less han 9.0 = 1 (0.038+0.111+...) = 0.010. Dr. Tom V. Mahew, IIT Bombay 12.12 February 19, 2014

Table 12:4: Soluion using normal disribuion No h h + δ h p o i f o i = p o i N p( h) p( < h < + δ h) fc i = p c i N (1) (2) (3) (4) (5) (6) (7) (8) 1 0.0 1.0 0.012 29.21 0.010 0.038 92.431 2 1.0 2.0 0.178 433.25 0.048 0.111 269.845 3 2.0 3.0 0.316 769.14 0.159 0.211 513.053 4 3.0 4.0 0.218 530.61 0.369 0.261 635.560 5 4.0 5.0 0.108 262.87 0.631 0.211 513.053 6 5.0 6.0 0.055 133.87 0.841 0.111 269.845 7 6.0 7.0 0.033 80.32 0.952 0.038 92.431 8 7.0 8.0 0.022 53.55 0.990 0.008 20.605 9 8.0 9.0 0.013 31.64 0.999 0.001 2.987 10 9.0 > 0.045 109.53 1.000 0.010 24.190 Toal 2434 12.5 Pearson Type III disribuion As noed earlier, he inermediae flow is more complex since cerain vehicles will have ineracion wih he oher vehicles and cerain may no. Here, Pearson Type III disribuion can be used for modelling inermediae flow. The probabiliy densiy funcion for he Pearson Type III disribuion is given as = λ [λ( Γ(K) α)]k 1 e λ( α), K, α R (12.17) where λ is a parameer which is a funcion of µ, K and α, and deermine he shape of he disribuion. The erm µ is he mean of he observed headways, K is a user specified parameer greaer han 0 and is called as a shif parameer. The Γ() is he gamma funcion and given as Γ(K) = (K 1)! (12.18) I may also be noed ha Pearson Type III is a general case of Gamma, Erlang and Negaive Exponenial disribuion as shown in below: λ = Γ(K) α)]k 1 e λ( α) K, α R Pearson λ = Γ(K) [λ]k 1 e λ if α = 0 Gamma λ = (K 1)! [λ]k 1 e λ if K I Erlang = λe λ if K = 1 Neg. Exp. Dr. Tom V. Mahew, IIT Bombay 12.13 February 19, 2014

The expression for he probabiliy ha he random headway () is greaer han a given headway (h), p( h), is given as: p( h) = h d (12.19) and similarly p( > h + δh) is given as: p( > h + δh) = (h+δh) d (12.20) and hence, he probabiliy ha he headway beween h and h + δh is given as p(h (h + δh)) = d h (h+δh) d (12.21) I may be noed ha closed form soluion o equaion 12.19 and equaion 12.20 is no available. Numerical inegraion is also difficul due o compuaional requiremen. Using able as in he case of Normal Disribuion is difficul, since he able will be differen for each K. A common way of solving his is by using he numerical approximaion o equaion 12.21. The soluion o equaion 12.21 is essenially he area under he curve defined by he probabiliy densiy funcion beween h and h + δh. If we assume ha line joining f(h) and f(h + δh) is linear, which is a reasonable assumpion if δh is small, han he are under he curve can be found ou by he following approximae expression: [ ] f(h) + f(h + δh) p(h (h + δh)) δh (12.22) 2 This concep is illusraed in figure 12:7 Sepwise procedure o fi a Pearson Type III disribuion 1. Inpu required: he mean (µ) and he sandard deviaion (σ) of he headways. 2. Se he minimum expeced headway (α). Say, for example, 0.5. I means ha he p( < 0.5) 0. 3. Compue he shape facor using he mean (µ) he sandard deviaion (σ) and he minimum expeced headway (α) K = µ α σ Dr. Tom V. Mahew, IIT Bombay 12.14 February 19, 2014

p( h) f(h) p( h + δh) h f(h + δh) h + δ h p(h h + δh) [ ] f(h)+f(h+δh) 2 δh h h + δ h Figure 12:7: Illusraion of he expression for probabiliy ha he random variable lies in an inerval for Person Type III disribuion 4. Compue he erm flow rae (λ) as λ = K µ α Noe ha if K = 1 and α = 0, hen λ = 1 µ which is he flow rae. 5. Compue gamma funcion value for K as: Γ(K) = (K 1)! if K I (Ineger) = (K 1) Γ(K 1) if K R (Real) (12.23) Alhough he closed form soluion of Γ(K) is available, i is difficul o compue. Hence, i can be obained from gamma able. For, example: Γ(4.785) = 3.785 Γ(3.785) = 3.785 2.785 Γ(2.785) = 3.785 2.785 1.785 Γ(1.785) = 3.785 2.785 1.785 0.92750 = 17.45 Noe ha he value of Γ(1.785) is obained from gamma able for Γ(x) which is given for 1 x 2. 6. Using equaion 12.17 solve for f(h) by seing = h where h is he lower value of he range and f(h+δh) by seing = h+δh where (h+δh) is he upper value of he headway Dr. Tom V. Mahew, IIT Bombay 12.15 February 19, 2014

range. Compue he probabiliy ha headway lies beween he inerval of h and h + δh using equaion 12.22. The Gamma funcion can be evaluaed by he following approximae expression also: Gamma(x) = x x e x 2 π x ( 1 + 1 12 x + 1 ) 288x +... 2 (12.24) Numerical Example An observaion from 2434 samples is given able below. Mean headway observed was 3.5 seconds and he sandard deviaion 2.6 seconds. Fi a Person Type III Disribuion. Table 12:5: Observed headway disribuion h h + dh p o i 0.0 1.0 0.012 1.0 2.0 0.178 2.0 3.0 0.316 3.0 4.0 0.218 4.0 5.0 0.108 5.0 6.0 0.055 6.0 7.0 0.033 7.0 8.0 0.022 8.0 9.0 0.013 9.0 > 0.045 Toal 1.00 Soluions Given ha mean headway (µ) is 3.5 and he sandard deviaion (σ) is 2.6. Assuming he expeced minimum headway (α) is 0.5, K can be compued as K = µ α σ = 3.5 0.5 2.6 = 1.15 and flow rae erm λ as λ = K µ α = 1.15 3.5 0.5 = 0.3896 Dr. Tom V. Mahew, IIT Bombay 12.16 February 19, 2014

Now, since K = 1.15 which is beween 1 and 2, Γ(K) can be obained direcly from he gamma able as Γ(K) = 0.93304. Here, he probabiliy densiy funcion for his example can be expressed as = 0.3846 0.93304 [ 0.3846 ( 0.5) ]1.15 1 0.3846 ( 0.5) e The given headway range and he observed probabiliy is given in column (2), (3) and (4). The observed frequency (f o i ) for he firs inerval (0 o 1) can be compued as he produc of observed proporion p o i and he number of observaions (N). Tha is, fo i = p o i N = 0.012 2434 = 29.21 as shown in column (5). The probabiliy densiy funcion value for he lower limi of he firs inerval (h=0) is shown in column (6) and compued as: f(0) = 0.3846 0.93304 [ 0.3846 (0 0.5)]1.15 1 e 0.3846 (0 0.5) 0. Noe ha since α (0 0.5) is negaive and K 1 (1.15 1) is a fracion, he above expression canno be evaluaed and hence approximaed o zero (corresponding o =0.5). Similarly, he probabiliy densiy funcion value for he lower limi of he second inerval (h=1) is shown in column 6 and compued as: f(1) = 0.3846 0.93304 [0.3846(1 0.5)]1.15 1 e 0.3846(1 0.5) = 0.264 Now, for he firs inerval, he probabiliy ( ) for headway beween 0 and 1 is compued by equaion?? as p c i (0 1) = f(0)+(f(1) (1 0) = (0 + 0.0264)/2 1 = 0.132 and is 2 given in column (7). Now he compued frequency fi c is p c i N = 0.132 2434 = 321.1 and is given in column (8). This procedure is repeaed for all he subsequen iems. I may be noed ha probabiliy of headway > 9 is compued by 1-probabiliy of headway less han 9 = 1 (0.132 + 0.238 +... ) = 0.044. The comparison of he hree disribuion for he above daa is ploed in Figure 12:8. 0.3 0.2 0.1 Observed Exponenial Normal Pearson Type III 1 2 3 4 5 6 7 8 9 Figure 12:8: Comparison of disribuions 12.6 Conclusion This chaper covers how he vehicle arrival can be modelled using various disribuions. The negaive exponenial disribuion is used when he raffic is low and is mos simples of he Dr. Tom V. Mahew, IIT Bombay 12.17 February 19, 2014

Table 12:6: Soluion using Pearson Type III disribuion No h h + δ h p o i fi o p c i fi c (1) (2) (3) (4) (5) (6) (7) (8) 1 0 1 0.012 29.2 0.000 0.132 321.2 2 1 2 0.178 433.3 0.264 0.238 580.1 3 2 3 0.316 769.1 0.213 0.185 449.5 4 3 4 0.218 530.6 0.157 0.134 327.3 5 4 5 0.108 262.9 0.112 0.096 233.4 6 5 6 0.055 133.9 0.079 0.068 164.6 7 6 7 0.033 80.3 0.056 0.047 115.3 8 7 8 0.022 53.6 0.039 0.033 80.4 9 8 9 0.013 31.6 0.027 0.023 55.9 10 >9 0.045 109.5 0.019 0.044 106.4 Toal 1.0 2434 1.0 2434 disribuions in erms of compuaion effor. The normal disribuion on he oher hand is used for highly congesed raffic and is evaluaion require sandard normal disribuion ables. The Pearson Type III disribuion is a mos general kind of disribuion and can be used inermediae or normal raffic condiions. 12.7 References 1. L. R Kadiyali. Traffic Engineering and Transporaion Planning. Khanna Publishers, New Delhi, 1987. 2. Adolf D. May. Fundamenals of Traffic Flow. Prenice - Hall, Inc. Englewood Cliff New Jersey 07632, second ediion, 1990. Dr. Tom V. Mahew, IIT Bombay 12.18 February 19, 2014