1 Ans

(a). Using Equation, N=[(z *a)/d?

95% confidence interval &d=+lseconds
g= 11.6 km/h, z 1.96 for

N=[1.96(1.6)/1|

N= 517 >517 observations

(b). For 99.7%, z - 3:

N=(3(11.6) / 1|

N= 1211.04 => 1211 observations.

2 Ans

Given the accident rate is 5200 vehicles per year, the accident rate per day is:

2=5200/365=14.25 per day

The Poisson probability mass function is given by

P(X=x) - (0ey/x!

Where,

P(X=x) is the probability of X events occurring

2is the average rate of occurrence (14.25 accidents per day)

X is the number of events we are interested in (5 accidents)

For x= 5 and 2. 14.25,

P(X=50) =(14.25°e425y/s! =0.0032 d

lo
The probability of exactly 15 vehicles arriving in one minute is
er
approximately 0.0032 or 0.32
lin

3 Ans
1:
(a) Given a normal
=4
distribution with a mean u=42, variance g 16, standard
20
deviation 25
30
Using the z-score formula to find the
corresponding speed 35
The 40
z-score formula is z= (X-u)/o
At X= 45, 45
50
 is z= (X-u)/o
The z-score formula

At X= 45,

Z= (45-42)y4=0.75
And at X-55,

Z= (55-42)/4-3.25

P(45<X<70) = P(0.75<2<3.25) = (area to the left of z = 3.25) -(area to the left of z = 0.75)
From the table we get,
P(45<X<70) =0.9994 -0.7734 =0.226
The Z-score for an 88% confidence interval is 1.55
N=55,6-4

Using Equation, N =
[(z *o)/e]

3e0.84
Mean +e=42+0.84= 42.84

Mean- e=42-0.84= 41.16

(b) Volume-Volume is
simplythe number of
a specific vehicles that
pointon the road during were actually counted
the hour of passin
Capacity- Capacity is the study.
the
that specific maximum number of vehicles
location during the that the road
how many hour, assuming could handlea
vehicles can pass ideal conditions.
It's the
Demand- Demand through the point. upper limit
represents the total
that point number of
during the study vehicles that
volume). plus all
hour. This
includes the wanted to travel past
the other vehicles that
congestion. These
vehicles that were were counted (the
"prevented" prevented from
Vehicles vehicles include: passing due to
stuck in a
queue waiting
Vehicles thattook to get to the
a different studypoint.
Vehicles that route to avoid
decided not to the
Vehicles that went travel at all congestion.
somewhere else because of the
because of the instead of
traffic.
So, demand is traffic. their
original
the destination
potential traffic
flow if there
were no
congestion.
4 Ans

Mid Frequency
fx f(x
Speed
range(km/ value(x) ()
h)

15-20 17.5

20-25 22.5 4 90 1530.37

25-30 27.5 247.5 1907.94

30-35 32.5 18 585 1645.08

35-40 37.5 35 1312.5 727.78

40-45 42.5 42 1785 8.13

45-50 47.5 32 1520 946.99

50-55 52.5 20 1050 2179.87

55-60 57.5 517.5 2145.54

60-65 62.5

f"X 7107.5, f 169

Mean (u)=Ef*x/f =42.06 km/h

Standard deviation
(o)=22-8.125km/h Ef-1

Where. =chi-squared stat istics

=fcqueney of ohservations

XISts
f-theotetical frequency
in spccd

tn speed group .
group

assuming that the assumed dstrbution

N=number of specd groups in the dstrbutton

Spee Observe Normal

d
Spee Probabilit Theoretic Combined
d probabilit Combined
low y of upper y of speed al groups (n)
upp frequen groups (f) (ni-f)?
er er limit group frequency
cy (ni)
limi limi
1
15 20
20 0.003
25 O.002 0.404
25 0.016 0.013 13
30 9 10.809
30 0.064 2.207
35 18 0.049
0.187 8.198
35 40 0.123
40 35 0.398 20.717 18 0.444
45 42 0.211 20.717
0.645 35.637 0.356
45 50 0.247 35
32 41.738 35.637
50 0.842 24 0.011
55 20 0.197 41.738
0.949 0.107
33.287 32 0.002
18.074 33.287
20 0.050
18.074 6.605
 0.949 0.107 18.074 20 18.074 6.605
50 55 20

60 0.988 0.040 6.680 9

60 8.359 0.049
65 0.998 0.010 1.680

H:Data doesnot follow nomal distribution.

=7.517

Degrees of freedom=7-3 =4 & Confidence level =95%=> a =0.05

Critical value ofy²
(0.05,4)
=9.488 =>< CTitLCal
> Accept Ho
Therefore, the given data is normally distributed.

5 Ans- Peak

Cumulative
Speed frequency
(km/h) Count (5)
14 3 10.00
23.33
18 6 43.33
20 4 56.67
22 4
70.00
24
83.33
26 4
83.33
andle 28 1 100.00
er limi

vel pat Histogram freauencydistribution Peakl

d (the

Vol

Volu

Peak
nation

Peak I

Therefo
 Curnulatiye Distribution Peak

70

60
50

40

20

10

14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
Speed kmuhi

Non- Peak

Cumulative
Speed frequency
(km/h) Count (%)
39 6.67

38 2 13.33

36 3 23.33
42 6 43.33
41 4 56.67

37 2 63.33

44 73.33

43 4 86.67

35 90.00

46 2 96.67

40 1 100.00

Peak)
frequency distribution(Non
-
Histogram

Speedtkmi

Graph Non Peak

Frequency
Cumulative

100
 Cumulative FrequericyGraph
100 Non Peak

-
90

80
(95)
70

60
Auanbay
50

40

3 30
angenung
20

10

34

Speed (emih)

Average speed- Zui / f

(a).
Non peak avg speed 1185/30=39.5 km/h =
Peak avg speed 610/30 20.33 km/h = =
(b). 85th percentile speed from cumulative
distribution
=
Non peak 41.87 km/h
plots

Peak =26 km/h

(C). 15th percentile speed from cumulative distribution plots
Non peak =36.17 km/h
Peak =14.75 km/h

(d). Mode from histograms

Non peak =38 km/h
Peak =
18km/h
(e). Median from cumulative distribution plots
Non peak =38.5 km/h
Peak 19 km/h
7Ans
6Ans Todetermine volume
1382 veh/h Ve(NaO-Prl
Volume during 16:00- 17:00 pm 332+390+410+250 values
Taking average 95.1.
N129.2. N
390+410+250+268-1318
P15PL6
veh/h
Volume during 16:15 - 17:15 pm
TE

+ 1.9- I5)60
17:30 pm 410+250 +268+315 -1243 veh/h V 95I
Volume during 16:30 1.l- L6)60
V(1292 +
--

250+268 +315+375 1208veh/h time

Volume during 16:45- 17:45 pm travel
To determine

18:00 prn 26843 15+375+305 -1263 veh/h Te Te f60(O-Py/V

Volume during 17:00- volume is I382 veh/h T
f60(I.9-15V27
10.58- T 023 m

Peak hour is between 16:00-17:00 pm and the corresponding Te 10.49 min

410 veh/h
we
in this period,
Vis For each vehicle,
Peak 1Smin volume 8Ans-
PHE (1382)(4x410)0.84 in ms:
Therefore, Speeds
kmh
VehiclA 20 Amh-
Vehicle B:25
30 kamh30
Vehicle C kmh22
2
Vehicle D:
Vehicle E:
35km/b35S
F: 28krm/h28
Vehicle
t
occupancy
Now calculate the

time (Vehicle
Opcupancy (Vehicle
B
tieme
Occupae(Vehicle
C
ime rVehic )
tinc (Vehicle E)
C
(Vehicle F)
cupancy time
 7 Ans
Todetermine
volume using the
moving vehicle technique,
usethe
V=(NwtOg-Pr)/(Tw+Te),Vw-(NgtOw- equations:
Taking average values Pw)/(Te+Tw)
= 129.2, Nw = 95.1,for times and vehicle count;
Op =1.9 & Ow =1|
N

Py=1.5, Pw=1.6,TE =
10.58 & Ty 10.15
V

Vw
(95.l
(129.2
+ 1.9-

+
1.5)60 /(10.58 + 10.15)
1.1- L.6)60/(10.58 +
> V 276.41 veh /h
10.15)=> Vw 372.50 veh
To determine time using the moving vehicle
travel
techniquc,use cquation:
TE= TE-[60(0;- PE) /Ve ], Tw- Tw -
eh/h (60(Ow Pw )/Vw] -
TE 10.58 - (60(1.9-1.5)276.41|, Tw-
10.15 - (60(1.1-1.6)/372.50]
T= 10.49 min Tw 10.23 min

8Ans- For each vehicle, we compute

will their occupancy times.

Speeds in m/s:

Vehicle A: 20 km/h -20/3.6-5.56 m/s

Vehicle B: 25 km/h =25/3.6-6.95m/s

Vehicle C: 30 km/h =30/3.6-8.34 m/s
Vehicle D: 22 km/h22/3.6-6.12 m/s
Vehicle E: 35 km/h35/3.6-9.73 m/s
Vehicle F: 28 km/h = 28/3.6-7.78m/s

occupancy time for cach vehicle:

Now calculate the

Occupancy time (Vehicle A) 5 m/5.56 m/s0.90 = scconds

m/s0.72 second
Occupancy tme (Vehicle B) = 5 m/6.95
m/s-0.60 seconds
Occupancy time (Vehicle C) = 5 m/8.34
D) 5 m/6.12 m/s-0.82 seconds
Occupancy time (Vehicle
5 m/9,73 m/s 0.51 seconds
Occupancy time (Vehicle E)
time (Vehicle F) 5 m/7.78 m/s 0.64 seconds
Occupancy

+0.82 +0.5 l +0.644.19 seconds

St 0.90+0.72+0.60

(4.19/60) *100= 6.98

%occupancy
360 veh/hr
Flow-(6*3600)/60
v
Flow, q=kx
=6(2+5) 10.02 m/scc36.07km/h
d/t
V 4.19

9,98 veh/km
K=g/v 360/36.07