Question

capacity of a 1-km section of northbound Highway 3 located on a 3.5% upgrade. The highway has three 3.5m-wide lanes in one direction with no lateral obstruction within 5m of the pavement edges. The ideal (base) free-flow speed is 100 km/hour. You observed that percentages of cars and truck are 90% and 10%, respectively (no RVs). A weekday peak hour volume is 4320 vehicles, with 1200 vehicles arriving in the most congested 15-minute period. Since a majority of drivers are local commuters and truck drivers, you assume that they are generally familiar with the highway. The interchange density is 0.75 interchange per km.
(a) Determine the level of service (LOS) for the highway. If the required LOS is “C”, is the current highway capacity enough to handle the traffic?
(b) If the number of lanes in one direction were increased to 4 lanes, would the LOS in part (a) be improved?

Answer 1

Peak hour flow = V = 4320 veh/hour

Peak 15mt flow = 1200 veh

So PHF = 4320/(4*1200) = 0.9

So peak hour flow V = 4320 veh/hr and PHF = 0.9

P_T = 10% =0.1, P_R = 0

Its 1km section on 3.5% upgrade so E_T = 2.5 and E_R = 2.0

Calculate Heavy Vehicle Factor

So we can calculate Heavy vehicle factor = f_HV and it is calculated using the formula

So f_HV = 1/(1+(P_T* 1.5) + (P_R *1) ) = 1/(1+ 0.1*1.5 + 0*1)=1/1.14 = 0.8696

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v_p is calculated by the formula

where

vp = 15 mt passenger car equivalent flow rate in pc/h

PHF = 0.9, N= number of lanes in each direction =3

f_HV = 0.8696

fp = driver population factor. Assume equal to 1 because all are familiar drivers

So v_p = 4320/ (0.9 * 3 * .8696 *1) = 4320/2.34792 = 1839.926 pc/hr/ln

Now determine the Freeflow speed FFS

Compute FFS using the roadway characteristics

FFS = BFFS - f_LW - f_LC -f_N -f_ID

where

DO all calcs in mph and then convert back to kmph at the last

we have 3 lanes in one direction and each lane is 3.5m (i.e 11.48 feet wide)

right shoulder lateral clearance = 5m = 16.4 feet

interchange density = 0.75 per km, and so it is 1.207 access points/mile

BFFS = base free flow speed = this is given as 100kph i.e 62.1372 mph

f_LW = adjustment for lane width = 1.04 mph (see HCM tables for 11.48ft )

f_LC = adjustment for right-shoulder lateral clearance = 0 because it says 16.4 feet (so zero)

f_N = adjusment for number of lanes = 3.0 (see HCM table corresponding to 3 lanes)

f_ID = adjustment for interchange density = 0 (see Table 4 assume we have 1.207 interchange per mile)

Therfore FFS = 62.1372 – 1.04 – 0 – 3 – 0 = 58.0972 mph = 93.498 kmph

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Density = v_p/FFS

= 1839.926/93.498 = 19.678 pc/km/ln

So based on LOS guideline tables, LOS = D

So IF REQUIRED LOS IS C, then current capacity CANNOT HANDLE THE TRAFFIC

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b) Suppose we have 4 lanes in each direction,

then your v_p value changes

So v_p = 4320/ (0.9 * 4 * .8696 *1) = 4320/3.13056 = 1379.945 pc/hr/ln

then the FFS will change

FFS = BFFS - f_LW - f_LC -f_N -f_ID

all other factors will remain the same, but f_N will become 1.5 instead of three

So FFS = 62.1372 – 1.04 – 0 – 1.5 – 0 = 59.5972 mph = 95.912 kmph

Density = v_p/FFS

= 1379.945/95.912 = 14.3876 pc/km/ln

So based on LOS guideline tables, LOS = C

So yes by increasing he # of lanes to 4, we are definitely able to improve the LOS from D to C.

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