Question

6-lane freeway (3 lanes in each direction) in level terrain has a current one-directional volume of 5000 veh/h; it is estimated that in 3 years, volume will have increased to 5600 veh/h. The traffic mix includes 10% trucks. The free-flow speed has been measured at 70 mph and the PHF is 0.95.

a) Determine the current LOS.

b) Determine the LOS in 3 years.

c) After the initial three years, traffic grows at 4% per year, when will capacity be reached?

Answer 1

one directional volume = V = 5000 veh/hour

Its level terrain so ET = 2.5 and ER = 2.0

Percentage Trucks = PT = .10 and Percentage RV's = 0

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

So fHV = 1/(1+(.1* 1.5) + (0*1) ) = 1/1.5 = 0.6667

-----------------------------------------------------------------------------------------------------------------------------------------

So

vp is calculated by the formula

where

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

= 5000/(.95 * 3 * .6667) = 5000/1.9 = 2632 pc/hr/lane

----------------------------------------------------------------------------------------------------------------------------------------------------

measured Free Flow Speed = FFS = 70 mph

So density = 2632/70 = 37.6 pc/mi/lane = 23.363 pc/km/ln

So based on following LOS threshold table LOS = D

------------------------------------------------------------------------------------------------------------------------------------------------------

b) after three years V = 5600 veh/hour

all other factors remain the same

So vp = 5600/(.95 * 3 * .6667) = 5600/1.9 = 2947.36 pc/hr/lane

So density = 2947.36/70 = 42.105 pc/mi/lane = 26.163 pc/km/ln.So LOS = E

so LOS worsens. it goes from D previously to D

--------------------------------------------------------------------------------------------------------------------------------------------------------

For capacity to be reached, LOS should touch E, so density should be 28 pc/km/ln i.e 45.06152 pc/mi/ln

so back calculate vp using this density

so 45.06152 = vp/70

so vp = 70 * 45.06152 = 3154.31 pc/hr/lane

We know

so vp = V/1.9

so V = vp * 1.9 = 5993.18 veh/hour

-------------------------------------------------------------------------------------------------------------------------

Traffic after n years i.e n is number of years after 3 years= Traffic at 3 year mark * ( 1+ rate)n

so 5993.18 = 5600 * (1.04)n

From this we find that n = 1.729

So basically capacity will be reached 1.729 years after the 3 year mark i.e 4.729 years from now i.e 57 months from today.