3 Suppose there is an accident on eastbound of I90 The traff

3. Suppose there is an accident on eastbound of I-90. The traffic operation center has to decide how to response to the accident. They are considering two alternative methods: (a) Completely shut the freeway off for 15 minutes, and clear accident and reopen the (b) Partially open the freeway at reduced capacity (shown in the following table), but In order to help traffic operators make decision to take, you are asked to analyze which freeway. the clear process requires 40 minutes, before reopening the freeway. alternative will generate shorter queue extension (the farthest point queue reaches)? See the following table for more details: Condition Description 2 4 Arrival flow Jammed flow Capacity flow Reduced capacity flow2100 q(veh/hr) k(veh/mi) v(mph) 3000 0 4200 50 240 100 175 60 42 12 k*

Solution

initial arrival flow= q_a =3000 veh/hr

final reduced flow= q_r =2100 veh/hr

jammed arrival flow= q_j =0 veh/hr

final reduced cocentration= k_r =175 veh/mi

final reduced velocity= v_r =12mi/hr

initial arrival concetration = k_a =50 veh/mi

initial arrival velocity= v_a =60 mi/hr

jammed arrival concentration= k_j =240veh/mi

jammed arrival velocity= v_j =0 mi/hr

velocity of shockwave = (q_a-q_j )/(k_a-k_b) = (3000-0)/(50-240) = -15.7895 mi/hr (negative toward upstream)

length of queue = velocity of shockwave * time = -15.7895 * 15//60 = -3.95 mi(negative toward upstream)

2.

When the highway is partially close for t₂=40 min,

velocity of shockwave = (q_r-q_a )/(k_r-k_a) = (2100-3000)/(175-50) = -7.2 mi/hr (negative toward upstream)

length of queue= velocity of shockwave*time= -7.2*40/60 = -4.8 mi(negative toward upstream)

From these calculation we conclude that completly shut the highway for 15 minutes resulting in a shorter queue.