Task Order 5.2 Benefits Analysis for the Georgia Department of Transportation NaviGAtor Program

APPENDIX A: FORMULA FOR INCIDENT DELAY COMPUTATIONS

A.1Total Delay and Time Delay in Queue for Incidents without Partial Clearance (Single Full Clearance)

Figure A. 1: Delay for Incident without Any Partial Clearance*

From basic trigonometry:

AC = l . t Q

AB = m . ( t Q - t R)

BC = m R . t R

Therefore

l.t Q = m R . t R + m . ( t Q - t R )

or   l.t Q = m R . t R + m . t Q-m . t R

or   l.t Q-m . t Q = m R . t R -m . t R

or   t Q (l. -m) = . t R (m R-m)

or   t Q ( m-l ) = t R ( m - m R )

or   t Q = t R . ( m-m R ) / (m-l)

Total delay is equal to the area of the shaded portion in the figure. Therefore:

TD    = ∆AYC - ∆AXB - ∆XYZ -∆BXZC

   = (1/2) . l . t Q . t Q- (1/2) . m . (t Q - t R ) . (t Q - t R )

- (1/2) . m R . t R . t R -m R . t R . (t Q - t R )

   = (1/2) [ l . t Q 2-m . t Q 2-m . t R 2 +2m . t Q . t R-m R . t R 2 + 2m R . t R 2- 2m R . t Q . t R]

   = (1/2) [(l-m). t Q 2 - (m-m R). t R 2 +2(m-m R) . t Q . t R]

Substituting

t Q ( m-l ) with t R ( m - m R ) in the RHS,

TD    = (1/2) [- (m-m R). t R. t Q - (m-m R). t R 2 +2(m-m R) . t Q . t R]

   = (1/2) (m-m R). t R. (t Q- t R)

Substituting

t Q with t R . ( m-m R ) / (m-l) in the RHS,

TD    = (1/2) (m-m R). t R. t R (( m-m R ) / (m-l) -1)

   = (1/2) (m-m R). t R. t R (l-m R) / (m-l)

   = (1/2) t R 2. (m-m R). (l-m R) / (m-l)

A.2Total Delay and Time Delay in Queue for Incidents with Intermediate Partial Clearance Leading to Reduction in the Number of Blocked Lanes

Figure A.2: Delay for Incident with Intermediate Partial Clearance*

From basic trigonometry:

AD = l . t Q

AB = m . ( t Q - t R1- t R2 )

BC = m R2 . t R2

CD = m R1 . t R1

Therefore

l.t Q = m R1 . t R1 + m R2 . t R2 + m . ( t Q - t R1- t R2 )

or   t Q ( m-l ) = m . ( t R1 + t R2 ) -m R1 . t R1 -m R2 . t R12

or   t Q = ( 1 / (m-l) ) . t R1 . ( m-m R1 ) + (1 / ( m-l ) ) . t R2 . ( m-m R2 )

The two terms on the Right Hand Side (RHS) can be separated as:

t Q1 = ( 1 / ( m-l ) ) . t R1 . ( m-m R1 )

and

t Q2 = ( 1 / ( m-l ) ) . t R2 . ( m-m R2 )

Therefore:

t Q = t Q2 + t Q2

Total delay is equal to the area of the shaded portion in the figure. Therefore:

TD =

(1/2) . l . t R1 . t R1- (1/2) . m R1 . t R1 . t R1

+ (1/2) . t R2 . ( l . t R1 + l . ( t R1 + t R2 ) ) - ( m R1 . t R1 . t R2 + (1/2) . t R2 . m R2 . t R2 )

+ (1/2) . ( t Q- ( t R1 + t R2 ) ) . ( l . ( t R1 + t R2 ) - ( m R1 . t R1 + m R2 . t R12 ) )

Substituting

-( m R1.t R1 + m R2 . t R12 ) with ( t Q (m-l) -m . ( t R1 + t R2 ) ) in the RHS,

and using some simplification:

TD =

(1/2) . (l-m R1 ). ( t R1 + t R1 ) 2

+ (1/2) . (m R1-m R2 ). t R2 2

+ (1/2) . (m-l ) . ( t Q- ( t R1 + t R2 ) ) 2

It can be easily verified that setting t R2 to 0 gives the equation for total delay for the incident that does not have an intermediate partial clearance.

* See Figure 3.2 in Chapter 3 of the report for detailed annotations for the figure