Greenshields Model

Example 1

Greenshields flow, speed and density — Greenshields Model

A freeway segment follows Greenshields with free-flow speed 56 mph and jam density 184 veh/mi/ln. Find the capacity, and the speed and flow at a density of 52 veh/mi/ln.

Given

$vf = 56 mph$
$kj = 184 veh/mi/\ln$
$k = 52 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 56(184)/4 = 2,576 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 92 veh/mi/\ln$
Speed at k
$v = 56(1 - 52/184) = 40.17 mph$
Flow
$q = vk = 40.17(52) = 2,089 veh/h/\ln$
Utilization
$q/qmax = 81.1%$

Answer:

qmax \approx 2,576 veh/h/\ln ; at k = 52, v \approx 40.2 mph and q \approx 2,089 veh/h/\ln

Why the other options are there

qmax = 10,304 veh/h/ln (factor of 4 omitted)
v = 15.8 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 2

Greenshields flow, speed and density — Greenshields Model (2)

A freeway segment follows Greenshields with free-flow speed 58 mph and jam density 151 veh/mi/ln. Find the capacity, and the speed and flow at a density of 81 veh/mi/ln.

Given

$vf = 58 mph$
$kj = 151 veh/mi/\ln$
$k = 81 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 58(151)/4 = 2,190 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 76 veh/mi/\ln$
Speed at k
$v = 58(1 - 81/151) = 26.89 mph$
Flow
$q = vk = 26.89(81) = 2,178 veh/h/\ln$
Utilization
$q/qmax = 99.5%$

Answer:

qmax \approx 2,190 veh/h/\ln ; at k = 81, v \approx 26.9 mph and q \approx 2,178 veh/h/\ln

Why the other options are there

qmax = 8,758 veh/h/ln (factor of 4 omitted)
v = 31.1 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 3

Greenshields flow, speed and density — Greenshields Model (3)

A freeway segment follows Greenshields with free-flow speed 53 mph and jam density 174 veh/mi/ln. Find the capacity, and the speed and flow at a density of 76 veh/mi/ln.

Given

$vf = 53 mph$
$kj = 174 veh/mi/\ln$
$k = 76 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 53(174)/4 = 2,306 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 87 veh/mi/\ln$
Speed at k
$v = 53(1 - 76/174) = 29.85 mph$
Flow
$q = vk = 29.85(76) = 2,269 veh/h/\ln$
Utilization
$q/qmax = 98.4%$

Answer:

qmax \approx 2,306 veh/h/\ln ; at k = 76, v \approx 29.9 mph and q \approx 2,269 veh/h/\ln

Why the other options are there

qmax = 9,222 veh/h/ln (factor of 4 omitted)
v = 23.1 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 4

Greenshields flow, speed and density — Greenshields Model (4)

A freeway segment follows Greenshields with free-flow speed 60 mph and jam density 183 veh/mi/ln. Find the capacity, and the speed and flow at a density of 47 veh/mi/ln.

Given

$vf = 60 mph$
$kj = 183 veh/mi/\ln$
$k = 47 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 60(183)/4 = 2,745 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 92 veh/mi/\ln$
Speed at k
$v = 60(1 - 47/183) = 44.59 mph$
Flow
$q = vk = 44.59(47) = 2,096 veh/h/\ln$
Utilization
$q/qmax = 76.3%$

Answer:

qmax \approx 2,745 veh/h/\ln ; at k = 47, v \approx 44.6 mph and q \approx 2,096 veh/h/\ln

Why the other options are there

qmax = 10,980 veh/h/ln (factor of 4 omitted)
v = 15.4 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 5

Greenshields flow, speed and density — Greenshields Model (5)

A freeway segment follows Greenshields with free-flow speed 62 mph and jam density 151 veh/mi/ln. Find the capacity, and the speed and flow at a density of 61 veh/mi/ln.

Given

$vf = 62 mph$
$kj = 151 veh/mi/\ln$
$k = 61 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 62(151)/4 = 2,341 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 76 veh/mi/\ln$
Speed at k
$v = 62(1 - 61/151) = 36.95 mph$
Flow
$q = vk = 36.95(61) = 2,254 veh/h/\ln$
Utilization
$q/qmax = 96.3%$

Answer:

qmax \approx 2,341 veh/h/\ln ; at k = 61, v \approx 37.0 mph and q \approx 2,254 veh/h/\ln

Why the other options are there

qmax = 9,362 veh/h/ln (factor of 4 omitted)
v = 25.0 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 6

Greenshields flow, speed and density — Greenshields Model (6)

A freeway segment follows Greenshields with free-flow speed 66 mph and jam density 203 veh/mi/ln. Find the capacity, and the speed and flow at a density of 39 veh/mi/ln.

Given

$vf = 66 mph$
$kj = 203 veh/mi/\ln$
$k = 39 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 66(203)/4 = 3,350 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 101.5 veh/mi/\ln$
Speed at k
$v = 66(1 - 39/203) = 53.32 mph$
Flow
$q = vk = 53.32(39) = 2,079 veh/h/\ln$
Utilization
$q/qmax = 62.1%$

Answer:

qmax \approx 3,350 veh/h/\ln ; at k = 39, v \approx 53.3 mph and q \approx 2,079 veh/h/\ln

Why the other options are there

qmax = 13,398 veh/h/ln (factor of 4 omitted)
v = 12.7 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 7

Greenshields flow, speed and density — Greenshields Model (7)

A freeway segment follows Greenshields with free-flow speed 52 mph and jam density 163 veh/mi/ln. Find the capacity, and the speed and flow at a density of 76 veh/mi/ln.

Given

$vf = 52 mph$
$kj = 163 veh/mi/\ln$
$k = 76 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 52(163)/4 = 2,119 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 82 veh/mi/\ln$
Speed at k
$v = 52(1 - 76/163) = 27.75 mph$
Flow
$q = vk = 27.75(76) = 2,109 veh/h/\ln$
Utilization
$q/qmax = 99.5%$

Answer:

qmax \approx 2,119 veh/h/\ln ; at k = 76, v \approx 27.8 mph and q \approx 2,109 veh/h/\ln

Why the other options are there

qmax = 8,476 veh/h/ln (factor of 4 omitted)
v = 24.2 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 8

Greenshields flow, speed and density — Greenshields Model (8)

A freeway segment follows Greenshields with free-flow speed 66 mph and jam density 159 veh/mi/ln. Find the capacity, and the speed and flow at a density of 37 veh/mi/ln.

Given

$vf = 66 mph$
$kj = 159 veh/mi/\ln$
$k = 37 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 66(159)/4 = 2,624 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 80 veh/mi/\ln$
Speed at k
$v = 66(1 - 37/159) = 50.64 mph$
Flow
$q = vk = 50.64(37) = 1,874 veh/h/\ln$
Utilization
$q/qmax = 71.4%$

Answer:

qmax \approx 2,624 veh/h/\ln ; at k = 37, v \approx 50.6 mph and q \approx 1,874 veh/h/\ln

Why the other options are there

qmax = 10,494 veh/h/ln (factor of 4 omitted)
v = 15.4 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 9

Greenshields flow, speed and density — Greenshields Model (9)

A freeway segment follows Greenshields with free-flow speed 70 mph and jam density 192 veh/mi/ln. Find the capacity, and the speed and flow at a density of 90 veh/mi/ln.

Given

$vf = 70 mph$
$kj = 192 veh/mi/\ln$
$k = 90 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 70(192)/4 = 3,360 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 96 veh/mi/\ln$
Speed at k
$v = 70(1 - 90/192) = 37.19 mph$
Flow
$q = vk = 37.19(90) = 3,347 veh/h/\ln$
Utilization
$q/qmax = 99.6%$

Answer:

qmax \approx 3,360 veh/h/\ln ; at k = 90, v \approx 37.2 mph and q \approx 3,347 veh/h/\ln

Why the other options are there

qmax = 13,440 veh/h/ln (factor of 4 omitted)
v = 32.8 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

Example 10

Greenshields flow, speed and density — Greenshields Model (10)

A freeway segment follows Greenshields with free-flow speed 66 mph and jam density 165 veh/mi/ln. Find the capacity, and the speed and flow at a density of 69 veh/mi/ln.

Given

$vf = 66 mph$
$kj = 165 veh/mi/\ln$
$k = 69 veh/mi/\ln$

Find

qmax, v(k) and q(k)

Start with the thinking

Greenshields is linear: v = vf(1 − k/kj).
Capacity occurs at k = kj/2, giving qmax = vf·kj/4.

Step-by-step solution

Capacity
$qmax = vf kj/4 = 66(165)/4 = 2,723 veh/h/\ln$
Optimum density
$k_{0} = kj/2 = 83 veh/mi/\ln$
Speed at k
$v = 66(1 - 69/165) = 38.40 mph$
Flow
$q = vk = 38.40(69) = 2,650 veh/h/\ln$
Utilization
$q/qmax = 97.3%$

Answer:

qmax \approx 2,723 veh/h/\ln ; at k = 69, v \approx 38.4 mph and q \approx 2,650 veh/h/\ln

Why the other options are there

qmax = 10,890 veh/h/ln (factor of 4 omitted)
v = 27.6 mph (relation inverted)

Reference: FE Reference Handbook — Transportation → Greenshields Model

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