Give step-by-step solution with explanation and final answer:
An intersection has a
3
-
timing stage signal with the movements allowed in each timing stage and corresponding analysis and saturation

Question

Give step-by-step solution with explanation and final answer:
 ﻿An intersection has a 
3
-
timing stage signal with the movements allowed in each timing stage and corresponding analysis and saturation flow rates shown below. Consider that the total lost time is 
1
2
 ﻿seconds and a critical v
/
c ratio of 
0
.
8
5
 ﻿was desired. The cycle length is 
6
0
 ﻿seconds. Calculate the delay and level of service for 
(
a
)
 ﻿the westbound left lane group, 
(
b
)
 ﻿the westbound approach and 
(
c
)
 ﻿the whole intersection.

VIPSolutions · Accepted Answer

# Signalized Delay and LOS Calculation

Let's break down the problem step by step:

## **Given Data**

- **3 timing stages** (movements and flow rates provided below)
- **Total lost time, $L = 12$ seconds**
- **Critical $v/c$ ratio desired: $.85$**
- **Cycle length, $C = 60$ seconds**

### **Movements and Saturation Flow Rates**

Suppose the table below is given (fill in with actual values if provided):

| Movement         | Lane Group         | Volume ($v$) | Saturation Flow ($s$) |
|------------------|-------------------|--------------|-----------------------|
| 1. WB Left       | Lane group 1      | $v_1$        | $s_1$                 |
| 2. WB Through    | Lane group 2      | $v_2$        | $s_2$                 |
| 3. WB Right      | Lane group 3      | $v_3$        | $s_3$                 |
| ...              | ...               | ...          | ...                   |

Let’s assume the following for illustration (replace with actual numbers if you have them):

| Lane Group           | Volume ($v$) | Saturation Flow ($s$) |
|----------------------|--------------|-----------------------|
| Westbound Left (a)   | 200 veh/h    | 160 veh/h            |
| Westbound Through    | 450 veh/h    | 180 veh/h            |
| Westbound Right      | 100 veh/h    | 150 veh/h            |
| **Whole Intersection** | **Total $v$** | **Total $s$**      |

---

## **Step 1: Calculate Lane Group Flow Ratios**

Flow ratio $y = v / (s 	imes 360/C)$

But often simplified as:

$$
y = \frac{v}{s}
$$

Calculate for each group (using example numbers):

### (a) **Westbound Left Lane Group**
$$
y_{WB\,Left} = \frac{200}{160} = .125
$$

### (b) **Westbound Approach**
$$
y_{WB\,Approach} = \frac{200 + 450 + 100}{160 + 180 + 150}
= \frac{750}{490} \approx .153
$$

### (c) **Whole Intersection**
Sum $v$ and $s$ for all approaches.

---

## **Step 2: Calculate Effective Green Time for Each Phase**

Effective green time per cycle:
$$
g = \left( C - L ight) 	imes \frac{y_{group}}{\sum y}
$$

Where:
- $C$ = cycle length (60 s)
- $L$ = total lost time (12 s)
- $y_{group}$ = flow ratio of group
- $\sum y$ = sum of critical flow ratios for all phases

Calculate $\sum y$ for all critical movements.

---

## **Step 3: Calculate Degree of Saturation ($X$)**

$$
X = \frac{v}{s \cdot (g/C)}
$$

For each lane group.

---

## **Step 4: Calculate Average Control Delay (HCM 200/201 Formula)**

Basic formula:
$$
d = \frac{.5 C (1 - g/C)^2}{1 - \min(X, 1) (g/C)} + \frac{900 T \left[(X - 1) + \sqrt{(X - 1)^2 + 8 k I X} ight]}{s (g/C)}
$$

But for $X < 1.$, and assuming random arrivals, the delay per vehicle is approximated as:

$$
d = \frac{.5 C (1 - g/C)^2}{1 - X (g/C)}
$$

Where:
- $d$ = control delay (s/veh)
- $X$ = degree of saturation
- $C$ = cycle length (s)
- $g$ = effective green (s)
- $s$ = saturation flow (veh/hr)

---

## **Step 5: Determine Level of Service (LOS)**

| **Delay (s/veh)** | **LOS** |
|-------------------|---------|
| -10              | A       |
| 10-20             | B       |
| 20-35             | C       |
| 35-55             | D       |
| 55-80             | E       |
| >80               | F       |

---

# **Sample Calculation (with Example Numbers)**

Let's proceed with the example for the **Westbound Left Lane Group**.

### 1. **Flow Ratio**
$$
y_{WB\,Left} = .125
$$

Assume other critical lane groups have $y_2 = .20$, $y_3 = .18$, so
$$
\sum y = .125 + .20 + .18 = .505
$$

### 2. **Effective Green for WB Left**
$$
g_{WB\,Left} = (60 - 12) 	imes \frac{.125}{.505} = 48 	imes .2475 \approx 11.88\,s
$$

### 3. **Degree of Saturation**
$$
X = \frac{200}{160 	imes (11.88/60)} = \frac{200}{160 	imes .198} = \frac{200}{316.8} \approx .631
$$

### 4. **Control Delay**
$$
g/C = 11.88/60 = .198
$$

$$
d = \frac{.5 	imes 60 	imes (1 - .198)^2}{1 - .631 	imes .198}
$$

- $(1 - .198) = .802$, squared = $.643$
- $.5 	imes 60 	imes .643 = 19.29$
- $1 - .125 = .875$
- $d = 19.29 / .875 \approx 22.05$ seconds/vehicle

### 5. **LOS**
- $d = 22.05$ s/veh → **LOS C**

---

# **Summary Table**

| Part | Lane Group/Approach | Delay (s/veh) | LOS   |
|------|---------------------|---------------|-------|
| a    | WB Left             | 22.1          | C     |
| b    | WB Approach         | Calculate as above |     |
| c    | Whole Intersection  | Weighted average |     |

---

# **Final Answers (with Steps)**

## **(a) Westbound Left Lane Group**
- **Delay:** $22.1$ s/veh
- **LOS:** C

## **(b) Westbound Approach**
- **Delay:** (Use sum of all WB lane groups' volumes and weighted delay)
- **LOS:** (Determine from delay)

## **(c) Whole Intersection**
- **Delay:** (Weighted average based on all approaches' flow and delay)
- **LOS:** (Determine from delay)

---

**Note:**  
If you provide the actual volumes and saturation flows for each movement, I can substitute and recalculate with your data.

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