Cruising Velocity and Operational Transit Cost Optimization

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Concept Explanation

1. Mathematical Formulation of the Total Trip Cost

For a transportation system traveling a fixed transit distance $D$ at a steady cruising speed $v$:

1. Transit Time:
   $$t(v) = \frac{D}{v}$$
2. Time-Dependent Operating Cost: Let $w$ denote the hourly cost of labor, driver wages, vehicle depreciation, and auxiliary life support systems ($\$ / \text{hour}$). The total temporal expenditure is:
   $$C_{\text{time}}(v) = w \cdot t(v) = \frac{w D}{v}$$
3. Speed-Dependent Energy/Fuel Cost: Aerodynamic drag force is given by $F_D = \frac{1}{2}\rho C_D A v^2$. The mechanical power required to overcome drag is $P_D = F_D \cdot v \propto v^3$.
   The fuel or electrical energy consumed per unit time is proportional to power, meaning hourly fuel cost is modeled by:
   $$R_{\text{fuel}}(v) = k v^n \quad (\text{where typically } n = 2 \text{ or } 3)$$
   Multiplying hourly fuel rate by total transit time gives total fuel cost over the trip:
   $$C_{\text{fuel}}(v) = R_{\text{fuel}}(v) \cdot t(v) = (k v^n) \cdot \left(\frac{D}{v}\right) = k D v^{n-1}$$
4. Total Objective Function:
   $$C(v) = C_{\text{time}}(v) + C_{\text{fuel}}(v) = \frac{w D}{v} + k D v^{n-1}$$

For quadratic hourly fuel models ($n = 2$, common in commercial truck logistics):

For cubic hourly fuel models ($n = 3$, common in marine naval architecture):

Cost ($)
  ^
  |        / Total Cost C(v) = A/v + Bv
  |       /
  |  \   /     * Global Min at v*
  |   \ /
  |    *-----------
  |   / \          \ Time Cost = A/v
  |  /   \          \
  | /     \          \
  +-----------------------------------> Velocity v
           v*

2. The Cost Equipartition Principle

When the hourly energy cost is quadratic ($n = 2$), the total trip cost is $C(v) = \frac{A}{v} + Bv$.
Differentiating with respect to velocity $v$:

Setting $C'(v) = 0$ yields the optimal cruising velocity:

Notice what occurs when we evaluate the two individual cost components at this optimal speed $v^*$:

3. Verification on Unbounded Domains (The Single Critical Point Rule)

Because velocity is physically positive ($v > 0$), the operational domain is often modeled on the open interval $(0, \infty)$ or a bounded legal range $[v_{\min}, v_{\max}]$.
Taking the second derivative:

Since $A, B, v > 0$, $C''(v) > 0$ strictly for all positive velocities. By the Single Critical Point Rule (formalized in First Derivative Test), the stationary point $v^*$ is guaranteed to be the unique absolute minimum across the entire domain.

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Engineering Applications & Context

In UNLV engineering disciplines, cruising velocity optimization governs mission planning and fleet routing:

• Mechanical & Automotive Engineering: Electric vehicle (EV) commercial delivery vans optimize highway cruising speeds to balance driver wages against battery discharge limits and recharge downtime.
• Aerospace Engineering & Aviation: Airline flight dispatchers program the Flight Management System (FMS) "Cost Index" ($CI = \frac{\text{Time Cost}}{\text{Fuel Cost}}$) to continuously calculate optimal cruising Mach numbers that minimize total direct operating costs per flight hour.
• Ocean & Marine Engineering: Container shipping lines implement "slow steaming" policies, mathematically reducing ship transit speed by $15\text{--}20\%$ across trans-oceanic legs to slash bunker fuel expenses by up to $50\%$.
• Autonomous Robotics: Autonomous Underwater Vehicles (AUVs) and long-endurance solar drones optimize forward velocity to maximize range and battery mission duration.

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Example Problems

Example Problem 1: Long-Haul Freight Truck Optimization

A long-haul transport truck operates on a $300\text{-mile}$ highway route. The driver earns a flat wage of $\$36/\text{hour}$. When cruising at a steady speed $v$ (in $\text{mph}$), diesel fuel costs $\frac{v^2}{100}$ dollars per hour. The truck operates within the legal speed interval $v \in [40, 75]\text{ mph}$. Find the cruising speed that minimizes the total trip cost, and calculate the cost at that speed.

Solution

Step 1: State governing formulas

• Total Trip Cost Function:
  $$C(v) = \text{Time} \times (\text{Hourly Wage} + \text{Hourly Fuel}) = \frac{D}{v} \left(w + \frac{v^2}{k}\right) = \frac{w D}{v} + \frac{D}{k} v$$
  (Defined in Cruising Velocity and Operational Transit Cost Optimization.)

Step 2: Substitute problem parameters
Given distance $D = 300\text{ miles}$, wage $w = \$36/\text{hr}$, fuel coefficient $k = 100$:

$$C(v) = \frac{300}{v}\left(36 + \frac{v^2}{100}\right) = \frac{10,800}{v} + 3v \quad \text{for } v \in [40, 75]$$

Step 3: Differentiate and find critical numbers

$$C'(v) = -\frac{10,800}{v^2} + 3$$

Set $C'(v) = 0$:

$$\frac{10,800}{v^2} = 3 \implies 3v^2 = 10,800 \implies v^2 = 3,600 \implies v = 60\text{ mph}$$

$v = 60\text{ mph}$ lies strictly within the legal interval $[40, 75]$.

Step 4: Execute Closed Interval Method

• At minimum speed ($v = 40\text{ mph}$):
  $$C(40) = \frac{10,800}{40} + 3(40) = 270 + 120 = \$390$$
• At maximum speed ($v = 75\text{ mph}$):
  $$C(75) = \frac{10,800}{75} + 3(75) = 144 + 225 = \$369$$
• At critical speed ($v = 60\text{ mph}$):
  $$C(60) = \frac{10,800}{60} + 3(60) = 180 + 180 = \$360$$

Conclusion:
The optimal cruising speed is $60\text{ mph}$, yielding a minimum total operating cost of $\$360$.
(Notice that labor cost $\$180$ exactly equals fuel cost $\$180$, confirming the Cost Equipartition Theorem.)

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Example Problem 2: Marine Container Cargo Transit

A commercial container ship navigates a $400\text{-nautical-mile}$ voyage. Crew and hotel electrical operations cost $\$500/\text{hour}$. Due to hydrodynamic hull resistance at sea, bunker fuel costs $0.25 v^3$ dollars per hour at cruising velocity $v$ (in knots). If navigational constraints require $v \in [6, 18]\text{ knots}$, find the velocity minimizing total voyage cost.

Solution

Step 1: State governing formulas

• Total Voyage Cost:
  $$C(v) = \frac{D}{v}(w + k v^3) = \frac{w D}{v} + k D v^2$$
  (Defined in Cruising Velocity and Operational Transit Cost Optimization.)

Step 2: Substitute parameters
$D = 400\text{ nmi}$, $w = \$500/\text{hr}$, $k = 0.25$:

$$C(v) = \frac{400}{v}(500 + 0.25 v^3) = \frac{200,000}{v} + 100 v^2 \quad \text{for } v \in [6, 18]$$

Step 3: Differentiate and find critical numbers

$$C'(v) = -\frac{200,000}{v^2} + 200 v$$

Set $C'(v) = 0$:

$$200 v = \frac{200,000}{v^2} \implies 200 v^3 = 200,000 \implies v^3 = 1,000 \implies v = 10\text{ knots}$$

Step 4: Verify absolute minimum

• At $v = 6\text{ knots}$: $C(6) = \frac{200,000}{6} + 100(36) \approx 33,333.33 + 3,600 = \$36,933.33$
• At $v = 18\text{ knots}$: $C(18) = \frac{200,000}{18} + 100(324) \approx 11,111.11 + 32,400 = \$43,511.11$
• At $v = 10\text{ knots}$: $C(10) = \frac{200,000}{10} + 100(100) = 20,000 + 10,000 = \$30,000$

The optimal cruising speed is $10\text{ knots}$, minimizing voyage operating expenses to $\$30,000$.

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Connections & References

• Parent Concepts:
  • Applied Optimization Modeling Framework
  • First Derivative Test
  • Extreme Value Theorem
• Related Optimization Techniques:
  • Fastest Path Across Media and Snell's Law
  • Squared-Distance Optimization Technique
• Course Links:
  • Calc-1 Session 13 - Critical Points and Extrema
  • Calc-1 Session 14 - Optimizing Areas and Volumes
  • Calc-1 Session 15 - Optimizing Distance and Time