Cruising Velocity and Operational Transit Cost Optimization

Core Definition

Cruising Velocity and Operational Transit Cost Optimization models the financial and energetic trade-offs inherent in moving physical vehicles across a fixed distance $D$ . Because elapsed travel time scales inversely with velocity ( $t = D/v$ ), fixed-rate temporal costs (e.g., driver wages, vehicle leasing, hotel overhead) decrease monotonically with speed as $C_t \propto 1/v$ . Conversely, aerodynamic and hydrodynamic drag forces scale non-linearly with velocity ( $F_{\text{drag}} \propto v^2 \implies P_{\text{drag}} \propto v^3$ ), causing energetic fuel/battery costs per mile to increase as $C_f \propto v$ or $v^2$ . Calculus identifies the unique velocity that balances these conflicting operational pressures.

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 $v$ :

Transit Time: $t(v) = \frac{D}{v}$
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: $C_{\text{time}}(v) = w \cdot t(v) = \frac{w D}{v}$
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: $C_{\text{fuel}}(v) = R_{\text{fuel}}(v) \cdot t(v) = (k v^n) \cdot \left(\frac{D}{v}\right) = k D v^{n-1}$
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):

C(v) = \frac{A}{v} + B v \quad \text{where } A = w D, \; B = k D

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

C(v) = \frac{A}{v} + B v^2 \quad \text{where } A = w D, \; B = k D

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 $v$ :

C'(v) = -\frac{A}{v^2} + B

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

\frac{A}{v^2} = B \implies v^2 = \frac{A}{B} \implies v^* = \sqrt{\frac{A}{B}}

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

C_{\text{time}}(v^*) = \frac{A}{v^*} = \frac{A}{\sqrt{A/B}} = \sqrt{AB}

C_{\text{fuel}}(v^*) = B v^* = B \sqrt{\frac{A}{B}} = \sqrt{AB}

The Cost Equipartition Theorem

For any transportation model where hourly propulsion expenditure scales quadratically with speed, the minimum total trip cost occurs at the precise velocity where total labor/time expenditure exactly equals total fuel/energy expenditure:

C_{\text{time}}(v^*) = C_{\text{fuel}}(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:

C''(v) = \frac{2A}{v^3} + 2B \quad (\text{or } \frac{2A}{v^3} > 0)

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.

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.

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.)

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.