Optimizing Truck Speed and Freight Load for Cost Efficiency on Interstate Travel

  • Context: MHB 
  • Thread starter Thread starter Ph4cm
  • Start date Start date
  • Tags Tags
    Calculus
Click For Summary

Discussion Overview

The discussion revolves around optimizing truck speed and freight load for cost efficiency during interstate travel, specifically between Adelaide and Sydney. Participants explore the implications of speed on fuel efficiency, cost modeling, and the relationship between freight size and travel speed. The conversation includes theoretical considerations and mathematical modeling.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose a cost model that factors in fuel efficiency, driver wages, and maintenance costs, with specific equations provided for total cost based on speed and freight load.
  • There is a suggestion that increasing speed beyond 110 km/h results in decreased fuel efficiency, with a proposed model indicating a loss of 0.1 km/L for each km/h increase.
  • Participants discuss the implications of carrying additional freight, noting that fuel efficiency decreases by 0.05 km/L for each extra container.
  • Different cost functions are presented, with some participants questioning the derivation of coefficients and the overall structure of the equations.
  • There are requests for clarification on how specific values in the cost functions were determined, particularly regarding the coefficients related to freight load.
  • One participant mentions a calculated minimum cost occurring at approximately 114 km/h without additional freight, while others present alternative formulations and results.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of the cost function and its parameters. There is no consensus on the optimal speed or the exact relationship between freight size and cost efficiency, indicating ongoing debate and refinement of ideas.

Contextual Notes

Some participants note the complexity of the cost functions and the assumptions underlying their models, including the impact of speed on fuel efficiency and the treatment of freight load. There are unresolved questions regarding the derivation of specific coefficients in the cost equations.

Who May Find This Useful

This discussion may be of interest to those involved in logistics, transportation economics, or mathematical modeling of cost efficiency in freight transport.

Ph4cm
Messages
6
Reaction score
0
A truck traveling interstate, driving at a constant speed of 110km/h, gets 7km/L efficiency and loses 0.1km/L in fuel efficiency for each km/h increase in speed. Costs include diesel ($1.49/L), truck drivers’ wage ($35/hour), and truck maintenance and repairs ($9.50/hour). This truck is mainly used for carrying freight between Adelaide and Sydney (1375km). Explain what would happen if the truck were able to maintain a constant speed over 250km/h?
 Is it reasonable to assume a constant speed for such a trip?
 (Find 2 sources)
 What factors affect the reasonableness of the model the most?
 Suppose that, for every extra freight container the truck carries, the fuel efficiency drops by 0.05km/L. Suggest what will happen to the optimum traveling speed as the load of the truck’s freight increases.
 Find a mathematical relationship between freight size and speed for which cost is a minimum. Discuss your findings.

I've figured out the cost model C= 1375*(801-2.96x)(x(18-0.1x)) (x is speed, C is total cost)
 
Physics news on Phys.org
ducduy said:
A truck traveling interstate, driving at a constant speed of 110km/h, gets 7km/L efficiency and loses 0.1km/L in fuel efficiency for each km/h increase in speed. Costs include diesel (\$1.49/L), truck drivers’ wage (\$35/hour), and truck maintenance and repairs (\$9.50/hour). This truck is mainly used for carrying freight between Adelaide and Sydney (1375km). Explain what would happen if the truck were able to maintain a constant speed over 250km/h?
 Is it reasonable to assume a constant speed for such a trip?
 (Find 2 sources)
 What factors affect the reasonableness of the model the most?
 Suppose that, for every extra freight container the truck carries, the fuel efficiency drops by 0.05km/L. Suggest what will happen to the optimum traveling speed as the load of the truck’s freight increases.
 Find a mathematical relationship between freight size and speed for which cost is a minimum. Discuss your findings.

I've figured out the cost model C= 1375*(801-2.96x)(x(18-0.1x)) (x is speed, C is total cost)

What are your units for speed, the $x$ in your cost function?

I get cost as a function of speed, $v$, in km/hr ...

$C(v) = 1375 \bigg[\dfrac{44.50}{v} + \dfrac{1.49}{7-0.1(v-110)} \bigg]$

unit analysis of my cost equation ...

$ \$ = km\bigg[\dfrac{\$ /hr}{km/hr} + \dfrac{\$/L}{km/L - \frac{km/L}{km/hr}(km/hr)}\bigg]$
 
Yes the x in the model represents in km/h.
Actually i really needed help in the last two questions. I've solved the first 3 questions. And don't mind the "find 2 sources" one it was my annotation. Would you mind to give me the direction to solve the last 2 questions ?
 
simplified the initial cost function ...

$C(v) = 1375\left(\dfrac{44.5}{v} + \dfrac{14.9}{180-v}\right)$

minimum cost occurs at $v \approx 114 \, km/hr$ with no additional freightcost as a function of speed and freight, where $n$ is the additional number of freight units

$C(v,n) = 1375 \left(\dfrac{44.5}{v} + \dfrac{29.8}{180-v-0.5n} \right)$

$\dfrac{\partial C}{\partial v} = 1375\left(-\dfrac{44.5}{v^2} + \dfrac{29.8}{(180-v-0.5n)^2}\right)$

$\dfrac{\partial C}{\partial v}=0 \implies v = \dfrac{180-0.5n}{1+A} \text{ where } A = \sqrt{\dfrac{149}{445}}$

attached is a calculated table of values for $v$ as a function of $n$ (in the table, $x$ represents $n$ and $Y_1$ is speed, $v$) ...
 

Attachments

  • n vs v.png
    n vs v.png
    576 bytes · Views: 139
Last edited by a moderator:
Could you explain more about how you've got the cost function C(v,n) ?
Cause I've got a different answer:
C= 1375∗(1.4918.05−0.1v−0.05n+44.5v)
 
ducduy said:
Could you explain more about how you've got the cost function C(v,n) ?
Cause I've got a different answer:
C= 1375∗(1.4918.05−0.1v−0.05n+44.5v)

I have a better idea ... explain how you derived your function.
 
skeeter said:
I have a better idea ... explain how you derived your function.
So the new fuel efficiency function is y(v,n)= 7-0.1(v-110)-0.05(n-1) = 18.05-0.1x-0.05n
Total cost for fuel: (1375*1.49)/(18.05-0.1x-0.05n)
Total cost for wages and repairs: 44.5*1375/x
Total cost for the trip: C(v,n) = 1375∗(1.49/18.05−0.1v−0.05n + 44.5/v)
Could you check if mine is correct ?
 
So the new fuel efficiency function is y(v,n)= 7-0.1(v-110)-0.05(n-1) = 18.05-0.1x-0.05n

what does n-1 represent?

in the equation I wrote in post #4, $n$ represents the number of additional freight loads
 
skeeter said:
what does n-1 represent?

in the equation I wrote in post #4, $n$ represents the number of additional freight loads
Sorry it was my mistake, it should only be n like in your equation at #4. But i can't understand how did you get the 29.8 in the equation at #4. Would you explain for me a little bit more please ? That will really helps me
 
  • #10
ducduy said:
Sorry it was my mistake, it should only be n like in your equation at #4. But i can't understand how did you get the 29.8 in the equation at #4. Would you explain for me a little bit more please ? That will really helps me

that's my mistake ... should be 14.9. I doubled the numerator and denominator to get the coefficient of $n$ to be 1 instead of 0.5. When I decided against it, I forgot to change the numerator back after I typed it.

$C(v) = 1375\left(\dfrac{44.5}{v} + \dfrac{14.9}{180-v}\right)$

minimum cost occurs at $v \approx 114 \, km/hr$ with no additional freight

corrected ...

cost as a function of speed and freight, where $n$ is the additional number of freight units

$C(v,n) = 1375 \left(\dfrac{44.5}{v} + \dfrac{{\color{red}14.9}}{180-v-0.5n} \right)$

$\dfrac{\partial C}{\partial v} = 1375\left(-\dfrac{44.5}{v^2} + \dfrac{{\color{red}14.9}}{(180-v-0.5n)^2}\right)$...this last equation was correct (see the 149 in the radical?)

$\dfrac{\partial C}{\partial v}=0 \implies v = \dfrac{180-0.5n}{1+A} \text{ where } A = \sqrt{\dfrac{149}{445}}$
 
  • #11
skeeter said:
that's my mistake ... should be 14.9. I doubled the numerator and denominator to get the coefficient of $n$ to be 1 instead of 0.5. When I decided against it, I forgot to change the numerator back after I typed it.

$C(v) = 1375\left(\dfrac{44.5}{v} + \dfrac{14.9}{180-v}\right)$

minimum cost occurs at $v \approx 114 \, km/hr$ with no additional freight

corrected ...

cost as a function of speed and freight, where $n$ is the additional number of freight units

$C(v,n) = 1375 \left(\dfrac{44.5}{v} + \dfrac{{\color{red}14.9}}{180-v-0.5n} \right)$

$\dfrac{\partial C}{\partial v} = 1375\left(-\dfrac{44.5}{v^2} + \dfrac{{\color{red}14.9}}{(180-v-0.5n)^2}\right)$...this last equation was correct (see the 149 in the radical?)

$\dfrac{\partial C}{\partial v}=0 \implies v = \dfrac{180-0.5n}{1+A} \text{ where } A = \sqrt{\dfrac{149}{445}}$
Thank you a lot for this help Skeeter, It really helped me with my work. You are a wonderful person.
 

Similar threads

MHB Optimization calculus question (Difficult)
  • · Replies 4 ·
Replies
4
Views
3K