driving slower

8/3/2019 Driving Slower

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On Driving Slower vs. FasterChuck Higgins

9th draftMarch 31, 2011

Driving more slowly will save on transportation costs, but is it always effective todo so? Transportation costs are a function of mileage, speed, price per gallon, and wagerates. Omitted here is depreciation associated with time in that depreciation would besame regardless. Wear and tear costs can be added to the price per gallon. Usagetraditionally in liters/kilometer is conceptually the reciprocal of the miles/gallon anddesirably lower and higher respectively.

Given M is mileage in m/g (or 1/L where L=lt/k) , P is price in $/g (or /lt), S isspeed in m/h (or k/h), and W is wage in $/h (or /h) with units in miles or kilometers,gallons or liters, dollars or euros, and hours. With K as total cost and D as distance:

1a) Ka = P(D/Ma) + W(D/Sa) or

Ka = PDLa + W(D/Sa) and1b) Kb = P(D/Mb) + W(D/Sb) or

Kb = PDLb + W(D/Sb)

for speeds Sa and Sb. Setting Ka equal to Kb, one gets:

2) P/Ma + W/Sa = P/Mb + W/Sb or PLa + W/Sa = PLb + W/Sb

3) W(1/Sa 1/Sb) = P(1/Mb 1/Ma) orW(1/Sa 1/Sb) = P(Lb La)

Given that when Sb > Sa that Mb < Ma (or Lb > La), then a positive solution is:

4) W = P([1/Mb 1/Ma]/[1/Sa 1/Sb]) orW = P([Lb La]/[1/Sa 1/Sb])

Scenario ASuppose at 60 m/h one gets 24m/g and at 80 m/h one gets 18m/g in miles, hours,

and gallons. A trip of 72 miles would require 3 gallons at 60 m/h and would take 1.2hours and at 80 m/h it would require 4 gallons and .9 hours. An equality of costs betweenthese speeds would be at:

5) 3P + 1.2W = 4P +.9W or

6) .3W = P or W = P/.3

where P is the price per gallon and W is the wage.If P is 4.5 dollars per gallon then the wage rate which makes a difference would

be 15 dollars per hour. Thus if one makes more than $15/h then it is more cost effective


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to drive at 80 m/h. However, is it significant? If one makes $50/h then 5) would be3x$4.5 + 1.2x$50 vs. 4x$4.5 + .9x$50 or $73.50 vs. $63 or a gain of 10.50 dollars todrive faster at a $50/h wage rate. But, if one makes $10/h then 5) would be 3x$4.5 +1.2x$10 vs. 4x$4.5 + .9x$10 or $25.50 vs. $27 or a gain of 1.50 dollars to drive slower ata $10/h wage rate.

Scenario BSuppose at 60 m/h one gets 40m/g and at 80 m/h gets 30m/g in miles, hours, and

gallons. A trip of 120 miles would require 3 gallons at 60 m/h and would take 2 hoursand at 80 m/h it would require 4 gallons and 1.5 hours. An equality of costs betweenthese speeds would be at:

7) 3P + 2W = 4P + 1.5W or

8) .5W = P or W = 2P

where P is the price per gallon and W is the wage.If P is 4.5 dollars per gallon then the wage rate that makes a difference would be 9

dollars per hour. Thus if one makes more than $9/h then it is more cost effective to driveat 80 m/h. However, is it significant? If one makes $50/h then 7) would be 3x$4.5 +2x$50 vs. 4x$4.5 + 1.5x$50 or $113.50 vs. $93 or a gain of 10.50 dollars to drive fasterat a $50/h wage rate. But if one makes $10/h then 7) would be 3x$4.5 + 2x$10 vs.4x$4.5 + 1.5x$10 or $33.50 vs. $33 or a gain of .50 dollars to drive still drive faster at$10/h wage rate.

Scenario C

Suppose at 100 k/h one gets .10 lt/k and at 120 k/h gets .12 lt/k in kilometers,hours, and liters. A trip of 100 kilometers would require 10 liters at 100 k/h and wouldtake 1 hour and at 120 k/h it would require 12 liters and .833 hours. An equality of costsbetween these speeds would be at:

9) 10P + 1W = 12P + .833W or

10) .166W = 2P or W = 12P

where P is the price per liter and W is the wage.If P is 1.5 euros per liter then the wage rate that makes a difference would be 18

euros per hour. Thus if one makes more than 18/h then it is more cost effective to driveat 120 k/h. However, is it significant? If one makes 40/h then 9) would be 10x1.5 +1x40 vs. 12x1.5 + .833x40 or 55 vs. 51.33 or a gain of 3.66 euros to drive faster ata 40/h wage rate. But, if one makes 10/h then 9) would be 10x1.5 + 1x10 vs.12x1.5 + .833x10 or 25 vs. 26.33 or a gain of 1.33 euros to drive slower at a 10/hwage rate.


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General CaseA general case can be examined with mileage as a diminishing function of speed,

or:11) M = A BS where mileage falls, or non similarly

L = C + ES where liter consumption rises

and substituting into a generalized version of 1a) or 1b), one gets:

12) K = PD/(A BS) + WD/S or non similarlyK = PD(C + ES) + WD/S

With 12) in liter/kilometers, the increase is linear with friction. However the dragrises to the cubic power of speed. Regardless, with P = 4.5, A=30, B=.20, and variousvalues for D, the optimal speeds associated with some various wage rates which achievedthe lowest total costs were:

Wage Speed

10 6015 6720 7325 7730 8035 8340 8645 8850 90

Graphically:

One can take the first derivative of the liter/kilometer version of 12):


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13) K/S = PDE WD/S2

and set equal to zero gives an optimal speed of:

14) S = (W/PE)1/2

An example from scenario C, would have E as .001 (lt/k)/(k/hr) and P as 1.5 /lt. With awage rate at 18 /h, the optimal speed would be about 109 kilometers per hour.

ConclusionIf ones wage rate (or the opportunity cost of time if on vacation instead of at

work) is high, then the cost savings of driving slower are swamped by the extra costassociated with longer trip times. Likewise, if ones wage rate is low, then it is worth itto slow down. As petroleum prices rise, the wage rate at the point of equality would riselinearly there from. As to the mileage, the wage rate at the point of equality rises per the

difference of the reciprocal mileage rates or difference of liter consumption. In the caseof Prius drivers, where mileage is a lesser consideration, then the conclusion would be tohave the Prii speed up!

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