on choosing to drive to a cheaper gas station

8/3/2019 On Choosing to Drive to a Cheaper Gas Station

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On choosing to drive to a cheaper gas station6th draft, April 25, 2007

Charles J. Higgins, Ph.D.

With automotive fuel prices rising, one may consider whether acheaper outlet is a better place to refuel. The tradeoff is whether theadditional cost of driving to a farther outlet is, or is not, outweighed

by the expected savings by proceeding to the farther outlet.Consider two outlets with a farther outlet which prices its fuel

at a lower price than a closer outlet, or:

=================A--------------------------------------- B --------------------PA------------D-------------PB

The cost of a full tank (or other quantity) at each station would be:

1) CA = PA Q say $3.50x15

2) CB = PB Q say $3.20x15,

where Q is the contemplated quantity to be purchased, and P is theprice and C is the cost of the purchase for outlets A and Brespectively. The nominal savings would thus be:

3) S = CA - CB = (PA- PB) Q or $.30x15 = $4.50.

However, the cost of traveling the additional distance would also haveto be borne, or:

4) T = 2DP / M say 2x1x$3.00/20 = $.30,

where D is the additional distance to be traveled in order to reachoutlet B, P is the previous purchased price of fuel, and M is the fuelmileage of the vehicle. Here one should note that if the trip is notlikely to be a round trip journey then the computation which considerstwice the distance (2D) may be inappropriate to the point that if outlet


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B is along an already contemplated route, then the marginal cost ofthe additional distance would be zero and thus would make theanalysis herein moot; or, if it is along the way it doesnt matter.

The question now leads to the consideration of whether S isgreater than T which would then prompt one to continue to outlet B (ifthe urgency to immediately refuel is not paramount). Thecomputation then is to compare the example herein of $.30x15 to2x1x$3.00/20 or $4.50 in savings versus $.30 in additionaltransportation cost, for a net savings of $4.20. If this would be aweekly consideration, the yearly savings could be $200.00.

The consideration of the value for depreciable wear and tearand for the time used in the additional distance has not yet beenconsidered. Such a consideration can be made by raising the value ofP; a rule of thumb would be to double the value of P to say $6.00

wherein the additional $3.00 represents an additional $.15 per milewith 20 miles per gallon for ownership costs of a $18,000 car acrosssay 120,000 miles (a $36,000 car would triple the rule of thumb justas a 60,000 mile expected usage would for a $18,000 car). In my

judgment I would not make an adjustment for insurance costs in thatthey usually do not ordinarily increase for additional mileage unlesssignificant; or in other words, insurance costs are more a function oftime regardless of marginal usage changes. In order to value theadditional time required, one might use the average speed and thevalue of a persons time for conversion to adjust the fuel mileage, or amore direct computation is to merely specify the value of additionaltime required. Using the latter approach, one would add V (the valueof ones time to travel to the farther outlet) to T, or

4a) T = 2DP / M + V or 2x1x$3.00/20 + $1.00 = $1.30.

In this example, now with V set to $1.00, the answer would changethe net savings to $3.20 for a yearly savings of about $150.00. With adoubled price for depreciable wear and tear now $6.00, the net

savings would lower to $2.90 with yearly savings now a bit lower.The value of ones additional travel time T can be made a

function of the additional distance D; here V would equal DW / Rwhere W is perhaps a wage rate per hour and R is the speed of travelin miles per hour. This would change equation 4a) to

4b) T = D (W/R + 2P/M) or 1x($60.00/30+2x$6.00/20) = $2.60.


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An interesting exercise is to set S equal to T and then solve forD and/or for PB. In solving for D, one gets:

5) D = [(PA- PB) Q - V] M / 2P or($.30x15-$1.00)x20/(2x$6.00) = 6.

Using equation 4b) instead, D becomes:

5a) D = [(PA- PB) Q] / (W/R+2P/M) or$.30x15/($60/30+2x$6/20) = 2

In solving for PB, one gets:

6) PB = PA- 2DP / MQ V/Q or

$3.50-2x1x$3/(20x15)-$1/15 = $3.39

Using equation 4b) instead, PBbecomes

6a) PB = PA- (D/Q) (W/R+2P/M) or$3.50-1/15x($60/30+2x$6/20) = $3.33,

An example using the above data with V having the value of $1.00and with P set at twice the rate for depreciable wear and tear wouldresult in a net savings breakeven distance using equation 5) of($.30x15 -$1.00)x20/(2x$6.00) or about 6 miles. Using equation 5a)requires a wage rate W of say $60.00/h and a speed R of say 30m/hwhich would now give a breakeven distance of ($.30x15) /($60/30+2x$6.00/20) or almost 2 miles. For equation 6) the necessaryfarther outlet price would need to less than the result of $3.50 -2x1x$6.00/(20x15) - $1/15 or $3.39. Using equation 6a) the fartheroutlet price would need to less than $3.50 (1/15)($60.00/30 +2x$6.00/20) or about $3.33. Because D was set at 1, it is clear that foreach additional mile farther, the price should be some $.17 cheaper;

this about a one cent per city block or two cents per suburban block; ifone values ones time at $30.00/hour then perhaps one cent for persuburban block.

on choosing to drive to a cheaper gas station

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