=(2d+10days×2r/day)(p2)/f2
=(2×10km+10days× 2 ×20km/day)(0.98$/liter)/(20km/liter)
=20.58$
∴ Money saved by Bao=Cost to Arwa- Cost to Bao
≈22.22$-20.58$
≈1.64$
Therefore we see that for the given set of values of p1,p2 & d, Bao saves 1.64$ and gets the better deal.
Is it possible that driving off route can result in money not being saved? Let’s investigate the effects of changing p1, p2 and d.
Before we do that Let’s find the range of d, p1 and p2.
If you get cheaper fuel on your way then why would you take expensive fuel
∴ Range of values of d are 0< d≤ 100km
In this world we get nothing for free.
It would make no sense for Bao to go out of his route for expensive fuel and hence p1>p2.
∴ Range of values of p1 and p2 are p1>p2>0
Now that we know the range of the values Let’s consider the sets of values.
- Set of values for d, when p1=1.00$ and p2=0.98$, are 10km, 30km, 50km, 70km and 90km.
- Set of values for p1, when p2=0.98$ and d=10km, are 1.00$, 1.10$,1.20$,1.30$ and 1.40$.
- Set of values of p2, when p1=1.00$ and d=10km, are 0.90$, 0.92$, 0.94$, 0.96$ and 0.98$
Now Let’s consider how to find the cost to Bao and Arwa per refuel.
Cost to Bao per refuel(US$)
= [(number of kilometers driven)/(fuel efficiency) ] × (cost of fuel)
=(2 × distance of detour + number of working days after which Bao refuels
× number of kilometers driven to work each day)(p2)/(f2)
=(2d+2rw/day)p2/f2
=(2d+2×20km×10day/day)p2/f2
=(2d+400km)p2/(20km liter-1)
Cost to Arwa per day (US$)
= [(number of kilometers driven)/(fuel efficiency) ] × (cost of fuel)
=(number of working days after which Arwa refuels× number of kilometers
driven to work each day)(p1)/(f1)
=(2r/day ×w)p1/(18km liter-1)
=(2×20km/day×10day)p1/(18km liter-1)
=(400km)p1/(18km liter-1)
∴Money saved by Bao (with respect to Arwa)
= Cost to Arwa per day- Cost to Bao per refuel
=(400km)p1/(18km liter-1)- (2d+400km)p2/(20km liter-1) Eq1
Now that we have the information we require Let’s start to investigate. First we shall see the values in table and then visualize them in the form of a graph in which money saved by Bao (with respect to Arwa) versus the variable changed i.e. p1,p2, or d. The equation of line will be the equation of money saved by Bao. ( which we can see is an inverse function if we replace d with x and the other variables p1 and p2 with constants)
Note: The costs to Arwa and Bao per day and the money saved by Bao per day are rounded to two decimal places in the tables
MS Excel was used to make the tables. It was used as it allows us to see the data in an organized manner, and there are options called formulas, which allow us to calculate the values easily. Geogebra was used to make the graphs as the GDC does not have good resolutions and the scale is not present on the graph.
Seeing the effect of changing p1 on the money saved by Bao (with respect to Arwa), when p2 and d are kept constant at 0.98$ and 10km respectively
Note: The equation of the line is the equation of money saved by Bao(with respect to Arwa) .Therefore it is y=(400)x/(18)- (2×10+400)×0.98/(20)
From the graph we can see that as p1 increases, the money saved by Bao (with respect to Arwa) increases. The reason for this is that as p1 increases the cost of fuel to Arwa increases and so Bao saves more money when compared to Arwa. The graph is linear; therefore money save by Bao (with respect to Arwa) is directly proportional to p1.
Seeing the effect of changing p2 on the money saved by Bao (with respect to Arwa), when p1 and d are kept constant at 1.00$ and 10km respectively
Note: The equation of the line is the equation of money saved by Bao(with respect to Arwa) .Therefore it is y=(400)×1/(18)- (2×10+400)x/(20)
From the graph we can see that as p2 increases, the money saved by Bao (with respect to Arwa) decreases. The reason for this is that as p2 increases the cost of fuel to Bao increases and so Bao saves less money when compared to Arwa. The graph is linear; therefore money save by Bao (with respect to Arwa) is directly proportional to negative of p2.
Seeing the effect of changing d on the money saved by Bao (with respect to Arwa), when p1 and p2 are kept constant at 1.00$ and 0.98$ respectively
Note: The equation of the line is the equation of money saved by Bao(with respect to Arwa) .Therefore it is y=(400)×1/(18)- (2×x+400)×0.98/(20)
From the graph we can see that as d increases, the money saved by Bao (with respect to Arwa) decreases. The reason for this is that as d increases Bao has to drive more and hence he consumes more fuel. The fuel costs money and hence Bao saves less money with respect to Arwa as Arwa is not affected by d. The graph is linear; therefore money save by Bao (with respect to Arwa) is directly proportional to negative of d.
Now that we have investigated the affects of p1,p2 and d on cost Let’s try deriving general formulas, which will allows motorist to decide which option is better for them, assuming:
- Every motorist buys fuel at regular intervals of time, and uses the car a fixed number of times only to go, to and fro the normal route once, during this interval.
- The only time they travel of this route, is when they go to buy cheaper petrol.
- Working days refers to the number of days the car is used between each refuel.
Let’s first define the cost E1$ of buying fuel from the normal route at relatively higher price, per working day.
E1 = Number of kilometers driven/fuel efficiency of car × price of fuel/liter
= Number of kilometers driven per day
/fuel efficiency of car ×price of fuel/l
=2r /f ×p1l
=2rp1/(f)l
Now Let’s define the cost E2$ of driving an extra distance out of the normal route to buy cheaper petrol per working day
E2 = Number of kilometers driven/fuel efficiency of car × price of fuel/liter
= (Number of kilometers driven per day (on normal route) + 2× distance of detour/number of working days) /efficiency of car ×price of fuel/l
=(2r +2d×day/w)/f×p2/ l
=2p2(r +/w)/f l
Now that we know the cost of the two options Let’s derive a formula to find which option is better.
Money saved by taking detour(on an average per working day)
= Cost of option1 –cost of option2
=E1-E2
=2rp1/(f l)- 2p2(r +d × day/w)/f l
If this is positive option of detour is cheaper and if it is negative then the option of buying expensive fuel on the way is better (as it is both cheaper and saves time).
Let's consider the limitations of these 2 equations. First of all the model doesn’t account for time, which is precious, on top of that we have assumed that drivers fuel up at regular intervals of time. But in reality, not all motorist fuel up at regular intervals of time. Many motorists, who do not have busy schedules, refuel only when their fuel almost gets over. By going off route and filling the entire tank, and coming once again when the fuel almost gets over, saves more time and money; hence many motorist prefer this option. Even though these equations are not perfect for all motorist it can be helpful to get a fair idea of the money saved or lost by taking detour.
Now that we have derived general equations for the two options let’s find the maximum distance of detour that Bao should drive to obtain a 2% saving.
Let’s assume w,p1,p2,r are constant at 10days,1.00$,0.80$ and 20km respectively.
There should be a minimum 2% saving by taking detour.
∴money saved by taking detour≥2%E1
∴E1-E2≥0.02E1
∴E2≤0.98×E1
∴2p2(r +d× day/w)/f l≤0.98×2rp1/(f l)
∴2×0.80$(20km+d/10) ≤0.98×2×1.00$×20km (multiplying f l to both sides, d>0 and hence there is no problem of 20km+d/10=0)
∴ 20km+d/10≤39.2$km/(2×0.80$)
∴20km+d/10≤24.5km
∴d/10≤2.5km
∴d≤25km
∴25km is the maximum distance of detour Bao should drive in order to get a 2% saving.
Let’s investigate the relationship between d and p2 when E2 is kept constant!
Let’s assume that f, r and w are constant at 20km/liter, 20km and 10days respectively.
For Arwa’s Vehicle
E2 =p2(r +d× day/w)/f l
=p2(20km+d/10)/(18km)
∴p2 =(E2×18km)/(20km+d/10)
y=p2 and x=d
∴The lines are of the form y=(E2×18)/(20+x/10), where E2 is a constant.
∴The line is an inverse function.
We observe that as d is increasing, p2 is decreasing and hence p2 is indirectly proportional to d when E2 is constant. The reason for this is that as d increases more fuel will be required, which means that E2 will increase. But as E2 is constant, for this to be possible the price will have to decrease.
From the graph we can see that 3 stations that provide Arwa with the same cost of 0.9$ are:
- A station 10km away and with price of fuel 0.77$/liter
- A station 15km away with price of fuel 0.75$/liter
- A station 20km away with price of fuel 0.74$/liter
How may this information be useful to Arwa?
This information may be useful to Arwa as a station may be closed one day, but if he knows other stations with the same cost he could go there, so as to save money (by not buying expensive fuel on the normal route). The information about distance can give Arwa the insight about the time it would take and hence Arwa could also consider whether the extra distance is worth the time, on that particular day. On top of that if he goes on the detour and is low on fuel and the station is closed he could use another one close by. (5km will take just a little more than a fourth of a liter) Even if the second one is closed, he can go to the third station. (10km will take only about half a liter).
Now Let’s see the relationship between p2 and d when E2 is kept constant for Bao.
For Bao’s Vehicle:
E2 =p2(r +d× day/w)/f l
=p2(20km+d/10)/(20km)
∴p2 =(E2×20km)/(20km+d/10)
y=p2 and x=d
∴The lines are of the form y=(E2×20)/(20+x/10), where E2 is a constant.
∴The line is an inverse function.
We observe that as d is increasing, p2 is decreasing and hence p2 is indirectly proportional to d when E2 is constant. The reason for this is that as d increases more fuel will be required, which means that E2 will increase. But as E2 is constant, for this to be possible the price will have to decrease.
For E2=1.00$ and p2=0.80$ Let’s find the maximum distance to work that each driver should drive and still save money by buying fuel on the normal route.
Buying fuel on the normal route is saving money
∴E1<E2
∴2rp1/(f)l<1.00$
Let’s assume p1 is constant at 1.00$
∴2r×1.00$/(f)l<1.00$
Let’s find the maximum distance for Arwa first.
∴2r×1.00$/18km<1.00$
∴r<9km
∴ Arwa can no longer save money by just buying fuel from his normal route after the distance to work becomes equal to or greater than 9km.
Now Let’s find the maximum distance for Bao
∴∴2r×1.00$/20km<1.00$
∴r<10km
∴ Bao can no longer save money by just buying fuel from his normal route after the distance to work becomes equal to or greater than 10km.
Arwa is a really business and wonders whether saving money is worth the time. Let’s help him decide.
Arwa must have a money value for time as he is considering saving money or saving time.
Let this be s dollar/hour.
Money saved by taking detour(on an average per working day)
=2rp1/(f l)- 2p2(r +d× day/w)/f l
Extra time taken per working day due to detour
=(2×distance of detour/working day)/speed
=(2d day/w)/(40km/hour) (Let’s assume the time taken per our in city is
constant at 40km/hour)
Therefore in money value time taken due to detour per day costs
= Extra time taken per working day due to detour × Arwa’s money value of time
=(2d day/w)/(40km/hour) ×s
Real benefit due to detour in money value
= Money saved by taking detour- in money value time taken due to detour per day costs
=2rp1/(f l)- 2p2(r +/w)/f l-(2d day/w)/(40km/hour) ×s
If the above equation is positive then Arwa should take the detour and if it is negative he shouldn’t.
Let’s consider the limitations of this model. This model also has the problems related to the assumption of refueling regularly. On top of that this model assume the speed in the city is constant at 40km/hour, which is not true. But there are so many uncontrolled variables, that a model cannot account for all of these variables. Even though these assumptions affect the accuracy of money saved, it can be used to get a fair idea of the benefits; hence I believe it is useful.