Fuel Economy; I understand that the distance traveled by the car using a liter of Petrol, can tell us how many liters it requires to travel the whole distance. This will also help us to eventually find Arwa's and Bao's spending on their cars and the total cost in the end. As the cars are of different sizes, this will make an important factor in the modeling of the equation.
Distance along Normal Route = S
Distance off Route for Fuel= d
Fuel consumed = F
Fuel Economy= Eco
Total Cost= C
Petrol Prices= P
But,
Therefore,
Here I try to determine who gets a better deal when P1= 1, P2 = 0.98, and d=10.
Arwa Spends US$1.00 per Liter for Fuel,
Therefore P1=1.
Bao Spends US$0.98 per Liter for Fuel,
Therefore P2=0.98.
The extra Distance that Bao travels is 10km,
Therefore d= 10
I assume that the distance they travel along their normal route is 100 Km as both the cars travel the same amount of distance. I will also assume here that although the cars are of different sizes, the fuel economy of these cars to be 10 Km/liter.
Arwa's Route:
Total Distance D = 100 Km
Price for Fuel P1= 1
Eco= 10 Km/l
Therefore, Arwa Spends 10$ on the fuel.
The amount Arwa spends per effective kilometer $
Bao's Route:
Total Distance D = 120 Km
Price for Fuel P2= 0.98
Eco= 10 Km/l
Therefore, Bao Spends 11.76$ on the fuel.
The amount Bao spends per effective kilometer
After all the Calculations done above, I can safely say that Arwa gets a better deal. But these calculations might differ when I account for the actual fuel economy of their cars.
Let Arwa be having a car with a fuel efficiency of F1 and Bao with one of F2.
If S is the fixed Distance of the Normal Route and P1 is the Price of Petrol in a petrol bunk on this route, whereas P2 be the price of Petrol on a Detour route of d kilometers out of this normal route.
Let L1 and L2 be the Effective Liters for each of the options of the routes.
Since Effective liters is a quantity of fuel utilized in traversing a normal rout S;
L1=
Likewise L2=
Effective Liters for route one E1 =P1 because all the amount of petrol used is utilized on the normal route.
Effective liters for route two E2=
Therefore
Considering S, F1, F2 of Fixed values, I shall try to derive a mathematical relationship between P2 (Fixing P1 at 1) and the maximum detour distance d with a economic perspective.
The Cost for Arwa along S=………………………………..(1)
The cost for Bao along S (But at Price P2) = P2…………………………(2)
The difference of these amounts is R$ =
With R$, Extra Liters can be obtained for which an additional distance of can be travelled.
Half of this Distance is our required maximum distance d.
i.e.: d=
Using this relation I prepare a table for the value the value of d for different values of P2
P1
P2
d
Constants
1
0.5
70
S=100
1
0.55
59.09091
F1=10
1
0.6
50
F2=12
1
0.65
42.30769
1
0.7
35.71429
1
0.75
30
1
0.8
25
1
0.85
20.58824
1
0.9
16.66667
1
0.95
13.15789
1
1
10
1
1.05
7.142857
1
1.1
4.545455
1
1.15
2.173913
1
1.2
0
1
1.25
-2
1
1.3
-3.84615
1
1.35
-5.55556
1
1.4
-7.14286
1
1.45
-8.62069
1
1.5
-10
When I plot P2 against d we get the following:
Repeating the above Process for different values of F2 I obtain the following family of curves:
P2
d when f2=12
d when f2=15
d when f2= 17
d when f2= 11
0.5
70
100
120
60
0.55
59.09090909
86.36363636
104.5454545
50
0.6
50
75
91.66666667
41.66666667
0.65
42.30769231
65.38461538
80.76923077
34.61538462
0.7
35.71428571
57.14285714
71.42857143
28.57142857
0.75
30
50
63.33333333
23.33333333
0.8
25
43.75
56.25
18.75
0.85
20.58823529
38.23529412
50
14.70588235
0.9
16.66666667
33.33333333
44.44444444
11.11111111
0.95
13.15789474
28.94736842
39.47368421
7.894736842
1
10
25
35
5
1.05
7.142857143
21.42857143
30.95238095
2.380952381
1.1
4.545454545
18.18181818
27.27272727
-8.88178E-15
1.15
2.173913043
15.2173913
23.91304348
-2.173913043
1.2
0
12.5
20.83333333
-4.166666667
1.25
-2
10
18
-6
1.3
-3.846153846
7.692307692
15.38461538
-7.692307692
1.35
-5.555555556
5.555555556
12.96296296
-9.259259259
1.4
-7.142857143
3.571428571
10.71428571
-10.71428571
1.45
-8.620689655
1.724137931
8.620689655
-12.06896552
1.5
-10
0
6.666666667
-13.33333333
Now I attempt to generate a model to help motorists to decide between L1 and L2 on economic grounds; for this I shall presume a fixed detour distance d at 10. The fuel economy we take constant F is 10 km per Liter we will also take distance along Normal route S to be 100 Km
The Total Cost for a Motorist taking Route 1 is C1=
The Total Cost for a Motorist taking Route 2 is C2=
I understand that there will be a breakeven point where C1 = C2
Therefore,
When I graph this model, for different values of P1 and P2 and without the loss of generality I will keep the difference between the prices constant at 0.03$
P1
P2
Price Along Option 1
Price along option 2
Difference in price
1
0.7
10
8.4
-1.6
1.1
0.8
11
9.6
-1.4
1.2
0.9
12
10.8
-1.2
1.3
1
13
12
-1
1.4
1.1
14
13.2
-0.8
1.5
1.2
15
14.4
-0.6
1.6
1.3
16
15.6
-0.4
1.7
1.4
17
16.8
-0.2
1.8
1.5
18
18
0
1.9
1.6
19
19.2
0.2
2
1.7
20
20.4
0.4
2.1
1.8
21
21.6
0.6
2.2
1.9
22
22.8
0.8
2.3
2
23
24
1
2.4
2.1
24
25.2
1.2
2.5
2.2
25
26.4
1.4
2.6
2.3
26
27.6
1.6
2.7
2.4
27
28.8
1.8
2.8
2.5
28
30
2
From this I can understand that if the Prices were P1 = 1.8$ and P2= 1.5$ the prices both the motorists would bear the same costs.
But if the prices P1 and P2 were any higher than 1.8$ and 1.5$ respectively, Route 1 would be a more viable option. But if P1 and P2 were higher lesser 1.8$ and 1.5$ respectively, Route 2 will be a more profitable option.
I can also use the model find the savings a motorist makes for choosing route 2 over route 1.
If I were to find Bao's savings, I will try to calculate the costs Bao would have to bear, if he took both the routes. In the following calculations Fuel economy F2 will be the same as I are only dealing with one car.
The cost Bao would have to bear if he took option 1 B1= P1
The cost Bao would have to bear if he were to obtain a saving of 2% if he took option 2 B2=
Cost B2 =
Therefore I get;
And hence I can derive d as
Therefore I can use this model to derive the farthest distance Bao needs to travel in order to get a price saving of 2%.
As an illustration I will take F2 as a constant at 10 km/l, S as 100 km, therefore L== 10
So I will plug-in P1 as 1.2$ and P2 as 0.87$, and use our model to find the farthest distance Bao would have to take a detour to get a price saving of 2%. I get d as 1.156. Hence Bao would have to take a detour of maximum 1.15 km to gain a price saving of 2%
When I keep the values of L1, P2 a constant in the model, I can understand the relationship between P1 and d for Bao's Vehicle.
P1
P2
d
1
0.6
33.33333
1
0.65
26.92308
1
0.7
21.42857
1
0.75
16.66667
1
0.8
12.5
1
0.85
8.823529
1
0.9
5.555556
1
0.95
2.631579
1
1
0
Here we take P1 as a constant and S as 100 km. and F2 as 10 km/l.
When we graph the results we get the following:
When we do the same, but for different values of P2 we get a family of curves:
P1
P2
d
P1
P2
d
P1
P2
d
1.2
0.6
50
1.4
0.6
66.66667
1.6
0.6
83.33333
1.2
0.65
42.30769
1.4
0.65
57.69231
1.6
0.65
73.07692
1.2
0.7
35.71429
1.4
0.7
50
1.6
0.7
64.28571
1.2
0.75
30
1.4
0.75
43.33333
1.6
0.75
56.66667
1.2
0.8
25
1.4
0.8
37.5
1.6
0.8
50
1.2
0.85
20.58824
1.4
0.85
32.35294
1.6
0.85
44.11765
1.2
0.9
16.66667
1.4
0.9
27.77778
1.6
0.9
38.88889
1.2
0.95
13.15789
1.4
0.95
23.68421
1.6
0.95
34.21053
1.2
1
10
1.4
1
20
1.6
1
30
When we represent these values on a graph we get a family of curves, which looks like this:
Now using E2 = = 0.9$
And keeping S/F2 a constant at 10
So the equation becomes 0.9=
Solving for d I get
Here I will plug in random values for P2 which will give me the corresponding value of d.
P2
d
0.75
0.75
0.8
0.5
0.85
0.25
E2 = =1$
When we keep P2 at a constant at 0.80$
1= =12.5 km
When we keep P2 at a constant at 0.80$
1= = 12.475
Therefore Arwa can afford to a maximum distance d of 12.475 km to be in a profitable position whereas for Bao the maximum distance is 12.5 Km. When the P2 = 0.8$
If we were to calculate the time spent on traveling both the routes, it can be possible if we know the average speed at which the vehicle is traveling. For this model we will take the average speed of the vehicle as V.
And the total distance traveled is S+2d.
Therefore the time taken to travel this route =
We can use this model to see the relation between time saved and the extra distance travel.
Extra time taken is t =
Therefore t= 2(
To illustrate, we will take the average speed V to be 20 km/h
S will be taken as 100 and the fuel economy F as 10 km/l
P1
P2
d
Extra time taken
1
0.6
33.33333
199.99998
1
0.65
26.92308
161.53848
1
0.7
21.42857
128.57142
1
0.75
16.66667
100.00002
1
0.8
12.5
75
1
0.85
8.823529
52.941174
1
0.9
5.555556
33.333336
1
0.95
2.631579
15.789474
