The document discusses two mathematical models for estimating the time spent driving to school based on starting time: 1) A difference equation model that represents the number of cars on different roads over time using a matrix equation. 2) A differential equation model that represents the rate of change in the number of cars on each road over time using another matrix equation. Both models aim to relate the number of cars on the road to the time spent driving, with the goal of estimating driving time for different departure times from school.

1. HOW MUCH TIME WILL BE USED FOR
DRIVING TO SCHOOL?
Ruo Yang
Winter 2015
Mathematical modelling

2. HOW DID I START?
1. Rush hours.
2. For different starting time t, it will costs a different time on the road
corresponding to t.
3. Different number of cars on the road at different starting time.
4. More cars costs more time for driving.

3. STRUCTURE
Input: Starting time
Road condition
(How many cars on the
road)
How many cars will
impacts my driving
Relation between
number of cars
who can impact
my drive and cost
of time
Output: cost of time

4. MODELLING
1. Difference equation model (Discrete in time).
2. Differential equation model (Continuous in time).
The core is finding the how many cars will impact driving for different starting time

5. THE ROAD TO UNIVERSITY

6. DIFFERENCE EQUATION
……….
S1
R1
F1
S2
F2
R2
R8
S9
F9
R9
I1 I2
I9

7. DIFFERENCE EQUATION
𝑆(𝑡)𝑖 > 0, 𝑖 = 1,2, … . . , 9 Represents how many cars on the road i at time t.
𝑅𝑖, 𝑖 = 1,2, … … , 9 Represents the percentage of cars go from ith road to i+1th road
after half minute
𝐹𝑖, 𝑖 = 1,2, … … , 9 Represents the percentage of cars stay in ith road after half
minute
𝐼(𝑡) > 0𝑖, 𝑖 = 1,2, … . . , 9, Represents how many cars will going to ith road from
outside every half minute.

8. DIFFERENCE EQUATION
∆𝑡 = ℎ𝑎𝑙𝑓 𝑚𝑖𝑛𝑢𝑡𝑒
𝑆 𝑡 + ∆𝑡 1 = 𝐼 𝑡 1 + 𝑅9 ∗ 𝑆 𝑡 9 + 𝐹1 ∗ 𝑆 𝑡 1
𝑆 𝑡 + ∆𝑡 2 = 𝐼 𝑡 2 + 𝑅1 ∗ 𝑆 𝑡 1 + 𝐹2 ∗ 𝑆 𝑡 2
𝑆 𝑡 + ∆𝑡 3 = 𝐼 𝑡 3 + 𝑅2 ∗ 𝑆 𝑡 2 + 𝐹3 ∗ 𝑆 𝑡 3
𝑆 𝑡 + ∆𝑡 4 = 𝐼 𝑡 4 + 𝑅3 ∗ 𝑆 𝑡 3 + 𝐹4 ∗ 𝑆 𝑡 4
𝑆 𝑡 + ∆𝑡 5 = 𝐼 𝑡 5 + 𝑅4 ∗ 𝑆 𝑡 4 + 𝐹5 ∗ 𝑆 𝑡 5
𝑆 𝑡 + ∆𝑡 6 = 𝐼 𝑡 6 + 𝑅5 ∗ 𝑆 𝑡 5 + 𝐹6 ∗ 𝑆 𝑡 6
𝑆 𝑡 + ∆𝑡 7 = 𝐼 𝑡 7 + 𝑅6 ∗ 𝑆 𝑡 6 + 𝐹7 ∗ 𝑆 𝑡 7
𝑆 𝑡 + ∆𝑡 8 = 𝐼 𝑡 8 + 𝑅7 ∗ 𝑆 𝑡 7 + 𝐹8 ∗ 𝑆 𝑡 8
𝑆 𝑡 + ∆𝑡 9 = 𝐼 𝑡 9 + 𝑅8 ∗ 𝑆 𝑡 8 + 𝐹9 ∗ 𝑆 𝑡 9

9. DIFFERENCE EQUATION
𝐹1 0 0 0 0 0 0 0 𝑅9
𝑅1 𝐹2 0 0 0 0 0 0 0
0 𝑅2 𝐹3 0 0 0 0 0 0
0 0 𝑅3 𝐹4 0 0 0 0 0
0 0 0 𝑅4 𝐹5 0 0 0 0
0 0 0 0 𝑅5 𝐹6 0 0 0
0 0 0 0 0 𝑅6 𝐹7 0 0
0 0 0 0 0 0 𝑅7 𝐹8 0
0 0 0 0 0 0 0 𝑅8 𝐹9
*
𝑆 𝑡 1
𝑆 𝑡 2
𝑆 𝑡 3
𝑆 𝑡 4
𝑆 𝑡 5
𝑆 𝑡 6
𝑆 𝑡 7
𝑆 𝑡 8
𝑆 𝑡 9
+
𝐼 𝑡 1
𝐼 𝑡 2
𝐼 𝑡 3
𝐼 𝑡 4
𝐼 𝑡 5
𝐼 𝑡 6
𝐼 𝑡 7
𝐼 𝑡 8
𝐼 𝑡 9
=
𝑆 𝑡 + ∆𝑡 1
𝑆 𝑡 + ∆𝑡 2
𝑆 𝑡 + ∆𝑡 3
𝑆 𝑡 + ∆𝑡 4
𝑆 𝑡 + ∆𝑡 5
𝑆 𝑡 + ∆𝑡 6
𝑆 𝑡 + ∆𝑡 7
𝑆 𝑡 + ∆𝑡 8
𝑆 𝑡 + ∆𝑡 9

10. DIFFERENCE EQUATION
Too many parameters, then reduce matrix
A=
𝐹𝐹1 0 𝑅𝑅3
𝑅𝑅1 𝐹𝐹2 0
0 𝑅𝑅2 𝐹𝐹3
, where
𝐹𝐹1 0 𝑅𝑅3
𝑅𝑅1 𝐹𝐹2 0
0 𝑅𝑅2 𝐹𝐹3
*
𝑆𝑆(𝑡)1
𝑆𝑆(𝑡)2
𝑆𝑆(𝑡)3
+
𝐼𝐼1
𝐼𝐼2
𝐼𝐼3
=
𝑆𝑆(𝑡 + ∆𝑡)1
𝑆𝑆(𝑡 + ∆𝑡)2
𝑆𝑆(𝑡 + ∆𝑡)3
.
To find the equilibrium points
1 − 𝐹𝐹1 0 −𝑅𝑅3
−𝑅𝑅1 1 − 𝐹𝐹2 0
0 −𝑅𝑅2 1 − 𝐹𝐹3
∗
𝑆𝑆(𝑡)1
𝑆𝑆(𝑡)2
𝑆𝑆(𝑡)3
=
𝐼𝐼1
𝐼𝐼2
𝐼𝐼3
.

11. DIFFERENCE EQUATION
𝑆𝑆(𝑡)1
𝑆𝑆(𝑡)2
𝑆𝑆(𝑡)3
=
−
𝐼𝐼1∗ 𝐹𝐹2−1 ∗ 𝐹𝐹3−1 +𝐼𝐼2∗𝑅𝑅1∗ 1−𝐹𝐹3 +𝐼𝐼3∗𝑅𝑅1∗𝑅𝑅2
𝐹𝐹1∗𝐹𝐹2∗𝐹𝐹3+𝑅𝑅1∗𝑅𝑅2∗𝑅𝑅3−𝐹𝐹1∗𝐹𝐹2−𝐹𝐹1∗𝐹𝐹3−𝐹𝐹3∗𝐹𝐹2+𝐹𝐹1+𝐹𝐹2+𝐹𝐹3−1
−
𝐼𝐼2∗ 𝐹𝐹1−1 ∗ 𝐹𝐹3−1 +𝐼𝐼3∗𝑅𝑅2∗ 1−𝐹𝐹1 +𝐼𝐼1∗𝑅𝑅3∗𝑅𝑅2
𝐹𝐹1∗𝐹𝐹2∗𝐹𝐹3+𝑅𝑅1∗𝑅𝑅2∗𝑅𝑅3−𝐹𝐹1∗𝐹𝐹2−𝐹𝐹1∗𝐹𝐹3−𝐹𝐹3∗𝐹𝐹2+𝐹𝐹1+𝐹𝐹2+𝐹𝐹3−1
−
𝐼𝐼3∗ 𝐹𝐹2−1 ∗ 𝐹𝐹1−1 +𝐼𝐼1∗𝑅𝑅3∗ 1−𝐹𝐹2 +𝐼𝐼2∗𝑅𝑅1∗𝑅𝑅3
𝐹𝐹1∗𝐹𝐹2∗𝐹𝐹3+𝑅𝑅1∗𝑅𝑅2∗𝑅𝑅3−𝐹𝐹1∗𝐹𝐹2−𝐹𝐹1∗𝐹𝐹3−𝐹𝐹3∗𝐹𝐹2+𝐹𝐹1+𝐹𝐹2+𝐹𝐹3−1

12. DIFFERENCE EQUATION
1. Nonnegative
2. Irreducible
So, primitive matrix
Perron-Frobenius Theorem
Positive E-value
𝑚𝑖𝑛𝑖 𝑗 𝑎𝑖𝑗 ≤ 𝜆 ≤ 𝑚𝑎𝑥𝑖 𝑗 𝑎𝑖𝑗
globally asymptotically stable.

13. DIFFERENTIAL EQUATION
………s1
r1
s2
r2 r8
s9
I1 I2 I9
O1 O2
O9
R
r9

14. DIFFERENTIAL EQUATION
𝑠(𝑡)𝑖, 𝑖 = 1,2, … . . , 9 Represents how many cars on the road i at time t.
𝐼(𝑡)𝑖, 𝑖 = 1,2, … … , 9 Represents how many cars will going to ith road from outside
every half minute.
𝑟𝑖, 𝑖 = 1,2, … … , 9 Represents the rate that how many cars leave from ith road to
i+1th road per half minute.
𝑂𝑖, 𝑖 = 1,2, … … , 9 Represents the rate that how many cars leave from ith road to
outside per half minute.

15. DIFFERENTIAL EQUATION
∆𝑡 = ℎ𝑎𝑙𝑓 𝑚𝑖𝑛𝑢𝑡𝑒
𝑑𝑠1
𝑑𝑡
= −𝑠(𝑡)1 ∗ 𝑂1 + 𝑠(𝑡)9 ∗ 𝑟9 + 𝐼(𝑡)1
𝑑𝑠2
𝑑𝑡
= −𝑠(𝑡)2 ∗ 𝑂2 + 𝑠(𝑡)1 ∗ 𝑟1 + 𝐼(𝑡)2
𝑑𝑠3
𝑑𝑡
= −𝑠(𝑡)3 ∗ 𝑂3 + 𝑠(𝑡)2 ∗ 𝑟2 + 𝐼(𝑡)3
𝑑𝑠4
𝑑𝑡
= −𝑠(𝑡)4 ∗ 𝑂4 + 𝑠(𝑡)3 ∗ 𝑟3 + 𝐼(𝑡)4
𝑑𝑠5
𝑑𝑡
= −𝑠(𝑡)5 ∗ 𝑂5 + 𝑠(𝑡)4 ∗ 𝑟4 + 𝐼(𝑡)5
𝑑𝑠6
𝑑𝑡
= −𝑠(𝑡)6 ∗ 𝑂6 + 𝑠(𝑡)5 ∗ 𝑟5 + 𝐼(𝑡)6
𝑑𝑠7
𝑑𝑡
= −𝑠(𝑡)7 ∗ 𝑂7 + 𝑠(𝑡)6 ∗ 𝑟6 + 𝐼(𝑡)7
𝑑𝑠8
𝑑𝑡
= −𝑠(𝑡)8 ∗ 𝑂8 + 𝑠(𝑡)7 ∗ 𝑟7 + 𝐼(𝑡)8
𝑑𝑠9
𝑑𝑡
= −𝑠(𝑡)9 ∗ 𝑂9 + 𝑠(𝑡)8 ∗ 𝑟8 + 𝐼(𝑡)9

16. DIFFERENTIAL EQUATION
A=
−𝑂1 0 0 0 0 0 0 0 𝑟9
𝑟1 −𝑂2 0 0 0 0 0 0 0
0 𝑟2 −𝑂3 0 0 0 0 0 0
0 0 𝑟3 −𝑂4 0 0 0 0 0
0 0 0 𝑟4 −𝑂5 0 0 0 0
0 0 0 0 𝑟5 −𝑂6 0 0 0
0 0 0 0 0 𝑟6 −𝑂7 0 0
0 0 0 0 0 0 𝑟7 −𝑂8 0
0 0 0 0 0 0 0 𝑟8 −𝑂9
𝐼(𝑡) =
𝐼 𝑡 1
𝐼 𝑡 2
𝐼 𝑡 3
𝐼 𝑡 4
𝐼 𝑡 5
𝐼 𝑡 6
𝐼 𝑡 7
𝐼 𝑡 8
𝐼 𝑡 9
, Then,
𝑑𝑠
𝑑𝑡
= 𝐴 ∗ 𝑠 𝑡 + 𝐼(𝑡).

17. DIFFERENTIAL EQUATION
Too many parameters, then reduce matrix
B =
−𝑂𝑂1 0 𝑟𝑟3
𝑟𝑟1 −𝑂𝑂2 0
0 𝑟𝑟2 −𝑂𝑂3
,
Where
−𝑂𝑂1 0 𝑟𝑟3
𝑟𝑟1 −𝑂𝑂2 0
0 𝑟𝑟2 −𝑂𝑂3
*
𝑠𝑠(𝑡)1
𝑠𝑠(𝑡)2
𝑠𝑠(𝑡)3
+
𝐼𝐼1
𝐼𝐼2
𝐼𝐼3
=
𝑑𝑠
𝑑𝑡
.
Nonhomogeneous system

18. DIFFERENTIAL EQUATION
𝑠𝑠(𝑡)1
𝑠𝑠(𝑡)2
𝑠𝑠(𝑡)3
=
𝐼𝐼1∗𝑂𝑂2∗𝑂𝑂3+𝐼𝐼2∗𝑂𝑂3∗𝑟𝑟1+𝐼𝐼3∗𝑟𝑟1∗𝑟𝑟2
𝑜𝑜1∗𝑜𝑜2∗𝑜𝑜3−𝑟𝑟1∗𝑟𝑟2∗𝑟𝑟3
𝐼𝐼2∗𝑂𝑂1∗𝑂𝑂3+𝐼𝐼3∗𝑂𝑂1∗𝑟𝑟2+𝐼𝐼1∗𝑟𝑟3∗𝑟𝑟2
𝑜𝑜1∗𝑜𝑜2∗𝑜𝑜3−𝑟𝑟1∗𝑟𝑟2∗𝑟𝑟3
𝐼𝐼3∗𝑂𝑂2∗𝑂𝑂1+𝐼𝐼1∗𝑂𝑂2∗𝑟𝑟3+𝐼𝐼2∗𝑟𝑟1∗𝑟𝑟3
𝑜𝑜1∗𝑜𝑜2∗𝑜𝑜3−𝑟𝑟1∗𝑟𝑟2∗𝑟𝑟3
𝜆3 + (𝑂𝑂1+𝑂𝑂2 + 𝑂𝑂3) ∗ 𝜆2+(𝑂𝑂1 ∗ 𝑂𝑂2 + 𝑂𝑂1 ∗ 𝑂𝑂3 + 𝑂𝑂2 ∗ 𝑂𝑂3) ∗ 𝜆 +
(𝑂𝑂1 ∗ 𝑂𝑂2 ∗ 𝑂𝑂3 − 𝑟𝑟1 ∗ 𝑟𝑟2 ∗ 𝑟𝑟3)
If unique equilibrium points exist then it is globally asymptotically stable.

19. NUMERICAL SIMULATION
By Recording the data everyday
8 10 12 14 16 18
4
6
8
10
12
14
timeo
fad
ay
number
o
f
c
ars
number_of_cars vs. time_of_a_day
untitled fit 1

20. NUMERICAL SIMULATION
For both models, using same initial condition for the road and input cars at each
traffic intersection.
6 8 10 12 14 16 18 20
0
2
4
6
8
10
12
14
Timeo
f the day
Number
of
cars
s1
s2
s3
s4
s5
s6
s7
s8
s9
6 8 10 12 14 16 18 20
0
2
4
6
8
10
12
14
time
Numberofcars
I1
I2
I3
I4
I5
I6
I7
I8
I9

21. NUMERICAL SIMULATION
0.1 0 0 0 0 0 0 0 0.01
0.8 0.1 0 0 0 0 0 0 0
0 0.8 0.1 0 0 0 0 0 0
0 0 0.55 0.1 0 0 0 0 0
0 0 0 0.02 0.1 0 0 0 0
0 0 0 0 0.85 0.1 0 0 0
0 0 0 0 0 0.85 0.1 0 0
0 0 0 0 0 0 0.5 0.4 0
0 0 0 0 0 0 0 0.5 0.1
for difference equation

22. NUMERICAL SIMULATION
Notice that: 2 models are using same ∆𝑡 = ℎ𝑎𝑙𝑓 𝑚𝑖𝑛𝑢𝑡𝑒
𝑅𝑖 + 𝐹𝑖 +𝑜𝑖=1 in difference equation.
𝑟𝑖=log(1+𝑅𝑖), 𝑂𝑖 = log(2 − (𝑅𝑖 + 𝐹𝑖)) in differential equation.
The
−0.09 0 0 0 0 0 0 0 0.01
0.59 −0.09 0 0 0 0 0 0 0
0 0.59 −0.30 0 0 0 0 0 0
0 0 0.44 −0.63 0 0 0 0 0
0 0 0 0.02 −0.05 0 0 0 0
0 0 0 0 0.62 −0.05 0 0 0
0 0 0 0 0 0.62 −0.37 0 0
0 0 0 0 0 0 0.41 −0.09 0
0 0 0 0 0 0 0 0.41 −0.64
is for differential equation

23. NUMERICAL SIMULATION
Average number of cars I will meet on the road at different starting time
This number is the number of cars can impact me during my driving.
6 8 10 12 14 16 18 20
30
40
50
60
70
80
90
100
110
120
130
Starting time
Averagenumberofcarswillbemet
Difference
Avarage of 2
Differential

24. NUMERICAL SIMULATION
The relation between time cost(without traffic light) and number of cars who can
impact my driving.
By recording data, transfer the starting time to corresponding average number of
cars met on the road.
40 50 60 70 80 90 100
8
9
10
11
12
13
14
15
Average number of cars met
Acturallycostoftime
t vs. sp
untitled fit 1

25. NUMERICAL SOLUTION
Starting time vs. cost time
Cost time = cost time (without traffic light) + random number from(0,3)* 1/3 seconds
6 8 10 12 14 16 18 20
9
9.5
10
10.5
11
11.5
12
12.5
13
13.5
Starting time
Timeofdrivingwithouttrafficlightsimpact

26. CONCLUSION
1. Different modelling method can provide different result.
2. A simple real life problem is harder than academic example.
3. Improvement :
1. Analyze more (5 by 5)
2. Change constant parameter to function respect to time.
3. PDE modelling

27. THANK YOU

28. QUESTIONS?