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PHYS5032 Techniques For Sustainability Analysis

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  • Course Code: PHYS5032
  • University: The University Of Sydney
  • Country: Australia


Part A: Passengers arriving at the platform
Assume that the platform is completely empty at the beginning of the observed timeframe. The modelers have measured the influx of people onto the platform in people per second. They observed that as the platform fills up, the rate of people entering the platform decreases (due to overcrowding). For peak use times (mornings and evenings), the modelers have concluded that the influx of people ?(?) (number of people arriving on the platform per second) can be described by the function ?(?) = 1
2 √?
Note that this function measures the people that arrive at the station at time ? in persons per second. The variable ? is given in seconds.
1. (A) (4 points) Calculate amount of people that are on the platform after five minutes. For this, assume that no people are on the platform at the beginning of the timeframe, and that the influx of people is perfectly described by the above function. Write a short paragraph explaining why you chose mathematical integration or numerical integration.
2. (A) (4 points) If you chose mathematical integration in the first question, use numerical integration with a step length of 1 second and compare the results. What is the percentage deviation between the mathematical and the numerical solution (do not submit detailed calculations for the numerical integration results, only submit the final result of the numerical integration and the percentage deviation).
Part B: Passengers leaving the platform by train
The station design is intended to allow one train to leave every five minutes which should transport all passengers that are waiting on the platform. As the train pulls up and opens its doors, the number of passengers that enter the train is modeled in two different steps. The rate of passengers entering the train is again measured in persons per second.
a) As the train opens its doors and the passengers enter, the rate of passengers entering the train increases for the first 6 seconds and is described by ?(?) = 10?* ∗ 1

b) After 6 seconds the level of people getting on the train stays constant at the level that is has reached at 6 seconds according to the function in a).
3. (A) (3 points) How many people are getting on the train within the first 6 seconds of the train being at the platform?
4. (A) (3 points) The train is scheduled to remain at the platform for 15 seconds in total. How many people can get on the train within this time? Is this stopping time long enough for all people that are waiting on the platform to get on the train?
5. (A) (2 points) How long does the train have to stay at the platform for all people to enter the train? Remember to round your results to integer numbers.
6. In your own words (A) (4 points) For what effects is integration the right tool to carry out your analysis? In your response, it is sufficient to discuss phenomena that span over a certain time frame (similar to what we discussed in the lecture). You can also use different examples if you wish, but this is not necessary. 



Home worksheet 4:1 

?(?) = √?

The reason I chose Mathematical integration is that it provides accurate results

We will integrate this to obtain the number of people that have arrived after 5 seconds is given by:

F(t)     =√?

= 300

=1/3* 300^1.5

=1732 ~1733 people 

The reason I chose Mathematical integration is that it provides accurate results compared to Numerical integration. Even though Numerical integration can integrate everything, it only gives approximation which sometimes is always slightly wrong answers. 


This is the formula for obtaining numerical integration


Where our s = 1

and n-1=300 seconds

We will use Excel to compute this

We obtain 1731.698 ~ 1732 people 

Percentage deviation from Mathematical Integration and Numerical Integration is obtained by:

= (1733-1732)/1733*100% = 0.0577% 


f(t) = 10t3e-t/2

F(t) =

We will use integration by parts to tackle this problem 

udv = uv-

let u = 10t^3 and dv = e^-t/2dt

du= 30t^2 dt and V= -1/2e^-t/2 

10t3e-t/2dt = -5t^3e^-t/2 - -15t^2e^-t/2dt


 = -5t^3e^-t/2 + 15t^2e^-t/2

=e^-t/2(-5t^3+ 15t^2)

People getting into the train after 6 seconds 

= e^-3(-5*6^3+ 15*6^2)

= 26 people 

For the first seconds =26

Remaining 9 seconds = 26*9=234

Total =260 people 

Total people 1733

After 6 seconds 26 people enter the train

People remaining = 1733-26 = 1707

Time taken = 1707/26 = 67 seconds

Time that the train will have to wait = 67 + 26 =93 seconds ~ 1 min 33 seconds

Despite the fact that Mathematical integration is hard to do, It's still the best method to tackle many Mathematical problems because it's so accurate. As we have seen in the above example it produces the best answer compared to the numerical integration . For example if you were to obtain a speed of a vehicle given an equation of the distance, it would be better to use Mathematical integration to tackle this problem.

Homework Sheet 5: Structural Path Analysis 

Cash spent at the pub = cash spent per hour at the pub * hours at the pub * how often you go out

In 2019

2000 =20 *5*20 = 20*100

In 2017

3000 = 50*4*15=50*60 

1st Option

(50-20)*60+ 20(60-100)

1800 +(-800)

= 1000 

2nd Option

(50-20*100+ 50(-40)

3000 +(-2000)


The estimate is accurate and therefore, the gap between the contributors is zero 

Average = (1800+3000)/2

    = 2700


(50+20)/2 = 35 

Total 2700+35 = 2735

 The difference between the contributors is not smaller but bigger than the difference between the pub expenditure. 


Structural Decomposition Analysis 6: Structural Path Analysis 


The CO2 footprint that my company causes are correlated with the level of contributors involved in the process. During the supply chain, the contributor with uppermost correlation becomes the highest while other contributors follow in the same manner. We will use this concept to arrange the top ten most intensive nodes. 

Assume a matrix

And q = (0:8 2:5 4 5) 

The total footprint is 1088 t CO2

This is the supply chain G  S C  C 

0.2$ *100 = 20$ are required in the second node of the path. (tier-1)

0.3$ * 20 = 6$ are required in the third node (tier-2)

0.5 $ * 6 = 3$ are required in the fourth node (tier-3)

0.2$ * 3 = 0.6$ are required in the fifth node (tier -4)

0.6$ * 5 =3t CO2

The path translates to = 3/1088*100 = 0.275% 

  1. (A) Top 10 paths in descending order

Solution: Check the graph

0.2$ *100 = 20$

0.3$ * 20 = 6$

0.5 $ * 6 = 3$

0.2$ * 3 = 0.6

0.6$* 0.1=0.06$

0.1$ * 0.06 =0.006$

0.3 $ 0.006 =0.0018$

0.4 $ * 0.0018 = 0.00072$

0.2$ * 0.00072 = 0.000144

0.2$ * 0.000144 =0.000288$

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