Solutions to Assignments

1) World Oil Reserves vs. Consumption

  1. Which share of countries consume 75% of oil_consumption()?

    >>> Effects(oil_consumption()).attain_effects(3/4)
0.167

  2. How much of co2() emissions account the biggest five emitting countries for?

    >>> Effects(co2()).attain_causes(5/len(co2()))
0.606

  3. Which number of countries is required to attain 75% of oil_reserves()?

    >>> Effects(oil_reserves()).attain_effects(0.75)*len(oil_reserves())
7.992

  4. How many years do 10% of the oil_reserves() last for 10% of oil_consumption() (note: reserves are in million barrels per year, consumption in thousand barrels per day)?

    >>> (Effects(oil_reserves()).attain_causes(0.1)*sum(oil_reserves()))/((Effects(oil_consumption()).attain_causes(0.1)*sum(oil_consumption()))/1000/365)
3295.5143003164517

2) Euro Coins & Bank Notes

  1. Is a pareto distribution present for euro_banknotes()?

    >>> Effects(euro_banknotes()).pareto
True

  2. If, which distribution is stronger, that for the euro_coins() or for the euro_banknotes()?

>>> coins = Effects(euro_coins())
>>> notes = Effects(euro_banknotes())
>>> if notes.pareto:
...    print(notes.ratio < coins.ratio)
False

As the ratio of euro_banknotes() is less than of the euro_coins(), the coins have the stronger pareto distribution. Remember: As the ratio depends on the entropy, a smaller ratio means less entropy, less entropy means stronger pareto distribution.

  1. How much Euros do the largest 10% of euro_banknotes() represent?

    >>> notes.attain_causes(0.1)*sum(euro_banknotes())/1000
461.80144999999993
462 billion Euros.

  1. How much Euros do the largest 50% of euro_coins() represent?

    >>> coins.attain_causes(0.5)*sum(euro_coins())/1000
23.701013600000003
24 billion Euros.

3) Swiss Railway Passenger Frequency

  1. Suppose Swiss railways classifies railway stations into A (60% of passenger volume), B (60 to 80%) and C (80 to 100% of volume). Which share of causes would each class represent?

    >>> class_a = passengers.attain_effects(0.6)                 # 0 to 60%
>>> class_b = passengers.attain_effects(0.8)-class_a         # 60 to 80%
>>> class_c = passengers.attain_effects(1.0)-class_a-class_b # 80 to 100%
>>> print('A: {}, B: {}, C: {}'.format(class_a, class_b, class_c))
A: 0.051, B: 0.093, C: 0.8119999999999999

    0 to 5% of stations make up 60% of passengers, the next 9.3% (5.1 to 14.4%) another 20% and the final 81% make up for only 20% of passengers.

  2. Following your results from 1., how many stations fall into the categories A, B and C?

    >>> [cls*len(swiss_rail_passengers()) for cls in [class_a, class_b, class_c]]
[37.647999999999996, 66.608, 619.744]

    A holds 37 stations, B 67 and C 588.

  3. Determine the corresponding threshold values of passenger frequency enabling you to allocate each single station to class A, B or C.

    >>> [passengers.separate_effects(threshold) for threshold in [0.6, 0.8]]
[(19200.0, 1, 1), (6000.0, 5, 5)]

    A station belongs to class A if it has 19200 or more passengers per work day. It falls into class B if it is the fifth of the five stations with 6000 passengers per day. Otherwise it belongs to class C.

4) Exoplanets

  1. How many exoplanets fall into the interval of 30 to 60% of total mass?

    >>> masses.interval_effects(0.3, 0.6)*len(exoplanets())
33.39
33 exoplanets fall into the interval of 30 to 60% of total mass.
  1. Which exoplanets are this namely?
>>> list(keys_in_effects_interval(eplanets, 'mass', 0.3, 0.6))
['2M 0122-24 b',
 '2M 2140+16 b',
 'Kepler-57 b',
 'Kepler-53 b',
 'HD 41004 B b',
 'Kepler-39 b',
 'CT Cha b',
 'HN Peg b',
 'Kepler-53 c',
 '1RXS 1609 b',
 'kappa And b',
 '2M 0219-39 b',
 'Kepler-27 c',
 'WISE 0458+6434 b',
 'XO-3 b',
 'Ross 458 (AB) c',
 'GU Psc b',
 'WASP-18 b',
 'Kepler-52 c',
 'CoRoT-27 b',
 'HR 8799 c',
 'HR 8799 d',
 'Kepler-448 b',
 'Kepler-75 b',
 'SWEEPS-11 ',
 'Kepler-27 b',
 'ROXs 42B b',
 'HAT-P-2 b',
 'Kepler-52 b',
 'Kepler-14 b',
 'CoRoT-14 b',
 'WASP-14 b']
  1. Find out the exoplanets that fall into the interval of 70 to 80% of total mass but not into the interval of 10 to 90% of volume.
>>> set(keys_in_effects_interval(eplanets, 'mass', 0.7, 0.8)).difference(set(keys_in_effects_interval(eplanets, 'volume', 0.1, 0.9)))
{'CoRoT-10 b',
 'CoRoT-17 b',
 'HATS-22 b',
 'HD 118203 b',
 'K2-33 b',
 'Kepler-419 b',
 'Kepler-58 c',
 'Kepler-59 b',
 'POTS-1 b',
 'WASP-43 b',
 'WASP-8 b'}

5) Fibonacci Sequences

  1. Check whether a pareto distribution can be observed for fibonacci series. If, all or just from/to a specific fibonacci series?

    >>> [Effects(fib_list(n)) for n in range(5, 20)]
[<pareto not present [1.04]: 1/1 causes => 1/1 effects [total ∆: 0.0 % points]>,
 <pareto not present [1.01]: 1/3 causes => 2/3 effects [total ∆: 0.0 % points]>,
 <pareto present [0.977]: 3/10 causes => 2/3 effects [total ∆: 3.1 % points]>,
 <pareto present [0.94]: 1/4 causes => 2/3 effects [total ∆: 3.0 % points]>,
 <pareto present [0.905]: 1/3 causes => 4/5 effects [total ∆: 2.2 % points]>,
 <pareto present [0.872]: 1/5 causes => 3/5 effects [total ∆: 2.5 % points]>,
 <pareto present [0.842]: 1/1 causes => 1/1 effects [total ∆: 0.0 % points]>,
 <pareto present [0.814]: 1/4 causes => 4/5 effects [total ∆: 3.3 % points]>,
 <pareto present [0.789]: 3/10 causes => 9/10 effects [total ∆: 5.1 % points]>,
 <pareto present [0.766]: 3/10 causes => 9/10 effects [total ∆: 5.9 % points]>,
 <pareto present [0.745]: 1/5 causes => 4/5 effects [total ∆: 3.5 % points]>,
 <pareto present [0.726]: 3/10 causes => 9/10 effects [total ∆: 2.3 % points]>,
 <pareto present [0.709]: 3/10 causes => 9/10 effects [total ∆: 1.6 % points]>,
 <pareto present [0.693]: 1/3 causes => 9/10 effects [total ∆: 4.4 % points]>,
 <pareto present [0.679]: 1/1 causes => 1/1 effects [total ∆: 0.0 % points]>]
A pareto distribution is present from the seventh fibonacci sequence on, that is [0, 1, 1, 2, 3, 5, 8].