101 lines
3 KiB
Python
101 lines
3 KiB
Python
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"""
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fish.py
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~~~~~~~
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This module solves the fish homework problem.
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Run with "python fish.py"
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Description:
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Every year (for 3 years) we decide whether to fish in a lake or not.
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If we fish, the population of fish in the lake remains the same.
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If we don’t fish, the population of fish in the lake doubles.
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Years of experience tell us that our profit is exactly 70% of the
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number of fish in the lake if we fish.
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If there are 10 fish in the lake, we make $7.
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The interest rate is 25%
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We need a contingency plan: no matter what situation we find
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ourselves in, we need to have an optimal strategy.
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We could try every strategy, but that quickly gets out of control:
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there are too many possibilities!
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Output:
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{0: {10: {'fish': False, 'value': 32.25600000000001}},
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1: {10: {'fish': False, 'value': 20.160000000000004},
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20: {'fish': False, 'value': 40.32000000000001}},
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2: {10: {'fish': True, 'value': 12.600000000000001},
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20: {'fish': True, 'value': 25.200000000000003},
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40: {'fish': True, 'value': 50.400000000000006}},
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3: {10: {'fish': True, 'value': 7.0},
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20: {'fish': True, 'value': 14.0},
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40: {'fish': True, 'value': 28.0},
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80: {'fish': True, 'value': 56.0}},
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4: {10: {'value': 0},
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20: {'value': 0},
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40: {'value': 0},
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80: {'value': 0},
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160: {'value': 0}}}
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Optimal strategy: do not fish -> do not fish -> fish -> fish
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"""
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from copy import deepcopy
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from pprint import pprint
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# Problem parameters
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initial_population = 10
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growth_factor = 2
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profit_rate = 0.7
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interest_rate = 0.25
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final_value = 0
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n_periods = 4
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# Initialize data structures to keep track of calculations
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tree = {
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0: {
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initial_population: {},
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},
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}
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max_population = initial_population
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# Construct the tree by forward calculation
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for i in range(n_periods):
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tree[i+1] = deepcopy(tree[i])
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max_population *= growth_factor
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tree[i+1][max_population] = {}
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# Set the values for all nodes in the last period to 0
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for key in tree[n_periods].keys():
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tree[n_periods][key]['value'] = 0
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# Value the tree by backward induction
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for i in reversed(range(n_periods)):
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for key in tree[i].keys():
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fish = profit_rate * key + (1 / (1 + interest_rate)) * tree[i+1][key]['value']
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no_fish = (1 / (1 + interest_rate)) * tree[i+1][key * growth_factor]['value']
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tree[i][key]['value'] = max(fish, no_fish)
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if fish > no_fish:
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tree[i][key]['fish'] = True
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else:
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tree[i][key]['fish'] = False
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# Print the decision rule
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result = []
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optimal_population = initial_population
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for i in range(n_periods):
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if tree[i][optimal_population]['fish']:
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result.append('fish')
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else:
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result.append('do not fish')
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optimal_population *= growth_factor
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pprint(tree)
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print('Optimal strategy: ' + ' -> '.join(result))
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