For Round 1’s algo portion, it’s identical to last year’s, so I reused my 2023 code. Interested friends can refer to last year’s posts.
Manual Part Analysis:
There’s a group of fish selling items with reserve prices uniformly distributed between 900 and 1000:
⦁ Minimum probability at 900
⦁ Maximum probability at 1000
⦁ Linear probability distribution in between
We need to set two bid prices. Any fish with reserve price below our bids will sell to us at the lowest bid that exceeds their reserve price. We can then resell all acquired items at 1000.
Example:
If I bid 930 and 970:
⦁ Fish with reserves 900-929 sell at 930
⦁ Fish with reserves 930-969 sell at 970
⦁ Fish with reserves 970-1000 don’t trade
Mathematical Optimization:
Assuming fish quantity follows an arithmetic sequence:
⦁ 1 fish at 900
⦁ 2 fish at 901
⦁ …
⦁ 101 fish at 1000
For lower bid X:
Number of willing fish = Σ from 900 to X-1 = (X-900)(X-901)/2
Profit E1 = (X-900)(X-901)/2 × (1000-X)
For higher bid Y:
Number of willing fish = Σ from X to Y-1 = (X+Y-1801)(Y-X)/2
Profit E2 = (X+Y-1801)(Y-X)/2 × (1000-Y)
Total profit E = E1 + E2
Maximizing E requires solving ∂E/∂X = 0 and ∂E/∂Y = 0
Optimal solution found at X≈952, Y≈978
