Brain-Teaser 350: Catering crisis
From The Sunday Times, 21st January 1968 [link]
The caterer at our Village Hall is in a quandary, as the Hall has been double-booked for next Saturday — by the Cricket Club and the Darts Club.
Unfortunately the matter cannot be resolved until the return of the Vicar on Saturday morning. He is quite unpredictable in such decisions, and the caterer must order the food by Friday.
The Cricketers want 100 sausage rolls and 60 meat pies; the Darts Club wants 50 sausage rolls and 90 meat pies.
The caterer is empowered to spend exactly £6.00. He buys sausage rolls at 3p and sells at 4p; meat pies cost him 5p and sell at 8p. Any left-overs are disposed of at a loss to a local prep school, which pays 2p per sausage roll and 3p a pie.
What should the caterer order so that he makes the best safe profit no matter which club gets the booking? And what will that profit be?
This puzzle is included in the book Sunday Times Brain Teasers (1974). The puzzle text is taken from the book.
[teaser350]
Frits, and 
Lise Andreasen are discussing. 
S2T2 
Jim Randell 8:59 am on 21 May 2024 Permalink |
This Python program looks at possible orders of sausage rolls and pies that cost the caterer exactly 600p, and then the amount made if these supplies are available for each club, and records the minimum profit made in each case. We then look for the largest of these profits to find the answer to the puzzle.
It runs in 58ms. (Internal runtime is 240µs).
Run: [ @replit ]
from enigma import (Accumulator, express, printf) # amount earned for supply <s>, demand <d>, price <p>, disposal price <x> def amount(s, d, p, x): return (d * p + (s - d) * x if s > d else s * p) # calculate amount earned for a supply (s, p) and a demand (S, P) # of sausage rolls and pies def earned(s, p, S, P): return amount(s, S, 4, 2) + amount(p, P, 8, 3) # find the maximum profit r = Accumulator(fn=max, collect=1) # the caterer spends 600p for (s, p) in express(600, [3, 5]): # calculate amount earned from each club C = earned(s, p, 100, 60) D = earned(s, p, 50, 90) # record profit r.accumulate_data(min(C, D), (s, p, C - 600, D - 600)) # output solution for (s, p, C, D) in r.data: printf("{s} sausages rolls + {p} pies -> cricket = {C} / darts = {D}")Solution: The caterer should order 80 sausage rolls and 72 pies. Whichever club arrives he will make a profit of 236p.
In the originally published puzzle pre-decimal currency was used and the numbers were different:
And the answer is: 160 sausage rolls, 144 pies. The profit is 472 pence = £1, 19s, 4d.
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Frits 10:35 am on 21 May 2024 Permalink |
This program doesn’t assume that the caterer will spend exactly 600p.
from enigma import SubstitutedExpression, Accumulator # invalid digit / symbol assignments d2i = dict() # try to spend close to 600p in order to maximize the profit for dgt in range(0, 101): vs = set() if dgt < 50: vs.update('S') if dgt < 60: vs.update('P') if dgt > 90: vs.update('P') d2i[dgt] = vs # the alphametic puzzle p = SubstitutedExpression( [ # expenditure close to the maximum "595 < S * 3 + P * 5 <= 600", # calculate profits for cricketers and darters "(pc := S + min(P, 60) * 3 - (P - 60) * 2 * (P > 60))", "(pd := min(S, 50) + P * 3 - (S - 50) * 1 * (S > 50))", ], answer="min(pc, pd), S, P", base=101, d2i=d2i, distinct="", reorder=0, verbose=0, # use 256 to see the generated code ) # find the maximum profit r = Accumulator(fn=max).accumulate_from(ans for (_, ans) in p.solve()) # output solution w, s, p = r.value print(f"he should order {s} sausages and {p} pies for a profit of {w}p")LikeLike
Frits 10:32 am on 22 May 2024 Permalink |
Assuming that the caterer will spend exactly 600p and that the caterer will order a sensible amount of sausages S (50, 55, 60, …, 95, 100) the profit (if the cricketers get the booking) can be expressed as: 2.2 * S + 60 and 460 – 2.8 * S if the darters get the booking.
As the first function is increasing and the latter is decreasing for S the safe profit will be maximized if 2.2 * S + 60 = 460 – 2.8 * S or S = 80.
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Lise Andreasen 2:17 am on 22 May 2024 Permalink |
Interesting.
I approached it differently and arrived at 100 sausage rolls and a profit of 280.
https://onlinephp.io/c/47cce
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Lise Andreasen 2:23 am on 22 May 2024 Permalink |
Oh, hang on. Read the puzzle incorrectly.
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Lise Andreasen 2:25 am on 22 May 2024 Permalink |
Yeah. I get the correct result now.
https://onlinephp.io/c/37baa
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Lise Andreasen 2:27 am on 22 May 2024 Permalink |
Read the numbers in an early version of this page?
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Jim Randell 12:44 pm on 24 May 2024 Permalink |
Sorry. For a short while the numbers were mixed up between the original puzzle and the version published in the book. The puzzle text now reflects the version from the book, rather than the originally published version.
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