Dafoe, The Lone Entrepreneur is a very ambitious man. He wants to make $1000 in the shortest amount of time! He starts out with $0 but is currently receiving $5 per day and he can purchase the following shops at any given time:

1-Shack. Cost: $10 Income: $5/day
2-Mall. Cost $25 Income $15/day
3-Shopping Center. Cost $150 Income $150/day

The money is always paid in at night and Dafoe can purchase as many new shops as his money allows in the mornings.

What would be the best sequence of shops to be purchased to to have a balance of $1000 on hand?

How many days would that take?

Money spent comes out of his balance and so is no longer part of the needed total balance of $1000.

Modifying the program to allow multiple purchases of possibly different types of property on a given day, we can't allow perpetual non-purchase, so the choices programmed were to buy as many shopping centers as possible, then, from half as many to as many malls as possible, then from half as many to as many shacks as possible, given what's left, we still do not get fewer than 13 days, but we maximize the amount of money, and give Dafoe lots of properties for future earning:

day $ at shk ma sc shk ma sc    $
    start  bought     held    remaining
 1    5         0   1  0  0    5
 2   10   1 00 00   2  0  0    0
 3   10   1 00 00   3  0  0    0
 4   15   1 00 00   4  0  0    5
 5   25      1 00   4  1  0    0
 6   35   1 01 00   5  2  0    0
 7   55      2 00   5  4  0    5
 8   90   3 01 00   8  5  0   35
 9  150         1   8  5  1    0
10  265   1 04 01   9  9  2    5
11  485      1 03   9 10  5   10
12  955   1 01 06  10 11 11   20
13 1885         0  10 11 11 1885

The blank-vs-zero difference is a result of using one displayed number represent all three values.

DECLARE SUB addOn ()
DEFDBL A-Z
CLEAR , , 10000
DIM SHARED lowest, shacks, malls, centers, amount, best
DIM SHARED p(40), dayNo, cost(3), hist(40), hs(40), hm(40), hc(40), hist2(40)
DATA 10,25,150
FOR i = 1 TO 3: READ cost(i): NEXT

lowest = 13
shacks = 1

addOn

SUB addOn
  STATIC ct

  dayNo = dayNo + 1
  aSave = amount
  amount = amount + shacks * 5 + malls * 15 + centers * 150
  hist(dayNo) = amount
  hist2(dayNo) = amount
  IF amount >= 1000 AND (dayNo < lowest OR dayNo = lowest AND amount > best) THEN
    lowest = dayNo
    best = amount
    ct = ct + 1
    p(dayNo) = choice
    hs(dayNo) = shacks
    hm(dayNo) = malls
    hc(dayNo) = centers
    PRINT dayNo
    FOR i = 1 TO dayNo
      PRINT USING "## #### ###### ## ## ## ####"; i; hist(i); p(i); hs(i); hm(i); hc(i); hist2(i)
    NEXT
    PRINT ct
  ELSE
   IF dayNo < lowest THEN
     pce = INT(amount / cost(3))
     amt2 = amount - pce * cost(3)
     FOR pma = INT(amt2 / (2 * cost(2))) TO amt2 / cost(2)
       amt3 = amt2 - pma * cost(2)
       FOR psh = INT(amt3 / (2 * cost(1))) TO amt3 / cost(1)
        amount = amt3 - psh * cost(1)
        hist2(dayNo) = amount
        shacks = shacks + psh
        malls = malls + pma
        centers = centers + pce
        hs(dayNo) = shacks
        hm(dayNo) = malls
        hc(dayNo) = centers
        p(dayNo) = psh * 10000 + pma * 100 + pce

        addOn

        shacks = shacks - psh
        malls = malls - pma
        centers = centers - pce
        amount = amt3 + psh * cost(1)
       NEXT psh
     NEXT pma
   END IF
  END IF
  amount = aSave
  dayNo = dayNo - 1
END SUB

 

Posted by Charlie on 2006-01-31 14:34:59