Question

6. Recall question 3a from Homework 2: Albert Cornman makes his living buying and selling corn. On January 1 he has 50 tons of corn and $15,000 cash in his bank account. The bank pays an interest of 0.5% per month for the money in Albert's account. On the first day of each month Albert can buy corn at the following prices per ton: Jan $300, Feb $350, Mar $350, Apr $500, May $500, Jun $500. On the last day of each month he can sell corn at the following prices per ton: Jan $250, Feb $400, Mar $350, Apr $550, May $550, Jun $600. Albert stores his corn in a warehouse that can hold at most 100 tons of corn. The lease on the warehouse expires at the end of June; hence he must sell all of his corn by the end of June. Each month corn loses 4% of its weight; that is, for instance, 50 tons of corn he owns on Jan 1, weighs 0.96 * 50 = 48 tons on the last day of Jan (and on the first day of Feb, for the purpose of modeling). Albert must pay cash for all corn at the time of purchase. Write a linear programming model to maximize Albert Cornman's cash on hand at the end of June. Solve the problem in AMPL. What is the optimal solution and the optimal objective value? Print and attached your AMPL code.

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(a) Write a linear programming model to maximize Albert Cornman's cash on hand at the end of June. Solution: Note that there are many different ways to model the situation, depending on your assumptions. One of the possible ways is presented here. Let us define the following notation. bt = Amount of corn bought at the beginning of month t = 1,2,3,4,5,6. St = Amount of corn sold at the end of month t = 1,2,3,4,5,6. it = inventory at the beginning of month t just before buying, t = 1,2,3,4,5,6,7. m+ = money at the beginning of month (before increase / decrease of loan and buying), t = 1,2,3,4,5,6,7. Objective is to maximize the money at the end of month 6. This can be written as: max m7 We now need to write the inventory balance constraints. Since we buy corn at the beginning of the month, it gets stored for one month, and so the weight loss happens for the amount it + 2+ for every month. Then, ii i2 iz (1) (2) (3) == 50 0.96(61 + ii) - $1 0.96(b2 + i2) - $2 0.96(b3 + i3) - $3 0.96(64 + 14) - 84 0.96(65 + is) - 55 0.96(66 +16) - S6 15 (5) (6) 16 = i7 (7) 17 = 0 (8) Where the last constraint enforces that we have no corn remaining at the end of june. Note also that the above constraints (and the nonegativity) imply that we can't sell more corn than the amount we have.

The following constraints enforce the warehouse capacity. ii + bi < 100 i2 + b2 < 100 i3 + b3 < 100 i4 + b4 < 100 i5 + 65 < 100 i6 + b6 < 100 (9) (10) (11) (12) (13) (14) The following constraints enforce that we cannot buy more corn than the corn we can afford. 300b1 < mi 350b2 < m2 35063 < m3 50064 < m4 50065 m5 50056 <m6 (15) (16) (17) (18) (19) (20) Finally, we have the following cash-flow constraints. We assume that all the revenue is deposited in the bank. mi m2 m3 m4 15000 1.005(mi - 30061) + 250sı 1.005(m2 - 350b2) + 40082 1.005(m3 - 35063) + 35083 1.005(m4 + -500b4) + 55054 1.005(m5 + - 50065) + 55085 1.005(m6 - 50066) + 60086 (21) (22) (23) (24) (25) (26) (27) m5 = mo m7 The objective function would be to maximize my subject to all of the above constraints, along with non-negativities for all variables.

(b) Solve part (a) in AMPL. What is the optimal solution and the optimal objective value? Print and attached your AMPL code. 100 Solution: The optimal objective value is: 63107. The optimal solution is: 1 'sellAmount [1]' 0 2 'sellAmount [2]' 92.16 3 sellAmount [3]' 0 4 'sellAmount [4]' 95.7568 5 'sellAmount [5] ' 96 6 'sellAmount [6]' 96 7 'buy Amount [1]' 50 8 'buy Amount [2]' o 9 'buy Amount [3] 100 10 'buy Amount [4]' 3.74664 11 'buy Amount [5]' 12 'buy Amount [6]' 100 13 'money [1]' 15000 14 'money [2]' 0 15 ’money [3]' 36864 16 'money [4] 1873.32 17 'money [5]' 52666.2 18 'money [6]' 55479.6 19 'money [7]' 63107 20 'inventory [1] 50 21 'inventory [2] 96 22 'inventory [3] 0 23 'inventory [4] 96 24 'inventory [5] ' 0 25 'inventory [6]' 0 26 'inventory [7]' 0