,A ship has two cargo holds, one fore and one aft. The fore cargo hold has a wei

Question

,A ship has two cargo holds, one fore and one aft. The fore cargo hold has a weight capacity of $7,000 tons and a volume capacity of 224,000 cubic feet. The aft hold has a weight capacity of 9,000 tons and a volume capacity of 288,000 cubic feet. The ship has to be balanced, therefore the two cargo holds weights have to be within 15 percent of each other,i.e. the aft cannot be heavier than the fore by more than 15 percent and vice versa.

The shipowner has contracted to carry bales of cotton and ingots of aluminum. Each bale of cotton weights 500 lbs and each ingot of aluminum is 20 lbs. One ton is equivalent to 2,000 lbs. and therefore a ton is 4 bales of cotton or 100 ingots of aluminum.

The total weight of the available cotton is 8,000 tons; the total weight of the available aluminum is 10,000 tons. The volume per mass of cotton is 1 ton of cotton takes 32 cubic feet, and the volume per mass for aluminum is one ton of aluminum takes 25 cubic feet.

The profit for transporting cargo is fixed $0.035 per pound, regardless of the commodity transported. The ship owner is free to accept all or part of the available cargo; he wants to know how much cotton and aluminium to accept in order to maximize profit.

Formulate as a linear programming problem.

Explanation / Answer

Let x1 and x2 be the number of balses of cotton in fore and aft respectively and y1 and y2 be the number of ingots of aluminium in fore and aft respectivly
Minimize
((x1+x2)/4+(y1+y2)/100)*2000*0.035
x1/4+y1/100<=7000
x2/4+y2/100<=9000
x1/4+y1/100<=1.15*(x2/4+y2/100)
x2/4+y2/100<=1.15*(x1/4+y1/100)
x1+x2<=8000*4
y1+y2<=10000*100

32*x1/4+25*y1/100<=224000
32*x2/4+25*y2/100<=288000

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