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Apr 17th, 2007, 09:55 AM
#1
Thread Starter
New Member
Another linear programming problem
The optimum daily intake of vitamins is 10 units of b1, 8 units of b2 and 12 units of b12.
But the vitamins cannot be obtained in pure form.
VIT tablets are £1.00 and each contain 1 b1, 1 b2, and 1 b12.
AL tablets cost £2.10 and each contain 3, 2, and 4 units of b1, b2, and b12.
ITY tablets cost £1.50 and each contain 3, 2 and 0 units og b1, b2 and b12.
I need to set up a linear program and then find the dual problem and solve this using the simplex algorithm. I have to find the daily recommend intake at minimum cost.
So i have done this:
Let x be the number of VIT tablets needed.
Let y be the number of AL tablets needed.
Let z be the number of ITY tablets needed.
Cost per VIT tablet = £1.00/100 = £0.01
Cost per AL tablet = £2.10/100 = £0.021
Cost per ITY tablet = £1.50/100 = £0.015
Objective function = 0.01x + 0.021y + 0.015z
x, y, z ≥ 0
number of units of b1 = 1x + 3y + 3z ≤ 10
number of units of b2 = 1x + 2y + 2z ≤ 8
number of units of b12 = 1x + 4y + 0z ≤ 12
[COULD BE THIS INSTEAD]
number of units of b1 = 1x + 3y + 3z ≥ 10
number of units of b2 = 1x + 2y + 2z ≥ 8
number of units of b12 = 1x + 4y + 0z ≥ 12
[COULD BE THIS INSTEAD]
Linear Programme
Minimise C= 0.01x + 0.021y + 0.015z
Subject to:
1x + 3y + 3z ≤ 10
1x + 2y + 2z ≤ 8
1x + 4y≤ 12
x, y, z ≥ 0
The Dual Problem:
Add in the variable u, w, v ≥ 0
1ux + 3uy + 3uz ≤ 10u
1wx + 2wy + 2wz ≤ 8w
1vx + 4vy ≤ 12v
(1u + 1w + 1v)x + (3u + 2w + 4v)u + (3u + 2w)z ≤ (10u + 8w + 12v)
Maximise: 10u + 8w + 12v
Subject to:
1u + 1w + 1v ≤ 0.01
3u + 2w + 4v ≤ 0.021
3u + 2w ≤ 0.015
but I am incountering problems when it come to the simplex algorithm as you cannot divide anything by 0.
Is there a chance the problem has been set up wrong.
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Apr 20th, 2007, 04:45 AM
#2
Addicted Member
Re: Another linear programming problem
According to the Simplex Method Tool (http://people.hofstra.edu/faculty/St...d/simplex.html)
Minimize p = 0.01x + 0.021y + 0.015z subject to
1x + 3y + 3z = 10
1x + 2y + 2z = 8
1x + 4y + 0z = 12
x >= 0
y >= 0
z >= 0
Optimal Solution: p = 41/500; x = 4, y = 2, z = 0
...este projecto dos Deuses que os homens teimam em arruinar...
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Apr 20th, 2007, 07:27 AM
#3
Thread Starter
New Member
Re: Another linear programming problem
But thats the original not the dual I have to solve the dual problem.
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