sci.op-research

This is the Mathematica result, copied as plain text.

It's much better formatted in Mathematica. I've added some spaces to make
it a little more readable.



Minimize[{x+y,a*x+y >=4,b*x+y>=7,x>=0,y>=0},{x,y}]

{\[Piecewise]

(7/8) b==8&&a>=32/7

1 b==8&&a==4

(7/2) b==8&&a==8/7

4 (b==7/4&&a>=1)||(b==8&&a==1)||(b==1/2 (2+Sqrt[7])&&a==1)||(b==1/4
(5+Sqrt[13])&&a==1)||(b==1/2 (5+Sqrt[21])&&a==1)||(b==1/4
(7+Sqrt[21])&&a==1)||(b==1/8 (11+Sqrt[57])&&a==1)||(b==1/8
(11+Sqrt[93])&&a==1)||(b>8&&a==1)||(7/4(2+Sqrt[7])(11+Sqrt[57])&&a==1)||(1/2 (5+Sqrt[21])(5+Sqrt[21])&&a==1)||(1/8 (11+Sqrt[57])(11+Sqrt[93])
(35/8) b==8&&a==0

(40/7) b==7/4&&a==0

7 (b==1&&a>=4/7)||(b==1&&0
(4/a) (b==8&&1(2+Sqrt[7])&&1(5+Sqrt[21])&&1(11+Sqrt[57])&&1b)/7)||(b>8&&1(2+Sqrt[7])(11+Sqrt[57])&&1(7+Sqrt[21])(2+Sqrt[7])&&1b)/7)

((-35+7 a)/(-8+a)) (b==8&&0
((-10+7 a)/(-(7/4)+a)) (b==7/4&&0
((-3+7 a-4 b)/(a-b)) (b==1/2 (2+Sqrt[7])&&0(2+Sqrt[7])&&a<0)||(b==1/4 (5+Sqrt[13])&&0(5+Sqrt[13])&&a<0)||(b==1/2 (5+Sqrt[21])&&0(5+Sqrt[21])&&a<0)||(b==1/4 (7+Sqrt[21])&&0(7+Sqrt[21])&&a<0)||(b==1/8 (11+Sqrt[57])&&0(11+Sqrt[57])&&a<0)||(b==1/8 (11+Sqrt[93])&&0(11+Sqrt[93])&&a<0)||(b>8&&a==(4
b)/7)||(b>8&&08&&a<0)||(1b)/7)||(1(5+Sqrt[13])&&0(11+Sqrt[93])&&a==(4 b)/7)||(1/2 (2+Sqrt[7])(11+Sqrt[93])&&0(5+Sqrt[13])(11+Sqrt[57])&&0(5+Sqrt[21])(5+Sqrt[21])b)/7)||(1/4 (7+Sqrt[21])(5+Sqrt[21])&&a<0)||(1/8 (11+Sqrt[57])(11+Sqrt[57])(2+Sqrt[7])&&a<0)||(1/8 (11+Sqrt[93])(11+Sqrt[93])(7+Sqrt[21])&&a<0)

(7/b) (b==1/2 (2+Sqrt[7])&&a>(4 b)/7)||(b==1/4 (5+Sqrt[13])&&a>(4
b)/7)||(b==1/2 (5+Sqrt[21])&&a>(4 b)/7)||(b==1/4 (7+Sqrt[21])&&a>(4
b)/7)||(b==1/8 (11+Sqrt[57])&&a>(4 b)/7)||(b==1/8 (11+Sqrt[93])&&a>(4
b)/7)||(b>8&&a>(4 b)/7)||(1(4 b)/7)||(7/4(4
b)/7)||(1/2 (2+Sqrt[7])(4 b)/7)||(1/4
(5+Sqrt[13])(4 b)/7)||(1/2 (5+Sqrt[21])(4
b)/7)||(1/4 (7+Sqrt[21])(4 b)/7)||(1/8
(11+Sqrt[57])(4 b)/7)||(1/8 (11+Sqrt[93])(7+Sqrt[21])&&a>(4 b)/7)

-((-3-4 b)/b) True

,{x->\[Piecewise

]0 b<1

(3/8) b==8&&a==0

(7/8) b==8&&a>=32/7

1 b==8&&a==4

(12/7) b==7/4&&a==0

(31/14) b==8&&a==1

(7/2) b==8&&a==8/7

4 b==7/4&&a>=1

5 b==1&&a>=4/7

(4/a) (b==8&&1(2+Sqrt[7])&&1(5+Sqrt[21])&&1(11+Sqrt[57])&&1b)/7)||(b>8&&1(2+Sqrt[7])(11+Sqrt[57])&&1(7+Sqrt[21])(2+Sqrt[7])&&1b)/7)

1/8 (7-(-32+7 a)/(-8+a)) (b==8&&0
4/7 (7-(-7+7 a)/(-(7/4)+a)) (b==7/4&&0
7+1/2 (-7-(-4+7 a)/(-1+a)) b==1&&a<=0

7+1/2 (-4-(-4+7 a)/(-1+a)) b==1&&0
4-(7-4 b)/(2 (1-b)) (b==1/2 (2+Sqrt[7])&&a==1)||(b==1/4
(5+Sqrt[13])&&a==1)||(b==1/2 (5+Sqrt[21])&&a==1)||(b==1/4
(7+Sqrt[21])&&a==1)||(b==1/8 (11+Sqrt[57])&&a==1)||(b==1/8
(11+Sqrt[93])&&a==1)||(b>8&&a==1)||(7/4(2+Sqrt[7])(11+Sqrt[57])&&a==1)||(1/2 (5+Sqrt[21])(5+Sqrt[21])&&a==1)||(1/8 (11+Sqrt[57])(11+Sqrt[93])
((4-(7 a-4 b)/(a-b))/a) (b==1/2 (2+Sqrt[7])&&0(2+Sqrt[7])&&a<0)||(b==1/4 (5+Sqrt[13])&&0(5+Sqrt[13])&&a<0)||(b==1/2 (5+Sqrt[21])&&0(5+Sqrt[21])&&a<0)||(b==1/4 (7+Sqrt[21])&&0(7+Sqrt[21])&&a<0)||(b==1/8 (11+Sqrt[57])&&0(11+Sqrt[57])&&a<0)||(b==1/8 (11+Sqrt[93])&&0(11+Sqrt[93])&&a<0)

(3/b) (b==1/2 (2+Sqrt[7])&&a==0)||(b==1/4 (5+Sqrt[13])&&a==0)||(b==1/2
(5+Sqrt[21])&&a==0)||(b==1/4 (7+Sqrt[21])&&a==0)||(b==1/8
(11+Sqrt[57])&&a==0)||(b==1/8
(11+Sqrt[93])&&a==0)||(b>8&&a==0)||(1(5+Sqrt[13])&&a==0)||(1/2 (2+Sqrt[7])(5+Sqrt[13])(7+Sqrt[21])(2+Sqrt[7])&&a==0)||(1/8 (11+Sqrt[93])
(7/b) (b==1/2 (2+Sqrt[7])&&a>(4 b)/7)||(b==1/4 (5+Sqrt[13])&&a>(4
b)/7)||(b==1/2 (5+Sqrt[21])&&a>(4 b)/7)||(b==1/4 (7+Sqrt[21])&&a>(4
b)/7)||(b==1/8 (11+Sqrt[57])&&a>(4 b)/7)||(b==1/8 (11+Sqrt[93])&&a>(4
b)/7)||(b>8&&a==(4 b)/7)||(b>8&&a>(4 b)/7)||(1b)/7)||(1(4 b)/7)||(7/4b)/7)||(7/4(4 b)/7)||(1/2 (2+Sqrt[7])(11+Sqrt[93])&&a==(4 b)/7)||(1/2 (2+Sqrt[7])(4
b)/7)||(1/4 (5+Sqrt[13])(5+Sqrt[13])(4 b)/7)||(1/2 (5+Sqrt[21])b)/7)||(1/2 (5+Sqrt[21])(4 b)/7)||(1/4 (7+Sqrt[21])(5+Sqrt[21])&&a==(4 b)/7)||(1/4 (7+Sqrt[21])(4
b)/7)||(1/8 (11+Sqrt[57])(11+Sqrt[57])(4 b)/7)||(1/8 (11+Sqrt[93])(7+Sqrt[21])&&a==(4 b)/7)||(1/8 (11+Sqrt[93])(4 b)/7)

((7-(7 a-4 b)/(a-b))/b) True

,y->\[Piecewise]

0
(b==7/4&&a>=1)||(b==8&&a==8/7)||(b==8&&a==4)||(b==8&&a>=32/7)||(b==8&&1(2+Sqrt[7])&&a>(4 b)/7)||(b==1/2 (2+Sqrt[7])&&1(5+Sqrt[13])&&a>(4 b)/7)||(b==1/4 (5+Sqrt[13])&&1(5+Sqrt[21])&&a>(4 b)/7)||(b==1/2 (5+Sqrt[21])&&1(7+Sqrt[21])&&a>(4 b)/7)||(b==1/4 (7+Sqrt[21])&&1(11+Sqrt[57])&&a>(4 b)/7)||(b==1/8 (11+Sqrt[57])&&1(11+Sqrt[93])&&a>(4 b)/7)||(b==1/8 (11+Sqrt[93])&&18&&a==(4
b)/7)||(b>8&&a>(4 b)/7)||(b>8&&1b)/7)||(1(4 b)/7)||(7/4b)/7)||(7/4(4 b)/7)||(7/4b)/7)||(1/2 (2+Sqrt[7])(2+Sqrt[7])(4 b)/7)||(1/2 (2+Sqrt[7])(11+Sqrt[93])&&1b)/7)||(1/4 (5+Sqrt[13])(4 b)/7)||(1/4
(5+Sqrt[13])b)/7)||(1/2 (5+Sqrt[21])(4 b)/7)||(1/2 (5+Sqrt[21])b)/7)||(1/4 (7+Sqrt[21])(7+Sqrt[21])(4 b)/7)||(1/4 (7+Sqrt[21])(5+Sqrt[21])&&1b)/7)||(1/8 (11+Sqrt[57])(4 b)/7)||(1/8
(11+Sqrt[57])(7+Sqrt[21])&&a==(4 b)/7)||(1/8 (11+Sqrt[93])(4
b)/7)||(1/8 (11+Sqrt[93])
(25/14) b==8&&a==1

2 b==1&&a>=4/7

4 (b==7/4&&a==0)||(b==8&&a==0)||(b==1/2 (2+Sqrt[7])&&a==0)||(b==1/4
(5+Sqrt[13])&&a==0)||(b==1/2 (5+Sqrt[21])&&a==0)||(b==1/4
(7+Sqrt[21])&&a==0)||(b==1/8 (11+Sqrt[57])&&a==0)||(b==1/8
(11+Sqrt[93])&&a==0)||(b>8&&a==0)||(1(5+Sqrt[13])&&a==0)||(1/2 (2+Sqrt[7])(5+Sqrt[13])(7+Sqrt[21])(2+Sqrt[7])&&a==0)||(1/8 (11+Sqrt[93])
7 b<1

((-32+7 a)/(-8+a)) (b==8&&0
((-7+7 a)/(-(7/4)+a)) (b==7/4&&0
1/2 (4+(-4+7 a)/(-1+a)) b==1&&0
1/2 (7+(-4+7 a)/(-1+a)) b==1&&a<=0

((7-4 b)/(2 (1-b))) (b==1/2 (2+Sqrt[7])&&a==1)||(b==1/4
(5+Sqrt[13])&&a==1)||(b==1/2 (5+Sqrt[21])&&a==1)||(b==1/4
(7+Sqrt[21])&&a==1)||(b==1/8 (11+Sqrt[57])&&a==1)||(b==1/8
(11+Sqrt[93])&&a==1)||(b>8&&a==1)||(7/4(2+Sqrt[7])(11+Sqrt[57])&&a==1)||(1/2 (5+Sqrt[21])(5+Sqrt[21])&&a==1)||(1/8 (11+Sqrt[57])(11+Sqrt[93])
((7 a-4 b)/(a-b)) True

}}

wrote in message
news:c3e6d341-edac-46c8-acf4-48dd762adc23@13g2000hsb.googlegroups.com...
Hi Frank!

Please, could you post the Mathematica 6's output for this problem?
I'm curious to see it!

Thanks in advance, Humberto.


On 12 mar, 20:40, "Frank Kampas" wrote:
> Mathematica 6 can solve the problem in a couple minutes. The solution is
> quite complex, having over 10 separate cases.
> The Mathematica input is simply:
> Minimize[{x+y,a*x+y >=4,b*x+y>=7,x>=0,y>=0},{x,y}]
>
> wrote in message
>
> news:1e336acc-cf9d-4753-a25e-02c9cf36088b@n36g2000hse.googlegroups.com...
>
>
>
> > Greetings!
>
> > Is there a mathematical software that solves symbolically aparametric
> >linearprogram, like that one:
>
> > min x + y
> > subject to ax + y >= 4
> > bx + y >= 7
> > x >= 0, y >=0
>
> > That is, I would like to see the solution in terms of the coefficients
> > a and b:
>
> > (x*, y*) = (3/(a - b), (4a - 7b)/(a - b)) if 7/a <= 4/b
> > (x*, y*) = (7/a, 0), otherwise.
>
> > Thanks in advance,Humberto.- Ocultar texto entre aspas -
>
> - Mostrar texto entre aspas -