1.
a)
First, we transform the variables:
(1) x1<= 5: let x1+ x1'= 5, x1'>=0
(2) x2 is free:
x2=x2'-x2'', x2' >= 0 and x2''>= 0
(3) -10 <= x3 <= 10:
it is the same as: 0 <=x3+10 <= 20
let x3'= x3 + 10, x3'>= 0
we get x3'<= 20,
let x3'+ x3''= 20, x3''>= 0
---------------------Then, We add x3'+ x3'' = 20 to the constraints,
and substitute x1, x2, x3 in the objective function and in the constraints by:
x1 = 5-x1',
x2 = x2'- x2''
x3 = x3'-10,
the objective function becomes:
max 15 * x1
= -min (-15
= -min [-15
= -min (-75
= 5 - min (
- 3 * x2 + 7 * x3
* x1 + 3 * x2 - 7 * x3)
* ( 5 - x1') + 3 * (x2'- x2'')- 7 * (x3'-10)]
+ 15 * x1'+ 3 * x2' - 3 * x2'' - 7 * x3' + 70)
15 * x1' + 3 * x2' -3 * x2''- 7 * x3')
For the constraint -10 * x1 + 6 * x2 - 2 * x3 >= 6,
we add a new slack vaiable x4>=0,
which makes -10 * x1 + 6 * x2 - 2 * x3 - x4 = 6,
and we did the substitution in the constraints.
-------------Finally,We get the standard form as
5 - min (15 * x1' + 3 * x2'
s.t.
-5 * x1'+ 3 * x2'- 3 * x2''
10 * x1'+ 6 * x2' - 6 * x2''
x3'+ x3''
=
x1', x2', x2'', x3', x3'',x4
x=
[x1' x2' x2'' x3' x3'' x4]'
c=
follows:
-3 * x2''- 7 * x3')
- 14 * x3'
- 2 * x3' -x4
20
>=0
=
=
-155
36
[15,3,-3,-7,0,0]'
b=
[-155,36,20]'
A=
[-5
|10
[ 0
3
6
0
-3
-6
0
-14
-2
1
0 0 ]
0 -1 |
1 0 ]
--------------(b) The dual problem is:
5 - max -155 * y1+ 36 * y2 + 20 * y3
s.t
-5 * y1 + 10 * y2
<= 15
3 * y1 + 6 * y2
<= 3
-3 * y1 - 6 * y2
<= -3
-14 * y1 - 2 * y2 + y3
<= -7
y3
<= 0
-y2
<= 0
---------------------------------------------------------------2.
(a) The dual problem is:
max
3y
s.t
ay <= 1
-by = 1
y is free
(b) Transform the problem into the standard form:
min
s.t.
x1 + x2' - x2''
ax1 - bx2' + bx2'' = 3
x1, x2', x2'' >= 0
The corresponding dual problem is:
max
s.t
(c) when
(d) when
(e) when
x1=0.
(f) when
(g) when
3y
ay <= 1
-by <= 1
by <= -1
a <= 0 and b = 0, the problem is infeasible
a >= 0 or b != 0, the problem is feasible
(a/b) >= -1, an optimal solution exists. The optimal value is -3/b when
b = 0 and a >= 0, the problem is feasible and unbounded.
a/b = -1, the problem is feasible and has multiple optimal solutions
-----------------------------------------------------------------3.(a)
objective:
find the minimum cost diet which meets all the nutrient requirements
variable:
x(i) be the weight of the i-th fruit used in the diet.
min 5 * x1 + 1 * x2 + 8 * x3 + 3 * x4 + 9 * x5
s.t.
2 * x1 + 2 * x3 +
x4 + 3 * x5 >= 41
3 * x2 + 4 * x3 + 2 * x4 + 2 * x5 >= 80
2<=x1<=10,
3<=x2<=6,
0<=x3<=7,
5<=x4<=20,
0<=x5<=5
--------------------------------(b)
2<=x1<=10 is the same as 0<=x1-2<=8
let x1-2=x1',x1'>=0.
we get x1'<=8,
let x1'+x1''=8, x1'>=0, x1''>=0
Similarly, we can get:
x2-3=x2', x2'>=0
x2'<=3,
let x2'+x2''=3, x2'>=0, x2''>=0
x4-5=x4', x4'>=0
x4'<=15,
let x4'+x4''=15, x4'>=0, x4''>=0
For x3 and x5, we can let x3+x3'=7, x3'>=0
x5+x5'=5,x5'>=0
-------------We add:
x1'+x1''=8,
x2'+x2''=3,
x3+x3'=7,
x4'+x4''=15,
x5+x5'=5
to the constaints;
keep x3 and x5;
substitue x1, x2, x4 in the objective function and in the constraints, :
x1=x1'+2,
x2=x2'+3,
x4=x4'+5
we get:
min 5 * (x1'+2) + 1 * (x2'+3) + 8 * x3 + 3 * (x4'+5) +9 * x5
s.t
2 * (x1'+2)+ 2 * x3 + (x4'+ 5)+ 3 * x5 >= 41
3 * (x2'+3)+ 4 * x3 + 2 * (x4'+5) + 2 * x5 >=80
x1'+x1''=8
x2'+x2''=3
x3 +x3' =7
x4'+x4''=15
x5 +x5' =5
x1',x1'',x2',x2'',x3, x3', x4', x4'', x5, x5' >=0
try to transform into the standard form:
28 + min 5 * x1'+ x2'+ 8 * x3 + 3 * x4'+ 9 * x5
s.t
2 * (x1'+2) + 2 * x3 + (x4'+5) + 3 * x5 - x6 = 41
3 * (x2'+3) + 4 * x3 + 2 * (x4'+5)+ 2 * x5 - x7 = 80
x1'+x1''=8
x2'+x2''=3
x3 +x3' =7
x4'+x4''=15
x5 +x5' =5
x1',x1'',x2',x2'',x3, x3', x4', x4'', x5, x5', x6, x7>=0
The standard form is:
min 5 * x1'+ x2'+ 8 * x3 + 3 * x4'+ 9 * x5
s.t
2 * x1'+ 2 * x3 +
x4'+ 3 * x5 - x6 = 32
3 * x2'+ 4 * x3 + 2 * x4'+ 2 * x5 - x7 = 61
x1'+x1''=8
x2'+x2''=3
x3 +x3' =7
x4'+x4''=15
x5 +x5' =5
x1',x1'',x2',x2'',x3, x3', x4', x4'', x5, x5',x6, x7>=0
-------------------------------------------** please remember:
(1) explain the objective and the variables!
(2) transform the problem into the standard form!