1. The Simplex Method. Restaurant owner problem. max 8 x + 6 y Subject to 5 x + 3 y ≤ 30 2 x + 3 y ≤ 24 1 x + 3 y ≤ 18 x , y ≥ 0. Seafoods available: 30 seaurchins
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max 8x + 6y
Subject to
5x + 3y ≤ 30
2x + 3y ≤ 24
1x + 3y ≤ 18
x,y ≥ 0
30 seaurchins
24 shrimps
18 oysters
$8 : including 5 seaurchins, 2 shrimps et 1 oyster
$6 : including 3 seaurchins, 3 shrimps et 3 oysters
owner in order to maximize his revenue according to the seafoods
available
max f(w)
Subject to
where f : X → R1.
or – f(w*) ≤ – f(w)
– f(w*) = min – f(w)
Subject to w X Rn
and w* is a point of X where the function – f(w) reaches its minimum.
solution w* .
f(w*) =max f(w) = – min – f(w) = – (–f(w*) )
or – f(w*) ≤ – f(w)
– f(w*) = min – f(w)
Subject to w X Rn
and w* is a point of X where the function – f(w) reaches its minimum.
solution w* .
Subject to
5x + 3y≤ 30
2x + 3y≤ 24
1x + 3y≤ 18
x,y ≥ 0
min–(8x + 6y)
Subject to
5x + 3y≤ 30
2x + 3y≤ 24
1x + 3y≤ 18
x,y ≥ 0
Restaurant owner problemmin z = –8x– 6y
Subject to
5x + 3y≤ 30
2x + 3y≤ 24
1x + 3y≤ 18
x,y ≥ 0
5x + 3y= 30
The set of points satisfying
the constraint
5x + 3y≤ 30
is under the line since the origin
satisfies the constraint
Feasible Domain2x + 3y= 24
The set of points satisfying
the constraint
2x + 3y≤ 24
is under the line since the origin
satisfies the constraint
Feasible Domainx + 3y= 18
The set of points satisfying
the constraint
x + 3y≤18
is under the line since the origin
satisfies the constraint
Feasible Domainz = –8x– 6y.
The more we move away on the right of the origin, the more the objective function decreases:
x = 0 and y = 0 => z = 0
Solving the problemz = –8x– 6y.
The more we move away on the right of the origin, the more the objective function decreases:
x = 0 and y = 0 => z = 0
x = 0 and y = 6 => z = – 36
Solving the problemz = –8x– 6y.
The more we move away on the right of the origin, the more the objective function decreases:
x = 0 and y = 0 => z = 0
x = 0 and y = 6 => z = – 36
x = 6 and y = 0 => z = – 48
Solving the problemz = –8x– 6y.
The more we move away on the right of the origin, the more the objective function decreases:
x = 0 and y = 0 => z = 0
x = 0 and y = 6 => z = – 36
x = 6 and y = 0 => z = – 48
x = 3 and y = 5 => z = – 54.
Cannot move further on the right without going out of the feasible domain.
Solving the problemOptimal solution:
x = 3 et y = 5
Optimal value:
z = – 54
ai1x1 + ai2x2 + … + ainxn ≤bi → ai1x1 + ai2x2 + … + ainxn+yi=bi
yi≥ 0
ai1x1 + ai2x2 + … + ainxn≥bi → ai1x1 + ai2x2 + … + ainxn–yi=bi
yi≥ 0
min z = – 8x– 6y min z = – 8x– 6y
s.t. s.t.
5x + 3y≤ 30 5x + 3y+ u =30
2x + 3y≤ 24 2x + 3y + p=24
1x + 3y≤ 18 1x + 3y + h= 18
x, y ≥ 0 x, y, u, p, h≥ 0
u = 30 – 5x– 3y
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 – 8x– 6y
u = 30 – 5x– 3y
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 – 8x– 6y
is the following
x = y = 0 => u = 30, p = 24, h = 18 et z = 0.
u = 30 – 5x– 3y≥ 0
p = 24 – 2x– 3y≥ 0
h = 18 – 1x– 3y ≥0
u = 30 – 5x≥ 0 x ≤ 30 / 5 = 6
p = 24 – 2x≥ 0 x ≤ 24 / 2 = 12
h = 18 – 1x≥0 x ≤ 18
x ≤ min {6, 12, 18} = 6.
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 – 8x– 6y
The solution remains feasible as long as
x ≤ min {6, 12, 18} = 6.
i.e., x = 6.
x = 6, y = 0 => u = 0, p = 12, h = 12 et z = – 48.
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 – 8x– 6y
x = 6, y = 0 => u = 0, p = 12, h = 12 et z = –48.
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 – 8x– 6y
=> x = 6 – 1/5u – 3/5y
p = 24 – 2x– 3y
h = 18 – 1x–3y
z = 0 – 8x– 6y
p = 24 – 2x– 3y
=> p = 24 – 2(6 – 1/5u – 3/5y) – 3y
=> p = 12 + 2/5u – 9/5y
h = 18 – 1x– 3y
z = 0 – 8x– 6y
p = 24 – 2x– 3y => p = 12 + 2/5u – 9/5y
h = 18 – 1x– 3y
=> h = 18 – (6 – 1/5u – 3/5y) – 3y
=> h = 12+ 1/5u – 12/5y
z = 0 – 8x– 6y
p = 24 – 2x– 3y => p = 12 + 2/5u – 9/5y
h = 18 – 1x– 3y => h = 12+ 1/5u – 12/5y
z = 0 – 8x– 6y
=> z = 0 – 8(6 – 1/5u – 3/5y) – 6y
=> z = – 48+ 8/5u – 6/5y
p = 24 – 2x– 3y => p = 12 + 2/5u – 9/5y
h = 18 – 1x– 3y => h = 12+ 1/5u – 12/5y
z = 0 – 8x– 6y => z = – 48+ 8/5u – 6/5y
x = 6 – 1/5u – 3/5y
p = 12 + 2/5u – 9/5y
h = 12+ 1/5u – 12/5y
z = – 48+ 8/5u – 6/5y
x = 6 – 1/5u – 3/5y ≥ 0
p = 12 + 2/5u – 9/5y≥0
h = 12+ 1/5u – 12/5y ≥ 0
x = 6 – 3/5y ≥ 0 y ≤ 10
p = 12 – 9/5y≥ 0 y ≤ 20/3
h = 12– 12/5y ≥0 y ≤ 5
y ≤ min {10, 20/3, 5} = 5.
p = 12 + 2/5u – 9/5y≥0
h = 12+ 1/5u – 12/5y ≥ 0
z = – 48+ 8/5u– 6/5y
The solution remains feasible as long as
y ≤ min {10, 20/3, 5} = 5.
y = 5, u = 0 => x = 3, p = 3, h = 0 et z = – 54.
h = 12+ 1/5u – 12/5y
x = 3 – 1/4u + 1/4h
p = 3 + 1/4u + 3/4h
y = 5+ 1/12u – 5/12h
z = – 54+ 3/2u + 1/2h
The initial solution is
x = y = 0 ( u = 30, p = 24, h = 18 ) and the value of z = 0
When increasing the value of x,
the solution becomes
x = 6, y = 0 (u = 0, p = 12, h = 12) and the value of z = – 48
When increasing the value of y,
the solution becomes
x = 3, y = 5(u = 0, p = 3, h = 0) and the value of z = – 54
Link with graphic resolution5x + 3y ≤ 30
5x + 3y + u =30
2x + 3y ≤ 24
2x + 3y + p =24
1x + 3y ≤ 18
1x + 3y + h = 18
It can be shown that this is not the case.
min
Sujet à
If ≥ 0, then the solution
Is optimal,
and the algorithm stops
If < 0, then the variable
xs becomes the entering varaiable.
We move to Step 2.
In this case, the algorithm stops indicating that the problem is not
bounded below
The solution remains feasible
The solution remains feasible
Consequently, the largest value of the
entering variable xs is
The solution remains feasible
Consequently, the largest value of the
entering variable xsis
The independent variable xr
limiting the increase of the entering
variable xs is the leaving variable.
becomes an independent variable with a 0 value
Use the rth equation to specify xsin terms of xm+1, …, xs1, xs+1, …, xn, xr
Replace xsspecified in terms of xm+1, …, xs1, xs+1, …, xn, xr, in each of the other equations
Replace xsspecified in terms of xm+1, …, xs1, xs+1, …, xn, xr, in each of the other equations
Replace xsspecified in terms of xm+1, …, xs1, xs+1, …, xn, xr, in each of the other equations
Replace xsspecified in terms of xm+1, …, xs1, xs+1, …, xn, xr, in each of the other equations
min z = –8x– 6y min z
Sujet à Sujet à
5x + 3y+ u =30 5x + 3y+ u =30
2x + 3y + p=24 2x + 3y + p=24
1x + 3y + h= 18 1x + 3y + h= 18
x, y, u, p, h≥ 0 –8x –6y –z = 0
x, y, u, p, h≥ 0
min z = –8x– 6y min z
Subject to Subject to
5x + 3y+ u =30 5x + 3y+ u =30
2x + 3y + p=24 2x + 3y + p=24
1x + 3y + h= 18 1x + 3y + h= 18
x, y, u, p, h≥ 0 –8x –6y –z = 0
x, y, u, p, h≥ 0
u = 30 – 5x– 3y
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 –8x– 6y
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 –8x– 6y
Step 1: Entering criterion
Determine the entering variable
by selecting the smallest element
in the last row of the tableau
min {–8, –6, 0, 0, 0} = –8.
x is then the entering variable
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 –8x– 6y
Step 2: leaving criterion entering variable
To identify the leaving variable
determine the min of the ratio
right hand side terms divided by the
corresponding elements in the
column of the entering variable
that are positive:
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 –8x– 6y
Step 2: leaving criterion entering variable
min {30/5, 24/2, 18} = 30/5 = 6
The corresponding variable u
becomes the leaving variable
p = 24 – 2x– 3y
h = 18 – 1x– 3y
z = 0 –8x– 6y
leaving variable entering variable
Step3 : Pivot
Transform the system or
the tableau
entering variable
RECALL: We use the equation including variable x and u to specify x in terms of u and y:
u = 30 – 5x– 3y => (5x = 30 – u – 3y) / 5
=> x = 6 – 1/5u – 3/5y
This is equivalent to
5x + 3y+ u =30
entering variable
RECALL: We use the equation including variable x and u to specify x in terms of u and y:
u = 30 – 5x– 3y => (5x = 30 – u – 3y) / 5
=> x = 6 – 1/5u – 3/5y
This is equivalent to
(5x + 3y+ u =30) / 5
entering variable
RECALL: We use the equation including variable x and u to specify x in terms of u and y:
u = 30 – 5x– 3y => (5x = 30 – u – 3y) / 5
=> x = 6 – 1/5u – 3/5y
This is equivalent to
(5x + 3y+ u =30) / 5 => x + 3/5y + 1/5u = 6
entering variable
This is equivalent to
(5x + 3y+ u =30) / 5 => x + 3/5y + 1/5u = 6
In the tableau, this is equivalent to divide the row including the leaving variable by the coefficient of the entering variable in this row
x = 6 – 1/5u– 3/5y
p = 24 – 2x– 3y
=> p = 24 – 2(6– 1/5u– 3/5y) – 3y
This is equivalent to: p = 24 – 2(6– 1/5u –3/5y)+2x– 2x – 3y
2x + 3y + p – 2 (x + 3/5y +1/5u) = 24 – 2(6)
2x + 3y + p – 2 (x +3/5y + 1/5u) = 24 – 2(6)
2x + 3y + p = 24
– 2 (x +3/5y + 1/5u = 6)
0x + 9/5y –2/5u + p = 12
second row
minus
2(the first row)
If ≥ 0, then the current
solution is optimal, and
the algorithm stops
Entering variable
If < 0, then xs is the
entering variable
–
If
the problem is not
bounded below,
and the alg. stops
Entering variable
If
then the sol. remains feasible
–
The pivot element is located at the intersection
of the column including the entering variable xs
and of the row including the leaving variable xr
Entering variable
Leaving variable
–
Devide row r by the pivot
element to obtain a
new line r.
Variable d’entrée
Variable de sortie
–
Devide line r by the pivot
element to obtain a
new line r.
Entering variable
Leaving variable
–
Multiply the new line r by ,
and substrack this from the line i. This induces that the coefficient of the entering variable xs to become equal to 0.
Entering variable
Leaving variable
–
Multiply the new line r by ,
and substrack this from the line i. This induces that the coefficient of the entering variable xs to become equal to 0.
Entering variable
Leaving variable
–
Multiply the new line r by ,
and substrack this from the line i. This induces that the coefficient of the entering variable xs to become equal to 0.
Entering variable
Leaving variable
–
Multiply the new line r by ,
and substrack this from the line i. This induces that the coefficient of the entering variable xs to become equal to 0.
Entering variable
Leaving variable
–
Denote also
of xR in the objective function
Note that the two problems are
equivalents since the second one is
obtained from the first one using
elementary operations based on a
non singular matrix B1
the dependent variables)
associated with the columns
of the basis B, are now denoted
basic variables
The variables xR(denoted
independent variables)
are now denoted
non basic variables
basis B associated to the relative costs of the variables:
basis B associated to the relative costs of the variables:
Denote the vector specified by
Then
or
where denotes the jth column of the contraint matrix A
Let B* beanoptimal basis, and the corresponding basic solution
having the optimal value
Δb1<0
the size of the feasible domain is reduced
5x + 3y≤ 30
2x + 3y≤ 24
1x + 3y≤ 18
Solving the problem graphiclyΔb1>0
the size of the feasible domain is increased
5x + 3y≤ 30
2x + 3y≤ 24
1x + 3y≤ 18
Solving the problem graphiclyΔb3<0
the size of the feasible domain is reduced
5x + 3y≤ 30
2x + 3y≤ 24
1x + 3y≤ 18
Solving the problem graphiclyΔb2<0
the size of the feasible domain remains the same
5x + 3y≤ 30
2x + 3y≤ 24
1x + 3y≤ 18
Solving the problem graphiclyProof: Without lost of generality, assume that the first m variables
x1, x2, …, xm are basic; i. e.,

The objective function is as follows
Consider another feasible solution ≥ 0 with the value
But the hypothesis induces that
The objective function is as follows
Consider another feasible solution ≥ 0 with the value
But the hypothesis induces that