**Unformatted text preview: **Linear Programming and Integer Programming
6.2-1. Mid-term Examination 2 (a)
Iteration 0: Since all coefficients are zero, at the current solution Ð!ß !Ñ, the three
Solution
resources (production time per week at plant 1, 2 and 3) are free goods. This means
increasing them does not improve the objective value.
Iteration 1: Ð!ß &Î#ß !Ñ. Now resource 2 has been entirely used up and contributes
&Î# to profit per unit of resource. Since this is positive, it is worthwhile to continue fully
6.1-5.
using this resource.
1.
(a) Iteration 2: Ð!ß $Î#ß "Ñ. Resources 2 and 3 are used up and contribute a positive
minimize
"&C" #&C#
amount to profit.to
ResourceC1 is a free good while resources 2 and 3 contribute $Î# and "
subject &
"
per unit of resource respectively. C %
#
(b)
Iteration 0: Ð$ß &Ñ. #Cactivities 1 and 2 (number of batches of product 1 and
C" Both # $
2 produced per week) can ß be initiated ! give a more profitable allocation of the
C" C# to
resources. The current contribution of the resources required to produce one batch of
‡ ‡
(b) Optimal Solution: ÐC" ß C# Ñ œ Ð&ß %Ñ, so shadow prices for resources 1 and 2 are & and
product 1 or 2 to the profit is smaller than the unit profit per batch of product 1 or 2
% respectively.
respectively.
Iteration 1: Ð$ß !Ñ. Again activity 1 can be initiated to give a more profitable use
of resources, but activity 2 is already being produced (or the resources are being used just
as well in other activities).
Iteration 2: Ð!ß !Ñ. Both activities are being produced (or the resources are being
used just as profitably elsewhere).
(c)
Iteration 1: Since activities 1 and 2 can be initiated to increase the profit (give the
same amount of resources), we choose to increase one of these. We choose activity 2 as
the entering activity (basic variable), since it increases the profit by & for every unit of
product 2 produced (as opposed to $ for product 1).
Iteration 2: Only activity 1 can be initiated for more profit, so we do so.
Iteration 3: Both activity 1 and 2 are being used. Furthermore, since the
coefficients for B$ , B% and B& are nonnegative, it is not worthwhile to cut back on the use
of(c) of the resources. Thus, we must be at the optimal solution.
any
2.
6.3-1.
(a) (a) minimize
subject to [ œ #!C" "!C#
&C" C# '
#C" #C# )
C" ß C# ! 1 (b) Primal: (b)(b) Primal: ÐB" ß B# Ñ œ Ð&Î#ß "&Î%Ñ is optimal with ^ œ %&. Infeasible corner point solutions are
Ð!ß "!Ñ and Ð"!ß !Ñ.
(x , x2 ) = (5/2, 15/4) is optimal with Z = 45. Infeasible corner point solutions are (0, 10)and(10, 0).
ÐB" ß B# Ñ 1 Ð&Î#ß "&Î%Ñ is optimal with ^ œ %&. Infeasible corner point solutions are
œ
Dual:
Ð!ß "!Ñ and Ð"!ß !Ñ.
Dual: ÐC" ß C#(y1œy2 ) = (1/2, 7/2) optimal with [ = 45. .Infeasible corner point solutions are (0, 4), (0, 0)and(6/5, 0).
Ñ , Ð"Î#ß (Î#Ñ is is optimal with W œ %& Infeasible corner point solutions are
Ð!ß %Ñ, Ð!ß !Ñ and Ð'Î&ß !Ñ.
ÐC" ß C# Ñ œ Ð"Î#ß (Î#Ñ is optimal with [ œ %&. Infeasible corner point solutions are
Ð!ß %Ñ, Ð!ß !Ñ and Ð'Î&ß !Ñ.
6-9
6-9 2 (c)(c)
Primal BS
Ð!ß &ß "!ß !Ñ
Ð!ß !ß #!ß "!Ñ
Ð%ß !ß !ß 'Ñ
Ð&Î#ß "&Î%ß !ß !Ñ
Ð!ß "!ß !ß "!Ñ
Ð"!ß !ß $!ß !Ñ Feasible?
Yes
Yes
Yes
Yes
No
No ^
%!
!
#%
%&
)!
'! Dual BS
Ð!ß %ß #ß !Ñ
Ð!ß !ß 'ß )Ñ
Ð'Î&ß !ß !ß #)Î&Ñ
Ð"Î#ß (Î#ß !ß !Ñ
Ð%ß !ß "%ß !Ñ
Ð!ß 'ß !ß %Ñ (d)
(d) Primal: Ð!ß !ß #!ß "!Ñ 10)
Primal:(0, 0, 20,
Dual: Dual:(0,'ß )Ñ
Ð!ß !ß 0, −6, −8) Primal: Ð!ß &ß "!ß !Ñ 0)
Primal:(0, 5, 10,
Dual: Dual:(0,#ß !Ñ0)
Ð!ß %ß 4, −2, Primal:(5/2, 15/4, 0,
Primal: Ð&Î#ß "&Î%ß !ß !Ñ 0)
Dual:(1/2, 7/2, 0, 0)
Dual: Ð"Î#ß (Î#ß !ß !Ñ 6.3-2.
(a) minimize [ œ )C" %C# subject to C" C# "
$C" C# #
C" ß C# !
3 6-10 Feasible?
No
No
No
Yes
Yes
Yes 3.
We denote by
x1 : the units of Product 1 produced.
x2 : the units of Product 2 produced.
yi = 1 if any Product i is produced and yi = 0 otherwise, for i = 1, 2.
Then, the integer programming model is:
max z = 2x1 + 5x2 − 10y1 − 20y2
s.t. 3x1 + 6x2 ≤ 120,
x1 ≤ 40y1 ,
x2 ≤ 20y2 ,
x1 ≥ 0, x2 ≥ 0;
x1 , x2 be integer numbers;
y1 , y2 = 0 or 1.
4.
We deﬁne: x1 : x :
2 x3 : x4 :
y1 = the
the
the
the number
number
number
number of
of
of
of soldiers;
tanks;
planes;
trains; 1, If any soldiers are manufactured,
0, otherwise. y2 = 1, If any tanks are manufactured,
0, otherwise. y3 = 1, If any planes are manufactured,
0, otherwise. y4 = 1, If any trains are manufactured,
0, otherwise. Then, the proﬁt is:
= (12x1 + 10x2 + 8x3 + 15x4 ) − (6x1 + 5x2 + 4x3 + 8x4 ) − (200y1 + 150y2 + 120y3 + 100y4 ) = z 6x1 + 5x2 + 4x3 + 7x4 − 200y1 − 150y2 − 120y3 − 100y4 . The resource constraints are:
3x1 + 5x2 + 2x3 + 6x4 ≤ 150;
4x1 + x2 + 3x3 + 4x4 ≤ 160;
The machine tool rentage constraints are:
xi ≤ M i yi , , i = 1, 2, 3, 4, 4 where Mi are suﬃciently large constants.
The either-or constraints for economical productions are:
xi ≤ Ni zi ;
i = 1, 2, 3, 4,
5 − xi ≤ Ni (1 − zi );
where Ni are suﬃciently large constants.
In addition, the requirements 1-3 can be fulﬁlled by: y1 + y2 + y3 + y4 ≤ 3; y = y3 ; 1 y3 + y4 ≤ 1.
Hence, the linear IP model is:
max z= 6x1 + 5x2 + 4x3 + 7x4 − 200y1 − 150y2 − 120y3 − 100y4
3x1 + 5x2 + 2x3 + 6x4 ≤ 150;
4x1 + x2 + 3x3 + 4x4 ≤ 160;
xi ≤ Ni zi ;
i = 1, 2, 3, 4,
5 − xi ≤ Ni (1 − zi ); y1 + y2 + y3 + y4 ≤ 3; y = y3 ; 1 y3 + y4 ≤ 1.
xi ≤ Mi yi , , i = 1, 2, 3, 4;
xi ≥ 0, integers;
yi ∈ {0, 1}, , i = 1, 2, 3, 4;
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