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2 The linear one-dimensional case.
2.1 Extremizers are not in the interior.
Let c!! be given and assume that c " 0. The collection of nontrivial continuous linear functions is given by fc :! #! defined by f c $^ x %& cx. All x!! are in the interior of!. The derivative f c ' $^ x %& c is never zero. It follows that there are no extremizers. If C (! and extremizers of fc : C #! are sought, then the only possibility are points in C that are not in the interior of C.
2.2 Linear constraints.
A standard linear constraint in one variable has the form ax ) b , where a b ,!! and a " 0. The corresponding constrained set is given by
/ if 0 ( , / ] if 0 / if 0 [ / ,+^ ) if^0
x x b a a b a a C x ax b (^) b a a x x b a a
The set C is a closed. The only x! C that is not in the interior of C is given by x & b / a. Let m! " be given. Write a x 1 ) b 1 (^) , #, am x ) bm when m linear constraints are imposed at once. Denote the corresponding sets by C 1 (^) , #, Cm. Imposing these constraints at once means x!! must satisfy a xj ) bj for every single j! *^ 1, $, m +. The constrained set is therefore given by C & C 1 (^) 5 # 5 Cm , which is a closed set.
2.3 The primal problem.
The primal problem deals only with maximizers of functions fc :! #! given by f c $^ x %& cx with c (^) " 0. It is assumed that x 3 0 , and a xj ) bj for j! * 1, $, m +and some given m! ".
The following notation is used: a 1 1
max when 0
m m
x b
cx (^) a x b x
With the help of the column matrices
1
m
a A a
&^88
1
m
b b b
& 8^8
this becomes
max when (^0)
Ax b cx (^) x
The inequality sign between two matrices of the same order indicates that each pair of corresponding entries satisfies the inequality. The collection of all x!! such that Ax ) b and x 3 0 is known as the feasible set of the primal problem. If aj. 0 and bj 2 0 for some j , then the constraints are inconsistent and the feasible set is empty. Suppose the constraints are consistent with a (^) j. 0 and bj & 0 for some j , then the case is trivial since x & 0 is the only
point in the feasible set. If aj 2 0 and b (^) j 3 0 for some j , then the constraint is redundant since it is subsumed by the constraint x 3 0. In order to have a nonempty nontrivial feasible set with no redundant constraints it must be that a (^) j. 0 and bj. 0 , or a (^) j 2 0 and bj 2 0 for all j.
Let! & max * b j / aj a j 2 0, bj 2 0 +, or! & 0 if there is no j such that a j 2 0 and bj 2 0.
Let " & min * bj / a j a j. 0, bj. 0 +, or " & <0 if there is no j such that a j. 0 and
b (^) j. 0. There are three possibilities. If_!. " , then the feasible set is empty. If!_ & " , then x &! & " is the only number in the feasible set. If_!_ 2 " , then the feasible set is equal to =! " , > when " 2 <0 , and [! , <0) if " & <0. If c. 0 , then x &! is a minimizer. If c 2 0 , then x &! is a maximizer. If " 2 <0 , then x & " is a maximizer if c. 0 , and a minimizer if c 2 0.
2.4 Simple primal problems.
The simplest case is given by max cx when x 3 0. If c. 0 , then there is no maximizer. If c 2 0 , then x & 0 is a maximizer. The second simplest case is given by ax b max cx when (^) x 0
,--^ )
If a. 0 and b 2 0 , then there is no maximizer. If a. 0 and b & 0 , then x & 0 is the only point in the feasible set and therefore a maximizer. If a. 0 and b. 0 , then the feasible set is given by = 0, b / a >. In this case c. 0 implies that x & b / a is a maximizer, and c 2 0 implies that x & 0 is a maximizer. If a 2 0 and b 3 0 , then the feasible set is equal to [0, <0). This time there is no maximizer if c. 0 , and x & 0 is a maximizer if c 2 0. Finally, if a 2 0 and b 2 0 , then the feasible set is given by [ / , b a <0). Again, there is no maximizer if c. 0 , and x & b / a is a maximizer if c 2 0.
2.5 Duality.
To get a concrete interpretation of the primal problem, assume x is the mass in kilograms of the product to be produced and ultimately sold. Let c be the selling price in dollars per kilogram. The efficiency at different stages j of the production is given by a (^) j in hours per kilogram. The
available resources at stage j are given by bj in hours. The problem is to maximize the income
y (^) j 3 0. Let
1
m
x x y y y
1 2
1
m m
a a A I a a
/
The constraints in the primal problem have the new form
x A I (^) y b x y
6 7 8^6 7 9 &
The dimension of the problem has increased from 1 to m < 1. In the dual problem
min when (^0)
T #T b A^ # c #
let $ & AT # / c be the one slack variable. The first constraint is satisfied if and only if $ 3 0. The constraints have the new form
A^ _T # $ c
$_
The dimension of the problem has increased from m to m < 1.
2.7 Tableau.
With the slack variables in place, the data of the primal problem is recorded in the following initial tableau 1 1 2 2
1 1
m m m m
a b a b
a b a b c
/ /
The data above the horizontal line corresponds to the equality constraints. The bottom row has the coefficients of the function to be maximized to the left of the vertical line, and the negative of the value of the function to the right. To find a solution that is not in the interior of the feasible set it suffices to let one of the variables be equal to zero. It is convenient to choose x to be zero so that s (^) j & bj for all j , because s (^) j & bj is readily seen in the tableaux. The columns corresponding to the slack variables each look like a standard basis vector. The corresponding
variable is referred to as a basic variable , and all other variables are non-basic variables. The non- basic variables will always have value zero.
2.8 Pivoting rule.
Suppose x is a non-basic variable in the feasible set. It follows that all bj are positive and it may be assumed that all aj are positive as well. Assume c. 0 so that it is clear that it is beneficial
to increase x. Using elementary row operations the x column will be transformed so that there is a single 1 in row j and 0 in all other rows. The next tableau is given by
(^0 1 0 10 01 )
0 1 0 0
1 1
0 0 1 0
0 0 0 0 1
0 0 0 0 0
j j j
j j j
m (^) m m j j j j j j
a (^) b a b a a
b a a
a b a b^ a a c (^) cb a a
The subsequent value of x is b (^) j / a (^) j. Examine the column to the right of the vertical line in the tableau. Since all variables are required to be greater than or equal to zero, it follows that it is
necessary to choose j so that bj / a j & min * ak / bk k! * 1, $, m ++. This choice is referred to as
the pivoting rule. The j th slack variable is equal to zero. The problem is transformed, and now the function
j (^) j j s cs ' / a
is to be maximized. Since sj 3 0 is required and / c / aj 2 0 , it follows that there is nothing to
be gained by increasing s (^) j from zero, which is the current value. The current solution x & bj / aj is optimal. The new tableau is in fact the final tableau.
2.9 Duality and the final tableau.
If #k & 0 for all k " j and #j & c / aj , then T j j
_b
b_ & c a and^ c^ &^ AT^ #^3 c. It follows that the
final tableau has produced a feasible # for the dual problem with the property that the value of the dual objective is equal to the value of the primal objective. It is shown below how this forces # to be optimal in the dual problem.
