Mathematical linear programming notes

Applied Mathematical
Programming Using
Algebraic Systems
Class Notes
Winter Semester 2008/2009
Uwe A. Schneider

Table of Contents
Linear Programming (LP) theory ______________________________________________ 4
Graphical illustration of LP ________________________________________________ 4
Normal solution illustration _______________________________________________ 4
Unwanted solutions______________________________________________________ 5
Unbounded solution ___________________________________________________ 5
Infeasible solution _____________________________________________________ 6
Multiple optima_______________________________________________________ 6
Degenerate solution____________________________________________________ 6
Algebraic Representation of LP_____________________________________________ 8
Constraints_____________________________________________________________ 8
Introducing slack variables ________________________________________________ 8
Matrix Algebra _________________________________________________________ 8
Matrix Formulation of the Linear Programming Problem ______________________ 8
Solving LP's by Matrix Algebra __________________________________________ 9
Primal-Dual Solution Inter-Relationships____________________________________ 12
Objective Function Interrelationships _____________________________________ 13
Constructing Dual Solutions ____________________________________________ 13
Complementary Slackness _____________________________________________ 14
Zero Profits _________________________________________________________ 15
Degeneracy and Shadow Prices ___________________________________________ 15
Solutions and Their Interpretation__________________________________________ 16
Fixing improper models _________________________________________________ 17
Infeasible models_____________________________________________________ 17
Unbounded models ___________________________________________________ 18
LP-Formulations __________________________________________________________ 18
Resource allocation ______________________________________________________ 18
Transportation__________________________________________________________ 19
Mixing_________________________________________________________________ 19
Multi-input, Multi-output_________________________________________________ 19
Spatial equilibrium (class example from GAMS block session) __________________ 19
Sequencing _____________________________________________________________ 20
Storage ________________________________________________________________ 21
Multi-objective programming _____________________________________________ 21
Lexicographic preferences _______________________________________________ 21
Weighted preferences ___________________________________________________ 21
Separable Programming (Non-linear, well-behaved functions) __________________ 22
Risk programming ______________________________________________________ 23
Dynamic linear programming _____________________________________________ 23
Disequilibrium - Known Life _____________________________________________ 23
Disequilibrium - Unknown Life ___________________________________________ 25
Comments __________________________________________________________ 25
Equilibrium - Known Life________________________________________________ 25
Comments __________________________________________________________ 27
2

Equilibrium - Unknown Life______________________________________________ 27
Overall Comments on Dynamic Linear Programming __________________________ 27
Recursive Programming__________________________________________________ 28
Comments on Recursive Programming______________________________________ 28
Integer programming _______________________________________________________ 28
Fixed Costs_____________________________________________________________ 28
Fixed Capacity__________________________________________________________ 29
Minimum Habitat Size ___________________________________________________ 29
Knapsack ______________________________________________________________ 30
No item repetition ____________________________________________________ 30
Item repetition _______________________________________________________ 30
Warehouse____________________________________________________________ 30
Logical Conditions _____________________________________________________ 30
Mutual exclusive products _____________________________________________ 30
Either-Or-Active constraints ____________________________________________ 31
Multiple Active Constraints. ____________________________________________ 31
Distinct Variable Values _________________________________________________ 31
Non-linear representation ________________________________________________ 31
Approximation of not well behaved non-linear functions _______________________ 32
3

Linear Programming (LP) theory
A linear program is the mathematical formulation of an optimization problem. It consists of n
nonnegative decision variables, 1 free objective variable, 1 objective function, and m
constraints, which restrict and/or interrelate the decision variables.
Graphical illustration of LP
Normal solution illustration
Linear Programming Example
Max 2*X1 + 3*X2 = z
s.t. X1 + 2*X2 ≤ 10
X1 , X2 ≥ 0
Feasibility Region
Max 2*X1 + 3*X2 = z
s.t. X1 + 2*X2 ≤ 10
X1 , X2 ≥ 0
X2
X1
10
5
X2 ≤ 5 – 0,5*X1
X1,X2 ≥ 0
The feasibility region of an LP is a convex set. A convex set is a set, where all convex
combinations of points pertaining to the set will also pertain to the set.
Feasibility Region
Convex Set
X2
X1
10
5
Objective Function Isoclines - 1
Max 2*X1 + 3*X2 = z
s.t. X1 + 2*X2 ≤ 10
X1 , X2 ≥ 0
X2
X1
10
5
X2 = z/3 – 2/3*X1
z=6X2=2–2/3*X1
Objective Function Isoclines - 2
Max 2*X1 + 3*X2 = z
s.t. X1 + 2*X2 ≤ 10
X1 , X2 ≥ 0
X2
X1
10
5
X2 = z/3 – 2/3 * X1
z=15
z=6
z=0
Graphical Solution
Max 2*X1 + 3*X2 = z
s.t. X1 + 2*X2 ≤ 10
X1 , X2 ≥ 0
X2
X1
10
5
X2 = z/3 – 2/3 * X1
z=20
4

The optimal LP solution occurs at so-called extreme points of the convex feasibility
regions. Extreme points are those which cannot be represented as a convex combination of
other points pertaining to the convex set.
Graphical Solution
X2
X1
10
5
z=20
Other Extreme Points
X2
X1
10
5
Optimal solution
occurs at extreme
point!
1 Constraint ⇒1 Non-Zero Variable
Max Objective
s.t. X1 + 2*X2 + S = 10
X1 , X2 , S ≥ 0
X2
X1
10
5
X1 10
X 2 0
S 0
=⎛ ⎞
⎜ ⎟
=⎜ ⎟
⎜ ⎟=⎝ ⎠⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
=
=
10S
02X
01X
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
=
=
0S
52X
01X
2 Constraints
X2
X1
10
5
5
10 Max Objective
s.t. X1 + 2*X2 + S1 = 10
2*X1 + X2 + S2 = 10
X1 , X2 , S1, S2 ≥ 0
⎟⎟
⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎜
⎝
⎛
=
=
=
=
02S
51S
02X
51X
Max Objective
s.t. X1 + 2*X2 + S1 = 10
2*X1 + X2 + S2 = 10
X1 , X2 , S1, S2 ≥ 0
X2
X1
10
5
⎟⎟
⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎜
⎝
⎛
=
=
=
=
102S
101S
02X
01X
⎟⎟
⎟
⎟
⎟
⎠
⎞
⎜⎜
⎜
⎜
⎜
⎝
⎛
=
=
=
=
52S
01S
52X
01X
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
=
=
=
=
02S
01S
3
102X
3
101X
10
5
2 Constraints ⇒ 2 Non-Zero Variables
Each extreme point has no more non-zero variables than it has constraints.
Unwanted solutions
Unbounded solution
The objective variable can reach plus infinity (maximization) or minus infinity
(minimization). It is caused by misspecified or missing constraints. In the example below,
there is no constraint to the right. Thus, the objective variable value can go to infinity as x1
and x2 go to infinity.
5

Objective
increases
x1
x2 1 2
1 2
1 2
Max x x z 0
1
s.t. x x 0
2
x , x 0
+ − =
− + ≤
≥
Infeasible solution
The feasibility region is empty. It is caused by misspecified constraints or an infeasible
problem context. Below is an example, where one restrictions forces the two non-negative
decision variables x1 and x2 to be in area A but a second restriction causes x1 and x2 to be in
area B. Since the two regions are exclusive, the solution space is empty.
x1
x2
A
B
Multiple optima
Multiple optima are possible if the objective function isocline with the highest feasible
objective level is parallel to a binding constraint. Any feasible point on this binding constraint
then represents an optimal solution. The number of multiple solution is infinite. In the
graphical illustration below, an infinite number of alternative solutions exists between P1 and
P2.
x1
x2
P1
P2
Isocline with
highest
objective
Degenerate solution
A degenerate solution is caused by redundant constraints and their implications. The
illustration below demonstrates degeneracy graphically. There are three constraints shown.
1 2
1 2
1 2
Max ...
s.t. 5x x 20
1
x x 3
2
x , x 0
+ ≤
− + ≤ −
≥
1 2
1 2
1 2
1 2
Max 10x x
s.t. 10x x 100
1
x x 5
2
x , x 0
+
+ ≤
− + ≤
≥
6

Suppose, we have a maximization problem, where the optimal solution would occur at the
maximum feasible level of x1, namely at point P1.
x1
x2 1 2
1
2
1 2
1 2
Max x x
s.t. x 100
x 50
1
x x 50
2
x , x 0
−
≤
≤
+ ≤
≥
P1
By removing the vertical constraint (see below), we would not alter the solution.
x1
x2
P1
Similarly, one could remove the diagonal constraint (see below) without altering the solution.
1 2
1 2
2
1 2
Max x x
1
s.t. x x 100
2
x 50
x , x 0
−
+ ≤
≤
≥
1 2
1
2
1 2
Max x x
s.t. x 100
x 50
x , x 0
−
≤
≤
≥
x2
x1
P1
Graphically, it seems that redundant constraints don’t cause harm. However, degenerate
solutions cause zero levels for some basic variables and zero shadow prices for some binding
constraints. This may cause an incorrect solution interpretation.
7

Algebraic Representation of LP
Constraints
Constraints are either
technical constraints, or
convenience / accounting constraints.
Constraints can have three different mathematical forms:
a) 1 1a X b≤
b) 2 2a X b≥
c) 3 3a X b=
where a1, a2, and a3 are coefficients, b1, b2, and b3 are right hand side constants, and X is a
nonnegative variable. Constraint type a) is conventionally used as standard type. Constraint
types b) and c) can be easily converted into type a). Particularly, out of constraint of type b),
we can form a constraint of type a) by multiplying both sides by –1 yielding .
Similarly, out of each constraint of type c) we can form two constraints of type a):
2 2a X b− ≤ −
3 3a X b≤
and .3 3a X b− ≤ −
Introducing slack variables
Suppose we have n non-negative variables X1 through Xn contained in m inequality
restrictions (standard type) and one objective function.
1 1 n n
11 1 1n n 1
m1 1 mn n m
1 n
Max c X ... c X
s.t a X ... a X b
...
a X ... a X b
X , X 0
+
+ ≤
≤
+ ≤
≥
=
This system can be converted to an equality system by adding a slack variable to all less than
or equal constraints and an unrestricted objective function variable Z to the objective function.
After this modification we have a total number of n+1+m variables and m+1 equations.
1 1 n n
11 1 1n n 1 1
m1 1 mn n m m
1 n 1 m
Max Z c X c X 0
s.t a X a X S b
a X a X S b
X , , X , S , , S 0
− + + =
+ + =
=
+ +
≥
Matrix Algebra
The text below is copied from Chapter 3 of McCarl and Spreen book.
Matrix Formulation of the Linear Programming Problem
The matrix version of the basic LP problem can be expressed as in the equations below.
Max CX
8

s.t. AX ≤ b
X ≥ 0
Here the term CX is maximized where C is an 1xN vector of profit contributions and X is an
Nx1 vector of decision variables. This maximization is subject to inequality constraints
involving M resources so that A is an MxN matrix giving resource use coefficients by the X's,
and b is an Mx1 vector of right hand side or resource endowments. We further constrain X to
be non-negative in all elements.
It is common to convert the LP inequality system to equalities by adding slack variables.
These variables account for the difference between the resource endowment (b) and the use of
resources by the variables (AX) at no cost to the objective function. Thus, define
S = b - AX
as the vector of slack variables. Each slack variable is restricted to be nonnegative thereby
insuring that resource use is always less than or equal to the resource endowment. One slack
variable is added for each constraint equation. Rewriting the constraints gives
AX + IS = b,
where I is an MX M identity matrix and S is a Mx1 vector. The slack variables appear in the
objective function with zero coefficients. Thus, we add an 1xM vector of zero's to the
objective function and conditions constraining the slack variables to be nonnegative. The
resultant augmented LP is
MAX CX + OS
s.t. AX + IS = b
X, S ≥ 0.
Throughout the rest of this section we redefine the X vector to contain both the original X's
and the slacks. Similarly, the new C vector will contain the original C along with the zeros
for the slacks, and the new A matrix will contain the original A matrix along with the identity
matrix for the slacks. The resultant problem is
MAX CX
s.t. AX = b
X ≥ 0
Solving LP's by Matrix Algebra
LP theory reveals that a solution to the LP problem will have a set of potentially nonzero
variables equal in number to the number of constraints. Such a solution is called a Basic
Solution and the associated variables are commonly called Basic Variables. The other
variables are set to zero and are called the nonbasic variables. Once the basic variables have
been chosen; the X vector may be partitioned into XB, denoting the vector of the basic
variables, and XNB, denoting the vector of the nonbasic variables. Subsequently, the problem
is partitioned to become
MAX CBXB + CNBXNB
s.t. BXB + ANBXNB = b
XB , XNB ≥ 0.
9

The matrix B is called the Basis Matrix, containing the coefficients of the basic variables as
they appear in the constraints. ANB contains the coefficients of the nonbasic variables.
Similarly CB and CNB are the objective function coefficients of the basic and nonbasic
variables.
Now suppose we address the solution of this problem via the simplex method. The simplex
solution approach relies on choosing an initial B matrix, and then interactively making
improvements. Thus, we need to identify how the solution changes when we change the B
matrix. First, let us look at how the basic solution variable values change. If we rewrite the
constraint equation as
BXB = b - ANBXNB.
Setting the nonbasic variables (XNB) to zero gives
BXB = b.
This equation may be solved by premultiplying both sides by the inverse of the basis matrix
(assuming non-singularity) to obtain the solution for the basic variables,
B-1
BXB = IXB = B-1
b or XB = B-1
b.
We may also examine what happens when the nonbasic variables are changed from zero.
Multiply both sides of the equation including the nonbasic variables by B-1
giving
XB = B-1
b - B-1
ANB XNB.
This expression gives the values of the basic variables in terms of the basic solution and the
nonbasic variables. This is one of the two fundamental equations of LP. Writing the second
term of the equation in summation form yields
XB = B-1
b - ∑∈NBj
jj
-1
xaB
where NB gives the set of nonbasic variables and aj the associated column vectors for the
nonbasic variables Xj from the original A matrix. This equation shows how the values of the
basic variables are altered as the value of nonbasic variables change. Namely, if all but one
(Xη ) of the nonbasic variables are left equal to zero then this equation becomes
XB = B-1
b - B-1
aη Xη
This gives a simultaneous system of equations showing how all of the basic variables are
affected by changes in the value of a nonbasic variable. Furthermore, since the basic
variables must remain non-negative the solution must satisfy
XBi* = (B-1
b)i* - (B-1
aη)i* Xη = 0
This equation permits the derivation of a bound on the maximum amount the nonbasic
variable Xη can be changed while the basic variables remain non-negative. Namely, Xη may
10

increase until one of the basic variables becomes zero. Suppose that the first element of XB to
become zero is XBi*. Solving for XBi* gives
XBi* = (B-1
b)i* - (B-1
aη )i* Xη = 0
where ( )i denotes the ith
element of the vector. Solving for Xη yields
Xη = (B-1
b)i*/(B-1
ai)i*, where (b-1
aη)i*
This shows the value of Xη which causes the i*th
basic variable to become zero. Now since
Xη must be nonnegative then we need only consider cases in which a basic variable is
decreased by increasing the nonbasic variable. This restricts attention to cases where (B-1
aη)i
is positive. Thus, to preserve non-negativity of all variables, the maximum value of Xηis
Xη = {(B-1
b)i/(B-1
ai )i } for all i where (B-1
a η )i > 0
The procedure is called the minimum ratio rule of linear programming. Given the
identification of a nonbasic variable, this rule gives the maximum value the entering variable
can take on. We also know that if i*
is the row where the minimum is attained then the basic
variable in that row will become zero. Consequently, that variable can leave the basis with Xη
inserted in its place. Note, if the minimum ratio rule reveals a tie, (i.e., the same minimum
ratio occurs in more than one row), then more than one basic variable reaches zero at the same
time. In turn, one of the rows where the tie exists is arbitrarily chosen as i*
and the new
solution has at least one zero basic variable and is degenerate1
. Also, note that if all the
coefficients of Xη are zero or negative (B-1
aη)i -- for all i -- then this would indicate an
unbounded solution, if increasing the value of the nonbasic variable increases the objective
function, since the variable does not decrease the value of any basic variables.
Another question is which nonbasic variable should be increased? Resolution of this question
requires consideration of the objective function. The objective function, partitioned between
the basic and nonbasic variables, is given by
Z = CBXB + CNBXNB
Substituting the XB equation (3.1) yields
Z = CB(B-1
b – B-1
ANBXNB) + CNBXNB
or
Z = CBB-1
b - CBB-1
ANBXNB + CNBXNB
or
Z = CBB-1
b - (CBB-1
ANB - CNB)XNB
This is the second fundamental equation of linear programming. Expressing the second term
in summation notation yields
1
A degenerate solution is defined to be one where at least one basic variable equals
zero.
11

( ) jX∑∈
=
NBj
jj
-1
B
-1
B c-aBC-bBCZ
This expression gives both the current value of the objective function for the basic solution
(CBB-1
b since all nonbasic Xj equal zero) and how the objective function changes given a
change in the value of nonbasic variables. Namely, when changing Xη
( ) ηηη Xc-aBC-bBCZ -1
B
-1
B=
Since the first term of the equation is equal to the value of the current objective function (Z),
then it can be rewritten as
( ) ηηη Xc-aBC-ZZ 1-
B=
For maximization problems, the objective value will increase for any entering nonbasic
variable if its term, CBB-1
aη - cη , is negative. Thus the criterion that is most commonly used
to determine which variable to enter is: select the nonbasic variable that increases the value
of objective function the most per unit of the variable entered. Namely, we choose the
variable to enter as that variable X such that the value of CBB-1
aη - cη , is most negative. This
is the simplex criterion rule of linear programming and the term CBB-1
aη - cη, is called the
reduced cost.
If there are no variables with negative values of CBB-1
aη - cη, then the solution cannot be
improved on and is optimal. However, if a variable is identified by this rule then it should be
entered into the basis. Since the basis always has a number of variables equal to the number
of constraints, then to put in a new variable one of the old basic variables must be removed.
The variable to remove is that basic variable which becomes zero first as determined by the
minimum ratio rule. This criteria guarantees the non-negativity condition is maintained
providing the initial basis is non-negative. These results give the fundamental equations
behind the most popular method for solving LP problems which is the simplex algorithm.
Primal-Dual Solution Inter-Relationships
Every LP problem formulation can be converted into a different yet equivalent formulation.
The two corresponding setups are called primal and dual. Several relationships exist between
primal and dual solutions which are fundamental to understanding duality and interpreting LP
solutions.
In index notation:
j j
j
ij j i
j
j
Primal
Max c X
s.t. a X b for all i
X 0 for al
≤
≥
∑
∑
l j
i i
i
i ij j
i
i
Dual
Min U b
s.t. U a c for all j
U 0 for all i
≥
≥
∑
∑
In matrix notation:
12

Primal Dual
Max CX Min U b
s.t. AX b s.t. U A C
X 0 U 0
′
′≤ ≥
≥ ≥
Objective Function Interrelationships
Suppose we have any two feasible primal and dual solutions X*
, U*
and we want to determine
the relationship between the objective functions CX*
and U*
’b. We know the feasible
solutions must satisfy
* *
* *
'
b and CU AAX
0 0X U
≤ ≥
≥ ≥
To determine the relationship, we take the above constraint inequalities (not the non-
negativity conditions) and premultiply the left one by while postmultiplying the right one
by X
U
*′
*
.
* **
* **
U bU AX
U CAX
'
X
′
′
≤
≥
Noting that the term U*
’AX*
is common to both inequalities we get
b*'U*AX*'U*CX ≤≤
b*'U*CX ≤
This shows that the dual objective function value is always greater than or equal to the primal
objective function value for any pair of feasible solutions.
Constructing Dual Solutions
We can construct an optimal dual solution from an optimal primal solution. Suppose an
optimal primal solution is given by XB
*
= B-1
b and XNB
*
= 0. This solution must be feasible;
i.e., XB
*
= B-1
b ¥ 0. It also must have XNB ¥ 0 and must satisfy nonnegative reduced cost for
the nonbasic variables CBB-1
ANB - CNB ¥ 0. Given this, suppose we try U*’
= CBB-1
as a
potential dual solution.
First, let us investigate whether this is a feasible solution in the dual constraints. To be
feasible, we must have U*’
A C and U*
¥ 0. If we set U*’
= CBB-1
, then we know U*’
ANB -
CNB ¥ 0 because at optimality this is exactly equivalent to the reduced cost criteria, i.e., CBB-
1
ANB - CNB ¥ 0. Further, we know for the basic variables the reduced cost is CBB-1
AB - CB =
CBB-1
B - CB = CB - CB = 0, so U*’
B = CB. By unifying these two statements, we know when
U*
= CBB-1
then the dual inequalities U*’
A ¥ C are satisfied.
Now we need to know if the dual non-negativity conditions U ¥ 0 are satisfied. We can look
at this by looking at the slacks in the problem. For the slacks, A contains an identity matrix
and the associated entries in C are all 0's. Thus, for the part of the UA ¥ C that covers the
slacks and since we know that where A*'
sU A C≥ s s and Cs are the portions of A and C
relevant to the slacks. Substituting in the known structure of As and Cs, i.e., As = I and Cs = 0
yields U*’
¥ 0 or U ¥ 0. So the U's are non-negative. Thus, *' 1
BU = C B−
is a feasible dual
solution.
Now the question becomes, is this choice optimal? In this case the primal objective function
Z equals CX*
= CBXB
*
+ CNBXNB
*
and since XNB
*
equals zero, then Z = CBB-1
b + CNB0 =
13

CBB-1
b. Simultaneously, the dual objective equals U*’
b = CBB-1
b which equals the primal
objective. Therefore, the primal and dual objectives are equal at optimality. Furthermore,
since the primal objective must be less than or equal to the dual objective for any feasible pair
of solutions and since they are equal, then clearly CX*
cannot get any larger nor can U*’
b get
any smaller, so they must both be optimal. Therefore, CBB-1
is an optimal dual solution. This
demonstration shows that given the solution from the primal the dual solution can simply be
computed without need to solve the dual problem.
In addition given the derivation in the last chapter we can establish the interpretation of the
dual variables. In particular, since the optimal dual variables equal CB B-1
(which are called
the primal shadow prices) then the dual variables are interpretable as the marginal value
product of the resources since we showed
*1
B
Z
= = .C UB
b
′−∂
∂
Complementary Slackness
Yet another, interrelationship between the primal and dual solutions is the so called
complementary slackness relation. This states that given optimal values U*
and X*
then
( )
( )
*' *
*' *
0U b AX
0U A C X
=−
=−
This result coupled with the primal and dual feasibility restrictions (U ¥ 0; UA ¥ C; AX § b;
X ¥ 0) implies that (in the absence of degeneracy) for each constraint, either the resource is
fully used (bi - (AX)i = 0) with the associated dual variable (Ui
*
) nonzero, or the dual variable
is zero with associated unused resources (bi - (AXj)) being nonzero. Alternatively, for each
variable (again ignoring degeneracy) at optimality, either the variable level (Xj) is non-zero
with zero reduced cost (U*’
A)j - cj = 0) or the variable is set to zero with a non-zero reduced
cost.
This result may be proven using matrix algebra. Given optimal primal (X*
) and dual (U*
)
solutions
( )
( )
*' *
*' *
Let = U b AX
= U A C X
α −
β −
Now suppose we add together α + β and examine the result. Taking α + β equals U*’
(b -
AX*
) + (U*’
A - C)X*
which equals U*’
b - U*’
AX*
+ U*’
AX*
- CX*
= U*’
b - CX*
.
We know this equals zero at optimality. Further, we know that both α and β will be
nonnegative, since AX*
§ b, U*
¥ 0, U*’
A - C ¥ 0 and X*
¥ 0, thus α + β can be equal to zero
if and only if α and β are both equal to zero. Thus, complementary slackness must hold at
optimality.
The complementary slackness conditions are interpretable economically. The U*’
(AX*
- b) =
0 condition implies that the resource will: a) be valued at zero if it is not fully utilized (i.e.,
AX*
< b → U*
= 0) or b) have a nonzero value only if it is fully utilized (i.e., AX = b → U > 0
[note a zero value could occur]). Thus, resources only take on value only when they have
been exhausted. The condition ( )'
U A C X 0− = implies that a good will only be produced if
its reduced cost is zero (i.e., X > 0 → A - C = 0) and that only zero X's can have a reduced
cost (i.e., ( )'
U A C 0− > → X = 0). This last result also shows the returns (C) to every n
variable are fully allocated to the shadow prices (U) times the amount of resources (A) used
producing that activity (i.e., U’A=C).
onzero
in
14

Zero Profits
We have noted that Ui is the imputed marginal value of resource i and bi is its endowment.
Thus, U*
bi is sometimes called the "payment" to resource i. When we sum over all m
resources, the dual objective function can be interpreted as the total imputed value of the
resource endowment. If the total imputed value of the resources is viewed as a "cost", then it
makes sense that firm should seek to find U1, U2,..., Um, which minimizes .
However, at optimality the dual objective equals that of the primal
i i
i
U b∑
.c=b jj
j
ii
i
XU ∑∑
Thus, total payments to the resources will equal the profit generated in the primal problem.
All profits are allocated to resource values, and the solution insures that the imputed rate of
the resources allocated by the primal problem are such that their total value equals total
profits. Thus, if the firm had to pay U for the resources, zero profits would result.
Finding the Dual Solution Information
When you have solved the dual, you have solved the primal. Thus given the optimal B-1
, the
optimal dual variables are the primal shadow prices CBB-1
without any need for solution. In
general, one can show that the following correspondence holds
Primal Solution
Item
Dual Solution Item
Corresponding Dual Solution Information
Objective function Objective function
Shadow prices Variable values
Slacks Reduced costs
Variable values Shadow prices
Reduced costs Slacks
For example, if one wants to know the optimal values of the dual slacks, those values are the
primal reduced costs.
Degeneracy and Shadow Prices
The above interpretations for the dual variables depend upon whether the basis still exists
after the change occurs. As shown in the previous chapter, there is a right hand side range
over which the basis remains optimal. When a basic primal variable equals zero, the dual has
alternative optimal solutions. The cause of this situation is generally that the primal
constraints are redundant at the solution point and the range of right hand sides is zero. This
redundancy means one does not need a full basic solution, so one of the basic variables is set
to zero with the other basic variables likely to be nonzero. The best way to explain the
implications of this situation is through an example.
Consider the following problem
0X,X
50X
50X
100XX
2X3XMax
21
2
1
21
21
≥
≤
≤
≤+
+
15

Notice that at the optimal solution, X1 = 50, X2 = 50, the constraints are redundant. Namely,
either the combination of the last two constraints or the first two constraints would yield the
same optimal solution which is X1 = X2 = 50. The simplex solution of this problem shows a
tie for the entering variable in the second pivot where one has the choice of placing X2 into
the solution replacing either the slack variable from the first or the third constraint. If the first
slack variable (S1) is chosen as basic then one gets X1 = 50, X2 = 50, and S1 = 0 while S1 is
basic. The associated shadow prices are 0 (constraint 1), 3 (constraint 2), and 2 (constraint 3).
On the other hand, if S3 were made basic one gets X1 = 50, X2 = 50, and S3 = 0 with the
shadow prices 2 (constraint 1), 1 (constraint 2), 0 (constraint 3). Thus, there are two
alternative sets of shadow prices, both of which are optimal. (Note, the dual objective
function value is the same as the optimal primal in each case.) The main difficulty with
degeneracy is in interpreting the shadow price information. The shadow prices are taking on
a direction. Note that if one were to increase the first right hand side from 100 to 101 this
would lead to a zero change in the objective function and X1 and X2 would remain at 50. On
the other hand if one were to decrease that right hand side from 100 to 99 then one would
obtain an objective function which is two units smaller because X2 would need to be reduced
from 50 to 49. This shows that the two alternative shadow prices for the first constraint (i.e.,
0 and 2) each hold in a direction. Similarly if the constraint on X1 was increased to 51, the
objective function increases by one dollar as one unit of X2 would be removed from the
solution in favor of X1; whereas, if the constraint was moved downward to 49, it would cost
three dollars because of the reduction in X1. Meanwhile, reducing the constraint on X2 would
cost two dollars, while increasing it would return to zero dollars. Thus in all three cases
shadow prices take on a direction and the value of that change is revealed in one of the two
dual solutions. This is quite common in degeneracy and may require one to do a study of the
shadow prices or try to avoid degeneracy using a priori degeneracy resolution scheme as
implemented automatically in OSL.
For more details see chapter 4 of McCarl and Spreen book.
Solutions and Their Interpretation
Note, the text below is copied from Chapter 3 of McCarl and Spreen book. The two
fundamental equations developed in section 3.1 [of the book] may be utilized to interpret the
LP solution information. The first (3.1) shows how the basic variables change as nonbasic
variables are changed,
∑∈
=
NBj
jj
-1-1
B xaB-bBX
and the second (3.2) give the associated change in the objective function when a nonbasic
variable is changed
( ) ,xc-aBC-bBCZ
NBj
jjj
-1
B
-1
B ∑∈
=
Suppose we assume that an optimal basic solution has been found and that B and B-1
are the
associated basis and basis inverse. Now suppose we consider changing the constraint right
hand sides. The implications of such a change for the solution information may be explored
using calculus. Differentiating the above equations with respect to the right hand side b
yields
16

1
1
−
−
=
∂
∂
=
∂
∂
BC
b
Z
B
b
X
B
B
These results indicate that CBB-1
is the expected rate of change in the objective function when
the right hand sides are changed. The values CBB-1
are called the shadow prices and give
estimates of the marginal values of the resources (later they will also be called the Dual
Variables or Dual Solution). Similarly, B-1
gives the expected rate of change in the basic
variables when resources are changed. Thus when the first right hand side is changed, the
basic variables change at the rate given by the first column within the basis inverse; i.e., the
first variable changes at rate (B-1
)11, the second at (B-1
)21 and so on.
Other results may be derived regarding changes in nonbasic variables. Partially
differentiating the objective function equation with respect to a nonbasic variable yields
NBj)c-aB(C-
x
Z
jj
1-
B
j
∈=
∂
∂
This shows that the expected marginal cost of increasing a nonbasic variable equals the
negative of CBB-1
aj - cj , a consequence the CBB-1
aj - cj term is usually called reduced cost.
The marginal effect of changes in the nonbasic variables on the basic variables is obtained by
differentiating. This yields
NBjaB-
x
X
j
1-
j
B
∈=
∂
∂
which shows that the marginal effect of the nonbasic variables on the basic variable is minus
B-1
aj. The B-1
constitutes a translation from the original resource use space (i.e., aj) into the
basic variables space and tells us how many units of each basic variable are removed with a
marginal change in the nonbasic variable. We can also use these results to further interpret
the
jx
Z
∂
∂ equation. The marginal revenue due to increasing a non basic variable is equal to
its direct revenue (cj the objective function coefficient) less the value of the basic variables
(CB) times the amount of the basic variables diverted (B-1
aj). Thus, this equation takes into
account both the direct effect from increasing Xj plus the substitution effect for the basic
variables.
Fixing improper models
Infeasible models
Infeasible models are models, whose restrictions cannot be satisfied. To fix infeasibilities one
must first make the model artificially infeasible. This is done by adding artificial variables to
all constraints, which cannot be satisfied by setting the contained variables equal to zero.
Thus, one need to add one artificial variables with a coefficient of “+1” to all “¥” constraints
with a positive right hand side variable. To all “=” constraints with a non-zero right hand side,
one must add two artificial variables: one with a “+1” and one with a “-1” coefficient. All
artificial variables are restricted to be non-negative and have a large undesirable objective
function coefficient. The second step in fixing infeasible models is to solve the modified
model and examine, which artificial variables are non-zero. The equations associated with
17

non-zero variables should used as a starting point to identify misspecifications or incorrect
data.
Example:
1 2
1 2
1 2
1
Max 50X 50X
s.t. X X 50
50X X 65
X 20
+
+ ≤
+ ≤
≥
The above model is infeasible because constraint three causes X1 to be at least 20. With X1
being 20 or more, however, constraint 2 cannot be satisfied.
1 2
1 2
1 2
1
Max 50X 50X - 10000A
X X 50
50X X 65
X A 20
+
+ ≤
+ ≤
≥
The model is now artificially feasible.
Infeasibilities become more difficult to detect and fix the more complex a model is. With
artificial variables, one can narrow down the search for possible causes potentially.
Unbounded models
Models are unbounded if the objective can reach plus infinity (maximization) or minus
infinity (minimization). To detect the cause of unboundedness, one has to impose artificial
bounds at unrealistically high levels (M) on all potentially unbounded variables. Then the
model is solved and one can look for variables reaching their artificial bound. From there one
can examine the constraints, in which the variables are present and identify misspecified or
missing constraints.
0X
MX
bAXs.t.
cXMax
≥
≤
≤
LP-Formulations
The following formulations are mostly taken from the McCarl and Spreen book Chapters 5
and higher. For more explanations see the McCarl and Spreen book.
Resource allocation
jallfor0X
iallforbXas.t.
XcMax
j
i
j
jij
j
jj
≥
≤∑
∑
18

X’s are production activities, a’s are resource requirements of production activities, c’s are
benefit coefficients, and b’s are resource endowments. There are i resources and j production
activities. The objective is to maximize total benefits from production subject to given
resource endowments.
Transportation
This formulation pertains to a cost minimization problem for a transportation company. What
is the cheapest way to transport goods from a number of supply sites to a number of demand
sites?
ji,allfor0
jallford
iallforss.t.
Min
X
X
X
Xc
ij
j
i
ij
i
j
ij
i j
ijij
≥
≥
≤
∑
∑
∑∑
Mixing
The purpose of this formulation is to find the cost-minimizing combinations of ingredients,
which satisfy all nutritional restrictions.
jallfor0F
1F
iallforLLFa
iallforULFas.t.
FcMin
j
j
j
i
j
jij
i
j
jij
j
jj
≥
=
≥
≤
∑
∑
∑
∑
Multi-input, Multi-output
This formulation solves for the optimal activity level of a company, which has several multi-
input and multi-output technologies available.
kj,p,allfor0Z,Y,X
mallforbYs
kallfor0Z-Yr
pallfor0Yq-Xs.t.
Ze-Yd-XcMax
kjp
m
j
jmj
k
j
jkj
j
jpjp
k
kk
j
jj
p
pp
≥
≤
≤
≤
∑
∑
∑
∑∑∑
Spatial equilibrium (class example from GAMS block session)
Y are sales variables, T are trade variables, and X are production variables. The indexes
denote regions (r, ), products (k), crops (j), management alternatives (u), and resources (i).r
19

The coefficients represent prices (p), costs (c), resource requirements (a), yields (y), resource
endowments (b), and minimum demands (d).
T X
r,k r,k r,r,k r,r,k r,j,u r,j,u
r,k r,r,k r,j,u
r,j,u,i r,j,u r,i
j
r,r,k
r
r,k r,j,u,k r,j,u
j,ur,r,k
r
r,k r,k
r,k r,j,u
Max p Y c T c X
s.t. a X b for allr andi
T
Y y X 0 for allr and k
T
Y d for allr and k
Y , T , X 0
− −
≤
+
− ≤
−
≥
≥
∑ ∑ ∑
∑
∑
∑
∑
The objective function maximizes revenue from selling products (summed over all products
and regions) minus transportation costs (summed over all transportation ways and products)
minus the costs of production (summed over all regions, crops, and management alternatives).
The first constraint restricts the amount of resources used in production (labor and land) to be
at or below given levels. The second constraint links sales of products to trade and production.
Particularly, the amount of product sold in a region cannot exceed the amount produced plus
the amount transported from other regions minus the amount transported into other regions.
The third constraints represents minimum demand restrictions for product k in region r. The
last line gives the non-negativity conditions.
Note that all restrictions except the third one would be feasible by setting all decision
variables to zero. Thus, the third restriction is a potential candidate for infeasibility and could
be augmented using the artificial variable technique.
Sequencing
For explanations of the formulations below, see the McCarl and Spreen book.
333
222
111
321321
2121
11
TfYcXWeek3
TeYbXWeek2
TdYaXWeek1
0YYYX-X-X-Week3
0YYX-X-Week2
0YX-Week1
≤+
≤+
≤+
≤+++
≤++
≤+
321st,ktjt
mt
s
sts
k
ktj
j
jtj
3
s tt
st
k tt
kt
2
k tt
kt
j tt
jt
s t
sts
k t
ktk
j t
jtj
t,t,ts,k,j,allfor0ZY,X
tm,allforgZfYbXa
tfor t0ZY
tfor t0YXs.t.
ZeYd-Xc-Max
22
3
3
2
2
2
2
1
1
3
3
2
2
1
1
≥
≤++
∈≤+−
∈≤+−
+
∑∑∑
∑∑∑∑
∑∑∑∑
∑∑∑∑∑∑
≺≺
≺≺
20

Storage
t t t t
t t
t T
1 1 0
t t-1 t
T T-1
t t
t t
t t
Max c X - cs H
s.t. X H s
X - H H 0 for all 1 t T
X - H 0
X U for all t
X L for all t
X , H 0 for all t
≠
+ ≤
+ ≤ <
≤
≤
≥
≥
∑ ∑
<
or allr i
or allr i
or allr
Multi-objective programming
Lexicographic preferences
i
r r r
r rj j
j
mj j m
j
r r
r
r
j
r
Min w
s.t. w gl T for allr
gl g X 0 for allr
a X b for allm
w w f
w f
w 0 f
X 0 for all j
gl unrestrictedfor allr
+ ≥
− =
≤
≤ <
≤ ∞ ≥
≥
≥
∑
∑
Weighted preferences
r r
r
rj j r
j
r r r
mj j m
j
r
j
r
Max c q
s.t. g X gl 0 for allr
N q gl 0 for allr
a X b for allm
q 0 for a
X 0 for all j
gl unrestrictedfor allr
− + =
− =
≤
≥
≥
∑
∑
∑
llr
21

r r r r
r r
rj j r
r r r r r
mj j m
j
r r
j
r
Max c q c q
s.t. g X gl 0 for allr
T q T q gl 0 for allr
a X b for allm
q , q 0 for allr
X 0 for al
gl unrestrictedforallr
+ + − −
+ −
+ −
+
− =
− + + =
≤
≥
≥
∑ ∑
∑
l j
Separable Programming (Non-linear, well-behaved functions)
Well behaved functions display non-decreasing costs, or equivalently non-increasing
revenues. Here costs refer to negative objective contributions and revenues refer to positive
objective contributions.
In the graphical example below, the per unit costs increase as the input level increases. We
can approximate this nonlinear function through a stepwise linear function as shown below.
First we separate the whole variable range for variable X into different classes (steps). For
each step, we define a new non-negative variable Si and compute the average per unit cost.
A1 A2 A3 A4
Cost
Input X
Particularly, S1 represents the range between 0 and A1 and the average per unit cost are at c1
(see below).
22

A1 A2 A3 A4
Cost
Input X
c1
c4
c3
c2
The formulation of the problem is straightforward. Each step variable cannot exceed a certain
size, which is given by di. These maximum sizes are d1 = A1 – 0, d2 = A2 – A1, d3 = A3 – A2,
and d4 = A4 – A3. The total amount of input used is the sum over all step variables. Each step
variable has a separate objective function contribution, representing different per unit cost
coefficients.
i i
i
i
i
i i
i
Min c S
s.t. S X 0
S d
S , X 0
− ≤
≤
≥
∑
∑
for alli
Risk programming
See chapter 14 of the McCarl and Spreen book.
Dynamic linear programming
This is copied from McCarl and Spreen book – chapter 8.
The most straight forward linear programming model of a dynamic situation contains
multi-year enterprises of known life with multiple years modeled. This is where our
discussion of dynamic models will begin. Subsequently, models with unknown life will be
discussed, followed by sections on the equilibrium, typical single period model with known
and unknown life. For economy of description, we will refer to multi-period models as
disequilibrium models, and typical period representations as equilibrium models.
Disequilibrium - Known Life
Consider a problem involving decisions on how to commit resources over a number of time
periods considering items that, once begun, will require resources for a known number of
periods. A formulation of this problem must consider resource availability, choice of
variables and continuing resource usage during the life of multi-period enterprises. Time
preference of income and initial and terminal conditions must also be incorporated. Assume
23

all enterprises have a K year life so that an enterprise undertaken in year t lasts until t+K. A
model of this situation, assuming resource use, yields and returns are independent of the year
in which the activity begins is
jjetj,
jje
j
eTj,
j
*
ej,ej,
it
j Ke
etj,ije
Ke
e
jeje
T
t j
tj,jt
t
Keandtj,allfor0I,X
jallandKeallfor0IX
jallandKeand0eallforXX
tandiallforbXAs.t.
IFr)(1XCr)(1Max
j
j
<≥
<=+−
<>=
≤
+++
∑
∑ ∑
∑ ∑∑∑
−
−−
<
−
<
−−
j
Here t is the year identifier and begins at zero for the first year; T is the last year in the model;
j identifies alternative variables; e indexes the elapsed age of an enterprise; Kj is the life of
enterprise j (i.e., the number of periods after starting enterprise j that resources are used); i
indexes resources; r is the discount rate; Cjt is the profit from initiating variable Xjt in year t;
Xjt is the number of units of enterprise j initiated in year t; Fje is the terminal value of the jth
enterprise when it is e years old after the explicit time period in the model lapses; Ije is the
number of units of the jth enterprise which are e years old at the model completion; Aije is the
usage of resource i by the jth alternative when it is e years old; bit is the endowment of
resource i in year t; and X*
j,-e is the initial condition giving the amount of Xj done e years
before the model begins.
The model maximizes net present value of activity from years 0 through T plus the terminal
value of in-process inventory. Annual profits are converted to present value by multiplying
through by the discount factor. The model assumes the profits (Cjt) are collapsed into the
period when the Xj variable is initiated. This is done by calculating the term Cjt as
t}T,tK{ewhere
h)r+(1=C
j
je
e)+(t
e
jt
−−≤
−
∑
where hje is the net profit from the jth enterprise when it is e years old. The inequality for e
accounts for cases where the life of the enterprise is both in process and possibly extends
beyond the time horizon. The second term of the objective function gives the terminal value
of enterprises not entirely completed during the years the model explicitly covers. For exam-
ple, enterprises begun in the last year explicitly modeled are valued at Fj1, since they will be
one year old in the period following the model. The Fje terms depict a future stream of profits
to activities lasting beyond the model horizon discounted back to year T.
The first model constraint requires resource use to be less than or equal to resource
availability in that year. Annual resource use depends upon enterprises begun in the current
time period as well as items that began up to Kj years in the past. Thus, the summation
subscripting adds all variables that are of age 0 to Kj years old in the current year (t). Re-
source use is assumed to depend only on the elapsed age (e) of the activity. Each alternative
(j) once committed will always exist for years t+0 to t+Kj using an a priori specified amount
of resources (Aije) each year. The next constraint set specifies the initial conditions: i.e., the
amount of enterprises undertaken before the model was started which use resources during the
time period covered by the model. The last constraint set adds up in process inventory at the
end of the explicit time horizon setting the variable Ije equal to the amount of the ith enterprise
which is e years old in the final model year and will extend into future years, thus the e=Kj
case is not included.
The formulation assumes equal length periods.
24

Disequilibrium - Unknown Life
A second possibility involves modeling of situations wherein the exact life of activities is to
be endogenously determined (although a maximum life is known) in a multi-period context.
A formulation of this problem must include the considerations in the known life model as well
as variables depicting the decision on whether to continue or terminate an activity. Adopting
the same assumptions as above, a general formulation of this model is
et,j,allfor0I,X
Keand0eandjt,allfor0X
Keandjallfor0IX
KeandjallforXX
tandiallforbXAs.t.
IFr)(1XCr)(1Max
jeet,j,
jet,j,1,1,
jjeeT,j,
j
*
ej,0,ej,0,
itet,j,
j Ke
ije
je
j Ke
je
T
et,j,
t j Ke
je
t
j
jj
≥
≤>≤+−
<=+−
<=
≤
+++
−−
≤
<
−
≤
−
∑ ∑
∑ ∑∑∑ ∑
etjX
Where t is the time index; T is the length of the total planning horizon; j is an index for
identifying alternative variables; e is an index identifying the elapsed age of a variable; Kj is
the maximum age of a variable; i is an index which identifies resources; r is the discount rate;
Cje is the per unit profit from variable j when it is of age e; Xj,t,e is the number of units of
alternative j on hand in period t which are of elapsed age e; Fje is the terminal value of
incomplete units of enterprise j which are of elapsed age e; Aije is the resource i usage by one
unit of the production represented of enterprise j when it is of elapsed age e; bit is the
endowment of resource i in period t; and Xj,o,e is the initial amount of enterprise j of elapsed
age e before the model begins (in time period 0).
This model has many common features with the known life model. In particular, the
objective function interpretation is virtually identical. However, there are several distinctive
features that should be brought out. Since the life of a production variable may be terminated,
the decision variable now is Xj,t,e which allows for the possibility of employing variable j in
year t when it is of elapsed age e. The constraint before the non-negativity conditions relates
the amount of e year old variable in year t to the amount of e-1 year old variable in year t-1.
The inequality on this constraint permits the predecessor to be nonzero and the successor to be
zero (thus depicting selections less than maximum life for the variable).
This model is probably more realistic than the earlier one, but this has a price. The number of
variables has been multiplied by the number of years a production activity could exist
and the number of constraints has expanded. This is potentially a much larger
model.
( J
J
K 1+∑ )
Comments
An alternative form of the disequilibrium unknown life model merges the concepts of the
known and unknown life models. In this model the linkage equations above are dropped and
the X variables are assumed to have known life wherein the alternative variables are formula-
ted for each possible activity age. Thus in our example, we would have first year variables for
strawberries planted in the first year and kept with a known life of 1, 2, 3, and 4 years (4
variables). Such an example will be illustrated in the equilibrium unknown life section.
Equilibrium - Known Life
Disequilibrium models are usually relatively large. Further, the specification of the initial and
terminal conditions is of key importance. However, such conditions can be difficult to
specify. Equilibrium models provide an alternative approach. In an equilibrium model, the
firm is assumed to operate in a steady state manner, repeatedly making identical decisions
25

year after year. In such a case, initial and final inventories of in-process goods will be equal;
thus, the initial and final conditions may simply be set equal and their optimal levels
determined by the model. This leads to a smaller model depicting a representative
equilibrium year. Equilibrium solutions, however, do not portray growth situations or time
paths of adjustment; only final equilibriums are created.
Equilibrium models are developed as follows: assume we have a variable with life of 4
periods and resource use, yield, etc., equal to Ae, where e is the elapsed age of the activity (0-
3). Let us (assuming we start with zero initial activity) portray the resource use over several
periods.
Begin Activity in Period
1 2 3 4 5 6 7
Period 1 Resource
Usage
A0
Period 2 Resource
Usage
A1 A0
Period 3 Resource
Usage
A2 A1 A0
Period 4 Resource
Usage
A3 A2 A1 A0
Period 5 Resource
Usage
A3 A2 A1 A0
Period 6 Resource
Usage
A3 A2 A1 A0
Period 7 Resource
Usage
A3 A2 A1 A0
Once the activity pattern enters period 4 we have resource use by each age of the activity. In
this and subsequent periods, resource use in any period may be written as
where Xt is the quantity of the activity begun in period t.
XA+XA+XA+XA=R t01t-12t-23t-3t
Under an equilibrium assumption the same thing is done each year in a continually repeated
sequence. Thus, Xt = Xt-1 = Xt-2 = Xt-3 and the resource use may be rewritten as
AX=X)A+A+A+A(=XA+XA+XA+XA=R e
e
tt3210t0t1t2t3t Σ
In an equilibrium model, then, the resource use for a variable is the sum of its resource usages
over its whole life. (At this juncture we should observe that these assumptions imply that
initial and final inventories are equal.)
Thus, a general formulation of the equilibrium model with known life is
jallfor0X
jallforbXAs.t.
XCMax
j
i
j
jij
j
jj
≥
≤∑
∑
where: Xj is the quantity of the jth enterprise produced in equilibrium.
Cj is the revenue per unit of Xj and equals the sum of the returns to the activity over
the periods (e) of its life, which is assumed to be known;
Aij is the use of resource i per unit of Xj and equals Aije or the sum of the resource
usages over the years of the enterprise=s life; and
26

bi is the amount of resource i available.
Comments
Several aspects of the equilibrium model require discussion. First when a model
reaches an equilibrium, the resources available must be in equilibrium. Thus, the same
amount of resources must be available on a continuing basis. Similarly, resource use of a
variable of a given age must be the same year after year. Such assumptions rule out the use of
this model when examining cases with depletable resources.
A second assumption involves the word repeatability. Namely, the plan at any point in time
must be repeatable. Given an alternative which lasts n periods the solution will have 1/n units
of the alternative in each stage of its life. In the wheat strawberry 1/3 of the acreage is in each
of the three years of the strawberries life. Further, in the corn-soybean example, no more than
1/2 the acreage will be devoted to corn following soybeans and an equal acreage must be
devoted to soybeans followed by corn.
A final line of comments relates to discounting. These authors have not examined
discounting at great length, but feel that discounting should not be included in this model
since the return in every year is identical.
Equilibrium - Unknown Life
A version of the equilibrium model may be formulated in which the multi- period activities
are of unknown life. The assumption is retained that the same thing is done each period.
Thus, resource use in a typical period is summed across the possible ages that an activity may
be kept. A general formulation of the model is
je je
j e
ije je i
j e
je j,e 1
je
Max C X
s.t. A X b for all i
X X 0 for all j and e > 0
X 0 for all j and e
−
≤
− ≤
≥
∑∑
∑∑
where: Xje is the quantity of the jth activity produced and kept until it is e periods old;
Cje is the per unit returns to Xje;
Aije is the per unit usage of resource i by Xje;
bi is the amount of resource i available;
The model maximizes profits subject to constraints on resource use and age sequencing. The
age sequencing constraints state that the amount of enterprise j of age e must be less than or
equal to the amount of that enterprise that was kept until age e-1.
Overall Comments on Dynamic Linear Programming
Dynamic LP may be used in many settings. The models, however, do involve numerous
assumptions and require different computational and data resources. The disequilibrium
models in particular become quite large when used and also require constructing "good"
terminal conditions. Equilibrium models are more compact, yet use assumptions which may
be unrealistic given a problem situation.
Further, both models suffer from a "curse of certainty"; clearly the optimal solution depends
upon the assumed current and future conditions. However, the model can include uncertainty
through the expected value variance and MOTAD formulations, multistage discrete stochastic
method, or other approaches.
27

Recursive Programming
Another LP formulation dealing with dynamic situations is known as recursive programming.
Basically, recursive programming models involve problems in which model coefficients are
functionally dependent upon earlier model solutions and an exogenously specified time path.
A recursive programming model consists of a constrained optimization model; and a data
generator. The data generator given the solution in period t, prepares the input for period t+1
including defining a set of constraints which relates the feasible values of current variables to
past values of variables and exogenous events. The model then is optimized for each time
period with the data generator updating the model to the next time period.
While recursive programming does not have to be cast as a LP, the LP version will be
presented. The general recursive LP sequentially maximizes an LP for each of a number of
periods.
kandjallfor0Y,X
iallforbYEXAs.t.
YdXCZtperiodforMax
ktjt
it
k
ktik
j
jtijt
k
ktk
j
jtjtt
≥
≤+
+=
∑∑
∑∑
where: Cjt are objective function parameters functionally dependent upon the previous
objective function parameters (Cj,t-1), lagged optimal decision variables (Xjt-1, Ykt-1), and
exogenous events; Xjt are the values of the decision variables at time t; Aijt are the resource i
usages by Xjt functionally dependent upon lagged values, dk are objective function
coefficients which are stable over time; Ykt are the optimal values of the kth Y variable in
time period t; Eik are the usages of resource i which do not change over time; bit are the
resource i limits, functionally dependent upon lagged phenomena.
The recursive model is not solved over all time periods simultaneously, rather an optimal
solution is constructed for each time period. There is no guarantee (or necessarily a desire)
that these solutions are optimal over all periods. Rather, the solution represents an adaptive
stream of decisions.
We have not specified the functional form of the lagged functions as they most often depend
upon the problem at hand.
Comments on Recursive Programming
Recursive programming models explicitly represent lagged adjustment. Several forms of
lagged adjustment have been commonly used. A frequently used form involves "flexibility
constraints" in which a variable is allowed to vary at most from its predecessor by a
percentage (as done in the example).
Integer programming
For more information, see the chapters 15 and 16 of the McCarl and Spreen book.
Pure (seldom) and mixed (frequently) problems
Fractional solutions are not always acceptable
Duality is not a well-defined subject in the context of IP
Allows to approximate not well behaved functions
Fixed Costs
28

Max CX - FY
s.t. X - MY 0
X 0
Y 0 o
≤
≥
= r 1
in
,1
⋅ ≤
Fixed Capacity
m m k k
m k
m km k
k
m k
Max C X - F Y
s.t. X - Cap Y 0 for all m
X 0, Y 0 or 1 for all k and m,
≤
≥ =
∑ ∑
∑
Minimum Habitat Size
The following formulation can be used to represent the relationship between habitats and
supported population. Particularly, habitats must have a minimum size to support a certain
population at all. Beyond that area size, the supported population is the product of area size
times species density per area unit.
( )
0 m
0 min
1 max min
1 min
0 1
HAB h
HAB h I 0
HAB h h I 0
POP d HAB d h I 0
POP, HAB , HAB 0
I 0
≤
− +
− − ⋅ ≤
− ⋅ − ⋅ ⋅ ≤
≥
∈
hmin
0 area
population
HAB0
HAB1
29

Knapsack
The formulation solves a item selection problem for a hiker who has limited space in his
backpack. It can be applied to a variety of similar problem contexts.
For more explanations of the formulations below, see the McCarl and Spreen book.
No item repetition
jallfor1or0X
WXds.t.
XvMax
j
j
jj
j
jj
=
≤∑
∑
Item repetition
j j
j
j j
j
j j
j
Max v X
s.t. d X W
X U
X for
≤
≤
∈ all j
∑
∑
Warehouse
kj,i,allfor0Z,Y,X1,or0V
mallforb
VA
kallfor0YVCAP
kallfor0YX
jallforDZY
iallforSZXs.t.
ZEYDXCVFMin
ijkjikk
m
k
kmk
j
kjkk
j
kj
i
ik
j
i
ij
k
kj
i
j
ij
k
ik
i j
ijij
k j
kjkj
i k
ikik
k
kk
≥=
≤∑
≤∑+−
≤∑+∑−
≥∑+∑
≤∑+∑
∑∑+∑∑+∑∑+∑
Logical Conditions
Mutual exclusive products
1
2
1 2
1 2
X MY 0
Z MY 0
Y Y 1
X, Z 0
Y , Y 0 or 1
− ≤
− ≤
+ ≤
≥
=
30

X and Z are mutually exclusive products. For X to be produced, Y1 needs to be 1. For Z to be
produced, Y2 needs to be 1. However, Y1 and Y2 cannot be one simultaneously. Thus, only
X or Z can be produced. Note the objective function is not shown here.
Either-Or-Active constraints
1 1
2 2
A X - MY b
A X + MY b M
X 0
Y 0 or 1
≤
≤ +
≥
=
M is a sufficiently large coefficient.
Y = 1: The first constraint is not active but the second is.
Y = 0: The first constraint is active but the second is not
Multiple Active Constraints.
iallfor1or0Y0,X
P-KY
bMY-XA
:
:
bMY-XA
bMY-XA
i
i
i
kkk
222
111
=≥
=∑
≤
≤
≤
Logical Restrictions
1or0Z,0Y,X
0RZ-Y
0MZ-X
ki
k
k
i
i
=≥
≥∑
≤∑
Distinct Variable Values
1 1 2 2 k k
1 2 k
k
X - VY - V Y ... -V Y 0
Y Y ... +Y 1
X 0 Y 0 or 1.
=
+ =
≥ =
V represent the distinct values. Y are binary variables of which only one can take on a value
of 1.
Non-linear representation
1or0,X,XZ,
0X-X-2Z
1XXZ-
21
21
21
=
≤
≤++
31

Approximation of not well behaved non-linear functions
Suppose we have to approximate a non-concave function. Let us approximate this function
through 4 grid points A1 through A4 (as shown in the picture below). Then we can
approximate any point on this function as a convex combination of two adjacent points.Cost
Input X
A1 A2 A3 A4
Let l1 through l4 represent the weights for the grid points A1 through A4.
1 2 3 4
1 1
2 2
3 3
4 4
1 2 3 4
1 3
1 4
2 4
1 2 3 4
1 2 3 4
1
- Z 0
- Z 0
- Z 0
- Z 0
Z Z Z Z 2
Z Z
Z Z
Z Z
, , ,
Z , Z , Z , Z = 0 or 1
λ + λ + λ + λ =
λ ≤
λ ≤
λ ≤
λ ≤
+ + + ≤
+ ≤
+ ≤
+ ≤
λ λ λ λ ≥
1
1
1
0
32

Mathematical linear programming notes

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