Linear Programming Problems.

1. Linear Programming Problems
Dr. Purnima Pandit
Assistant Professor
Department of Applied Mathematics
Faculty of Technology and Engg.
The Maharaja Sayajirao University of Baroda

2. Different Kinds of
Optimization

3. Elements
 Variables
 Objective function
 Constraints
Obtain values of the variables
that optimizes the objective function
( satisfying the constraints )

4. Linear Programming (LP)
LP is a mathematical modeling
technique used to achieve an objective,
subject to restrictions called constraints

5. Types of LP

6. LP Model Formulation
 Decision variables
 mathematical symbols representing levels of activity of an
operation
 Objective function
 a linear relationship reflecting the objective of an operation
 most frequent objective of business firms is to maximize profit
 most frequent objective of individual operational units (such as
a production or packaging department) is to minimize cost
 Constraint
 a linear relationship representing a restriction on decision
making

7. LP Model Formulation (cont.)
Max/min z = c1x1 + c2x2 + ... + cnxn
subject to:
a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
:
am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
xj = decision variables
bi = constraint levels
cj = objective function coefficients
aij = constraint coefficients

8. LP Model: Example
RESOURCE REQUIREMENTS
Labor Clay Revenue
PRODUCT (hr/unit) (lb/unit) ($/unit)
Bowl 1 4 40
Mug 2 3 50
There are 40 hours of labor and 120 pounds of clay
available each day
Decision variables
x1 = number of bowls to produce
x2 = number of mugs to produce

9. LP Formulation: Example
Maximize Z = $40 x1 + 50 x2
Subject to
x1 + 2x2 40 hr (labor constraint)
4x1 + 3x2 120 lb (clay constraint)
x1 , x2 0
Solution is x1 = 24 bowls x2 = 8 mugs
Revenue = $1,360

10. Graphical Solution Method
1. Plot model constraint on a set of coordinates
in a plane
2. Identify the feasible solution space on the
graph where all constraints are satisfied
simultaneously
3. Plot objective function to find the point on
boundary of this space that maximizes (or
minimizes) value of objective function

11. Convex Hull
 A set C is convex if
every point on the line
segment connecting x
and y is in C.
 The convex hull for a
set of points X is the
minimal convex set
containing X.

12. Graphical Solution: Example
x2
50 –
40 –
4 x1 + 3 x2 120 lb
30 –
20 – Area common to
both constraints
10 –
x1 + 2 x2 40 hr
0– | | | | | |
10 20 30 40 50 60 x1

13. Computing Optimal Values
x2 x1 + 2x2 = 40
40 –
4x1 + 3x2 = 120
4 x1 + 3 x2 120 lb
4x1 + 8x2 = 160
30 – -4x1 - 3x2 = -120
5x2 = 40
20 –
x1 + 2 x2 40 hr x2 = 8
10 – 8
x1 + 2(8) = 40
0– | | 24 | | x1
x1 = 24
10 20 30 40
Z = $50(24) + $50(8) = $1,360

14. Extreme Corner Points
x1 = 0 bowls
x2 x2 =20 mugs
x1 = 224 bowls
Z = $1,000
40 – x2 =8 mugs
Z = $1,360 x1 = 30 bowls
30 – x2 =0 mugs
20 – A
Z = $1,200
10 –
B
0– | | | C|
10 20 30 40 x1

15. Objective Function
x2
40 –
4x1 + 3x2 120 lb
30 – Z = 70x1 + 20x2
Optimal point:
x1 = 30 bowls
20 –A
x2 =0 mugs
Z = $2,100
10 – B
x1 + 2x2 40 hr
0– | | | C |
10 20 30 40 x1

16. Minimization Problem
CHEMICAL CONTRIBUTION
Brand Nitrogen (lb/bag) Phosphate (lb/bag)
Gro-plus 2 4
Crop-fast 4 3
Minimize Z = $6x1 + $3x2
subject to
2x1 + 4x2  16 lb of nitrogen
4x1 + 3x2  24 lb of phosphate
x 1, x 2  0

17. Graphical Solution
x2
14 –
x1 = 0 bags of Gro-plus
12 – x = 8 bags of Crop-fast
2
10 –
Z = $24
8–A
Z = 6x1 + 3x2
6–
4–
B
2–
C
0– | | | | | | |
2 4 6 8 10 12 14 x1

18. Two type of Constraints
Binding and Nonbinding constraints:
A constraint is binding if the left-hand and right-
hand side of the constraint are equal when the
optimal values of the decision variables are
substituted into the constraint.
A constraint is nonbinding if the left-hand side
and the right-hand side of the constraint are
unequal when the optimal values of the decision
variables are substituted into the constraint. 18

19. Special Cases of Graphical Solution
• Some LPs have an infinite number of solutions
(alternative or multiple optimal solutions).
• Some LPs have no feasible solution (infeasible
LPs).
• Some LPs are unbounded: There are points in
the feasible region with arbitrarily large (in a
maximization problem) z-values.
19

20. Alternative or Multiple Solutions
X2
60
B
D
50
Any point (solution) Feasible Region
falling on line segment
40
AE will yield an optimal
solution with the same E
30
objective value z = 100
z = 120
20
z = 60
10
F
A C
10 20 30 40 50 20
X1

21. No feasible solution
X2
60
No Feasible Region
50
x1 >= 0
40
Some LPs have no
x2 >=0
solution. Consider
30
the following
formulation:
20
10
No feasible region exists
10 20 30 40 50 21
X1

22. Unbounded LP
X2
Feasible Region
6 D
The constraints are
5 B
satisfied by all points z=4
bounded by the x2 axis 4
and on or above AB and 3
CD.
2
There are points in the
feasible region which 1
z=6
will produce arbitrarily
A C
large z-values 1 2 3 4 5 6 X1
(unbounded LP). 22

23. Higher Dimensional Problems
Graphical Method is useful for two
dimensional Problems.
Simplex method for 2 and higher
dimensional problems

24. Cannonical form
 Standard form is a basic way of describing a LP
problem.
 It consists of 3 parts:
A linear function to be maximized
maximize c1x1 + c2x2 + … + cnxn
Problem constraints
subject toa11x1 + a12x2 + … + a1nxn < b1
a21x1 + a22x2 + … + a2nxn < b2
…
am1x1 + am2x2 + … + amnxn < bm
Non-negative variables
x1, x2 ,…, xn > 0

25. Obtain Standard Form
IMPORTANT: Simplex only deals with equalities
General Simplex LP model:
min (or max) z =  ci xi
s.t.
Ax=b
x0
In order to get and maintain this form, use
 slack, if x  b, then x + slack = b
 surplus, if x  b, then x - surplus = b
 artificial variables (sometimes need to be added to
ensure all variables  0 )

26. Different "components" of a LP model
 LP model can always be split into a basic
and a non-basic part.
 “Transformed” or “reduced” model is
another good way to show this.
 This can be represented in mathematical
terms as well as in a LP or simplex
tableau.

27. Simplex
 A simplex or n-simplex is
the convex hull of a set of
(n +1) . A simplex is an n-
dimensional analogue of a
triangle.
 Example:
 a 1-simplex is a line segment
 a 2-simplex is a triangle
 a 3-simplex is a tetrahedron
 a 4-simplex is a pentatope

28. Movement to Adjacent
Extreme Point
Given any basis we move to an adjacent extreme point
(another basic feasible solution) of the solution space by
exchanging one of the columns that is in the basis
for a column that is not in the basis.
Two things to determine:
1) which (nonbasic) column of A should be brought
into the basis so that the solution improves?
2) which column can be removed from the basis
such that the solution stays feasible?

29. Entering and Departing Vector
(Variable) Rules
General rules:
 The one non-basic variable to come in is the one
which provides the highest reduction in the objective
function.
 The one basic variable to leave is the one which is
expected to go infeasible first.
NOTE: THESE ARE HEURISTICS!!
Variations on these rules exist, but are rare.

30. Example Problem
Maximize Z = 5x1 + 2x2 + x3
subject to
x1 + 3x2 - x3 ≤ 6,
x2 + x3 ≤ 4,
3x1 + x2 ≤ 7,
x1, x2, x3 ≥ 0.

31. Simplex and Example Problem
Step 1. Convert to Standard Form
a11 x1 + a12 x2 + ••• + a1n xn ≤ b1, a11 x1 + a12 x2 + ••• + a1n xn + xn+1 = b1,
a21 x1 + a22 x2 + ••• + a2n xn ≥ b2, a21 x1 + a22 x2 + ••• + a2n xn - xn+2 = b2,

am1 x1 + am2 x2 + ••• + amn xn ≤ bm, am1 x1 + am2 x2 + ••• + amn xn + xn+k = bm,
In our example problem:
x1 + 3x2 - x3 ≤ 6, x1 + 3x2 - x3 + x4 = 6,
x2 + x3 ≤ 4, x2 + x3 + x5 = 4,
3x1 + x2 ≤ 7, 3x1 + x2 + x6 = 7,
x1, x2, x3 ≥ 0. x1, x2, x3, x4, x5, x6 ≥ 0.

32. Simplex: Step 2
Step 2. Start with an initial basic feasible solution (b.f.s.) and set up the
initial tableau.
In our example
Maximize Z = 5x1 + 2x2 + x3
x1 + 3x2 - x3 + x4 = 6,
x2 + x3 + x5 = 4,
3x1 + x2 + x6 = 7, cB Basis cj Constants
5 2 1 0 0 0
x1, x2, x3, x4, x5, x6 ≥ 0. x1 x2 x3 x4 x5 x6
0 x4 1 3 -1 1 0 0 6
0 x5 0 1 1 0 1 0 4
0 x6 3 1 0 0 0 1 7
c row 5 2 1 0 0 0 Z=0

33. Step 2: Explanation
Adjacent Basic Feasible Solution
If we bring a nonbasic variable xs into the basis, our system changes from the
basis, xb, to the following :
x1 + ā1sxs= b1 x = b a for i =1, …, m
i i is
xr + ārsxr= b xs = 1
 r
xj = 0 for j=m+1, ..., n and js
xm + āmsxs= b
 s
The new value of the objective function becomes:
m
Z  c (b  a
i 1
i i is )  c s
Thus the change in the value of Z per unit increase in xs is
cs = new value of Z - old m
m
value of Z
=  c (b  a
i 1
i
m
i is )  c s  c b
i 1
i i
This is the Inner Product rule
= c  c a
s i is
i 1

34. Simplex: Step 3
Use the inner product rule to find the relative profit coefficients
cB Basis cj Constants
5 2 1 0 0 0
x1 x2 x3 x4 x5 x6
0 x4 1 3 -1 1 0 0 6
0 x5 0 1 1 0 1 0 4
0 x6 3 1 0 0 0 1 7
c row 5 2 1 0 0 0 Z=0
c j  c j  cB Pj
c1 = 5 - 0(1) - 0(0) - 0(3) = 5 -> largest positive
c2 = ….
c3 = ….
Step 4: Is this an optimal basic feasible solution?

35. Simplex: Step 5
Apply the minimum ratio rule to determine the basic variable to leave the basis.
The new values of the basis variables:
xi = bi  a is x s for i = 1, ..., m
 bi 
max xs  min 
a is  0 a is
 
In our example:
cB Basis cj Constants
5 2 1 0 0 0 Row Basic Variable Ratio
x1 x2 x3 x4 x5 x6 1 x4 6
0 x4 1 3 -1 1 0 0 6 2 x5 -
0 x5 0 1 1 0 1 0 4 3 x6 7/3
0 x6 3 1 0 0 0 1 7
c row 5 2 1 0 0 0 Z=0

36. Simplex: Step 6
Perform the pivot operation to get the new tableau and the b.f.s.
cB Basis cj Constants
5 2 1 0 0 0
x1 x2 x3 x4 x5 x6
0 x4 1 3 -1 1 0 0 6
0 x5 0 1 1 0 1 0 4
0 x6 3 1 0 0 0 1 7
c row 5 2 1 0 0 0 Z=0
New iteration:
find entering cB Basis cj Constants
variable: 5 2 1 0 0 0
c j  c j  c B Pj x1 x2 x3 x4 x5 x6
0 x4 0 8/3 -1 1 0 0 11/3
cB = (0 0 5) 0 x5 0 1 1 0 1 0 4
5 x1 1 1/3 0 0 0 1/3 7/3
c2 = 2 - (0) 8/3 - (0) 1 - (5) 1/3 = 1/3 c row 0 1/3 1 0 0 -5/3 Z=35/3
c3 = 1 - (0) (-1) - (0) 1 - (5) 0 = 1
c6 = 0 - (0) 0 - (0) 0 - (5) 1/3 = -5/3

37. Final Tableau
cB Basis cj Constants x3 enters basis,
5 2 1 0 0 0 x5 leaves basis
x1 x2 x3 x4 x5 x6
0 x4 0 8/3 -1 1 0 0 11/3
0 x5 0 1 1 0 1 0 4
5 x1 1 1/3 0 0 0 1/3 7/3
c row 0 1/3 1 0 0 -5/3 Z=35/3
cB Basis cj Constants
5 2 1 0 0 0
x1 x2 x3 x4 x5 x6
0 x4 0 11/3 0 1 1 0 23/3
1 x3 0 1 1 0 1 0 4
5 x1 1 1/3 0 0 0 1/3 7/3
c row 0 -2/3 0 0 -1 -5/3 Z=47/3

38. Computational Considerations
 Unrestricted variables (unboundedness)
 Redundancy (linear dependency, modeling
errors)
 Degeneracy (some basic variables = 0)
 Round-off errors

39. Simplex Variations
Various variations on the simplex method exist:
 "regular" simplex
 two-phase method: Phase I for feasibility
and Phase II for optimality
 Condensed / reduced / revised method: only
use the non-basic columns to work with
 (revised) dual simplex, etc.

40. Limitations of Simplex
1. Inability to deal with multiple objectives
2. Inability to handle problems with integer variables
Problem 1 is solved using Multiplex
Problem 2 has resulted in:
 Cutting plane algorithms (Gomory, 1958)
 Branch and Bound (Land and Doig, 1960)
However,
solution methods to LP problems with integer or
Boolean variables are still far less efficient than those
which include continuous variables only

41. Primal and Dual LP Problems
•Economic theory indicates that scarce (limited) resources
have value. In LP models, limited resources are allocated, so
they should be, valued.
•Whenever we solve an LP problem, we implicitly solve two
problems: the primal resource allocation problem, and the
dual resource valuation problem.
•Here we cover the resource valuation, or as it is commonly
called, the Dual LP.

42. Max c X j j
Primal s.t. a X
j
ij j  bi for all i
j
Xj  0 for all j
Min b Y i i
Dual s.t.
i
a Y
i
ji i  c j for all j
Yi  0 for all i