2. UNIT II:LINEAR PROGRAMMING MODELS
Upon completion of the learning unit, the student will be able to
apply sensitivity and duality analysis to determine the parameter
values for which the solution of the linear programming model
remains valid. Will be able to solve linear programming models with
integer constraint for the decision variables. Will be able to
understand the concept of network models, as well as to use them to
solve transportation problems.
3. EVIDENCES:
EC1: Concept and operation of
sensitivity analysis
EC2: Concept and performance of
duality
EC3: Concept and operation of integer
programming models
4. The Simplex method was developed by Dantzig in 1947 and is an
algebraic method of geometric concepts. It is an iterative method
(repeats a series of fixed steps) that starts by finding an initial
feasible solution.
The method evaluates how much each potential solution improves the
objective function(Z) and chooses to move to the one with the best
improvement rate.
To simplify the original idea, a variant of the method called the
revised Simplex Method was developed that accommodates computer
processing. This variant uses matrix operations.
5. Linear Programming in its fundamentals has mainly four assumptions:
Proportionality
Additivity
Divisibility
Certainty
We have not repaired in these properties or assumptions of the LP since
in the reality of the problems these are not fulfilled. However, it is
necessary for the Simplex method that there are no large deviations from
them.
6. The application of the Simplex method allows us to perform future
post-optimality tasks with sensitivity analysis and parametric linear
programming. All this is facilitated by the computer programs
available today, which makes the task easier.
This also allows me to make decisions on what I can admit or if I
should improve the information to further refine the parameters.
7. SENSITIVITY ANALYSIS OR
POST-OPTIMALITY ANALYSIS
In general it is assumed that all model
parameters are known, but it happens that
in reality in most cases they are estimates.
For all these reasons, sensitivity analysis
becomes an important part of the decision
making process. Sensitivity analysis allows
us to see what effect changing parameter
values have on the optimal solution.
8. In general it is necessary to:
1. Determine the parameters that if changed will produce changes
in the optimal solution
2. 2. Determine the interval of values so that the optimal solution
does not change. (permissible interval)
3. 3. Determine the range of values for the solution to remain
feasible. feasible.
Ultimately I am interested in knowing how sensitive the optimal
solution is to inaccurate data.
9. DUALITY OPERATION - SHADOW
PRICES
Also called Lagrange multipliers in nonlinear programming.
The shadow price of a constraint gives the rate of change of the
objective function as the right-hand side (LD) of that constraint
increases.
In other words, it shows how the objective function (Z) varies with
unit changes in the resource at the optimum.
10. For example, a shadow price equal to zero for a constraint would
indicate that an increase in resources has no effect on the current
optimal solution. Resources has no effect on the current optimal
solution.
Therefore we could say:
Shadow price = Marginal value of the resource.
A constraint of type > will have a negative shadow price.
A constraint of type < will have a positive shadow price.
A constraint of type = will have a positive, negative or zero shadow
price.
11. It is the meaning of the Lagrange multiplier, which represents the
variation of a given objective when an additional unit of a certain
constrained resource is available.
The Shadow Price of a resource (associated to the availability limit
of that resource or Right Side of the Constraint) is the amount
corresponding to the change in the objective function as a
consequence of a unit increase in the availability of the resource
(i.e. in the value of the Right Side of the constraint).
12. Example SENSITIVITY AND SHADOW PRICE ANALYSIS:
The manufacturing company Omega discontinued production of a certain
unprofitable product line. This move created considerable excess production
capacity. Management wants to dedicate this capacity to one or more of three
products, called 1, 2 and 3. The following table summarizes the available
capacity of each machine that can limit production.
The number of machine-hours required to manufacture each unit of the
respective products is:
Type of machine Time available (in machine-hours per
week)
Milling machine 500
Lathe 350
Grinding machine 150
13. Productivity ratio (in machine hours per unit)
The sales department indicates that the potential sales of products 1 and 2
exceed the maximum production rate and that the potential sales of product 3
are 20 units per week. The unit profit would be $50, $20, and $25, for products 1,
2, and 3, respectively. The objective is to determine how many products of each
type the company must produce to maximize profit.
a) Formulate a linear programming model for this problem.
b) Use a computer to solve this model using the simplex method.
Type of
machine
Product 1 Product 2 Product 3
Milling machine 9 3 5
Lathe 5 4 0
Grinding
machine
3 0 3
15. Set all the variables as seen in the previous method, remember:
We put the
objective function
with the model
equation in terms
of equation
The constraints are
put in terms of each
equation in separate
cells with their least
equal time available
We have in here empty cell for that the solver
put the results automatically
19. Manufacture zero products of product
1, 87 and a half pieces of product 2
and 47 and a half pieces of product 3.
With a profit of $2938.5 pesos
Using
500=500 all milling time.
350=350 all lathe time
150 ≠ 142.5 and with a surplus of
grinding machine time.
20. Interpretation of the sensitivity analysis:
For the model to be fulfilled and give us the gain
we predict in the previous point I can only make
some allowable adjustments such that for product
1 I can increase 1 product and I can decrease as
much as I want up to 1x1030.
For product two I can increase as much as I want
up to 1x1030 of production, however I can only
reduce 1 product.
For product three I can increase up to 8.3 pieces
and decrease only 1.19.
The shadow price (duality) is the profit factor per unit
increased with increase restrictions, for example for
the first resource I have a limit of 12.5 hrs in the
increase part, however in the reduction part I can only
reduce 237.5 hrs and so on with each assigned
resource. In the right side restriction we would be left
with an allowable increase to gain 5 pesos more than
512.5 hrs.
IF THE LIMITS OF THE MODEL ARE
NOT RESPECTED. THE MODEL WILL
GO OUT OF THE GAIN RANGE AND
WILL NO LONGER BE VALID FOR OUR
PROBLEM.
21. INTEGER PROGRAMMING MODELS
An integer programming model is a
model that contains constraints and an
objective function identical to those
formulated by linear planning. The only
difference is that one or more of the
decision variables must take an integer
value in the final solution.
There are three types of integer
programming models:
INTEGER PROGRAMMING
MODELS
MIXED
BINARY
PURE
22. PURE INTEGER PROGRAMMING
A pure integer model (PLE) is, as the name implies, a problem in which all decision
variables are required to have integer values. For example:
Min 6X1 + 5X2 + 4X3
Restrictions
108X1 + 92X2 + 58X3 >= 576
7X1 + 18X2 + 22X3 >= 83
X1 ,X2, X3 >= 0 and integers
It is a pure integer model. Without the additional constraints that x1, x2, x3 are integers (i.e.
the integrality conditions) it would be a linear programming problem.
23. MIXED INTEGER PROGRAMMING
Some of the decision variables have integer values. The others comply with the
divisibility assumption.
A problem in which only some variables are required to have integer values while
others can assume any non-negative number (i.e., any continuous value) is called
mixed integer linear programming (MLEP). For example, suppose that in the
problem only X1 and X2 must be integers and X3 must not.
24. BINARY INTEGER PROGRAMMING
There are two methods to generate the special constraints that force the optimal
solution of the problem towards the desired integer optimal solution:
■ Branch and bound method.
■ Cutting plane method.
Here it helps us make decision factors such as yes or no.
Unfortunately, neither method is effective in solving integer linear programming
problems.
26. A blacksmith with 80 kg of steel and 120 kg of aluminum wants to make touring
and mountain bicycles that he wants to sell, respectively, at 20,000 and 15,000
pesos each in order to make the maximum profit. For the touring bike, he will use
1 kg of steel and 3 kg of aluminum. How many touring bicycles and mountain
bicycles will he have to make to maximize profits?
Steel Aluminum Selling Price
Touring bicycle
(x)
1kg 3kg $20,000
Mountain bike
(y)
2kg 2kg $15,000
Availability 80kg 120kg
30. Target
function
Restrictions
Minimum value that the variable can
take
Maximum value that the variable can
take M is a default value and
means: a VERY large
number. Type of
variable
Right-Hand side:
The right-hand side of a constraint is usually a
constant value that represents the maximum or
minimum available resource, in this case our Steel
availability is 80 kg, therefore it is our maximum
available resource, and the result of the equation
that governs this constraint cannot exceed it.
31.
32.
33. Interpreting each of the values of the solution table is as important as obtaining the
optimal solution, since from this interpretation we will be able to extract a good
sensitivity analysis:
Solution value: Solution value is the value taken by the decision variables in our
optimal solution, in this case it indicates that 20 touring bicycles and 30 mountain
bicycles should be produced.
Unit Cost or Profit: The unit cost or contribution is the value assigned to the
variables by us in the objective function.
Total Contribution: The total contribution to the objective solution is the product of
the solution value * unit cost or contribution.
Basic Status: After the problem is solved, this represents whether the variable is a
base variable, at the lower bound, or at the upper bound in the final simplex table.
Allowable MIN, MAX C(j): For a particular objective function coefficient. This is the
range in which the current basis of the solution remains the same.
34. Objective Function: Shows us the result of our objective function, in this case the
optimal solution has an objective function (utility) of $ 850,000.
Left Hand Side: On the left hand side, it is the value that the equation of each
constraint takes after replacing the variables that compose it by the solution values.
For example, the equation of the Steel constraint which is x + 2y <= 80, when
replacing the solution values it will be: (20) + 2(30) <= 80, the value of the left hand
side will then be 80.
Right Hand Side: On the right hand side, it is the value assigned by us to the
restrictions as maximum or minimum available resource.
Slack or Surplus: When the constraint in question has the operator <=, it corresponds
to a slack, i.e., it can be interpreted as the unused resource. When the constraint in
question has the operator >=, it corresponds to a surplus, i.e., it can be interpreted
as the resource used above the minimum usage constraint.
Shadow Price: The shadow price of a constraint is the marginal change of the
objective function when the value of the right-hand side of the constraint increases
by one unit. In our example it would be like this: for each additional kg of steel we
have available, the objective function will increase by $1250.
35. Problem Integer linear programming:
Three medical services have 10, 6 and 4
physicians respectively, with each
physician attending a maximum of 10
patients. The cost of each patient in
Service 1 is 10 euros per day, in Service 2
20 euros per day and in Service 3 25 euros
per day, and the total daily budget of the
three services is 2400 euros. In addition,
the first two services must treat at least
twice as many patients as Service 3.
10𝑆1 + 20𝑆2 + 25𝑆3 = 2400 𝐹. 𝑂.to win we must transform this
equation to an inequation
f.o. s1+s2+s3
RESTRICTIONS
10*10S1=100S1 S1≤100
10*6S2=60S2 S2≤ 60
10*4S3=40S3 S3≤ 40
10s1+20s2+25s3 ≤2400
S1+S2≥ 2𝑆3 twice as many patients as s3
S1, S2, S3 ≥0 and whole (half a patient cannot be attended)
Service
1
Service
2
Service
3
Total
10 6 4 10 Patients
10E 20E 25E Cost 2400
S3 S1+S2≥ 𝑆3
36. Interpretation:
For service 1 we will serve 100
people, service 2 is allocated 60
people and service 3 is allocated 8
people with a maximum service of
168 people.
We have the shadow price given in
the table and our maximum allowed
prices.
37. Evaluation of the examination and exposition of the problems
Mixed
Binary
By team 1 problem research it from the internet using the methodology of
your choice solver or winqsb and present it in class.
Teams of 5 people 1 of each 1
