April 9, 2020
Lec-1 Introduction to Linear Programming Formulations

Lec-1 Introduction to Linear Programming Formulations


This course is titled in Fundamentals of Operations
Research. In this course, you learn various tools of
operation research such as Linear Programming, Transportation, Assignment problems and so
on. Before you begin let us see what operations
research is. Operation Research is also called OR for short
and it is a scientific approach to decision making which seeks to determine how best to
design and operate a system under conditions requiring allocation of scarce resources. Operations research as a field, primarily
has a set or collection of algorithms which act as tools for problems solving in chosen
application areas. OR has extensive applications in engineering
business and public systems and is also used by manufacturing and service industries to
solve their day to day problems. The history of the OR as a field as goes up
to the Second World War. In fact this field operations research started
during the Second World War when the British military asked scientists to analyze military
problems. In fact Second World War was perhaps the first
time when people realized that resources were scarce and had to be used effectively and
allocated efficiently. The application of mathematics and scientific
methods to military applications was called operations research to begin with. But today it has a different definition it
is also called management science. A general term Management Science also includes
Operation Research. In fact these two terms are used interchangeably. A set of tools that are used to solve problems
is also called Management Science. Today OR is defined as a scientific approach
to decision making that seeks to determine how best to operate a system under conditions
or allocating scarce resources. In fact the most important thing in operations
research is the fact that resources are scarce and these scarce resources had to be used
efficiently. In this course we will primarily handle the
above topics. We started with Linear Programming. We introduced the formulations or problems
to formulate it in Linear Programming. We will do four examples there.We will also
spend lot of time on solving linear programming problems and understand the various issues
associated with the solution. We then get into a specific title called duality
and sensitivity analysis where we explore this linear programming topic further. Then we move on to two important problems
called the Transportation problem and the Assignment problem and when we do Dynamic
Programming and then we also spend some time on Deterministic Inventory Models. So these will be the topics that will be covered
in this course. Linear programming was first conceived by
Dantzig, around 1947 at the end of the Second World War. Very historically, the work of a Russian mathematician
first had taken place in 1939 but since it was published in 1959, Dantzig was still credited
with starting linear programming. In fact Dantzig did not use the term linear
programming. His first paper was titled ‘Programming in
Linear Structure’. Much later, the term ‘Linear Programming’
was coined by Koopmans. The Simplex method which is the most popular
and powerful tool to solve linear programming problems, was published by Dantzig in 1949. So this is the brief history of this field
called Linear Programming. The first title is called linear programming
formulations where we try to introduce to you how to formulate real life problems as
linear programming problems and to understand the various terminologies that is used in
formulating linear programming problems. We begin linear programming with an example. So let us consider a small manufacturer making
two products called A and B. Two resources are required which we call as R1 and R2 to
make these products. Now each unit of product A requires one unit
of R1 and 3 units of R2. For example we have these two products A and
B. Two resources are need R one and R2. A requires one unit of R1 and three units
of R two. B requires one unit of R one and two units
of R two. Manufacturer has 5 units of R one available
and 12 units of R two available. The manufacturer also makes a profit of rupees
6 per unit of A sold and rupees 5 per unit of B of sold. So this is the problem setting that we are
looking at. Now what are the issues that this manufacturer
faces? One, the manufacturer would like to produce
in such a way that the profit is maximized. How does the manufacturer go around formulating
his problem?The first thing is that the manufacturer has to decide on is the number of units of
A and B that is to be produced. It is also reasonable to assume that the manufacturer
produces this A and B in such a way to maximize the profit. So first thing that the manufacturer has to
do is to decide or determine how many units of A, and how many units of B, he or she is
going is to produce. So we first call these as X and Y. So let X be the number of units of A produced
and Y be number of units of B produced. So if the manufacturer produces X units of
A and Y units of B then the money that the person makes or the profit that the person
makes, which we call as Z, will now be 6X + 5Y. Now the manufacturer would ideally like to
produce as much of A and as much of B possible. But what happens is the availability of these
resources will now try and restrict the amount of A and B i.e., X and Y that the manufacturer
can produce. So we look at the resource requirement to
produce X of A and Y of B. So since each unit of A requires one of R1
and each unit of B requires one of R1, the total amount of resource R1 required is X
+ X + Y. What is available is 5. This X + Y should not exceed 5 or X + Y is
less than or equal to 5. It cannot be more than 5. Similarly 1 unit of A requires 3 of R2 and
1 unit of B requires two of R2. So the requirement to produce X units of A
and Y units of B is given by 3X + 5Y. This has to be 3X + 2Y and this has to be
less than or equal to 12 this cannot be greater than 12. Also we need to explicitly state that this
manufacturer will make X and Y which are greater than or equal to zero that is the person cannot
produce negative quantities of both X and Y. And you can see this here in this power point
slide.And you can see it here in this power point slide. Also since the person wants to maximize the
profit, the profit function Z is to be maximized. So the person would like to find out X and
Y in such a way that the function 6X + 5 Y is maximized and then X and Y has to satisfy
these conditions, (i) X + Y is less than or equal to 5 and
(ii) 3X + 2 Y less than or equal to 12 And explicitly stating that X and Y has to
be greater than or equal to zero which is what is shown here.This is called formulating
a linear programming problem. What we have done now is we have converted
the descriptive portion that was shown here earlier into a mathematical form and this
mathematical form is called as formulation. Now let us define some amount of terminologies
and we will try to maintain these terminologies consistently in all our formulations. The first thing that we define is called decision
variables. The two variables that we solve this problem
X and Y are called Decision variables and they represent the output. Whatever comes out as a solution is going
to imply that we make some amount of A and some amount of B, X amount of A and Y amount
of B. The values of X and Y is the output or the outcome after having solved this problem. So the variables that we want to solve are
called Decision Variables and then we have a function which represents what we are solving
this problem for. And as far as this problem is concerned, we
want to solve this problem to maximize the profit or to get maximum profit and this is
the objective with which the manufacturer works. This function 6X + 5Y represents the profit
function which is to be maximized and this is called the Objective Function. So the second term that is used is called
Objective Function which is precisely the purpose of working on this problem.The third
set of things is the Condition. The conditions actually represent the resources
that are available and the resources that have to be used efficiently. So let these conditions that we have to satisfy
namely, (i) X + Y less than or equal to 5 and
(ii) 3X + 2Y less than or equal to 12 are called constraints because they constrain
the decision variable from taking as larger value as we would ideally like the variable
to take. Because if we simply want to maximize 6X +
5Y without these constraints, then X and Y can take as higher value as possible. Now these constraints restrict the decision
variables form taking an unlimited or a very high value and so they are important and we
have to explicitly state in all linear programming problems (we will define what a linear programming
problem is formally) that these decision variables will have to be greater than or equal to zero. So the four terms that we have tried to introduce
here as you can see some there are (i) The decision variables
(ii) The objective function (iii) Constraints and
(iv) The non–negativity restriction So in a problem in where we define all the
four and write it in the mathematical form it is called formulating a linear programming
problem or formulating an operation research problem. (We will come to the definition a little later). Now as far as this problem is concerned, X
and Y are the Decision variables. The functions 6X + 5Y that is to be maximized
is called the Objective function. X + Y less than or equal to 5 and
3X + 2Y less than or equal to 12 are the Constraints. X, Y greater than or equal to 0 are the non
negativity restriction So the problem formulation has four steps. i) Identifying the decision variables
ii) Writing the objective function iii) Writing the constraints and
iv) Explicitly stating the non negativity restriction on the variables. Next, in this formulation, we realize that
the objective function is a linear function of the decision variables and all the constraints
are linear functions of the decision variables. So if you write a formulation such that the
objective function is linear and the constraints are linear then such a formulation is called
a Linear Programming Formulation. So the important requirements of a linear
programming formulation are that the objective function is linear, the constraints are linear
and we explicitly state that the non negativity restriction is held. So these are the three important aspects in
a linear programming problem. If either the objective function or the constraints
are Non linear or the case maybe that this does not exist then it is not a linear programming
problem. Now shall look into some more examples to
understand the linear programming situation. So this is the simple formulation where we
have looked at two variables, two constraints, and a maximization objective function.Now
we shall look at more formulations to understand various other aspects of problem formulation
in Linear Programming problems. Let us look at the second example which is
called as Production Planning Problem and the problem is as follows: Let us consider
a company making a single product demand of 1000, 800, 1200 and 900 respectively for 4
months. Now the company wants to meet the demand for
the product in the next 4 months. The company can use two modes of production. There is something called as Regular Time
Production and Overtime Production. Now the regular time capacity is 800/month
and overtime capacity is 200/ month. In order to produce one item in regular time,
it costs Rs 20 and to produce overtime it costs Rs 25. The company can also produce more in a particular
month and carry the excess to the next month. Such a carrying cost is rupees 3. What we call the inventory cost or carrying
cost is Rs 3 per unit per month.The important condition is that the demand has to be met
every month. We cannot afford to have shortages. Now let’s try to formulate a problem that
represents this situation and also try to understand a few more things about problem
formulation as we proceed. Now the first thing we do here is that we
define X1, X2, X3, and X4. Or in general Xj. as quantity produced using
regular time production in month j and Yj as quantity produce using overtime in month
j. Now these are our decision variables, Xj and
Yj (quantities that are produced). Now let us first write the constraints. In fact it is important first to define the
decision variables then to write the objective function & the constraints. Sometimes the constraints can be written first
and then the objective function or sometimes the Objective function & then the constraints
and then explicitly state the non negativity. So in this example we will try the write the
constraints first. Now as far as the first month is concerned,
we produce X1 + Y1. We assume that we do not have any initial
inventory or any stock to begin with. So whatever is produced in the first month
is X 1+ Y1. We have to meet the first month’s demand which
is 1000. So we write X1+ Y1 is greater than or equal
to 1000. We need to produce more than the first month’s
demand of 1000. In fact another alternative is that we can
either state X1 + Y1 is greater than or equal to 1000, which implies that X1 + Y1 is exactly
1000 or more than 1000. If X1+ Y1 is more than 1000 then the balance
is carried to the next month. So we can define another term called I1 which
is the quantity that is carried to the next month. Therefore we can say X1 + Y1 + I1 are equal
to 1000. Now for the second month we have already carried
an I1 here So we start with an I1 and then we produce
X2 in regular time. X1 + Y1=1000 + I1 The balance is carried
i.e., so X1 + Y1-1000 is carried so that is I1. So we begin with I1. We produce X2 by regular time in the second
month and Y2 by overtime in the second month and that has to be equal to the second month’s
demand of 800. Plus, what we carry at the end of the second
month into the third month. So, if this left hand side which is the quantity
available at the end of the second month, is more than the demand which is 800, and
if it exceeds 800 then the balance is carried to the third month. So here, we have this I1 and I2 which are
the quantities that are carried at the end of the month to the next month. Ij is the quantity that is carried at the
end of month j to meet the demand of month j + 1. Similarly for the third one, we begin with
I2 to produce X3 + Y3 which will now be equal to 1200 which we have to meet and if it exceeds
1200, the balance I3 is carried to the fourth month. Now for the fourth month, we begin with I3
to produce X4 + Y4 and that is equal to 900. We are not using an I4 here because the problem
stops at the end of the fourth month and therefore we would adjust these quantities in such a
way that we are able to get exactly 900 which is required to meet the demand of the fourth
month. So we do not use an I4 here. Moreover, we also have a situation here wherein
all XY Xj Yj and Ij are greater than or equal to 0.We have to write the objective function. We have not written the objective function
as such. The objective function will be equal to the
cost of regular time production. The cost of the regular time production we
have is 20. So 20 into sigma Xj. Sigma Xj represent the total quantities produced
in the four months using the regular time. So 20 into X1 + X2 + X3 + X4 + 25 into the
overtime production Yj, if not, you could say J equal to one to four here and j equal
to one to four here. In addition, whatever quantity that is carried
to the next month, we have an additional Rs 3 per unit per month we have + 3 into sigma
Ij, J equal to 1 to 3 and we don’t have I4 in this problem.Plus, we have Xj, Yj and Ij
greater than are equal to 0. Now what is that we want to do? This is the cost function. And this being a cost function, we would like
to minimize this cost function. We like to minimize this cost function.So
now this formulation is complete and this formulation finds out Xj. Xj, which is the quantity producing using
a regular time in month j & Yj which is the quantity produce using overtime in month j. Constraints are given here and the objective
function now tries to minimize the cost function. So this is the formulation corresponding to
the production planning problem As we go back, as far as the problem is concerned, this has
actually 11decision variables (we have left out the I4). We need to write some more constraints which
have not been written here. The regular time capacity and the overtime
capacity are also given here. So we need to write these constraints. These constraints will now be
(i) Xj is less than or equal to 800 for every j
(ii) Yj is less than or equal to two hundred for every j
So we have four constraints here we have four more constraints here. We also have 4 constraints for the 4 months. So we have 12 constraints and we have eleven
variables, 4 variables (X1 to X4) and another four variables (Y1 to Y4) and another three
variables (I1 to I3). So this formulation has a minimization objective
which is different from the maximization objective that we saw in the earlier example. It has twelve constraints and eleven variables. It also has some equations here in the earlier
formulation where we did not have any equation. We also realize that there are constraints
which are of the less than or equal to type here in the earlier formulation as well. We have some equations which are here. We can do a few more things. From the same formulation we can do a few
more things. The first thing is that we can eliminate this
I1, I2 and I3 from this which can easily be done.For example we can eliminate this I1
by saying that X1 + Y1 should be greater than or equal to 1000. Whatever is produced in the first month should
be more than or equal to the demand from that month and if there is an excess it is carried
to the second month.So what could possibly be the excess? The excess will be X1 + Y1 -1000. So this becomes X1 + Y1 – 1000 + X2 + Y2 should
be greater than or equal to 800. So we eliminate this I2 as well. Now this quantity is, what is available for
the second month this is the demand for the second month. So if this quantity exceeds 800, then the
balance is carried to the third month. So this I2 is nothing but X1+ Y1 + X2 + Y2
-1000-800 will be greater than or equal to 1200 and we eliminate this three because I2
is nothing but this left hand side – 800. So we substitute for I2 here to get this and
we also eliminate this I3. We now go back to write what is I3, I3 is
nothing but this left hand side minus 1200. So we eliminate this I3 here and write this
as X1 + X2+ X3 + Y1 + Y2+ Y3 -1000-800-1200=900 and also we can eliminate these If’s. Now we have the same number of constraints
here. We have few more variables.We have to rewrite
the objective function. Also here this objective function has these
If’s, so we have to write this Ij very carefully So this now becomes + three times (as far
as the first month is concerned) X1 + Y1- 800 + three times X1 + Y1 – X2 + Y2- 1800
& (as far as the third month is concerned) X1 + Y1 + X2 + Y2
+ X3 + Y3 – 3000.3000 is the sum of 1800 and 1200. So this is what is being carried as I3 which
comes in. Right now we have not introduced an I4 yet. We have an equation here. We can even keep this as greater than or equal
to just to be consistent with the other three constraints and if we choose to do that then
what will happen is we have another term that comes in. You will have a fourth term which is this
X1+ Y1 + X2 + Y2 + X3 + Y3 + X4 + Y4 – 3900. We will continue to have these two constraints
Xj less than or equal to 800 and Yj less than or equal to 200. So you will continue to have these two constraints
as well, but the only thing we have done is we have added another term into the objective
function We have replaced this equation by an inequality
just to be consistent with these four so we now have a second formulation for this problem. We can even have a discussion on this 900. We have a second formulation for this problem
which is the same as what we saw in the first formulation. The only difference is in this formulation,
we have fewer variables. We have 12 or even variables in the earlier
one. We now have 8 variables in this formulation. We had equations in the first one and we now
have all inequalities here and we also introduced the greater than or equal to inequity to this
problem which we did not see in the earlier formulations also so this slightly different
formulation but it also represents exactly the same problem that we are trying to model. Now let us come back to this 900. Now there are two ways of addressing this
900. One is to put an equation here which will
automatically correct the values of X4 and Y4 such that we do not have anything excess
and if we put an equation here.We can then you can leave out this term. If you don’t put an equation here and put
an inequality instead, it means you are allowing an I4 to exist and if such an I4 exists then
it is carried into the objective function,with the cost of this. Now both are correct. You can put an equation and leave out this
term. You can keep this as an inequality here and
then add this. Both are correct simply because any I4 that
is carried from here comes to increase your cost. So the solution will be such that you will
not carry an I4.And so the I4 will automatically be 0. So in the formulation stage, for the purpose
of consistency you can still keep all these four as inequalities. Then you can write this term. Also, it is just that the objective function
has an additional term that comes into play in this. This is shown in this slide. The same formulation is shown in this slide. The requirements for the four months are the
four constraints, except that the right hand side is simplified. You can see that these two terms are taken
to the right hand side (as you see the formulation there) and then you have the same objective
function which is shown here including all the four terms,3900 and you will also find
that all of them are inequalities. So this is the second way of formulating this
problem. Now when compared to the first, the question
arises as to which one is better? Which one is superior? There are some very simple guidelines. For the same number of constraints (both these
problems have twelve constraints) you can say that a formulation which has fewer variables
is superior.So this is seen to be the superior one when compared to the earlier. One, both has the same number of constraints
but the number of variables is much less in this. Also in general, a formulation that has inequalities
is better than formulations that have equations. Most of us are used to solving equations though. Later we see that we convert these inequalities
into equations and so on. But as a very general idea, one may assume
that the formulation which has inequalities is better and preferred over formulations
that have equations. So this is better than the earlier one. In this formulation, we have learnt a few
things like, minimization objective function. The earlier formulation had a maximization
of objective function.In this formulation we learnt to introduce greater than or equal
to type inequalities, which we did not see in the earlier formulation. Now let us look at a third way to solve the
same problem just to show you that the same problem can have different formulations depending
on how you think about the problem.So let us look at the third way to formulate the
same problem again. Now we introduced a slightly different set
of decision variables. Now we are going to introduce Ink as the quantity
produced in month I to meet the demand of month j using production
type K. Now we look at four months. So month I can take I equal to one to four,
j can also take j equal to 1 to 4 and K has two types of productions, Regular time and
Overtime. So K takes values 1 or 2. Now with this type of definition of decision
variable, let us try to write the constraints and the objective function corresponding to
this problem. As for as the first month is concerned, first
month we produced X111 indicates producing in regular time to meet the first month’s
demand using regular time production i.e., producing in the first month i.e., I equal
to one to meet the first month’s demand, j equal to one using regular time production. It is X111 + X produced in the first month
to meet the first month’s demand using overtime. So these are the only two variables that represent
the production to meet the demand of the first month. So X111 + X112 should be equal to the first
month’s demand which is thousand 1000.Here we are not looking at carrying anything. You see the way the variables are defined. Whatever quantity that is carried is also
defined as a decision variable. Now we try to meet the second month’s demand. The second month’s demand can be met into
two ways: (i) by production in the second month
(ii) by production in the first month In this, we also make an assumption that you
cannot meet the first month’s demand by producing in the later months to be consistent with
the earlier assumptions that we made, while you can actually produce in a month that is
early, carry and then meet the demand of subsequent month. So, the second month’s demand can be met by
production in first as well as second. So you have X121 produced in the first month
to meet the demand of the second month by regular time plus produced in the first month
to meet the second month’s demand by overtime plus produced in the second month to meet
the second month’s demand by regular time produced in the second month to meet the second
month’s demand by overtime So these are the four ways by which you can
meet the demands of the second month and these four should add to the second month’s demand
which is 800. Now as far as the third month is concerned,
we can meet the third month’s demand by six ways. (i) Producing in the first month, regular
time and overtime (ii) Producing in second month, regular time
and overtime and (iii) Producing in the third month, regular
time and overtime. So you end up writing X131 produced in the
first month to meet the third month’s demand with regular time + X132 produced in the first
month to meet the third month’s demand overtime, produced in the second month to meet the third
month’s demand regular time produced in the second month, meet the third month’s demand
overtime produce in the third month meet third month’s demand regular time produce in the
third month to meet third month’s demand overtime. This should be exactly equal to 1200 which
is the third month’s demand. Now you extend it and write it for the fourth
month. Now this can be done in 8 ways. You can produce in the first month by two
ways, second month by two ways and third month and fourth month two ways. So you have X141 + X142. This represents producing in the first month
to meet the fourth month’s demand by two types. X241 + X 242 + X 341 + X 342 + X 441 + X 442
are equal to 900. So we have written the four constraints corresponding
to the requirement of the four months. Now we need to look at the constraints on
the capacities. For the first month, the regular time, what
we do is, X111, produced in the first month to meet the demand of the first month by regular
time + X121. I produce in the first month to meet to the
demand of the second month, once again by regular time. So when we are looking at the first month’s
regular time production quantity, I will be 1 and k will be 1 but j can be 1, 2, 3, and
4. So you will have X111 + X121 +X131 + X141. Now these four represent the quantity produced
in the first month, using regular time, to meet the demand of 1, 2, 3 and 4 months. So this should be less than or equal to the
regular time capacity of 800. Now as far as the second month’s regular time
is concerned, you have X produced in second month to meet the second month’s demand by
regular time. So in this case I will be 2 because you are
producing in the second month. k will be 1, since you are using regular time
production. j will be 2,3 or 4 so you will have X221 +
X 231 + X 241 should be less than or equal to 800. So as far as the third month is concerned
you will get X331 + X341 because I is equal to 3 represents producing in the third month
j can be 3 or 4. You can produce in the third month to meet
the third month’s requirement or the fourth month’s requirement and k is 1 because you
are using regular time So this should be less than or equal to 800 and as far as the fourth
month is concerned you will have X441 to be less than or equal to 800. You have to write four more for the overtime
production. The only difference being the third subscript
will become 2. So the first month’s overtime will now look
like this X112 + X122 + X132 + X142 is less than or equal to 200. X222 + X232 + X242 is less than or equal to
200. X332 + X342 is less than or equal to 200 and
X442=200. 200 is the capacity that we have. So this will also be 200. So you have these capacity constraints written. Now let us go back and write the objective
function. The objective function will look like this. I want to minimize the total cost of regular
time, overtime and the quantity that is carried. So in the first month 20 is the cost of the
regular time production. So I produce X111 + X121 + X131 + X141. Because I can produce by regular time to meet
the demand of all the four. In the second month I produce X221 + X231
+ X241 because in the second month, I can produce only to meet the demands of months
2, 3 and 4. So I have three terms now. I have X331 + X341 which is the thirds month’s
production to meet the demand of either 3 or 4 and X441. So these are my regular time production quantities. So it is 20 for all these. Now I have to write a similar term for the
overtime production. The only difference is the third subscript
k will become 2 instead of 1. So I will have + 25 into X112 + X122 + X132
+ X142. These four terms represent overtime production
to meet the demands of months 1, 2, 3 and 4. k=implies overtime. Now for the second month, I will have X222
+ X232 + X242. Once again, as we saw earlier in the second
month, you can produce to meet the demand of months 2, 3 and 4 by overtime. You will have two terms for the third month
X332 + X342 and you will have one term for the fourth month which is X442. Now we also need to write the cost term corresponding
to the quantities that we are carrying. So what we do is + 3. 3 is the cost that we incur to carry a unit
to one month. So all the quantities that are carried from
the first month to second month will come here and the one produced in second month
carried to third month will come here and produced in third month carried to the fourth
month will come here. For example we will have X121; this is carried
by one month. If j -I=1, it means it is produced in a
particular month and used in the next month. So it will incur 3. So X121 + X122 are all carried by one month
+ X231 + X232 produced in the second month to meet the third month produced in the second
month to meet the third month by overtime and X341 + X342. So these are the quantities that are carried
one month. There are quantities that are carried for
2 months. For example, X131 + X132, produced in the
first month to meet only the third months demand so carried to two months with a cost
of 6. Similarly you will have X241 + X242 and there
are quantities which are carried to 3 months because you can produce in the first month
and use it in the fourth month. So you will have another two terms that come
in X141 + X142. So this is the objective function. Plus of course, all I or relevant Xijk is
greater than or equal to 0. So this is the third type of formulating the
same problem. You can see that formulation here. You will also observe that this formulation
has 20 variables, a little more than the previous. Ink would allow you 4 into 4 into 2=32. But you have only 20 active variables (the
other 12 are not active) 12 constraints, you also have some equations. So by the same discussion that we had had,
you realize that this is not a very efficient formulation compared to the second one. This is because this has more variables.This
also has some equations.Nevertheless this is also a formulation that we can have for
the same problem. We have learnt a few things in the second
example which is called the Production Planning example. The first thing that we learnt is that, there
are types of problems which can be formulated as Minimization problems. The first example was a maximization problem. Second example is the minimization problem. Objective functions are of two types, Maximization
and Minimization. Both have linear objective functions. So we looked at the second type of objective
function, which is minimization in this case. We also looked at the problem where we introduce
the greater than or equal to constraints. We do not see them here. You could introduce greater than or equal
to constraints when we eliminated the inventory that is carried. So constraints can be of three types,
(I) less than or equal to type that you see here
(II) Equations that you see here and (III) Greater than or equal to type. So we have seen three types of constraints
in this. Decision variables of course, are also included. We have also learnt that a formulation is
superior if it has fewer constraints, fewer variables. It happened that all the three types of formulation
had the same number of constraints so we couldn’t make out something superior based on fewer
constraints. But in general a formulation that has fewer
constraints is superior. For the same number of constraints a formulation
that has fewer variables is superior. We also saw that the same problem can be formulated
in more than one way depending on how the person formulating looks at the problem. Now we did see 3 different types of formulation
and all the 3 formulations will give effectively the same solution if they are correct. Some of them may have more variables. Some of them may have fewer variables. So we have come across 3 formulations. We also saw the non negativity restriction
that comes in. So in summary in these two examples that we
have seen, we have seen how to formulate a linear programming problem, terminology in
terms of Decision variable, Objective function, Constraints, Non negativity, different types
of objectives, different types of constraints and different types of variables. We later go through two more formulations
to understand some aspects of problem formulation that need to be covered which have not been
covered in these two.

100 thoughts on “Lec-1 Introduction to Linear Programming Formulations

  1. dear, Proffesor G.SRINIVASAN
    i would like to thank you from the bottom of my heart for your amazing way of teaching you are gifted proffesor you helped me a lot i am sorry because i dont have any other way of thanking you but leaving a comment .

  2. so much of my thanks to you Prof. it gives me joy that we have people like this to help us think Logically and objectively for optimal solutions. Greetings from Nigeria…..

  3. Prof. Srinivasan you are the best operations research teacher !! Thank You so much sir and nptel for uploading these lectures !!

  4. thanks a lot sir for videos , lucky are those who get the opportunity to study from teachers like you . i am a student of amity business school , and lecturer there named JK SHARMA only dictates whats pre written in the slides and we pay for such classes hefty amount like 15 lakhs .

  5. I am an Indonesian student, and wow! Your explanation in English makes me understand this subject better than my own lecturer's explanation, which was delivered in my own language…

  6. Sir ,You did a great job. These lectures are very good.. Even i suggest you to make your own channel. I need not to touch any book after watching your lectures. Thanxx a lot

  7. What if the problem has non linear constraints and linear objective function ? Is there any particular method to solve this kinda scenario ?

  8. good morning sir
    i have learned main topics from your lecture videos
    and i have a small drought please clarify me sir:
    A project proposal needs to be prepared for a mixed used development towards housing and retail area. The housing area would consist of efficiency apartments, duplex and single family homes. Maximum demand by potential renters is estimated to be of 600 efficiency apartments, 400 duplexes and 350 single family homes, but the number of duplexes must equal at least 60% of the number of efficiency apartments and single homes. Retail space is proportionate to the number of home units at the rates of at least 12 sqft, 18 sqft and 20 sqft for efficiency, duplex, and single family units respectively. However, land availability limits, rental space to no more than 10000sft. The monthly rental income is estimated at Rs.25000, Rs.35500 and Rs.50000 for efficiency, duplex and single family units respectively. The retail space rents for Rs.5500 per sft.
    a. Please write the Objective Function
    b. Identify the constraints
    c. Arrive at an optimal solution for the LP model developed using simplex method to determine the optimal retail space area and the number of residences
    d. Write the Dual Problem for this LP Problem
    e. Using Sensitivity Analysis comment on the change in optimal solution if the local statutory authorities insist only 6500 sqft of space be made available towards retail.

  9. Thanks a lot for explaining things so lucidly sir!! One of the best initiatives of MHRD for making quality higher education accessible to all with internet connection and device.

  10. Hi,
    with refer to 23:48,
    In my opinion, the cost function should have -ve sign for the Inventory cost. Can anyone please check and let me know?
    Many many Thanks in advance.

  11. Thank you sir when i see this topic i was confused but after seeing your video the concept of linear programming is clear.

  12. Sir, it was a very nice lecture. I have one doubt. You said that while writing constraints, inequalities are better than equations. Can you please clarify why is it so?

  13. There are no words for me to explain your patience in writing and delivering quality with your speech.

  14. The company can not produce more than 1000 (RT+OT). In that case what is the use of I1 in constraint 1…..

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