In the last lecture we looked at the complementary
slackness condition and we also said that the simplex method also gives the solution
to the dual at the optimum. We also wanted to find out how the Cj - Zj's
in the intermediate iterations represent something or do they even represent anything? To consider that further, we will take the
familiar example and try to interpolate the Cj - Zj's corresponding to the intermediate
iterations of this simplex algorithm. So the example is maximize 6X1 + 5X2 subject
to X1 + X2 less than or equal to 5 3X1 + 2X2 less than or equal to 12; X1 X2 greater than
or equal to 0 Very quickly performing this simplex iteration,
we get X1, X2, X3, and X4 at the right hand side. We start with the X3, X4 1 1 1 0 5, 3 2 0
1 12, 6 5 0 0. Now variable X1 with the largest positive
Cj - Zj enters. Corresponding theta values are 5 and 4 is
a minimum theta. Variable X1 enters to get X3, X1 and Cj - Zj,
Dividing by the pivot element, we would get 1, 2/3, 0, 1/3, 4. This minus this would give us 0, 1/3, 1 - 1/3,
1 is t = 24. 0 0 6 into 2/ 3 is 4. 5 - 4 is = 1. 6 into 1/3 is = 2. We get - 2 here. Variable X2 enters the basis so this is 1
divided by 3 is = 3; 4 divided by 2/ 3 is = 6. This leaves pivot element. X2 replaces X3 in the basis, so the final
table will look like this. Dividing by the pivot element we get 0, 1,
3 - 1, 3. This - 2/3 times this will give 1, 0 - 2,
1/3 + 2/3 which is = 1; 4 - 2/3 into 3 is 2. Cj - Zj values are 5 into 3 = 15 - 12 = 3;
I get a - 3 - 5 and 6 which will give us - 1 and 27. In the last lecture we saw that this - 3 and
- 1 when multiplied with the - 1 again would give us 3 and 1 which are the values of the
dual at the optimum. We also said that amongst X1, X2, X3, and
X4 now X3 and X4 are our u1 and u2 which are the primal slack variables. We replace them by u1 and u2 here. X3 also becomes u1. From the complementary slackness we understand
that there is a relationship between X and V and y and u. Whatever comes under X, we can write it as
V1 and V2. Primal decision variables have a relationship
with the dual slack. Primal slack have a relationship with the
dual decision variables. So we say that with the minus sign, y1 = - 3y2
= - 1 represents the value here y1 = + 3y2 = + 1 represents that the values are dual
at the optimum. Now let us look at an intermediate iteration
here. In this, intermediate iteration, solution
to the primal is X1 = 4; u1 = 1. Let us apply the complementary slackness and
see what happens. When we apply complementary slackness to this,
X1 basic indicates V1 = 0. u1 basic indicates y1 = 0 and non basic. Here X2 and u2 are non basic indicating that
V2 and y2 are basic for the equivalent solution to the dual if we apply complementary slackness
condition. We are writing the dual for this problem. Dual is to minimize 5y1 + 12y2, subject to
y1 + 3y2 - V1 = 6; y2 + 2y2 - V2 = 5; y1, y2, V1, 1 V2 greater than or equal to 0. When we apply complementary slackness to this
we have to solve for V2 and y2 which are the basic variables. This solution will become 3y2 = 6 from which
y2 = 2 and from here we have y2 - V2 = 5 from which V2 = - 3. When we apply complementary slackness conditions
two and intermediate iteration of the simplex algorithm and evaluate the corresponding dual
as we have done here, the corresponding primal solution is u1 = 1; X1 = 4; Z = 24. When we apply complementary slackness and
solve we get y2 = 2; V2 = - 3; Z or W = 5y1 + 12y2 W is also = 24. Let us go back to this intermediate iteration
and see whether we have this solution reflected somewhere in this Cj - Zj. If we go back now we have y2 = 2; we realize
this is y2 so the value here under the Cj - Zj once again multiplied with the - 1 gives
us the value of y2 = 3 and from this and I have y1 + 2y2 - V2. So 2y2 - V2 is = 5; V2 is - 1; V2 is here
which is shown under X2 and we realize the negative of this number is - 1. What we observe here is if we take an intermediate
iteration of the simplex algorithm, write the complementary slackness corresponding
to that and then evaluate a corresponding dual, we realize that the solution to that
dual is also seen under the Cj - Zj of the intermediate iteration. Simplex satisfy complementary slackness conditions
at every iteration of the algorithm not only at the optimum. We see some more interesting things happening. It satisfies the complementary slackness conditions. It gives an equivalent solution to the dual
after the complementary slackness conditions are applied. We get the same value of the objective function
for the primal and dual respectively. We also realize that the dual is infeasible
here because V2 is - 1 we would want y1, y2, V1 and V2 all greater than or equal to 0. Now V2 is - 1. The infeasibility of the dual which is V2
= - 1 is now reflected in the corresponding non optimality of the primal. You can see that V2 = - 1 is actually represented
as a C2 - Z2 = + 1 which enters the basis to get the next iteration. Whichever dual variable is infeasible or whichever
dual constraints are infeasible. After all V2 = - 1 comes from the fact that
y1 + 2y2 is not greater than or equal to 5. This V2 is nothing but the extent to which
y1 + 2y2 is greater than 5 because the actual constraint is y 1 + 2y2 greater than or equal
to 5. If y1 + 2y2 is strictly greater than 5 then
V1 will take a positive value. In this case because y1 + 2y2 is less than
5; V1, V2 takes a negative value. V2 taking a negative value indicates that
the second constraint is violated. A dual is infeasible when either the decision
variable takes negative value or the slack variables take negative value. Slack variable taking negative value represent
the extent of not satisfying a particular constraint so the extent of infeasibility
of the dual represents the extent of non optimality of the primal. If the second dual constraint is violated
by one unit then it implies that in the simplex algorithm, the corresponding digestion variable
(if this slack variables take a negative value) which is non basic because of the complementary
slackness condition will now try to enter the basis and will indicate the rate of increase
of the objective function which is 1 which is the same - 1 which we see here Simplex
not only satisfies the complementary slackness conditions at the optimum. At every iteration of this simplex algorithm,
the complementary slackness conditions are satisfied. In every intermediate iteration a non optimal
iteration with respect to the primal, is an infeasible solution to the dual and the extent
to which the entering variable in the intermediate iteration corresponds to the infeasible dual
slack if a decision variable or a dual decision variable or slack variable enters. The rate of the Cj - Zj corresponding to the
entering variable is actually the extent to which the dual is infeasible. Non optimality of the primal which is represented
by an intermediate iteration represents infeasibility to the dual. There is a different way of looking at this
Cj - Zj. What this simplex algorithm does is remember
that the right hand side is always greater than or equal to 0. The rule for the leaving variable will ensure
that the right hand side will never become negative. A simplex as an algorithm will always have
a primal solution which is basic feasible. It will not have an infeasible solution at
all. Now this represents the feasibility of the
primal now this represents either the optimality of the primal or the feasibility of the dual. This plus value here indicates that the corresponding
dual is infeasible with the negative. What simplex tries to do is it starts with
a feasible primal and an infeasible dual. This would mean both X1, V1 and V2 are negative. So it starts with a feasible primal. It has an infeasible dual and it tries to
make the dual feasible. The extent of infeasibility of the dual is
given by the Cj - Zj of the corresponding value. It tries to take that variable which is least
feasible by picking up this 6 here which is nothing but the maximum rate at which I can
increase objective function for the primal. In successive iteration it tries to make the
dual feasible. Here you have two dual variables that are
infeasible. We have one dual variable that is infeasible. It tries to make the dual also feasible by
applying complementary slackness. In the end when both the primal and the dual
are feasible, then the optimal is reached after applying the complementary slackness. The optimality can be looked at in this way. We can go back and try to even interpolate
all the duality theorems. One, you have a basic feasible solution here
with an objective function value of 27 that is feasible to the primal. You also have a feasible solution to the dual
with a same value of the objective of function therefore it is optimal to both primal and
dual respectively. A simplex as an algorithm is always feasible
to the primal. It applies complementary slackness and tries
to evaluate the corresponding dual in every iteration. The moment it finds that the corresponding
dual is also feasible then it is optimum. Simplex can be seen in an algorithm which
actually solves both the primal and the dual, but it keeps a basic feasible solution to
the primal and works towards getting a feasible solution to the dual there/making it optimum. Now simplex comes under a category of what
are called primal algorithms. A primal algorithm always has a feasible solution
to the primal. It satisfies complementary slackness or applies
complementary slackness and terminates when the corresponding dual is feasible indicating
that both primal and dual are feasible and optimum. We do not need to solve a dual at all. We now realize that when we solve the primal
we are automatically solving the dual and we can always get the solution to the dual
of the problem that we are solving from the primal iterations. This is the primal that we have solved. Solution to the primal is here. Solution to the dual of this problem is here
to the y1 = 3; y2 = 1; Z is = 27. Let us go back and see this. The optimum solution to the dual can be read
from the optimum tableau of the primal in the simplex algorithm and we need not solve
the dual explicitly at all. Now we look at the intermediate iteration. Whatever we have tried to show here is what
is explained here. This solution is infeasible to the dual because
variable V2 takes a negative value which is here. We have to remember constantly that we need
to multiply this with the - 1. This is actually the negative dual variable
not this. It just shows us that this constraint is violated
by the same quantity of one which is seen here and the optimum when the complementary
slackness conditions are applied resultant solutions is feasible to the dual and hence
optimum. Simplex can be seen as one that evaluates
basic feasible solutions to the primal and applies complementary slackness conditions
and evaluates a corresponding dual. When the primal basic feasible solution is
non optimal as in an intermediate iteration, a dual will be infeasible and as and when
the dual becomes feasible, we get the optimal solution for both the primal as well as a
dual. We next look at what is called as dual simplex
algorithm and try to see another version of simplex which is quite interesting and different
from the version that we have seen. We take an example. Minimize 4X1 + 7X2 subject to 2X1 + 3X2 greater
than or equal to 5; X1 + 7X2 greater than or equal to 9; X1, X2 greater than or equal
to 0. We add slack variables because of the greater
than or equal to. We have negative slack
So we have - X3 = 5; - X4 = 9; X3, X4 greater than or equal to 0. Normally we would added two artificial variables
a1and a2 to get the basic feasible solution because X3 and X4 by themselves are not capable
of giving us a starting basic feasible solution because X3 = - 5; X4 = - 9 is infeasible but
now let us see what happens if we still start with X3 and X4. Now we first thing we do is we convert this
problem to a maximization problem. We have X1, X2, X3 and X4. We are not going to add artificial variables
at all. The problem becomes maximize - 4X1, - 7X2,
0X3 and 0X4. We now still keep X3 and X4 as basic variables. We multiply this because this is equal to
5. We multiply this with the - 1 to get - 2X1
- 3X 2 + X3 = - 5. Hence, - 2X1 - 3X2 + X3 + 0X4 = - 5
- X1 - 7X2 - 1X1 - 7X2, 0, 1, X4 = - 9. We have 0 and 0
Now we have violated a very important assumption in simplex that the right hand sides should
be greater than or equal to 0. Now we have a solution which is not basic
feasible but has a completely negative right hand side values. But let us look at Cj - Zj values. Cj - Zj for this would give us a - 4 here
- 7 here 0 0. We have 0s here and they do not contribute
at all Cj - Zj would simply become Cj so we have this. Because of the problem we will now come back
to what is peculiar about this problem. But we definitely observe that the optimality
condition is satisfied. The optimality condition is satisfied, feasibility
condition is not satisfied or just extending whatever we had seen, the dual corresponding
to the solution X3 = - 5; X4 = 9 is feasible because I have a completely negative Cj - Zj. The dual is feasible. So we have a simplex iteration where the primal
is infeasible but the dual is feasible in all our earlier simplex versions. In any first iteration or intermediate iteration
the primal will be feasible. Dual will be infeasible. Now we have exactly the opposite happening. The primal is infeasible. The dual will feasible. Now we can still workout simplex. If we can maintain the feasibility of the
dual and slowly make this feasible then it will become optimum. In a normal simplex what we did is we maintained
this feasibility. We tried to bring the dual feasible and we
got the optimum. Now here if we can keep the dual feasible
consistent and then make this one feasible then it will be become optimum. Let us do that. Now we can have an entering variable here
because all the variables are with a negative Cj - Zj. The first thing we need to do is we need to
somehow make this feasible which means we need a positive value or a non negative value
on the right hand side. The most negative of this will leave first
so we first find out the leaving variable and leave out the variable with the most negative
value of the right hand side. So this will leave. We are trying to do some steps. Some part of the simplex which is the opposite
and some part of the simplex which is common to the earlier one. Here first find out the leaning variable. We need a variable that can enter and substitute
this X4. In order to do that, we compute a theta but
theta now comes in the row because we have to find out an entering variable. Next, to do that, we do something very similar. Take this value and divide here. - 4 divided by - 1 is = 4; - 7/- 7 is = 1;
we have 0. We can leave this because these are the basic
variables. We want only one of the non basic variables
to enter. We do not have to evaluate the theta for what
are presently the basic variables. You will evaluate the theta only to represent
non basic variables. These are the theta values.Once again the
smallest theta will enter so we have this. Now what happens, as a result of this, in
this algorithm the Cj - Zj's will always be negative and we need a negative pivot because
when we do the simplex iteration next, we divide this by the pivot element we will then
get a positive value here. In this case we would need a pivot that is
negative and this value will always be negative. So you will compute thetas only for negative
values in the row corresponding to the leaving variable. If there is an element with the positive value
in the row corresponding to the leaving variable you will not compute the theta even if that
is a non basic variable. For example if this had been a + 1 we would
not have computed this theta at all which is very similar to situations where we do
not compute theta in the earlier version of this simplex. (in the earlier version of the
simplex if that number is negative you would not compute theta or if that number is 0 you
would not compute theta). Here if that number is positive or 0 you will
not compute theta. So you will not compute theta here because
of this 0. In any case this is a basic variable so you
won't do that. Now variable X2 enters and variable X4 leaves
I have X3, X2. This is - 7 this is 0, so divide by the pivot
element to get 1/7, 1, 0, - 1/7, 9/7. We need a 0 here. This + 3 times this would give a 0. This + 3 times this will give a 0 - 2 + 3/7
is = - 11/7, 0, 1 this + 3 times this would give us - 3/7. This + 3 times this - 5 + 27/7 is - 8/7. The - 8/7, 9/7, 9/7 and we have Z is = - 9
here. Now we need to find out Cj - Zj, X3 and X2
are basic variables. We get 0 here. This 0 into - 11/7 + (- 7) into 1/7 - 1 - 4
- (- 1) is = - 3. Here I have a + 1
So 0 - (+ 1) will gives me = - 1. Once again I have Cj - Zj less than or equal
to 0. The optimality condition is satisfied. How is it satisfied? It is satisfied because of the minimum theta
rule that we followed to define the entering variable. We made sure that the optimality condition
is satisfied. The feasibility condition is not satisfied
because this as a negative right hand side. This has to go now. We need to find out the entering variable
corresponding to this leaving variable. How do we do this? We compute theta again. - 3 divided by - 11/7 is = 21/11. - 1 divided by - 3/7 is 7/3; 21/11 is smaller
than 7/3. Variable X1 enters and this is the pivot.Remember
that the pivot element has to be negative in this case. Only when the pivot element is negative, the
right hand side will become non negative in the next iteration. We continue with this simplex. The simplex iteration part of it is same. It does not change. The row operations are all the same. We will have X1, X2. X1 you get - 4; X2 you get - 7. So dividing by the pivot element or multiplied
by - 7/11 which is 1, 0, - 7/11 + 3/11. This is 8/11, this - 1/7 times this is = 0,
so 1/7 - this is = 0. This is 1. This - 1/7 times gives me + 1/11 - 1/7 - 1/7
into 3/11. So - 1/7 - 3/77 is - 14/77 which is - 2/11,
9/7 - 1/7 into 8/11. 9/7 - 8/77, so 99 - 8 is 91/77 which is 13/11.So
we get 13/11 here and then the value here will be 32 + 91 is = 123/11 with the minus
sign, 123/11. Cj - Zj values are all 0 and this is + 28/11
- 7/11 is + 21/11. 0 - (+ 21/11) is = - 21/11 - 12/11 + 14/11
is = 2/11. So 0 - 2/11 is - 2/11. Once again the Cj - Zj values are negative
indicating that the optimality condition is satisfied. Now we realize that the solution is feasible. Feasibility condition is also satisfied therefore
this is optimum. You can also show that, for example (21/11
into 5) + (2/11 into 9) will also gives us 123/11, 105 + 18 will give 123. So the optimal solution to the primal is X1
= 8/11; X2 = 13/11; Z = + 123/11. The minus comes because we have converted
a minimization problem into maximization problem by multiplying with the - 1. Optimum solution to the dual will be Y1 = 21/11;
Y2 = 2/11; W = 123/11. This algorithm is called the dual simplex
algorithm. Now let us go back to this. Both X1 and X2 are feasible. Right from the first iteration, the optimality
condition is satisfied. We do not have a leaving variable. The algorithm terminates to give us the optimum. The above algorithm is called the dual simplex
algorithm. What is special about the dual simplex algorithm? A dual simplex algorithm is very well suited
when you have all greater than or equal to constraints. We are not unduly worried about the objective. But it is very well suited when you have a
minimization problem with all positive coefficients here. So that the equivalent maximization will have
all negatives and the first Cj - Zj will have all negatives. It is very well suited for a minimization
problem with all positive coefficients in the objective function and all constraints
of the greater than or equal to type and with a non negative right hand side. A non negative right hand side is granted
when we formulate the problem. So in all greater than or equal to constraints
with non negative right hand side minimization function, all non negative or positive coefficients,
this algorithm is very well suited. We do not need to introduce artificial variables
at all. We do not need to do the big M method. We can still use the dual simplex and so on. The condition for the dual simplex is we will
have to have the Cj - Zj's negative to begin with. This will also be the negative indicating
infeasibility. We will retain the optimality condition by
a careful choice of the entering variable and theta by following a similar minimum theta
rule by making the pivot element negative or by pivoting on a negative element corresponding
to the leaving variable. We make the corresponding right hand side
non negative. We ensure that we are converting in every
iteration at least one negative value to a non negative value. Once again it is not absolutely necessary
that within two iterations this has to converge. We may encounter a situation where for example
here by doing one iteration, we are making sure that we have non negative number here. But this could turn out to be negative and
then it could go on. But if the algorithm has a solution then it
will terminate. In the dual simplex algorithm, we have to
make sure that the pivot element is negative. Only when the pivot is negative we can get
a positive here. Therefore the theta is computed in such a
way that it is computed only for those non basic variables which have negative coefficients
or are not completely negative coefficients in the leaving row. Then the minimum theta will ensure that the
optimality condition is satisfied. Feasibility is not satisfied by the starting
solution. If the solution is feasible we say that the
optimum is reached. It is called dual simplex for a very specific
reason. There are two ways of looking at it. So far we are used to having a maximization
problem with less than or equal to as our primal. Then we might be tempted to believe that the
dual of the primal by the way we have defined is a problem that fits into this structure
which is correct. The dual of the problem that we defined as
primal through fits into this structure. We might be tempted to call this dual simplex,
because it can solve the dual of our primal while that is not the real reason. The real reason is whatever problem you chose
to solve, moment you start the dual simplex algorithm, if you look at this carefully the
optimality condition being satisfied indicates that solution to the dual of the problem that
you are solving is feasible. So right through the dual simplex algorithm
you have a dual feasible solution, primal infeasible and the moment primal become feasible
it is optimum. It is called dual simplex because at every
iteration of this simplex, that dual of the problem that you are solving is feasible whereas
in the regular simplex which you may now call as primal simplex, the primal problem that
you are solving is feasible in all iterations. At the moment its dual becomes feasible and
it is optimum. We need to understand that it is call dual
simplex because at every iteration of this simplex algorithm, the dual to the problem
that you are solving is feasible and represented by the optimality condition being satisfied. Therefore it is called a dual simplex algorithm. This comes under the category of what are
called dual algorithms. Now dual simplex is an example of a dual algorithm. A dual algorithm will have a feasible dual
always or dual to the problem that you are solving is always feasible, applies complementary
slackness and the moment primal becomes feasible, it is optimal to both primal and dual. The regular simplex also called primal simplex
is a primal algorithm. Dual simplex is an example of a dual algorithm. Let us go to another type of problem. We have taken a problem which has mixed type
of constraint. You have a greater than or equal to type constraint. You have less than or equal to type constraint. You have all these. More importantly you have an objective function
though it has maximized it does not have all positive coefficients. So far we have not looked at a problem that
had a negative coefficient in the objective function. We looked at only after we convert it to a
maximization problem, you could get negatives but we never had a mixed set of objective
function coefficients. We never add a mixed set of right hand sides. If we have a problem like this then what do
we do is, the simplest and the easiest thing to do is the second constraint
X1 + X2 less than or equal to 3 is the kind of constraint that we want because the slack
variable + X4 will automatically qualify to be a basic variable. The first constraint being a greater than
or equal to type would now give us a negative slack which would not qualify normally. We would have added an artificial variable
a1 there and it is easy to start the simplex table with a1 and X4, use the big M method
and solve it. After having learned the dual simplex which
helps us to solve problems without introducing artificial variables, can we apply dual simplex
to this problem? Right now we cannot apply dual simplex to
this problem as it is because, when we when we write a - X3 here and a + X4 here, the
combination X3, X4 has an infeasible right hand side as you can see here Right hand sides will be 1 0 0 1 with a - 1
and 3. When we write this Cj - Zj, we write it and
convert it. We have a maximization problem. If we start with X3 and X4, we have a - 1
and 5. So we encounter situation where neither the
primal is feasible nor the dual is feasible. This 5 indicates that the dual is infeasible. A positive Cj - Zj would indicate that this
variable can enter which means that the corresponding dual when I apply complementary slackness
is infeasible. When we applied the dual simplex algorithm,
we made sure that all these were less than or equal to 0 and the dual was feasible in
all the iterations. Now we come into situation where the dual
is infeasible but we can still do this. So let us go back and see. You can do two things. You can leave this variable, treat this as
a leaving variable first and then perform a dual simplex iteration, or you can enter
this as your variable and perform a simplex iteration. You can do either. So what we are trying to show now is we have
a negative value for variable X3 as well as a positive value for Cj - Zj indicating that
both the primal and the corresponding dual are infeasible. So you cannot entirely apply the simplex algorithm
or entirely apply the dual simplex algorithm. If you want to solve this problem without
artificial variable, you have to judiciously mix simplex iteration and a dual simplex iteration
and proceed till the optimum. We show that through this example. You can actually do a simplex iteration by
entering variable X2 here with the 5 or you can do a dual simplex iteration by leaving
out this. So we choose the simplex iteration. In fact there is a general thumb rule that
when both the simplex and dual simplex iterations are possible, the thumb rule is, do the simplex
iteration first. We enter variable X2 with 5 and perform a
simplex iteration. For this you cannot have theta because in
a simplex iteration you would want the feasibility to be maintained. You will not compute the theta for this. You will compute a theta only for this. You have right hand side 3, you have X2 = 1,
so the theta value is 3 and variable X4 will leave the basis. This is 3 instead of 2 and variable X4 will
leave the basis. We do the next iteration. Next iteration is with X3 and X2. When we do simplex iteration, you now get
into situation where, your Cj - Zj's are negative which means that dual is feasible but then
the primal is infeasible because X3 now has a negative value. Now in this iteration you can do a dual simplex
iteration because your optimality condition is satisfied and the feasibility condition
is violated. You can do a dual simplex iteration which
is the only thing possible. Simplex iteration is not possible here because
you do not see an obvious entering variable whereas you see an obvious leaving variable. So you do a dual simplex iteration by leaving
out variable X3. That is what is written here. The primal is not feasible but the dual is
feasible. You can only do a dual simplex iteration with
X3 as a leaving variable. When we tried to compute the theta you will
get 6/5 and 5/3, 6/5 being smaller. Variable X1 will now enter. You will do a dual simplex iteration with
variable X1 replacing variable X3 and that is shown in the next iteration. Now you have X1 = 2; X2 = 1 with Z = 3 which
is your feasible solution to the primal. The feasible solution to the dual will be
Y1 = 6/5; Y2 = 7/5 and W = 3. You have a situation where your right hand
sides are non negative. Primal is feasible. We have optimality condition satisfied. So dual is feasible. You have now reached the optimum to both primal
as well as dual. The simplex method can also be used to solve
situations where we can have mixed type of constraints. When have mixed type of constraints and mixed
coefficients in the objective function, as we saw in this example, you could have a positive
and negative here. You could have a greater than or equal to
constraint and less than or equal to type constraints. When you have an equation you may still be
forced to use an artificial variable if necessary, because splitting the equation into two constraints
complicates further because you are increasing the number of constraints and the effort will
increase. We wouldn't choose to do that if you have
only inequalities and if you have a mixed type of constraints, you can still use a judicious
mix of simplex and dual simplex iterations and solve the problem. The only assumption that we have made here
is, the problem has a solution and we have got the optimal solution but we could get
into certain situations if for example a problem with mixed type of constraints with positive
and negative coefficients in the objective, if the problem is unbounded or infeasible
then we need a way to understand whether the problem is unbounded or infeasible. The usual conditions will still apply. For example even when you mix a simplex and
a dual simplex iteration or even within a dual simplex iteration you might get into
a situation where I can find a leaving variable, I may not be able to get an entering variable
that can happen. Now that does not directly ensure unboundedness
whereas those things are very well defined in a simplex iteration. In a dual simplex iteration if we get into
a situation where I have a leaving variable and I do not have an entering variable. That can happen. Such a thing would actually indicate infeasibility
of the problem. We have to go back and define how we look
at unboundedness and infeasibility with respect to the dual simplex or we look at unboundedness
and infeasibility with respect to applying simplex and dual simplex alternatively as
in the case of problems with different types of constraints and different positive and
negative coefficients in the objective function. Now both the dual simplex algorithm as such
and the example that we saw here where we solve problems with mixed type of constraints,
we have just tried to show that particularly in the second case when you have problems
with mixed constraints, where it is possible to solve using a judicious mix of simplex
and dual simplex iterations but then one has to be careful. If there is an optimum, we will definitely
get it but if the problem exhibits other things like unboundedness or infeasibility then we
need to carefully define unboundedness and infeasibility rules for those iterations. Here again we need to be little careful to
define the unboundedness and infeasibility but the equivalent unboundedness rule actually
represents infeasibility in the dual simplex iteration. Let us spend a couple of minutes on understanding
the simplex and dual simplex, the differences and so on. The regular problem that we had solved, i.e.,
the maximization problem which has all less than or equal to constraints and all greater
than or equal to variables is very ideal for a straight simplex application because slack
variables will qualify to be the basic variables. We can solve it entirely by the simplex algorithm. A minimization problem with all positive or
non negative coefficients in the objective with greater than or equal to constraints
is a very ideal situation or ideal problem to use the dual simplex. If we have problems with a mixture of constraints
and types of coefficients and so on, one would take the risk of solving it by judiciously
applying simplex or dual simplex or one would follow a very safe way of solving it only
by simplex algorithm by adding artificial variables. Both simplex and dual simplex algorithms solve
linear programming problems extremely efficiently. We have also seen the relationships between
the primal and dual that one is the primal algorithm and the other is a dual algorithm
and so on. The next thing that we have to see is, if
we can efficiently represent the simplex algorithm or the dual simplex algorithm. For example if we have to write a computer
program for the simplex algorithm, we will use the tabular form or are there better ways
of solving the simplex algorithm other than using the standard tabular form that we have
seen? We will see that in the next lecture
