 So, last class we have considered that solve to solve a LP problem using in tabular form. We have considered an example and that example we could not complete at that day. So, let us see from the tabular table 2 and table 2 70 we have considered in pivot element again. How pivot element is selected? First we will see the cost function coefficient in the cost function coefficient which is the most negative coefficient that will be treated as pivot column. So, pivot column is selected in other words x 2 is the our non-basic variable which will enter as a basic variable that is why the x 2 is called entering basic variables and how the pivot row is selected. So, once pivot column is selected find out the ratio of v divided by x 2 with positive sign of x 2 coefficient. So, in this situation this 51 by 51.6 divided by 7.332 ratio which is got 10.4 and another is your minus this minus sign will not help us to select the pivot element if you select the pivot element minus of the p 8.33 divided by minus 3. This will not help us to what is called reduce the function value if you consider x 1 is the living basic variable it will not have. So, minus sign will be not will not be considered in order to select the pivot element sorry pivot row. So, this is the only that this is the pivot row we got it. Once you got the pivot out of this two ratio minimum ratio is that that one 6.33 and that will be treated as a your pivot row. So, pivot row and pivot column will decide the pivot element. So, 70 is the pivot element. So, that that equation 3 is to be divided by 70 in order to make the coefficient of x 2 equal to 1. So, other coefficients equation number 1 and equation number 2 this coefficient must be 0 coefficient x 2 associated with x 2 and x 2 in equation number 1 and 2 must be 0 and similarly in cost function the coefficient associated with x 2 must be 0. This things we can do it by elementary row operations. Once you normalize this one by 70 we will get it this equation eliminate x 2 coefficient in order to eliminate the x 2 coefficient from equation number 1 2 and 4 that I have to multiplied by equation number 3 with coefficient 7.332 and subtract it from equation number 1. So, I am written 1 minus 3 multiplied by 7.332 if you subtract that will come 0 0 1 minus 0.28 and minus 0 147 and this last class we have written 20.9 point this is mistake we have to write 4.546. Similarly, this second equation x 2 we can remove it by multiplying the equation 3 equation 3 multiplying by this equation you have to after normalizing this equation you have to multiplied by 0.337 and added equation number 2. So, 2 plus 3 again then 0.337 this added if you add it you will get it this equation last class we have written 10 to the power minus actually it will be 10 to the power minus 3. So, similarly x 4 equation cost function equation the x 2 we can eliminate by multiplying the equation this equation by 1 2 3 and add with equation number 4 this equation number 4. When you are talking equation number 3 correspond to this equation. So, this we are doing so after doing this one look at the last row of the cost function look last row of the cost function this cost function all coefficients are positive. That means, further we cannot reduce the cost function value by changing one of the basic variable as a non basic variable and one of the basic non basic variable as a basic variable. So, we will stop our iteration of here and directly you can see what is the cost what is the function value basic variable function value. So, x 1 into 0 plus x 2 into 0 plus x 3 into 1 x 3 plus x 4 into this quantity, but x 4 is a non basic variable this value is 0 the x 5 into this since x 5 is a non basic variable this is 0 x 5 value is 0. So, x 1 directly we will get it 4.56. Similarly, x 1 we are getting directly is 10.47 then x 2 non basic variable value x 2 is directly 4.62 and what is the cost function value x 1 multiplied by 0 plus x 2 multiplied by 0 plus x 3 multiplied by 0 0 x 4 is a non basic variable this value is 0 0 multiplied by this non 0 quantity 0 this multiplied by 0 this equal to this quantity. So, this quantity is equal to 0 means f is equal to this one. So, our objective function value minimum value of the objective function is that one if you see this one and what we are supposed to find out you see our problem was maximize our problem if you see the maximize f of x that minimize value of x we got it. So, maximize values of that one will be f of x will be that is not here last page this one the maximum value of f will be what is this f max value will be minus of minus 210. So, it will be 2140.644 that is our maximum value of that one will get it and what is this our x 1 x 2 value x 1 is 10.4 x 2 is 6.48 these are two values are their design variables x 1 and x 2. So, these values are the design variables values are that one this is the design variables this and this agree and you see this is actually this will be z not f this is z. So, z is equal to this one and that f of x maximum value is that quantity. So, this is the way we will solve the LP problem using tabular form. Now, let us see what is two phase method the solving LP linear programming problem using two phase method a two phase two phase simplex method for solution of LP problem. So, let us see what is this our problem is maximize z is equal to y 1 plus twice y 2 subject to 3 y 1 plus 2 y 2 is less than equal to 12 and then 2 y 1 plus 3 y 2 is greater than equal to 6. Now, you see one inequality is less than equal to thing another equality is greater than equal to 6. So, this type of inequality when it is there the greater than equal to some constant quantity. Then we have to solve this problem by two phase simplex method what is this and it is also mentioned that y 1 is greater than equal to 0 and y 2 is unrestricted in sign that means is value can be negative positive anything 0 also. So, now this because when we will solve this problem using the standard LP method by simplex method you have to convert into standard LP problem. Then how will you solve because y must be greater than equal to 0. So, y we can redefine define y 2 as a y 3 minus y 4 and y 3 is greater than equal to 0 y 4 is also greater than equal to 0. Now, if you replace y 3 by 2 new variable y 3 y 2 by 2 new variable y 3 minus y 4 and both are greater than equal to 0. So, this quantity depending upon the value of y 4 or y 3 this will be y 2 may be positive negative and 0. So, this problem we can redefine into a what is called new variable form. If you define x 1 y 1 is equal to x 1 x 1 I have defined by y 1 then x 2 I defined y 3 and x 4 sorry x 3 x 2 I have defined by y 4. So, our original problem the original problem given this one we can rewrite into this form maximize z y 1 plus y 2 y 1 plus 2 y 2 y 1 value is what x 1 plus twice y 2 value is what if you say y 2 value you have y 3 minus y 4 y 3 value you have considered x 2 and y 4 consider x 3. That is we have defined here agree. So, this is our objective problem in terms of new variable x 1 x 2 x 3 subject to f of x that is maximize subject to what is this 3 y 1 3 y 1 value I will write x 1 2 y 2 means x 2 minus x 3 is less than 0. So, I will write 3 x 1 plus 2 x 2 minus x 3 is less than equal to 12. Then twice the second equation inequality equation twice x 1 plus 3 x 2 minus x 3 is greater than equal to 0 2 x 1 plus 3 x 2 minus x 3 is greater than equal to 6. Now, we are writing x 1 is greater than equal to 0 for I is equal to 1 2 3. So, what is our now problem is there you convert into a then our solution starts from here convert into standard LP problem. And standard LP problem you know this is minimize f of x what is f of x minus z what is this one minus x 1 minus 2 x 2 minus x 3. And we have to convert all inequality conditions in a equality conditions with right hand side is a positive constant quantity. So, let us say this one this one will write 3 x 1 plus 2 x 2 minus x 3 and this is less than 12. That means something we have to add with this one let us call the new variable x 4 is equal to 12. And right hand side you see positive quantity is that we have to make it right hand side that equality condition equality equation positive quantity it is already positive. So, x 4 is thus our slack variables the way we do it we do the what is called standard LP problem by adding slack variables. And that slack variables value greater than equal to 0. And second equation you see second equation 2 x 1 plus 3 x 2 minus x 3 and that is greater than equal to something we have to subtract some variable x 5. Suppose if you make it x 5 is equal to that 6 then you will see when you solve this equation x 1 x 3 this when it is 0 x 1 x 2 is 0 that values are coming minus. So, that creates a problem while you will solving the LP problem. So, in addition to this one I am adding with another new variables that is called artificial variables this is called artificial variable. This is called artificial variable and that artificial variable value is greater than equal to 0. And this is also it is called surplus variable this variable x 5 with minus sign it is called surplus variable. So, this is called surplus. So, when this type of inequality is there we have a two variables we have to introduce one is surplus variable another is artificial variable. If you just introduce surplus variable you see that this x 5 value will after solving you may get it negative quantity and this is not our constant is all are greater than equal to 0. So, this will violate our solution. So, in order to avoid that one we use another artificial variable that is called the what is called x 6. So, now it is converted into our standard LP problems once it is converted into standard LP problem we can solve is either in matrix form or we can solve by using tabular form. Let us see that how it is solved by this is solved by using the what is called tabular form. So, two phase method two phase simplex method you can write is an approach two phase simplex method is an approach to handle the artificial variable to handle the artificial variable when they are added to the system equations. So, two phase simplex method is an approach to handle the artificial variable or t whenever they are. Now, first phase that is called two phase first phase we will eliminate artificial variable from the problem. That means how the artificial variable is going to be formed. So, the all artificial variable all are artificial variable is considered as an artificial function artificial cost function. Suppose you have more than one artificial variables are there you have to add all are artificial variable together and considered as a artificial objective function or cost function and that cost function you have to minimize. In other words that cost function value artificial cost function value must be 0 some of the all artificial variables must be 0. So, one phase you have to eliminate w from the mathematical expressions. So, let us see how one can do this one phase one removal of artificial variable. So, this is here let us define artificial variable. So, this is agree so insult a artificial variable. So, how to remove this one. Suppose if you have a equation of this type of equation more than one let us go to the how many artificial variable will be there two artificial variable in each equation one are Maori fish is there, and two equation two are two artificial variables. two artificial variable you add together and treat as a objective function in addition to the original objective function. So, we have to do simultaneously minimize we have to minimize two objective function first you minimize the artificial objective function once we have minimize this one then you start to minimize in phase two you minimize the what is called objective function of the original problems. So, that is our steps to be followed. So, let us see in this example what is our artificial function where artificial variable x 6. So, that x 6 the minimize w and w I have considered x 6 and what is this x 6 expression see from this one x 6 is equal to I can write 6 minus 2 x minus 3 into x minus 2 x plus x 5. So, our 6 minus 2 x minus 3 x 2 plus 3 x 3 if you bracket you open the bracket then plus x 5 is our artificial objective function this is the artificial objective function minimize this one subject to what subject to you see this two constraint equality constraints. So, this two equality constraint 3 x 1 3 x 1 3 x 1 plus 2 x 2 minus 3 x 3 plus x 4 3 x 1 plus 2 x 2 minus 2 x 3 plus x 4 is equal to 12. Then second equation you see this equation 2 x 1 plus 3 x 2 minus 3 x 3 2 x 1 plus 3 x 2 minus 3 x 3 minus x 5 plus x 6 is equal to 6. So, our problem is minimize this one first phase minimize this one subject to this one. So, if you write it this one in tabular form this will look like this way. So, table one basic variables variables is the basic variables in these directions then how many variables are there if you see in this equation how many variables x 1 x 2 dot x 6. So, 6 variables are there. So, you write it x 1 x 2 x 3 dot x 4 x 5 x 6 then b then ratio. So, I will write first this equality constraints 3 into x 1. So, it will come 3 next is your see this one next is your 2 into x 2 minus 2 into x 3 plus x 1. So, under 2 under x 2 2 will be there under x 3 column minus 2 is there under x 4 column 1 is there x 5 x 6 no coefficients associated with x 5 x 6 is 0. So, I can write it that 1 2 minus 2 1 0 0 that right hand side is equal to 12. So, I will write this is equal to 12 similarly the second equation under x 1 2 I will write under x 2 3 under x 3 minus 3 under x 4 minus 1 x 5 minus 1 no x 4 is there under x 4 will be 0 under x 6 coefficient is 1 1. So, I can write it 2 3 minus 3 0 minus 1 1 6. So, we have a 2 objective function we have to give in phase 1 this is the phase 1 you can say this is the phase 1. Phase 1 job is what it will optimize the optimize means in this way minimize the function value W means artificial objective function value you have to minimize. In other words you have to make it 0 this one by minimizing this optimization problems LP problem linear optimization problem. So, our first is cost function or normal original cost function next is our artificial artificial cost function. So, let us say our original cost function is what if you see the our original cost function that minimize f of x that minus x 1 minus x 2 bracket if you open plus x 3. So, under x 1 is minus 1 the original cost function under x 2 is minus 2 under x 3 is 0 2 and then there is no x 3 x 4 is there if you see x 4 x 5 x 6 is not there. So, these all are 0 0 0 and what is this you right hand side of this one this value is this value is f of x. So, you will write simply that f of x this dot what is the function value of this one. Next is artificial variable you see artificial variable objective function what we have written it artificial objective function value expression just now I mentioned yes W is a 2 x 1 minus 3 x 2 plus 3 x 3 plus x 5. Since W is constant quantity you can make it with the W minus W minus 6 is equal to minus 2 x 1 minus 3 x 2 plus 3 x 3 plus x 5. So, if I write it this one accordingly I will write under x 2 is minus 2 x 1 then minus 3 this is you see this one if you take this one is a minus 3 x 3. So, minus 3 then plus 3 then x there is no coefficient with here there is no coefficient x 4 x 5 is 1 coefficient. So, x 5 1 coefficient x 6 there is no coefficient then what is the function value because it is 6 is constant we can take it left inside W minus 6 is the objective function value agree. So, minus 2 minus 3 0 1 0. So, now start your table of process of this one now you see first you have to identify which one is the basic variables. Now, if you consider x 4 x 4 is exist in first equation also x 1 variable belongs in equation number 1 it is does not belongs in equation number 2 agree, but x 5 you see that x 5 does not contain in equation number 1, but it is a with minus sign. So, you know to make minus plus then this will be minus sign. So, this we cannot take in as a basic variable, but x 6 contains only equation number 2 with positive and it does not contain in equation number 1. So, our basic variables are basic variables are x 4 and x 6 is a basic variable which is our remaining things are our non basic variables. Because we have a 2 equations are there 2 basic variable and we have a 6 variables are there 6 minus 4 6 minus 2 is a 4 non basic variable. So, our non basic variable generally denote with an arrow x 1 x 2 x 3 and x 5 is a non basic variables. Now, you see what is the basic variables value immediately because non basic variable value is 0. So, if you see this equation number 1 immediately we get x 4 is equal to 12 agree and x 6 is equal to 6 and remaining x 1 is equal to x 2 is equal to x 3 is equal to x 5 is equal to 0 agree. Now, this are the basic variables value values and this is the non basic variables agree. Now, which non basic variable will act as a basic variables. That you have to find out by choosing the pivot column and pivot row. Now, since I have to in phase 1 we have to optimize the artificial objective function or cost function. So, you have to concentrate only with the based on this coefficient reduction reduced coefficient objective coefficient of the artificial cost functions. Now, see this coefficient is minus this coefficient also minus we have to consider the most negative coefficient and most negative negative coefficient is that one agree this one. So, this is our pivot column because we have to minimize first w then we have to minimize that our cost function. So, we have to concentrate only this last row which is artificial cost function. So, this is the pivot column that means x 2 entering as a basic variables entering basic variable is which one x 2 and which one is non basic variable will enter as a non basic variables. So, again you have to see this ratio that we have to show 12 by 2 is equal to 12 by 2 is equal to your and 6 by 3 is equal to 12 by 2 is equal to 6. Which one is your what is called minimum ratio this one is your minimum ratio. So, then what will do it this is the minimum ratio. So, then what will do it this is the pivot row if you see the pivot row of this one that means this is our pivot element this is our pivot element. So, what will do you normalize this one you normalize this thing by a 3 that means what I am doing and after that you eliminate x 2 from equation number 1 equation number 3 cost function similarly equation number 4. So, what we can do it here if you normalize this one I am writing 1 equation number 1 minus equation number 2 this equation number 2 you have to normalize first then multiplied by 2. So, you have write equation number 2 normalized equation number 2 normalized then multiplied by 2 and stands for normalized equation number 2 normalized by 3 or you can write equation number 2 divided by 3 into 2. Then it subtract from equation number 1. So, if you do this one this is for equation number 2 and this after normalization whatever the equation you got it that you multiplied by 2 added with this one. So, x 2 will be eliminated similarly after normalization this you multiplied by 3 add with this one then x 2 also it eliminate that means I can write it if you see this one that equation number 3 plus equation number 2 normalized by 3 this one multiplied by what this you have to multiplied by 2 added then you will get it x 2 eliminate here x 2 also eliminate here. In this last equation that equation number 4 you add you normalize this one by 3 then multiplied by 3. So, equation number 2 normalized multiplied by 3 then add with equation number 4 if you do this operation then ultimately you will get the table like this way. So, we can say here our with this one pivot element selection x entering basic variable is our x 2 entering basic variable is x 1 is our x 2 and entering living basic variable is your x 6 x 6. So, now you will see that table table 2 basic variables basic variable x 1 x 2 x 3 basic variable in this direction x 3 x 4 x 5 x 6 b ratio. So, whatever the operation I asked you to do this one that this equation you divided by 3 then it will be coming the second equation and our basic variable says here what is this our basic variable x 4 and not changing x 6 is replaced by x 2 this is our basic variable just we have explained it here then second equation I mention it divided by normalize by 3. So, this will be a 2 by 3 this will be 1 this will be minus 1 this will be 0 then x 5 coefficient will be minus one third x 6 coefficient will be x 6 coefficient will be x 6 coefficient will be 1 by 3 then x coefficient will be 2 that b coefficient will be 2. Now, whatever the operation I did it to do if you do this one then this will be coming 5 by 3 what I did it to do this one you see the equation number 2 after normal normalization multiplied by 2. So, you multiply by this 2 that means what I will get it multiplied by 2 that means 4 by 3 4 by 3 3 minus see 3 minus 3 minus this one 3 minus 4 by 4 by 3. So, this one is coming if you see this one that I divided by this 2 by 3. So, I multiplied by this thing is 2 2 by 3 4 by 3. So, 4 by this was here 3 4 by 3 this is nothing but a 9 by 9 minus 4 by 3 is a 5 by 3. So, this is 5 by 3 similarly you have to do all this thing. So, 0 0 1 this will be a 3 by 2 2 by 3 it will be a point or you can write 2 by 3 if you like 2 by 3 this is 2 by 3 this is minus 2 by 3 then this is 8. Similarly, the cost function what I did it that equation number 2 this after normalization you multiply it by this thing by 2 multiplied by 2 and add with equation number 3. So, if you do this one you will get one third 0 0 0 0 0. So, you will get 0 minus 2 third plus 2 third and this will be f plus 4 and this is the artificial cost function artificial artificial cost function. And what I did it for this one you see this after normalization this is the this equation and this equation you multiplied by 3 that is what I am ready to multiply by 3 add with 4 add with this equation add with this equation sorry 4 with this equation. Then you will get it that one 0 0 0 0 0 last is 1 then this is w. Now, you see the last because our job is to minimize this first in phase 1 phase 1 this phase 1 first one to minimize that our w means our cost artificial cost function. You see last of the artificial cost function the reduced coefficient all are positive this all are positive. That means we cannot further reduce the value of the artificial cost function. Then immediately I can write the artificial cost function artificial cost function value w is equal to 0 how x 1 into 0 plus x 2 into 0 x 3 into 0 our basic variables are what x 4 and x 2 others are non basic variables. See x 1 sorry not x 2 x 2 is basic variable x 3 x 5 and x 6. So, now x 6 is non basic variable this into 1 means x 6 is 0. So, w is equal to 0 we got it. So, at this stage from the table 2 we can find from the table 2 this one we can find that x 1 is our non basic variable is 0 x 2 our basic variable x 2 our basic variable x 2 value is what we will get it see x 2 value is 2 directly you can see it is 2 a this into this 0. Similarly, this into this x 2 this is non basic variable 0 this basic variable 0 into this. So, it is a 2 and then x 1 x 2 x 1 is 0 x 2 is 0 then x 3 is our non basic variable 0 x 4 is basic variable state wave from this expression is 8. So, these are the basic variables this and this sorry 2 is the basic variables agree these are the basic variables values and x 4 x 5 is equal to 0 x 6 is a non basic variable 0 x 1 x 3 x 5 x 6 is 0. So, w x 2 is non basic variable this. So, after the end of the phase 1 that means we optimize the artificial objective function value which is equal to 0 then even though we added x 6 x 6 value is 0 this artificial variable. So, at the end of this one our basic solutions we got it this one agree since w is 0. So, this values x 6 value will not be considered in second phase 2 from now on and so phase 2 simplex method phase 2 simplex method phase 2. Since, w 6 w is 0 means x 6 values is 0 we got it this w is equal to x 6 we have considered this is. So, from this the new table what will form it x 6 you can omit or you cannot keep it in the table, but ignore it any operation you do further from onwards agree. So, I will take the table like this way basic variables see from this equation. So, our this optimization value is 0. So, now our only left is x 6 is 0 agree x 6 is 0 that is, but in turn during the phase 1 this our table of equation is now change it this one. Let us see this one x 1 x 2 x 3 x 4 x 5 x 6 will not include in the table even if you include in the table it ignore it is all operation from now onwards. So, our last table if you see this one I am reproducing here the our last table omitting the last column as well as the artificial cost function. Then the our problem is minimizing the our cost function. So, I am writing only this portion if you see from here up to his up to this and including that one this column also this thing I am reproducing now. So, now our say 5 by this is 5 by 3 now 5 by 2 agree. So, we are this is 5 by 3 is 5 by 3. Now see this one we are considering this table now this we have to now optimize that one this. Now, this cost coefficient you see whether we will be further we will be able to minimize the cost function value or not that we have to see it how we will see it that one that you see the reduced cost coefficients this is plus this is plus this is plus this is plus this is plus only this is minus. So, this quantity this that negative most negative coefficient of the cost of objective function is along the x 5 variables. So, this is our pivot column this is our pivot column if you see this column is our pivot column this column again now you find out the ratio that means 8 by multiplied by 8 multiplied by 3 by 2 that it comes 12. Now, this is minus ignored that means this if you consider this is our pivot row then this will not reduce our cost functions agree this will go beyond our what is called our region feasible region if you consider this one. So, this is our pivot this is the pivot row. So, our pivot element is that one. So, what will consider that x 5 is entering basic variable and x 4 is leaving basic variables. So, we can write it now here that our before entering the phase 2 our entering basic entering basic variable is x 5 leaving basic variables leaving basic variables is your that x 4. So, what you have to do here in phase 2 you divide the equation this that means normalize this one 2 by 3. So, if you divide by 2 by 3 this will be 1 and this is will be 3 by 2 minus 1 and this is will be 3 by 2 will be 5 by 2 that one this is 3 this is 3 this will be 5 by 3. So, if you do this one now our basic variables is which one say our basic variable x is leaving as a non basic variable entering x 5 is entering is a basic variable. So, x 5 x 5 and x 2 is our basic variables and remaining are non basic variables x 1 x 3 and x 4 is a non basic variables. So, look this is I divided by what is this I divided by 2 by 3 the normalize by 2 by 3 and you have to eliminate x 5 from equation 2 and 3 that means cost function. Then what will do it I have to multiply by after normalization I have to multiply it by this one third 1 by 3 and add with this equation. So, x 5 is eliminated similarly I multiplied by this 2 by 3 add with this one. So, I am writing here that equation 2 equation 2 add equation 2 you multiplied by this that equation 1 you normalize by 2 by 3 then multiplied by one third whatever you will get it add with equation number 2. If you do this one first you will normalize this one that will give you if you normalize this one that will give you 5 by 2 5 by 2 0 3 by 2 1 and b and ratio 3 by 2 1 minus 1 minus 1 then it will get half 0 and this is 6 and this is a 12 cost function 2 0 0 1 0 f plus 12. You see what I did it this equation number 1 you normalize by 2 by 3 normalize by 2 by 3. So, x 5 coefficient will be 1 x 5 coefficient I normalize by 2 by 3 this will be 3 by 2 you see this will be 3 by 2 this is 0 this is 0 this will be 5 by 2. So, this then what I did multiplied by the this is one third this equation I multiplied by this equation I multiplied by one third and add with this equation then you will get it this things. Now, look whether is there any possibility to reduce the cost function further by looking the reduced cost coefficient and reduced cost coefficient of the cost function objective function all are positive. So, there is no chance of reducing the function value further. So, we will stop our that iteration in phase 2 here suppose you got some of the coefficient in negative then you find out the pivot column pivot element that pivot row and then pivot element proceed in the similar manner. So, from this table so, from this table we get x 5 is equal to 12 x 2 is equal to 6 similar manner agree this is the basic variables and what are the non basic variable x 1 is equal to x 3 is equal to x 4 is equal to 0 is a non basic variables. Then what is our function value function value is minus 12 and our problem if you see the earlier our problem this problem what we have to solve it here you see that one maximize z. So, z value is what z value is minus of minus of this one. So, it will be a 12 plus and what is our corresponding y 1 y 2 all this things y 1 we have considered x 1. So, y 1 value is same as the x 1 whose value is 0 and similarly y 2 value is what y 2 value is nothing but a y 3 value is nothing but a x 2 minus x 3 and x 2 value is what we got it 6 and x 3 value is 0 x 2 value is 6. So, 6 minus 0 is equal to 6 that y 2 value is 6 and the other values x 2 value which we got it x 3 and y 1 y 2 value you got it. So, y 2 value is 6 and y 1 value is from this one y 1 value is this. So, graphically if you see this one if you see this figure in graphically in original problem you will see this one that our y 1 is greater than if you see this one that y 1 is greater than 0 y 1 if you consider in this directions and y 2 is in this directions agree. And y 2 can be positive and negative that means this will be on the line. This one and this region this region is y 1 greater than equal to 0 agree and this region that what is called and this region not this region you do not know that either top side or bottom side is y 2 if it is bottom top side it is y 2 greater than 0 and bottom side is y 1 less than y 2 is less than 0. So, our if you draw the equation number this and this and this you will see you will get it something like this equation. So, when y 2 is 0 that what we are getting the y 2 is 0 x y 1 is 4 y 1 is 4 then 1 y 1 is 0 y y 2 is 6 though it is not drawn to the scale. Let us call forget this is I am drawing to the scale is like this way forget about this one this is a 4. So, this is our which side is this one. So, this is our this expression this expression is 3 y 1 plus 3 y 2 2 y 2 2 y 2 is less than equal to 12 agree and another equation if you see that one this will be a it will go it can go extend like this way and this values is and that equation is twice y 1 plus 3 y 2 is greater than equal to 6 and this is what I am doing. So, next class I will discuss the figure more clearly we will stop it here.