in this video we are going to see linear programming problem using bigm method the bigger method is a modified version of Simplex method only so we can use the same Simplex algorithm to solve this bigm method there is no much difference between Simplex method and bigm method in Simplex method it contains lesser than or equal to constants so in that case we need to add slack variables that is plus s this we have already discussed in the Simplex method in a separate video now we are going to see bigm method which contains greater than or equal to constraints so in case of greater than or equal to we need to do two things first thing we need to subtract Surplus variables that is minus s and we need to add artificial variables that is plus a let us see the problem using bigam method see the problem solve the following linear programming problem using bigm method minimization Z is equal to 7 X1 + 15 X2 + 20 X3 subject to 2 X1 + 4 X2 + 6 X3 greater than or equal to 24 and second one 3 X1 + 9 X2 + 6 X3 greater than or equal to 30 X1 X2 X3 greater than or equal to0 so these are the two constraints this one is objective function so this problem is minimization problem using bigm method the first step is we need to subtract Surplus variables and we need to add artificial variables in the two constraints before that we need to write the objective function that is minimization z is equal to 7 X1 + 15 X2 + 20 X3 subject to the constraint subject to the first constraint is 2 X1 + 4 X2 + 6 X3 so in order to balance the equation we need to subtract Surplus variable 1 and add artificial variable 1 in the first constraint is equal to 24 in the same way in the next constraint is 3 X1 + 9 X2 + 6 X3 so here also greater than or equal to so in order to make it equal just subtract Surplus variable 2 and add artificial variable 2 is equal to 30 and what are the variables are there X1 comma X2 comma X3 comma we have subtracted two Surplus variable no so S1 comma S2 and we have added artificial variables to comma A1 and A2 greater than or equal to Z okay so here we have done two things first thing is we have subtracted Surplus variable and we have added artificial variable the same thing need to be added in the objective function okay so the objective function is minimization z is equal to 7 X1 + 15 X2 + 20 X3 and we need to add that is the for Surplus variable we need to give positive 0 okay the value is positive 0 so 0 S1 + 0 S2 for artificial variable we need to give maximum positive value that is m m A1 + m A2 M denotes maximum positive value see these artificial variable are included in the model just to solve the model okay and in the final solution these artificial variabl should not be available let us see the first initial Simplex table let us see the initial table format BV stands for basic variables that is in this problem we have these variables X1 X2 X3 X1 S2 A1 and A2 so these are the basic variables CJ stands for coefficient of objective function and CB stands for coefficient of basic variables so this one is Row one and second one row two here we need to write solution and finally ratio so now let us see how I'm framing these things so so BB is basic variables are these are the basic variables that we have already written CJ stands for coefficient of objective function this we can write from the objective function look at the objective function 7 X1 so X1 is there no right 7 15 X2 X2 is there right 15 next one 20 X3 20 0 S1 0 0 S2 0 and for artificial variable the most positive value that is M for A1 M for A2 also M so these are the objective function that is coefficient of objective function these are the basic variables in this column we need to write the two artificial constraints that is A1 and A2 so write A1 and A2 what is the coefficient of the basic variables look at the objective function for A1 M for A2 also M so write M for A1 and M for A2 also so now write the first row values so look at the first constraint 2 X1 4 X2 6 x3 - S1 + A1 is = 24 2 X1 4 X2 6 X3 S1 that is - S1 so S1 is there no right min-1 so in the first constraint there is no S2 put zero and in the first constraint A1 is there so write + 1 no A2 so write 0 is equal to 24 in the same way fill all these things for the row two for row two 3 X1 3 9 X2 for X2 9 for X3 6 and there is no S1 so put zero S2 is there Min -1 A1 is not there so 0 A2 is there so + 1 is equal to 30 so this is the way to frame initial table so now we need to proceed further the next step is we need to find ZJ okay so in order to find the ZJ this is the formula to find ZJ so instead of using this formula let me explain the simple method so first we need to take coefficient of basic variable of R1 into the first element plus coefficient of basic variable of R2 into first element then you'll be getting ZJ take the coefficient of basic variable into the first element plus the second row coefficient of basic variable into first element we'll be getting ZJ that is M into 2 + m into 3 in the same same way for the next value M into 4 + m into 9 M into 6 + m into 6 for the fourth value M into -1 + m into 0 in this way we need to find ZJ let us see the calculation for the first one M into 2 2 m + m into 3 3 m 2 + 3 5 m for the next value M into 4 4 m + m into 9 9 m so totally 13 M for the next one M into 6 + m into 6 so 12 M next value M into -1 - M + m into 0 0 so value - M next one M into 0 0 + m into -1 - M next one M into 1 m + m into 0 0 so Plus M next one M into 0 0 + m into 1 + m next one M into 24 24 m + m into 30 30 m totally 54 M okay so this is the way to find ZJ so after finding ZJ the next step is CJ minus Z J so we need to compare CJ and ZJ and we need to find the difference that is the formula is CJ minus ZJ so look at the first value CJ is 7 minus ZJ is 5 m so answer 7 - 5 m next value CJ is 15 ZJ is 13 M so cjus ZJ is equal to 15 - 13 M next value 20 - 12m next one 0 - M already minus is there so 0 + m so answer is M next 0 -- M so-- plus plus M next one M - M 0 m - M 0 after finding cjus Z J we need to check the optimality so what is optimality condition for minimization problem this is the optimality condition that is CJ minus ZJ must be greater than or equal to Z so all the cjus ZJ value must be 0 or positive value that is greater than or equal to 0 look at the answer CJ minus ZJ here we have got negative value for three things the first one is 7 - 5m here 15 - 13 M and here 20 - 12 M so we did not reach the optimality so in order to reach the optimality we need to proceed further in order to proceed further the first step is we need to find the key column for key column we need to find most negative value from the CJ minus ZJ so look at these three values which one is most negative minus 13 M that is 15 minus 13 m is the most negative value so this is the key column so this column is a key column that is X2 column is a key column okay so after finding key column the next one is we need to find key Row for that we need to find the ratio in ratio column for 24 ID by 4 is equal to 6 for the second row the solution is 30 no so 30 divid by 9 is equal to 10 by 3 in order to find the key row we need to select least positive value so which one is least positive value 10 by 3 is the least positive value than the six so this row is a key row this row is a key row okay so now we have got key column X2 is a key column and A2 is a key row and the intersecting point is called key element so the key element is nine and this one is key row and this one is key column so after finding key row and key column and key element we need to prepare the first iteration in the first iteration X2 will be the entering variable and A2 will be the leaving variable in the first iteration BV and CJ remains same okay so now we have to fill Row three and row four okay so in the previous initial table so which variable is entering variable and which variable is leaving variable X2 is the entering variable and A2 is the leaving variable okay in the first iteration instead of A2 you need to write X2 X2 what is the cost for X2 15 write 15 here okay and A2 is a leaving variable no so in the A2 column you need not find anything so put Dash okay now we need to find the new value in the first iteration okay in order to find the new value just take the old value divided by key element okay for the leaving variable okay you need to find the old value divided by the key element that is 3x 9 9 by 9 6 by 9 0 by 9 - 1 by 9 0 by 9 1X 9 and 30 by 9 in this way you can find a new value for row four so first one 3x 9 that is 1x 3 9 by 9 1 6 by 9 is = 2x 3 0 by 9 is = 0 - 1 by 9 is = - 1 by 9 0 by 9 is = 0 okay and 30 by 9 is equal to 10 by 3 okay we have got the new value for row four okay this is the way to find new value for row four now we need to find the values for row three in order to find values for Row three there is a Formula the formula is old value minus key column value into key row value divided by key value okay so look at the table initial table for example for the first Value Old value is 2 okay so 2 minus corresponding column value is 4 okay into correspondent key row value is three divided by key value is 9 okay this is the formula to find the new value so instead of this formula you can find this formula that is old value minus key column value into row four value okay this we have already applied in the previous step that is uh key row value divided by key value okay this is denotes R4 so you can simply take the R4 value let me explain with the problem old value minus key column value into row four value look at the initial table old value is 2 minus key column value is 4 into row four value is 1 by 3 again I repeat old value is 2 so 2 minus key column Valu is 4 so 4 into R 4 value 1 by 3 2us 4 into 1 by 3 is equal to 2 - 4X 3 = 3 6 - 4 is = 2x 3 so the new value is 2x 3 2 by 3 in the same way for the next Value Old value is four corresponding column value is also four and the new value row four value is 1 so formula is old value 4 minus correspondent column value 4 into our four value is 1 so 4 - 4 is = 0 0 in the same way for the next value is old value 6 so 6 minus corresponding column value is 4 into R4 value 2 by 3 6 - 4 into 2 by 3 so 6 - 4 2 8 8 by 3 LCM 18 - 8 is = 10 by 3 new value is 10 by 3 10 by 3 okay for the next Value Old value is - 1 minus key column value 4 into row value row value is 0er -1 - 4 into 0 = -1 - 1 for the next Value Old value is 0 so 0 - 4 into - 1 by 9 is equal to + 4 by 9 for the next Value Old value is 1 1 minus column value 4 into new row value 0 1 - 4 into 0 is = to 1 1 okay for the solution old values 24 4 minus column value 4 into new row value 10 by 3 so 24 - 4 into 10 40 by 3 so LCM 3 3 into 24 72 - 40 is = 32 by 3 32 by 3 see here here the value is the basic variable remain same and the coefficient of basic variable also remain same so this is the way to frame first iteration table so after framing all these things the next step is you need to find ZJ okay in order to find the ZJ you can follow the same procedure what we did in the initial table for the first value M into 2x3 + 15 into 1X 3 the answer is 2x3 m + 5 okay for the second value M into 0 0 + 15 into 1 15 next value M into 10x 3 10x 3 m + 15 into 2x 3 + 10 next value M into -1 + 15 into 0 so - M next value M into 4X 9 so 4X 9 m + 15 into - 1 by 9 so - 5x3 next value M into 1 m + 15 into 0 0 the answer is M okay in the same way for solution M into 32 by 3 + 15 into 10 by 3 the answer is 32 by 3 m + 50 after finding ZJ the next step is we need to find CJ minus ZJ so look at the first value the formula is CJ minus ZJ so first value is 7 7 - 2x 3 m + 5 so you cannot subtract this value from 7 because it's a here we have M value so we can subtract only five from 7 okay the formula is CJ minus Z J so we'll be getting - 2x3 m + 2 that is 7 - 5 we'll be getting 2 for the next value 15 - 15 will'll be getting 0 for the next value 20 - 10x 3 m + 10 the same way we cannot subtract this value okay from 20 we can subtract only 10 okay the formula is CJ minus ZJ no so the answer is - 10 by 3 m + 10 that is 20 - 10 we'll be getting 10 the next values 0 - - M so we'll be getting + m next value 0 - 4x 9 M - 5x3 the same thing is applicable here also because we have M value here so the answer is - 4 by9 m + 5 by3 okay again I repeat 0 - 4x 9 M - 5 by3 so minus into minus we'll be getting plus value the next value M - M 0 after finding CJ minus ZJ we need to check the optimality condition what is the optimality condition for minimization problem all the CJ minus ZJ value must be greater than or equal to zero which means we need to get either zero or positive value but in this iteration we have got negative values for three things so we didn't attain the optimality so we need to proceed further in order to proceed further we need to find key column key row and key element for the key column we need to find most negative value so which one is most negative value this one is most negative value so this column is a key column okay after finding key column then we need to find the ratio in order to find key Row for the first row 32 by 3 divid by 10 by 3 which means 3x 10 the answer is 6 16 by 5 okay for the next value 10x3 / 2x 3 that is 3 by 2 the answer is 5 in order to find the key row we need to select least positive value so which one is least positive value so this value is the least positive value when compared to 5 so 16 by 5 is the least positive when compared to five so this row is a key row after finding key row and key column this intersection point is the key element so X3 is a key column and A1 is a key row and 10x3 is the key element okay so according to this iteration X3 is the entering variable and A1 is the leaving variable so these are the two things is very important in order to frame iteration 2 see the iteration two basic variable and CJ values are remain same now we need to frame r five that is row five okay so in the previous iteration so which one is entering and which one is leaving variable X3 is entering variable and A1 is a leaving variable so instead of A1 we need to write X3 and X3 value is 20 here we need to write 20 and here we need to write X3 for these value take the old value divided by key element that is 2x 3/ 10x3 0 ID 10x3 10x 3id by 10x 3 - 1 / 10x3 4X 9/ 10x3 for A1 you need not write anything why because A1 is the leaving variable you can put dash for solution 32x 3 / 10x3 so this is the procedure to find the row five values for row five entering variable is X3 okay the cost is 20 okay for the new value you can apply the simple formula that is old value divided by key value 2x 3id 10x 3 is = 1x 5 then 0 ID 10x3 0 then 10x 3 10x 3 1 - 1 / 10x 3 - 3x 10 then 4X 9 / 10x 3 2x 15 for A1 A2 dash for solution 32 by 3 / 10 by 3 is = 16 by five okay so now we have got row five values now we need to find row six values so for row six basic variable is X2 as it is the coefficient of X2 is 15 so X2 15 now we need to find new value for row six so there is a formula to find out row six look at the formula row 6 value is equal to Old value minus key column value into row five value so for the first value old value is 1x3 minus key column value 2x3 into Row 5 value 1X 5 1X 3 - 2x3 into 1X 5 answer is 1 by 3 - 5 3 is 15 2 into 1 2 so LCM 15 5 * 1 into 5 5 5 - 2 is = 3x 15 is = 1 by 5 so new value is 1 by 5 for the next Value Old value is 1 1 minus corresponding column value same thing 2x 3 into R 5 value that is 0 1 - 2x 3 into 0 so 2x 3 into 0 you get 0 so the answer is 1 so new value 1 for the next Value Old Valu is 2x3 minus corresponding column value same thing 2x3 into Row 5 value is 1 2x 3 - 2x3 into 1 so 2x3 into 1 2x 3 2x3 - 2x3 is equal to 0 okay for the next Value Old value 0 minus corresponding column value 2x3 into Row 5 value - 3x 10 old value 0 minus column value 2x3 into Row 5 value - 3 by 10 so 3 3 cancel 1X 5 1X 5 next one old value is - 1 by 9 minus column value 2x3 into Row 5 value 2X 15- 1X 9 - 2x3 into 2x 15 so - 1 by 9 is = 3 into 15 45 - 45 2 into 2 4 so LCM 45 5 * so 1 into 5 5 - 5 - 4 - 99 by 45 is equal to - 1 by 5 the answer is -1 by 5 okay for A1 A2 need not calculate because these two variables are left A1 A2 put dash for solution old value is 10x 3 minus column value 2x 3 into Row 5 value 16x 5 10x 3 minus column value 2x3 into Row 5 value 16 by 5 is equal to 10 by 3 - 5 into 3 15 6 into 2 32 so LCM 15 5 * 5 into 10 50 - 32 15 - 32 is equal to 18 by 15 or 6 by 5 answer is 6 by 5 now we have got row five values and row six values now we need to find ZJ so in order to find ZJ just multiply coefficient of basic variable into each element plus coefficient of next row variable into each element so 20 into 1X 5 + 15 into 1 by 5 answer is 7 next 20 into 0 0 + 15 into 1 15 20 into 1 20 + 15 into 0 0 answer 20 and 20 into - 3x 10 + 15 into 1X 5 answer -3 then 20 into 2x 15 + 15 into - 1 by 5 answer is -1 by 3 the next value is 20 into 16x 5 + 15 into 6X 5 answer 82 after finding ZJ the next step is we need to find CJ minus Z J okay CJ is 7 7 - 7 0 15 - 15 0 20 - 20 0 0 - - 3 + 3 0 - - 1X 3 + 1X 3 that's all okay now check the optimality so what is the optimality condition for minimization problem all the CJ minus Z J must be greater than or equal to0 which means we need to get 0 or positive value so in this iteration we have got 0 0 0 3 and 1x3 so we have reached the optimality so this is the final iteration we have got the optimality solution the solution is 82 okay so now we need to write the X1 value X2 value and X3 value okay so the minimization minimization Z is equal to the final answer is solution is 82 okay now just apply we have got X2 value X2 value is 6X 5 so 6X 5 X3 value is 16x 5 16 by 5 so X1 value take zero there is no X1 we have got only X2 and X3 so put zero for X1 now apply these value in the objective function in order to end Ure the solution look at the objective function minimization Z is equal to 7 X1 + 15 X2 + 20 X3 7 X1 is 0 + 15 X2 is 6X 5 + 20 X3 value is 16 by 5 is equal to 82 we need to get 82 by applying these things just find 7 into 0 0 + 15 into 6X 5 so 6 into 3 18 + 4 into 16 64 so answer is 82 so we have attained the optimality okay this is the way to prepare lpp model using bigm method hope you all understood this concept thank you