2.  LP graphical solution is always associated with a corner point of the solution space.  The transition from the geometric corner point solution to the simplex method entails a computational procedure that determines the corner points algebraically.  This is accomplished by first converting all the inequalities constraint into equation and then manipulating the resulting equation, in a systematic manner.  A main feature of the simplex method is that it solve the LP in iterations.  Each iterations move the solution to a new corner point that has the potential to improve the value of objective function.

3.  LP solution space in equation form:  For the sake of standardization, the algebraic representation of the LP solution space is made under two condition:- A. All the constraints [ with exception of non negativity restriction] are equations with a non negative right-hand side. B. All the variables are non negative.

4. * To convert a (≤) inequality to an equation a non negative slack variable is added to the left-hand side of the constraints. Example :- 6x1 + 4x2 ≤ 24 - Defining s1 as the slack or unused amount of m1 and the constraint can be converted to the following equation 6x1 + 4x2 + s1 =24,s1 ≥ 0 * The converting from (≥) to (=) is achieved by subtracting a non negative surplus variable from the left- hand side of the inequality. Example :- x1 + x2 ≥ 800 - Defining s1 as surplus variables the constraint can be converted to the following equation: x1 + x2 – s1 = 800 s1 ≥ 0

5. * The only remaining requirement is for the right-hand side of the resulting equation to be non negative. * The condition can always be satisfied by multiplying both side of the resulting equation by -1. Example :- The constraint -x1 + x2 ≤ -3 is equivalent to the equation -x1 + x2 + s1 = -3, s1 ≥ 0 Now, multiplying both sides by -1 gives: x1 - x2 - s1 = 3

6.  Convert the following OR model into equation format? Maximize z = 5 x1 + 4 x2 Subject to 6 x1 + 4 x2 ≤ 24 x1 + 2 x2 ≤ 6 -x1 + x2 ≤ 1 x2 ≤ 2 x1, x2 ≥ 0

7.  in the graphical method, the solution space is delineated by the half-spaces representing the constraints and in the simplex method the solution space is represented by M simultaneous linear equations and N non negative variables.  In set of M x N equations (M < N), if we set (N-M) variables equal to zero and then solve the M equation, for the remaining M variables, the resulting solution, if unique, must correspond to a corner point of the solution space.

8.  The following example demonstrates the procedure:- Example:- Consider the following LP model with two variables. maximize z = 2x1 + 3x2 subject to:- 1) 2x1 +x2 ≤ 4 2) x1 + 2x2 ≤ 5 x1, x2 ≥ 0  The following figure provides the graphical solution space for the problem :- 1) 2x1+x2 = 4 => x1 = 0 → x2 = 4 (0,4) X2 = 0 → x1 = 4/2 = 2 (2,0) 2)x1 + 2x2 = 5 => x1 = 0 → x2 = 5/2 = 2.5 (0,2.5) x2 = 0 → x1 = 5 (5,0)

9.  Algebraically the solution space of the LP is represented as :- maximize z = 2x1 + 3x2 subject to : 2x1 + x2 +s1 =4 x1 + 2x2 + s2 =5 x1,x2,s1,s2 ≥ 0 The system has m=2 equation and n=4 variables.  The corner point can be determined algebraically by setting N-M = 4 - 2 = 2 variables equal to zero and then solving for the remaining m=2 variables.

10. Example:- If we set x1=0 and x2=0 the equation provides the solution s1=4 and s2=5, this solution corresponds to point A. Another point can be determined by setting s1=0 and s2=0 and solving the two equation: 2x1 + x2 =4 x1 + 2x2 =5 This yield x1=1 and x2=2 which is point c. x1 = 5-2x2 2x1 +x2 =4 2(5-2x2)+x2=4 10 -4x2+x2=4 10-4=3x2 6=3x2 x2=6/3 =2 X1=5-2x2= 5-4 =1 Table Method: Consider all combinations in which n-m variables are set to zero and solve the resulting equation, once done the optimum solution is the feasible corner point that yield the best objection function value.

11.  The N-M variables that are set to zero are known as non basic variables.  If the remaining M variables have a unique solution, then they are called basic variables and their solution is referred to as basic solution.  The following table provide all basic and non basic solution of the current example:- Objective value feasibleAssociated corner point Basic solution Basic variables Non basic variables (zero) 0YesA(4,5)(s1,s2)(x1,x2) -NoF(4,-3)(x2,s2)(x1,s1) 7.5YesD(2.5,1.5)(x2,s1)(x1,s2) 4YesB(2,3)(x1,s2)(x2,s1) -NoE(5,-6)(x1,s1)(x2,s2) optimum 8 YesC(1,2)(x1,x2)(s1,s2)

12. Ex: Find the optimal solution for the following OR model using the table method?  Maximizez = x 1 + x 2 Subject tox 2 – x 1 ≤ 0 x 2 ≤ 3 x 1, x 2 ≥ 0

13.  The vehicle of explanation is a numerical example. Example:- We use the Sadoline model to explain the details, of the simplex method.  The problem is expressed in equation from as:- maximize z=5x1+4x2+0s1+0s2+0s3+0s4 subject to 6x1+4x2+s1=24 x1+2x2+s2=6 -x1+x2+s3=1 x2+s4=2 x1,x2,s1,s2,s3,s4 ≥ 0

14.  The variables s1,s2,s3 and s4 are the slacks associated with the respective constraints.  We express the objective equation as : z - 5x1 – 4x2 = 0  In this manner, the starting simplex tableau can be representing as follow: rowsolutions4s3s2s1x2x1zbasic Z-row00000-4-51z S1-row240001460s1 S2-row60010210s2 S3-row1010010s3 S4-row21000100s4

15.  The design of the tableau specifies the set of basic and non basic variables as well as provides the solution associated with the starting iteration.  The simplex iteration, start at the origin (x1,x2) =(0,0), thus, the associated set of non basic variables and basic variables are defined as non basic variables(x1,x2) basic variables(s1,s2,s3,s4)  The following solution is immediately available (without further computation) Z=0 S1=24 S2=6 S3=1 S4=2

16.  Is the starting (current) solution optimal?  To determine that we check if all the objective coefficients happen to be ≥ 0, no further improvements in z is possible, signifying that the optimum has been reached so the current solution is not optimal.  Determine the entering variable: - The entering variable is x1 because that it has the most negative coefficient in the (maximization) objective function. - To determine the leaving variable directly from the tableau, we compute the intercepts of all the constraints with non negative direction of the x1- axis (recall that x1is the entering variable). - These intercepts are the ratios of the right-hand side of the equations (solution-column) to the corresponding constraint coefficients under the entering variable, x1 as the following table shows:

17. Entering Ratio x1= 24/6 =4 minimum x1= 6/1 =6 x1= 1/-1 = -1 (Ignore) x1= 2/0 = ∞ (Ignore) - The minimum non negative ratio corresponds to basic s1, signifying that s1 is the leaving variable (its value is zero in the new iteration) - The value of the entering variable x1 in the new solution also equals the minimum ratio (x1=4). solutionx1basic 246s1 61s2 1s3 20s4

18. - The corresponding increase in the value of objective z is 5*4 =20. - The end result of “swapping” the entering and the leaving variables is that non basic and basic variable at the new solution point will be given as: non basic (zero) variables :( s1,x2) basic variables : (x1,s2,s3,s4) - We now need to manipulate the equations in the last tableau so that the basic-column and the solution –column will identify the new solution. - The process called the Guss-Jordon row operations. - The following is a replica of the starting tableau, it associates the pivot column and the pivot row with the entering and the leaving variables, respectively. - The intersection of the pivot column and the pivot row is called the pivot element

19. solutions4s3s2s1x2x1ZBasic 00000-4-51z 240001460s1 60010210s2 1010010s3 21000100s4 - The Guss-Jordan computation needed to produce the new basic solution include two types: A)Pivot row: new pivot row = current pivot row /pivot element. B)All other rows including z: new row = (current row)-(its pivot column coefficient * (new pivot row). - These computations are applied to the preceding tableau in the following manner:

20.  New pivot s1-row = current s1-row ÷ 6 =( 0 6 4 1 0 0 0 24) ÷ 6 =( 0 1 2/3 1/6 0 0 0 4) New z row = current z row – (-5)*new pivot row =( 1 -5 -4 0 0 0 0 0)+5 * (0 1 2/3 1/6 0 0 0 4) =(1 0 -2/3 5/6 0 0 0 20) New s2 row=current s2 row – (1)*new pivot row =(0 1 2 0 1 0 0 6 ) – (0 1 2/3 1/6 0 0 0 4) =(0 0 4/3 -1/6 1 0 0 2) New s3 row=current s3 row –(-1) * new pivot row =(0 -1 1 0 0 1 0 1 ) + (0 1 2/3 1/6 0 0 0 4) =(0 0 5/3 1/6 0 1 0 5) New s4 row=current s4 row –(0) * new pivot row =(0 0 1 0 0 0 1 2 ) – (0) * new pivot =(0 0 1 0 0 0 1 2 )

21. - The new tableau corresponding to the new basic solution (x1,s2,s3,s4):- - The solution-column automatically yields new basic solution (x1=4, s2=2, s3=5, s4=2). - The corresponding new objective value: z=20. - An examination of the tableau shows that it is not optimal because the non basic variable x2 has a negative coefficient in the z-row, thus x2 is the entering variable. solutions4s3s2s1x2x1zbasic 200005/6-2/301z 40001/62/310x1 2001-1/64/300s2 50101/65/300s3 21000100s4

22. - Next, the ratio computation in the next table show that s2 is the leaving variable. Ratio x2= 4÷2/3 = 6 x2= 2÷4/3 = 1.5 x2= 5÷5/3 = 3 x2= 2÷1= 2 - The computations show that x2=1.5 and the corresponding increase in z 2/3*1.5=1 yielding new z=20+1=21. - Given x2 and s2 as the entering and leaving variables, need to apply the following Guss- Jordan row operation to produce the next tableau.  New pivot s2-row=current s2 row ÷4/3  New z row=current z row –(-2/3)*new pivot row.  New x1 row=current x1 row –(2/3)*new pivot row.  New s3 row=current s3 row –(5/3)*new pivot row.  New s4 row=current s4 row –(1)*new pivot row. solutionEntering x2 basic 42/3x1 24/3s2 55/3s3 21s4

23. - The computations produce the following tableau: - Because non of the z row coefficient associated with the non basic variables s1 and s2 are negative, the last tableau is optimal. - The optimum solution can be read for the simplex tableau as follows: x1=3, x2=3/2, z=21. solutions4s3s2s1x2x1zbasic 21001/23/4001z 300-1/21/4010x1 3/2003/4-1/8100x2 5/201-5/43/8000s3 1/210-3/41/8000s4

24.  Simplex method for minimize objective function in minimization the selection of the leaving variables is the same as in the maximization case (minimum nonnegative ratio ) The minimization case selects the entering variable as the non basic variable with the most positive objective coefficient. Minimum z is attained when all the z row coefficient are non positive. Example :- minimize z= 5x1- 4x2+ 6x3- 8x4 Subject : x1+ 2x2- 2x3 +4x4 ≤ 40 2x1 -x2+ 1x3+ 2x4 ≤ 8 4x1- 2x2 + x3 – x4 ≤ 10 x1, x2, x3, x4 ≥ 0

25. - Iteration 1: Minimize z - 5x1+ 4x2- 6x3+ 8x4 =0 subject to : x1+ 2x2- 2x3+ 4x4+ s1 = 40 2x1- x2+ x3+ 2x4+ s2 = 8 4x1- 2x2+ x3- x4+ s3 =10 New pivot row = ( 0, 2, -1, 1, 2, 0, 1, 0, 8 ) ÷ 2 = ( 0, 1, -0.5, 0.5, 1,0, 0.5, 0, 4) solutions3s2s1x4x3x2x1zbasic 00008-64-51z 400014-2210s1 80102120s2 101001-240s3

26.  New z=current z- (8) * (0, 1, -0.5, 0.5, 1, 0, 0.5, 0, 4) = (1, -5, 4, -6, 8, 0, 0, 0, 0) + (0, -8, 4, -4, -8, 0, -4, 0, -32) = (1, -13, 8, -10, 0, 0, -4, 0, -32)  New s1 =current s1 - (4) * (0, 1, -0.5, 0.5, 1, 0, 0.5, 0, 4) = (0, 1, 2, -2, 4, 1, 0, 0, 40) + (0, -4, 2, -2, -4, 0, -2, 0, -16) = (0, -3, 4, -4, 0, 1, -2, 0, 24)  New s3 =current s3 - (-1) * (0, 1, -0.5, 0.5, 1, 0, 0.5, 0, 4) = (0, 9, -2, 1, -1, 0, 0, 1, 10) + (0, 1, -0.5, 0.5, 1, 0, 0.5, 0, 4) = (0, 5, -2.5, 1.5, 0, 0, 0.5, 1, 14) - Iteration 2:- solutions3s2s1x4x3x2x1zbasic -320-400-108-131z 240-210-44-30s1 400.501 -0.510x4 1410.5001.5-2.550s3

27. - Iteration 3:- - We will stop because non of the non basic variables has positive coefficient. Z= -80, x1=0, x2=6, x3=0, x4=7. solutions3s2s1x4x3x2x1zbasic -8000-20 0-71z 60-0.50.2501-0.750x2 700.250.131000.630x4 291-0.750.63003.130s3

28.  Using the simplex method find the solution for the following OR model? minimize z= -0.3x1 + 0.9x2 subject to: x1 +x2 ≥ 800 0.21 x1 – 0.3 x2 ≤ 0 0.03 x1 – 0.01 x2 ≥0 x1, x2 ≥0