Please show how to do it like in the following example so I can understand it using this method.

Problem 3 Use the two-phase method to solve the following linear program: mins.t.z=15x1+20x2x1+2x2102x13x26x1+x26x1,x20 2 The Two-phase Method The two-phase method adopts a similar idea as the big- M method, which is to drive the artificial variables to 0 as fast as possible. The algorithm is operated in two phases. In phase one, we maximize an objective function with only the artificial variables. We expect that the variables will be optimized to 0 at the end of phase one. Then, we switch to the original objective, and continue the simplex method until optimality is reached. We will show the two-phase method using the same example: maxsubjecttoz=3x1x2x3+0x4+0x5Mx6Mx7x12x2+x3+x4=11,4x1+x2+2x3x5+x6=32x1+x3+x7=1x1,x2,x3,x4,x5,x6,x70. We will optimize the following objective in phase one: maxz=x6x7, where only the artificial variables are involved. Note that although x4 is in the basis with x6 and x7,x4 is a slack variable, so it is not involved in the objective. Also note that we set the coefficients to 1, because of the maximizing objective, which drives x6 and x7 to 0 . Coefficient 1 would make the linear program unbounded. If the objective is to minimize, the coefficients should be 1 instead of 1. Similar to big- M, we need to adjust the initial objective to form the basis using x4, x6, and x7 due to the 1 coefficient. Table 5 shows the initial iteration. Table 5: Phase 1, iteration 0 of the two-phase method