Patrick ReinagelIE 4161/14/2019Assignment #2Pg. 63, #3ChemicalProductionProcess 1ProductionProcess 2DemandA 3 1 10B 1 1 5C 1 0 3ProductionCost$4 $1Times toRunProductionProcessP1 P2Decision Variables:1. P1 = Number of times to run Production Process 12. P1 = Number of times to run Production Process 2(O.F)Zmin = 4*P1 + 1*P2(S.T.)1. 3*P1 + P2 > 10 A must meet demand of at least 10 units each day2. P1 + P2 > 5 B must meet demand of at least 5 units each day3. P1 > 3 C must meet demand of at least 3 units each dayAll variables >=0Equation:Z-4P1 + 1P2 =03P1 + 1P2 – e1 + a1 = 101P1 + 1P2 – e2 + a2 = 51P1 – e3 + a3 = 3Matrix: By Ms. Siswanto below:Patrick ReinagelIE 4161/14/2019First Simplex Tableau: (This should be updated as explained below but it is not expected from you.)Z P1 P2 e1 e2 e3 a1 a2 a3 RHS1 -4 -1 0 0 0 0 0 0 00 3 1 -1 0 0 1 0 0 100 1 1 0 -1 0 0 1 0 50 1 0 0 0 -1 0 0 1 3We should consider Big M method to handle artificial variables (make sure they will not appear as BV in final solution.) and then apply modified Simplex method for minimization. I did not cover these methods in class so they are not expected in exam.Just for your general information I have added below how we use these methods.Big M method: Add Mai to OF then the standard form will look like:Z-4P1 + 1P2 -Ma1 –Ma2 –Ma3 = 0Multiply each constraint that has ai and add to OF above, we will have:Z + (5M-4)P1 + (2M+1)P2 -Me1 –Me2 –Me3 = 18MThen new first Simplex tableau will be as:Modified Simplex method for minimization:There are two possible modifications. I select the one that selects the “most positive coefficient of OF” for entering variable. This will lead to selecting P1. Ratio testing will lead to select the last row. Then perform ero until this iteration is over. We will get the following tableau.The next tableau will be:Patrick ReinagelIE