Question: Problem 1: Suppose you have eight binary decision variables X1, X2, . ..., Xg in an integer-linear program (ILP). Please explain how you would incorporate

Problem 1: Suppose you have eight binary decision variables X1, X2, . ..., Xg in an integer-linear program (ILP). Please explain how you would incorporate the following restrictions into an ILP. It is acceptable to add variables and/or constraints. Part a: If x5 = 1, then x2 = 0. = Part b: If X4 = 1, then X1 = O or X5 = 1. = Part c: It is unacceptable to have both x2 = 0 and x5 = 1. = = Part d: At least five out of eight decision variables must take the value of zero. Part e: If either x3 = 0 or Xo = 1, then x2 = 0. = Part f: If x3 = 1 and X5 = 1, then neither x2 nor X7 can take the value of 1. = = Part g: If x1 = x2 = x3 = x4 = 1, then x6 = 1. = = = Part h (adapted from Hillier and Lieberman, 2005): [X1 + x2 + x3 + x4 x5 X6 X, X8] = 0, 5, 12 or 15. (Hint: Let Zk denote a binary variable that equals 1 if (x1 + x2 + x3 + x4 x5 X6 X, xg) takes its k-th possible value.) Part i: At least one of the following inequalities is satisfied: X1 + X5 > 1 Problem 1: Suppose you have eight binary decision variables X1, X2, . ..., Xg in an integer-linear program (ILP). Please explain how you would incorporate the following restrictions into an ILP. It is acceptable to add variables and/or constraints. Part a: If x5 = 1, then x2 = 0. = Part b: If X4 = 1, then X1 = O or X5 = 1. = Part c: It is unacceptable to have both x2 = 0 and x5 = 1. = = Part d: At least five out of eight decision variables must take the value of zero. Part e: If either x3 = 0 or Xo = 1, then x2 = 0. = Part f: If x3 = 1 and X5 = 1, then neither x2 nor X7 can take the value of 1. = = Part g: If x1 = x2 = x3 = x4 = 1, then x6 = 1. = = = Part h (adapted from Hillier and Lieberman, 2005): [X1 + x2 + x3 + x4 x5 X6 X, X8] = 0, 5, 12 or 15. (Hint: Let Zk denote a binary variable that equals 1 if (x1 + x2 + x3 + x4 x5 X6 X, xg) takes its k-th possible value.) Part i: At least one of the following inequalities is satisfied: X1 + X5 > 1