(Chapter 12] Integer Linear Optimization Models Fairfield Investment Group is considering investing in six new projects. The required capital at the present time and the expected net present value (NPV) for each project are given in the table below. Project Required Capital ($) Expected NPV ($) 1 2,000 7,000 2 4,000 9,000 3 3,000 8,000 4 7,000 18,000 5 6,000 15,000 6 5,000 14,000 At present, a budget of $24,000 is available for investment. Fairfield has some specific requirements (as detailed in the Constraints section below). Please help Fairfield develop an investment plan by formulating an integer linear optimization model. Write your final answer only (without intermediate steps) for the fill-in-the-blank questions. 1. Define the decision variables. (Note: This step is done for you. Please use these decision variables hereafter.) X1 = 1 if project 1 is selected for investment; 0 otherwise. X2 = 1 if project 2 is selected for investment; O otherwise. X3 = 1 if project 3 is selected for investment; O otherwise. X4 = 1 if project 4 is selected for investment; 0 otherwise. X5 = 1 if project 5 is selected for investment: 0 otherwise. X6 = 1 if project 6 is selected for investment; 0 otherwise. Constraint on the budget. 7,000 X + 9,000 X2 + 8,000 X3 + 18,000 X + 15,000 X5 + 14,000 0% 2 24,000 OX1 + X2 + X3 + X4 + X5 + X6 $ 24,000 2,000 X1 + 4,000 X2 +3,000 X3 + 7,000 X4 + 6,000 X5 + 5,000 X6 > 24,000 2,000 X1 + 4,000 X2 +3,000 X3 + 7,000 X4 + 6,000 X5 + 5,000 X6 $ 24,000 O2,000 X + 4,000 X + 3,000 X + 7,000 X + 6,000 X5 + 5,000 X = 24,000 7,000 X1 + 9,000 X2 + 8,000 X3 + 18,000 X4 + 15,000 X5 + 14,000 X6 = 24,000 (7,000 - 2,000) X1 + (9,000 - 4,000) X2 + (8,000 - 3,000) X3 + (18,000 - 7,000) X4 + (15,000 - 6,000) X5 + (14,000 - 5,000) X6 x 24,000 7,000 X1 + 9,000 X2 +8,000 X3 + 18,000 X4 +15,000 X5 + 14,000 X6 = 24,000 If project 3 is selected, projects 1 or 2 must be selected. OX3 s X1 + X2 OX; sX1, X3 s X2 O Xz = X + X2 OX3 2 X1 + X2 OX1 + X2 + X3 2 1 OX3 X1 + X2 If project 2 is selected, project 3 cannot be selected. O X+Xz51 X2 + X3 > 1 O X sXz O = X2 2 X3 OX2 + X3 = 1 X2 + X3 2 1 OX > Xz ) X + Xz 0 or 1 OX + X2 + X3 + X + Xs + X6 2 0 OX1, X2, X3, X4, X5, X6 2 0 or 1 X + X2 + X3 + X4 + Xs + X