Consider the following optimization problem: min:s.t.:8x1+12x2+16x320x424x528x6x1+3x2+3x3=5x4+x5+x6100xi9i=1,,6 where the variables of the multivariable optimization problem are integers. Use Extension-2 type approach, (i.e., Treat the multiple variables together as one structure or as one vector). Answer the following: (a) Let [4,5,3,1,1,1]T be the current solution. Ignore Constraints (2) \& (3), and write all the possible feasible immediate neighbors (star neighbors or unit neighbors) of the current solution. (b) Let [4,5,3,1,1,1]T be the current solution. Ignore Constraints (2) \& (3), and write 3 possible feasible extended neighbors of the current solution. (c) Execute one full iteration of the greedy search with the immediate/star neighborhood. Use starting solution as [3,0,1,1,1,1]T. Handle Constraint(2) \& (3) by creating a penalized objective function. Assume all the penalty coefficients are equal to 1000 . (d) Execute 3 full iterations of the random-walk search with the extended/expanded neighborhood. Use starting solution as [3,0,1,1,1,1]T. Handle Constraint(2) \& (3) by creating a penalized objective function. Assume all the penalty coefficients are equal to 1000 . Use random numbers from the random number table. See explanation at the end of this HW for generating random numbers. (e) Execute 4 iterations of the simulated annealing with following parameters: Initial temperature be 1000 , and starting solution be [3,0,1,1,1,1]T. Neighborhood type = Extended neighborhood, Move type = Random walk, Pool size =1, Max # tries =4. Cooling mechanism = After 2 iterations (irrespective of success or failure in the iteration), reduce the temperature to 500 , and continue with the remaining iterations. Handle Constraint(2) \& (3) by creating a penalized objective function. Assume all the penalty coefficients are equal to 1000 . (f) Execute one next iteration of the tabu search with following parameters: Current solution = [3,0,1,1,1,1]T. Neighborhood type = Immediate neighborhood, Move type = Greedy move, Tenure period =6, and the current tabu list is: {[4,0,1,1,1,1]T,[3,0,2,1,1,1]T,[3,2,1,1,1,1]T,[3,0,1,2,1,1]T,[3,0,1,2,1,1]T,[3,0,1,1,1,2]T} The Constraint(2) \& (3) were handled by creating a penalized objective function, where all the penalty coefficients are equal to 1000 . For each problem ( 1&2 ) or its part where random numbers are needed, obtain them from the consecutive random digits from Random Number Table as follows. Start from a seed location. Assume for this HW the seed location is first eolumn \& first row of the table (i.e-, seed at (1,1) ). Now read the five-digits of from the seed location and form a random numbers by placing a decimal point in front. From seed (1,1) the first random number is 0.13962 , and the second random number is 0.70902 . Once you reach the end of the first row, start from the first column of the next row. Always restart the generation of random numbers for each new problem (or any sub-problem). For this HW the seed is (1,1). If any random number that you use results in infeasibility, then discard it and move to the next random number. An example for Problem-1 is given below. Extended Neighbors For generating an extended neighbor, you need two random variables R1 and R2, such that 0 current solution. Use second random variable (R2) to get the value of the selected element. To select the variable or position to update, use the following formula: position=1+R1+M where M is the total number of variables, U is the floor function. If the position is infeasible, then repeat the above position calculation with the next random number until we get a feasible position. To get the value of the selected position, use the following formula. value=oldvalue+(R20.5)(UL) where U and L are upper and lower bounds on the selected variable. If the value is infeasible, then repeat the above value caleulation with the next random number until we get a feasible value. Examples Say the seed is at (1,1) and let [9,4,1,1,3,1]T be the current solution. Since seed is at (1,1),R1=0.1396. Now, based on the position formula, we get position =1+0.1396+6=1+0.8376=1+0=1. Therefore, first variable is updated, i.e., the extended neighbor will be [,4,1,1,3,1]2 of this form. Now the value for the position is obtained by using the next random number (i.e., R2=0.70992 ). So, the value =9+ (0.709920.5)(90)=1+1.88928=9+1=10. The value is infeasible for Problem-1, so we repeat with the next random number, which is R2=0.65172. Now using the formula, we get value =10 (again infeasible). Naw, the next in line random number, R2=0.28053, will be used. This will result in value =9+(0.280530.5)+(0)= 1+1.97523=92=7 Thus, the randomly selected extended neighbor will be [7,4,1,5,3,1]T