] In the derivation of the spectral graph partitioning algorithm, we relax a combinatorial optimization problem to a continuous optimization problem. This relaxation has the following effects.  The combinatorial problem requires an exact bisection of the graph, but the continuous algorithm can produce (after rounding) partitions that aren’t perfectly balanced  The combinatorial problem cannot be modified to accommodate vertices that have different masses, whereas the continuous problem can  The combinatorial problem requires finding eigenvectors, whereas the continuous problem requires only matrix multiplication  The combinatorial problem is NP-hard, but the continuous problem can be solved in polynomial time

Solve the following exercise using jupyter notebook for Python, to find the objective function, variables, constraint matrix and print the graph with the optimal solution. Use linear programming methods to solve this problem. Represent the constraints and the objective function mathematically in code. Use Python and the SciPy and Matplotlib libraries to solve the problem and visualize the optimal solution on a graph. Clearly state the optimal solution and any assumptions made while solving the problem.   A farm specializes in the production of a special cattle feed, which is a mixture of corn and soybeans. The nutritional composition of these ingredients and their costs are as follows: - Corn contains 0.09 g of protein and 0.02 g of fiber per gram, with a cost of.$0.30 per gram.- Soybeans contain 0.60 g of protein and 0.06 g of fiber per gram, at a cost of $0.90 per gram.0.90 per gram. The dietary needs of the specialty food require a minimum of 30% protein and a maximum of 5% fiber.…

W1 a W2 w6 w7 6. -1 4 W3 W5 e W4 Figure 1: Weighted, undirected graph G. Weights and Source Node All problems use weights wi through w7. To determine these weights, look at your 7 digit ID. The weight wi corresponds to the first digit of your ID, the weight w2 corresponds to the second digit of your ID, and so on. For problems 2 and 4, we use a source node or starting node s, determined as follows: find the digit in your ID that has the highest value. If it is the first digit then s = a, if it is the second digit, then s = b, and so on. If there is more than one digit with the same highest value, choose the first occurrence of that digit. For example, if your ID is 4526767, then the highest value is 7, and you choose the first 7, i.e., digit 5, and set s = е.

We know that if the heuristic function in A* is good enough, then A* can always find a shortest weighted path between two vertices, and is generally much faster than Dijkstra. Assume we are using a graph where a good heuristic function is well defined for A*, such that A* can always find the same shortest paths as Dijkstra. Briefly explain when you should choose Dijkstra over A* in this case.