09NN01 OPTIMIZATION TECHNIQUES

09NN01 OPTIMIZATION TECHNIQUES 3 1 0 4   LINEAR PROGRAMMING: Linear Programming: Graphical method, Simplex method, Revised simplex method, Duality in linear programming (LP), Sensitivity analysis, other algorithms for   solving LP problems, Transportation, assignment and other applications.                                                                                                                                                                (9)   NON-LINEAR PROGRAMMING: Non Linear Programming: Unconstrained optimization techniques, Direct search methods, Descent methods, constrained optimization.                                                                                                               (9)   INTEGER AND DYNAMIC PROGRAMMING: Formulation of Integer Programming Problems, Gomory’s cutting plane methods, Branch and Bound Techniques, Characteristics of Dynamic programming, Bellman’s principle of optimality, Concepts of dynamic method of solution.                                        (11)   PERT/CPM: Network Construction-Computation of earliest start time, latest start time,  total, free and independent float time  -Crashing – Computation of optimistic, most likely, pessimistic and expected time- Resource analysis in Network scheduling.                                                                                                                                                                          (5)                                                                                                                                                    NON TRADITIONAL TECHNIQUES: Genetic Algorithm, Simulated Annealing, Tabu Search and Neural Networks.       (8) Total 42 REFERENCES: Rao S S, “Engineering Optimization: Theory and Practice”, New Age International, New Delhi, 2006. Trivedi K S, “Probability and Statistics with Reliability, Queuing and Computer Applications”, Prentice Hall, New Delhi, 2006. Taha H A, “Operations Research: An Introduction”, Pearson Education, New Delhi, 2006.   Alberto Leon–Garcia, “Probability and Random Processes for Electrical Engineering”, Pearson Education, New Delhi, 2007. Jang J S R, Sun C T and Mizutani, “Neuro-Fuzzy and Soft Computing: A Computational Approach to Learning and Machine Intelligence”, Pearson Education, New Delhi, 2005.       09NN02 DATA STRUCTURES AND ALGORITHMS   3 1 0 4   INTRODUCTION: Primitive Data Types – Abstract Data types – Algorithm Analysis – Time and Space Complexity.        (3)   LISTS: Arrays – Linked Lists – Stacks and Queues – Applications – Implementation of Recursive Functions.                    (5) TREES: Binary Trees – Tree Traversals – Binary Search Trees – Balanced Search Trees – AVL – Red-Black…

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4 digit HEX number in MIPS

4 digit HEX number in MIPS         The first one is an hex number it is straightforward to convert it into binary. Every HEX digit can be converted in 4 binary digit as in the following example     Ex $$_7=2^2+2^1+2^0=0111_$$ For your hex number 0x71014802 is     0x71014802 => 0111 0001 0000 0001 0100 1000 0000 0010 7    1    0    1    4    8    0    2 Regarding to a conversion from decimal to binary I suggest you converting the decimal in hex and then hex to binary, let’s do an example with a decimal (I choose 71014802 DEC)           71014802 => 0x43B9992 => 0000 0100 0011 1011 1001 1001 1001 0010 0    4    3    B    9    9    9    2 another way to convert a decimal number into a binary one is by dividing by two and picking the remainder… but it is a longer procedure (you can find it here) Anyway if the opcode is in a fixed position (26-31) the opcode can be simply obtained through a mask and shift right as follow:     opcode=(reg>>26)&0x3F An algorithm to print a binary number can be like this: void print_binary(int n) { while (n) { char bit = n & 0x1; putchar (bit+’0′); n >>= 1; } putchar(‘\n’); }   char bits = (num >> (sizeof(num)*8 – 6)) & 0x3F; 0x3F = 00111111 that’s six bits pushed all the way to the left, giving you the result you’re looking for. EDIT: An more detailed explanation. What we are trying to do is to get the six leftmost bits of an integer. To determine if a bit is flipped or not (i.e. 1 or 0) we and it with 1. Getting six bits of a number is done by constructing an integer with six flipped bits, 00111111, or 0x3F, and anding…

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