ASSIGNMENT-FALL 2013
Name:

__Narinder Kumar_________________________

Registration No:

__511225739______________________...
Ans:- The application of Operations research methods helps in making decisions in such
complicated situations. Evidently t...
When the scale parameter equals 2, the distribution simplifies to the chi-squared distribution
with 2k degrees of freedom....
Step – IV: a1x1 + a2x2= b divides the whole plane into two half planes, which are a1x1+
a2x2 <= b (one side) and a1x1+ a2x...
Notations: Notations used are different and are given below:
The queue discipline in a finite queuing process can be:
i)
F...
E(w) = λ/( µ(µ - λ))
= 1 minute
b. Spent receiving or waiting for service
W(v) = 1/( µ - λ)
= 2 minutes
b) Find the mean n...
The mini-max is
Upcoming SlideShare
Loading in …5
×

MC0079 SMU 2013 Falll Session

591 views
557 views

Published on

Published in: Education, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
591
On SlideShare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
31
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

MC0079 SMU 2013 Falll Session

  1. 1. ASSIGNMENT-FALL 2013 Name: __Narinder Kumar_________________________ Registration No: __511225739_____________________________ Learning Center: __ Artex Informatic ________________________ Learning Center Code: __1688__________________________________ Course: __ MCA _________________________________ Subject: COMPUTER BASED OPTIMIZATION METHODS Semester: __IV____________________________________ Subject Code: __MC0079_______________________________ Date of submission: __8 December 2013________________________ Marks awarded: ________________________________________ Directorate of Distance Education Sikkim Manipal University II Floor, Syndicate House Manipal – 576 104 Signature of Coordinator Signature of Center Signature of Evaluator Q1:- Discuss the various application domains of Operations Research.
  2. 2. Ans:- The application of Operations research methods helps in making decisions in such complicated situations. Evidently the main objective of Operations research is to provide a scientific basis to the decision-makers for solving the problems involving the interaction of various components of organization, by employing a team of scientists from different disciplines, all working together for finding a solution which is the best in the interest of the organization as a whole. The solution thus obtained is known as optimal solution or decision. A few examples of applications in which operations research is currently used include: � Designing the layout of a factory for efficient flow of materials. � Constructing a telecommunications network at low cost while still guaranteeing QoS (quality of service). � Road traffic management and 'one way' street allocations i.e. allocation problems. � Designing the layout of a computer chip to reduce manufacturing time and hence reducing cost. � Managing the flow of raw materials and products in a supply chain based on uncertain demand for the finished products. � Roboticizing human-driven operations processes. � Globalizing operations processes in order to take advantage of cheaper materials, labor, land or other productivity inputs. � Scheduling: - personnel staffing - manufacturing steps - Network data traffic: these are known as queueing systems. - Sports events and their television coverage. � Blending of raw materials in oil refineries. � Operations research is also used extensively by various governments in Defence Operations, Industry, Planning, Agriculture, Hospitals, Transport, Research and Development etc. Q2:- Explain Erlang family of distributions of service times. Ans:- The Erlang distribution is a continuous probability distribution with wide applicability primarily due to its relation to the exponential and Gamma distributions. The Erlang distribution was developed by A. K. Erlang to examine the number of telephone calls which might be made at the same time to the operators of the switching stations. This work on telephone traffic engineering has been expanded to consider waiting times in queuing systems in general. The distribution is now used in the fields of stochastic processes and of biomathematics. Probability density function The probability density function of the Erlang distribution is The parameter is called the shape parameter and the parameter is called the rate parameter. An alternative, but equivalent, parameterizations (gamma distribution) uses the scale parameter which is the reciprocal of the rate parameter (i.e., ):
  3. 3. When the scale parameter equals 2, the distribution simplifies to the chi-squared distribution with 2k degrees of freedom. It can therefore be regarded as a generalized chi-squared distribution, for even degrees of freedom. Because of the factorial function in the denominator, the Erlang distribution is only defined when the parameter k is a positive integer. In fact, this distribution is sometimes called the Erlang-k distribution (e.g., an Erlang-2 distribution is an Erlang distribution with k=2). The Gamma distribution generalizes the Erlang by allowing to be any real number, using the gamma function instead of the factorial function. Cumulative distribution function (CDF) The cumulative distribution function of the Erlang distribution is: where is the lower incomplete gamma function. The CDF may also be expressed as Q3:- Explain the algorithm for solving a linear programming problem by graphical method. Ans:- Linear programming (LP) is a mathematical method for determining a way to achieve the best outcome (such as maximum profit or lowest cost) in a given mathematical model for some list of requirements represented as linear equations. More formally, linear programming is a technique for the optimization of a linear objective function, subject to linear equality and linear inequality constraints. Given a polytope and a real-valued affine function defined on this polytope, a linear programming method will find a point on the polytope where this function has the smallest (or largest) value if such point exists, by searching through the polytope vertices. An Algorithm for solving a linear programming problem by Graphical Method:(This algorithm can be applied only for problems with two variables). Step – I: Formulate the linear programming problem with two variables (if the given problem has more than two variables, then we cannot solve it by graphical method). Step – II: Consider a given inequality. Suppose it is in the form a1x1 + a2x2 <= b (or a1x1 + a2x2 >= b). Then consider the relation a1x1+ a2x2= b. Find two distinct points (k, l), (c, d) that lie on the straight line a1x1+ a2x2= b. This can be found easily: If x1= 0, then x2 = b / a2.If x2=0, then x1 = b / a1. Therefore (k, l) = (0, b / a2) and (c, d) = (b / a1, 0) are two points on the straight line a1x1+a2x2= b. Step – III: Represent these two points (k, l), (c, d) on the graph which denotes X–Y-axis plane. Join these two points and extend this line to get the straight line which represents a1x1+ a2x2= b.
  4. 4. Step – IV: a1x1 + a2x2= b divides the whole plane into two half planes, which are a1x1+ a2x2 <= b (one side) and a1x1+ a2x2 >= b (another side). Find the half plane that is related to the given inequality. Step – V: Do step-II to step-IV for all the inequalities given in the problem. The intersection of the half-planes related to all the inequalities and x1 >= 0,x2 >= 0 , is called the feasible region (or feasible solution space). Now find this feasible region. Step – VI: The feasible region is a multisided figure with corner points A, B,C, … (say). Find the co-ordinates for all these corner points. These corner points are called as extreme points. Step – VII: Find the values of the objective function at all these corner/extreme points. Step – VIII: If the problem is a maximization (minimization) problem, then the maximum (minimum) value of z among the values of z at the corner/extreme points of the feasible region is the optimal value of z. If the optimal value exists at the corner/extreme point, say A (u, v), then we say that the solution x1= u and x2= v is an optimal feasible solution. Step – IX: Write the conclusion (that include the optimum value of z, and the co-ordinates of the corner point at which the optimum value of z exists). Q4:- Explain the use of finite queuing tables. Ans:- There will be cases, where the possible number of arrivals is limited and is relatively small. In a production shop, if the machines are considered as customers requiring service from repair crews or operators, the population is restricted to the total number of machines in the shop. In a hospital ward, the probability of the doctors or nurses being called for service is governed by the number of beds in the ward. Similarly, in an aircraft the number of seats is finite and the number of stewardesses provided by the airlines will be based on the consideration of the maximum number of passengers who can demand service. As in the case of a queuing system with infinite population, the efficiency of the system can be improved in tens of reducing the average length of queues, average waiting time and time spent by the customer in the system by increasing the number of service channels. However, such increases mean additional cost and will have to be balanced with the benefits likely to accrue. If the queuing system in a machine shop is under study, the cost of providing additional maintenance crews or operators can be compared with the value of additional production possible due to reduced downtime of the machines` In cases where it is not possible to quantify the benefits, the management will have to base its decisions on the desired standards for customer service The queue discipline in a finite queuing process can be: i) First come-first served ii) Priority e.g.: Machines of high cost may be given priority for maintenance while others may be kept waiting even if they had broken down before. iii) Random e.g.; in a machine shop if a single operator is attending to several machines and several machines call for his attention at a time, he may attend first to the one nearest to him. The analysis of Finite Queuing Models is more complex than those with infinite population although the approach is similar.
  5. 5. Notations: Notations used are different and are given below: The queue discipline in a finite queuing process can be: i) First come-first served ii) Priority e.g.: Machines of high cost may be given priority for maintenance while others may be kept waiting even if they had broken down before. iii) Random e.g.; in a machine shop if a single operator is attending to several machines and several machines call for his attention at a time, he may attend first to the one nearest to him. The analysis of Finite Queuing Models is more complex than those with infinite population although the approach is similar. Notations: Notations used are different and are given below: (i) Find mean service timeTand mean running tirne U. (ii) Compute the service factor (iii) Select the table corresponding to the population N. (iv) For the given population, locate the service factor value. (v) Read off from tables, values of D and F for the number of service crews M. If necessary, these values may be interpolated between relevant values of X. (vi) Calculate the other measures L W, H, and J from the formulae given. The overall efficiency F of the system will increase with the number of service channels (M) provided. As mentioned earlier, addition of service crews involves cost, which should be justified by the increase in the efficiency of the system i.e. additional running time of machines possible. However it will be seen from the tables that as M increase the rate 0f increase in efficiency decreases. The practical significance is that beyond a certain value of M, it is not worthwhile increasing M as there would be no appreciable increase in the efficiency of the system. Q5:- Customers arrive at a small post office at the rate of 30 per hour. Service by the clerk on duty takes an average of 1 minute per customer 1) Calculate the mean customer time. a. Spent waiting in line b. Spent receiving or waiting for service. 2) Find the mean number of persons a. in line (ii) Receiving or waiting for service Ans:- Mean arrival rate λ = 30 customers per hour = ½ customer per minute. Mean service rate λ = 1 per minute Traffic intensity P = λ/µ =½ a) Mean Customer Time a. Spent waiting in line
  6. 6. E(w) = λ/( µ(µ - λ)) = 1 minute b. Spent receiving or waiting for service W(v) = 1/( µ - λ) = 2 minutes b) Find the mean number of persons a. in line E(m) = λ2/( µ(µ - λ)) = ½ Customer b. Receiving or waiting for service E(n) = λ /( µ - λ) = 1 Customer Q6:- Explain the terms I. Saddle point II. Max-min and min-max principle Ans:- I. Saddle point: For any game, if the maxi-min and the mini-max are equal, then such games are said to have a saddle point Steps to detect a Saddle Point Step (1): At the right of each row, write the row minimum and ring the largest of them. Step (2): At the bottom of each column, write the column maximum and ring the smallest of them. Step (3): If these two elements are same, the cell where the corresponding row and column meet is a saddle point and the element in that cell is the value of the game. Step (4): If the two ringed elements ate unequal, there is no saddle point, and the value of the game lies between these two values. Step (5): If there are more than one saddle points then there will be more than one solution, each solution corresponding to each saddle point. Max-min and min-max principle: Suppose player A and player B are to play a game without knowing what the other player would do. However, player A would like to maximize his profit and player B would like to minimize his loss. And thus, each player would expect his opponent to be calculative. Suppose player A plays A1. Then, his gain would be a11, a12, � a1n according as B �s choice is B1, B2, �. Bn. Let ?1 = min {a11, a12,�a1n}. Then, ?1 is the minimum gain of A when he plays A1. (Here, ?1 is the minimum pay-off in the first row.) Similarly, if A plays A2, his minimum gain is ?2 which is the least pay-off in the second row. Thus proceeding, we find that corresponding to A�s play A1, A2, �.. Am, the minimum gains are the row minimums . Suppose A chooses that course of which is maximum. This maxi-mum of the row minimum in the pay-off matrix is called maxi-min. The maxi-min is Similarly, by taking the minimum of the column maximums in the pay-off matrix is called mini-max.
  7. 7. The mini-max is

×