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TYBMS Semester 6 Operation Research (Q.P. April 2019 with Solution)

 paper/ subject code 86001/ operation research, tybms semester 6 operation research , (q.p. april 2019 with solution), 1) april 2019 q.p. with   solution (pdf) , 2) november 2019 q.p. with  solution (pdf), 4 ) april 2023 q.p. with  solution (pdf).

Please check whether you have got the right question paper.

1. All questions are compulsory. (Subject to Internal Choice)

2. Figures to the right indicate full marks

3. Use of non-programmable calculator is allowed and mobile phones are not allowed.

4. Normal distribution table is printed on the last page for reference. Support your answers with diagrams / illustrations, wherever necessary

6. Graph papers will be supplied on request.

___________________________________________________________________________

Q1.A) Multiple choice questions (Attempt Any 8)                                                  (8)

I) ln linear programming, unbounded solution means ____________. .

a) infeasible solution   b) Degenerate solution

c) Infinite solutions     d) Unique solution

  Ans:  c) Infinite solutions

Ii) if m + n-1 = number of allocations in transportation, it means _________..

(Where 'M' is number of rows and N' is number of columns)

a) There is no degeneracy       b) Problem is unbalanced

c) Problem is degenerate        d) Solution is optimal

 Ans:  a) There is no degeneracy   

Ijü) floats for critical activities will be always _________________..

a) One             b) Zero

c) Highest        d) Same ás duration of the activity

 Ans:  b) Zero

Iv) the total time required to complete all the jobs-in a job sequencing problem is known as __________,.

a) Idle time                                 b) Processing time

c)Elapsed time                        d) Processing order

 Ans:  c)Elapsed time

V) in linear programming, _____________ represents mathematical equation of the limitations imposed by the problem..

a) Objective function              b) Decision variable

c)Redundancy                         d) Constraints

Ans:    d) Constraints

Vi) if in an assignment problem, number of rows is not equal to number of columns then ____________..

a) Problem is degenerate                                    b) Problem is ‘unbalanced

c) It is a maximization problem                      d) Optimal solution is not possible

Ans:    b ) Problem is ‘unbalanced

 vii) the maximum time in which an activity will be completed assuming all possible delays and postponements is termed as __________..

a) Optimistic time                   b) Most likely time

c) Pessimistic time                  d) Expected time

Ans:  c) Pessimistic time

Viii) the various alternatives or courses of actions available to each player in a game are called as __________..

a) Saddle points                      b) Strategies

c) Pay off                                d) 'n' player game

Ans:  b) Strategies

Ix) in simplex, a maximization problem is optimal when all delta j, i.e., cj-zj values are.

a) Either zero or positive                       b) Either zero or negative

c) Only positive                                  d) Only negative

Ans:     b) Either zero or negative

X) which of the following considers difference between two least costs for each row and column while finding initial basic feasible solution in transportation.

a) North west corner rule                    b) Least cost method

c) Vogel's Approximation method       d) Row minima method

Ans:    b) Least cost method

Q1.B) True or false (Attempt Any 7)                                               (7)

i) Probability of a project completing in its expected time (Te) will be always 100%

Ii) lf saddle point is available in a game, it is called as pure strategy game., iii) slack represents unutilized resources., iv) if in a transportation problem, number of rows is not equal to number of columns, then the problem is unbalanced., v) lf we introduce an unnecessary dummy activity, the error is termed as redundancy., vi) job sequencing problems are solved to ensure that, both, the total time to complete all jobs and idle time of each machine are maximum., vii) when more than one optimal solution is possible in a linear programming problem, it is termed as 'unique solution'., viii) regret matrix is made to convert a maximization problem into minimization problem in assignment., ix) critical path method (cpm) considers the three-time estimates: most likely, optimistic and pessimistic time estimates., x) in solving a job sequencing problem, it is assumed that all jobs require the same sequence of operations., q.2 a) a company produces 2 products a and b: x 1 and x 2 , are the quantities manufactured of products a and b respectively. the following objective function along with constraints is given to you: .

Max Z = 8x 1 + 16x 2 ,

Subject to constraints:

x 1 + x 2 < 200

x 2 < 125

x 1 + 2x 2 < 300

x 1 , x 2 > 0

Find how many units of Product A and Product B should be produced by the company so that the profit is maximized. Ls it a case of multiple optimal solutions? Use graphical method to solve the LPP.

Video solution.

x 1 + x 2 < 200   ---- (i)

x 2 < 125           ---- (ii)

x 1 + 2x 2 < 300 ---- (iii)

x 1 , x 2 > 0  

x 1 + x 2 = 200

Let X 1 = 0

0 + x 2 = 200

Therefore x 2 = 200                  (x 1 , x 2 ) = (0, 200)

Let X 2 = 0

 x 1 + 0 = 200

Therefore x 1 = 200                      (x 1 , x 2 ) = (200, 0)

  x 2 < 125    ---- (ii)

Let x 2 = 0

x 2 = 125                 (x 1 , x 2 ) = (0, 125)

x 1 + 2x 2 < 300     ---- (iii)

x 1 + 2x 2 = 300

Let x 1 = 0

 0 + 2x 2 = 300

            x 2 = 300 / 2                             Therefore x 2 = 150             (x 1 , x 2 ) = (0, 150)

x 1 + 2x 2 = 300 Let x 2 = 0 x 1 + 2(0) = 300 x 1 = 300                     (x 1 , x 2 ) = (300, 0)

Corner Point Method:

Optimal Profit Maximization is Rs. 2,400

Q2.b) you are given the per unit cost of transporting goods from 3 factories to 4 customers. the 3 factories a b and c have capacity to supply 500,300 and 200 units respectively. the 4 customers p, q, r and s require 180,150,350 and 320 units respectively., i) you are required to find the initial basic feasible solution using vogel's approximation method                                                                                                                 (5) ii) find the total cost of transportation schedule obtained using vam          (2).

Total Demand = 1000

Total Supply = 1000

Total Demand = Total Supply

Cost data is given

Minimization Problem

RIM Condition:

M + N – 1 = No. of Allocation

3 + 4 – 1 = 6

Therefore, RIM Condition is satisfied.

Optimal Solution is Non-Degeneracy

(ii) Total Cost of Transportation :

            = (150 x 10) + (230 x 12) + (120 x 13) + (180 x 7) + (120 x 8) + (200 x 7)

            = Rs. 9, 440

Q.2 C) You are given the following details for a project consisting of 8 activities:

(i) Draw the network diagram and identify the critical path.                    (3) (ii) Find earliest start time, earliest finish time, latest start time and latest finish time for each activity.                                                                                                      (4) (iii) Find free float for activity B.                                                              (1)

  Solution:    [Video Explanation ]

(i)                  Network Diagram

Paths in Network:

(1) A-E-F = 4 + 20 + 10 = 34 days

(2) A-D-G = 4 + 5 + 6 = 15 days

(3) B-G = 6 + 6 = 12 days

(4) C- H = 13 + 16 = 29

Projection Completion Time : 34 days

(ii)               Free Float for Activity B:

= Ej – Ei – t

= 9 – 0 – 6

= 15  

Q.2 D) There are 6 jobs to be performed in a factory and each would go through 2 machines A and B in the order AB. The processing time (in hours) is given for each job in each machine.

(i) determine the sequence of performing the jobs that would minimize the total time of completing all the jobs:                                                                                 (2) (ii) find the elapsed time.                                                                               (3) (iii) find idle time for both the machines.                                                    (2).

Solution :    [Video Explanation ]

(i)                  Optimal Sequence

A                                                                                                                           B

(ii)                 Elapsed Time Calculation :

  Total Min. Elapsed Time = 36 Hrs.

(iii)            Idle time for Machinery A    = 36 – 35 = 1 Hrs

Idle time for Machinery A    = 36 – 28 = 8 Hrs

Q.3. B) You are given the following details for a project with 8 activities.

(i) Draw the network diagram.                                                                                     (3)

(ii) Find the expected time of project completion along with standard deviation.       (2)

(iii) What is the probability of the project completing in 55 days?                               (2)

Ans:    [Video Explanation]

Network Diagram:

Paths in the Network:

1)       1-2-3-5-7-8  = 47 days

2)       1-2-4-6-7-8 = 33 days

Expected Project Completion Time : 47 days

Variance of critical path = 16/36 + 100/36 + 256/36 + 64/36 + 4/36 = 440/36 =12.22

SD of critical path = √variance

SD = √ 12.22   = 3.495                        = 3.5 days

Time = 55 days = te

Z = t – CP / SD

= 55 – 47 / 3.5 = 2.29 

Table value = 0.4890

P(55 days) = 0.5 + 0.4890      = 0.9890                      = 98.90% 

Q.3. C) You are given information about the cost (in Rs. Thousands) of performing different jobs by different persons. PI cannot perform J3. P3 cannot perform J4

(i)Obtain optimal assignmnt and find cost of such assignment.

(i)Is it a case of alternative optimal solution?

Ans:      [Video Explanation]

Prohibited Assignment = X (infinity)

The problem is unbalanced.

Taking P5 as Dummy

Cost is given.

Hence Minimization Problem.

Row Minimization

Column Minimization Not Required.

Min. No. of Lines = 4

Size of Matrix = 5 x 5

Therefore Not optimal

Change = Min. open Value = 4

Optimal Cost = 66,000

Q3.D) Two firms, Lacko textiles and Rayon textiles have 3 strategies each to select from. The 3 strategies are no advertisement, using moderate advertising and using heavy advertising. You are given the pay off matrix from view point of Lacko textiles, showing its market share under several combinations of strategies:

(    (i) Find the saddle point and value of game.      (ii) Comment on the strategy to be selected by both the companies.

Value of the Game = 50 x 10000 =     5,00,000

Q. 4 a) you are given a solution for a transportation cost problem. figures in each cell represent per unit transportation cost. figures in circle within each cell represent number of units allocated for transportation. x, y and z. are the 3 factories and a, b, c and d are the 4 customers..

(i) You are required to check the above solution for optimality.

(ii) If it is not optimal, use modified distribution method to obtain optimal solution.

(iii) Find optimal transportation cost.

Ans:  [Video Explanation]

No negative ∆, Hence Optimal Solution.

(iii) Optimal transportation Cost:

= (200 x 7) + (120 x 18) + (100 x 15) + (280 x 7) + (180 x 11) + (12 x 5)

= RS. 9,600

B) You are given the following information, for a project with 8 activities:

The cost of completing the eight activities in normal time is Rs.6,500.

Indirect cost is Rs 160 per day. The contract includes a penalty of Rs. 100 per day for every day of delay more than 17 days.

  (i)   Draw the network diagram and find critical path.

(i     (ii)  Crash the projèct duration to find the total 'cost of completing the project in 17 days (4)

  Ans:  [Video Explanation]

Q.4. C) A company produces 2 products x and xX) using three resources S1, S2 and S. Product x1 gives profit of Rs.30 per unit and product x2 gives profít of Rs.40 per unit. The 3 resources Si. S and S; are available to the extent of 200 units, 600 units and 500 units respectiveiy The following objective function and constraints are given to you:

Max Z = 30x1 + 40x2

Subject to constraints;

x 1 + 2x 2 < 200

8x 1 + 5x 2 < 600

3x 1 + 4x 2 < 500

X1 >0; x2 >0

You are given the following simplex solution to the above problem:

  A) With reference to the above table answer the following:

i) Check if the above solution is optimal or not.

ii) lf it is not optimal, find optimal solution.

B) With reference to the optimal simplex table in the above problem obtained by you, answer the following:

iii) Find the optimal product mix and optimal profit

iv) Which resources are scarce and which are, unutilized?

v) Is it a case of alternative solution? Justify your answer

vi) What are the shadow prices of the resources? Justify

Ans:       [Video Explanation]

Testing the given solution for optimality ;

Positive ∆

Hence, solution not optimal.

(ii) Revising the Solution for Optimality:

No positive ∆

Therefore, Optimal Solution

(iii) Optimal Product Mix;

No. of units of X1 = 200/11        = 18.18

No. of units of X1 = 1000/11      = 90.90

Optimal Profit:

Max Z = [30 x 200/11] + [40 x 1000/11]

          = Rs. 46,000 /11

          = Rs. 4,181.81

(iv) Abundant Resource:

S3 = 900/11 ----- in optimal solution

Therefore Unutilized Capacity of S3 = 900 / 11

S3 is Abundant resource.

S1 and S2 are non-basis variable in optimal solution.

 Unutilized Capacity of S1 and S2 are 0

(i)                 ∆ of non basis variable :

∆ of S1 = - 170 / 11

∆ of S2 = -20/11

No zero ∆ value.

Hence, this is a case of Unique Solution.

There is no alternate optimal solution.

(ii)              Zj value of slack variables:

Q.5 A) Explain the concepts : Total float, free float, Independent float and Interfering float.           (8) 

(a) Total float:

lt is the amount of time by which the completion of an activity can be delayed beyond the earliest possible finishing time without affecting total project completion time.

 Total Float = (LST - EST) or (LFT - EFT)

.:: Total float = LST - EST -[;-t]-E Total Float = [Lj- E -t

(b) Free float:  It is the amount of time by which the completion of an activity can be delayed beyond the earliest finishing time without affecting the earliest possible start of a subsequent activity.

Free float = Earliest starting time of subsequent activity - Earliest finishing time of activity ij = Ej- [E, + t;] . Free Float = Ej- E-tj

(c) Independent float:  It is the amount of time by which the start of an activity can be delayed without affecting the earliest start of a subsequent activity, assuming that the preceding activity has finished at its latest finishing time. .

:. Independent float = Earliest starting time of subsequent activity - Latest finishing time of preceding activity - ij

:. Independent Float = [Ej-Li]-tij;

(d) Interfering float:  It is that part of total float which reduces the float of a subsequent activity. It is the amount of time by which the earliest possible start of a subsequent activity will be delayed if activity ÷ finishes on latest finishing time.

:. Interfering float = Latest finishing time of activity ij - Earliest starting time of subsequent activity. :. Interfering Float= Lj-Ej

B) Discuss any 5 areas where techniques of operation Research can be applied.    (7) 

OR techniques are applied to a variety of business problems. Some examples are:

(A) Production management:

(1) To calculate loss of time due to waiting time, queuing time etc.

(2) To decide optimum allocation of jobs and optimum sequence in which jobs should be sequenced.

(B) Personnel management:

(1) To study labour turnover. (2) To do human resource planning. (3) To decide number of personnel required to be kept on standby in case of demand for higher manpower.

(C) Inventory management: (1) To study economic lot size to be ordered.

(D) Marketing management: (1) To decide optimal product mix for maximum profit. (2) Media selection for advertising for maximum reach. (3) Sales forecasting.

(E) Transportation management: (1) To determine transportation schedule for minimum cost or minimum time.

(F) Project management (1) To identify critical and non-critical activities of a project (2) To determine minimum project completion time. (3) To determine optimal project cost.

(G) Financial management: project. (1) To decide investment portfolio to maximize return on investment. of such nature that it is impossible to find a feasible solution that can satis1y a the constraints. 

c) Answer any 3 of the following :            (15) 

I) explain the terms: redundant constraints and infeasible solution in linear programming. .

Ans:  A redundant constraint is the one which does not affect the optimal solution. Even if that constraint is not considered, we can still obtain the solution to the problem. Since the constraint is not required for obtaining the optimal solution, it is called redundant constraint cost.

Following sketch explains redundancy (redundant constraint):

There is no common feasible region for line AB and line CD

Hence, solution is infeasible.

A redundarnt constraint is the one which does not affect the optimal solution. Even if that constraint is not considered, we can still obtain the solution to the problem. Since the constraint is not required for obtaining the optimal solution, it is called redundant constraint. cost.

The feasible region for the above problem is OABC. The 3rd constraint does not affect the feasible region. Hence, the constraint [X1 S 16] is redundant constraint.

ii) what do you means by alternative optimal solution in transportation? How do you find that alternative solution? 

Alternate optimal solution or multiple optimal solutions mean there are two sets of solutions which provide the same optimal cost or optimal profit:

(a) In the optimal solution to a transportation problem, if there is an empty cell with zero A value, it means there is alternate optimal solution.

(b) To find alternate optimal solution, we should construct a closed loop from the empty cell with zero A. The new table we obtain after looping is the alternate optimal solution.

iii) Explain the time coat trade off in project crashing. 

The execution of a project involves two types of costs, direct cost and indirect cost.

(a) Direct costs: These include costs of materials, machinery, tools, manhours etc. When we estimate the duration of various activities in the project, it is assumed that normal amount of labour (manhours) & machines will be required to complete these activities. The estimation of direct costs is done based on the normal amount of resource required. This estimation gives us minimum possible direct cost required to complete the project. When we want to complete the project in a shorter period than critical path, we will need to employ more resources. Hence, direct cost will increase.

(b) Indirect costs: These include rent, overheads, administrative costs etc. Indirect costs vary with time. They are expressed on per day (or per week etc.) basis. Hence, when we shorten the project completion time, total indirect cost reduces.

Total P'roject Cost = Direct Cost + Indirect Cost

iV) Discuss the signification of theory of game. Briefly discuss the terms. Player and pay off.

Ans: Every one business party. The situation (also called competitive situation) involves more parties involved could be corporate. individuals, small firms or A number of outcome alternatives are available to each party for decision making. The (i.e. profit or loss) of the business decision does not depend on the alternative selected by one party alone, but on the interaction between the decisions of all the parties involved in the situation. Each party will try to select the alternative which maximizes gains for it, at the cost of its competitors. "Games Theory" deals with such business situations or problems where multiple parties are involved in the conflict through interaction of their alternatives or decisions.

Players: The various participants or decision markers in the game are called "Players". A game must have minimum two players (or competitors). A game having two competitors opposing each other is called 'two person game' and a game having more than 2 players is called 'n-person game'.

Payoff: Payoff is the outcome of the interaction of selected strategies of opponents in the game. Positive payoff is gain and negative payoff is loss. In a 'two person game', when one party gains, the other party loses.

v) Distinguished between PERT and CPM. 

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Operations Research Proceedings 2023

Selected Papers of the Annual International Conference of the German Operations Research Society (GOR), Germany, August 29 – September 1, 2023

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• Malte Fliedner 1 ,
• Knut Haase 2 ,
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• Kai Hoberg 4 ,
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Universität Hamburg, Hamburg, Germany

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Kühne Logistics University, Hamburg, Germany

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About this book

This book gathers a selection of peer-reviewed papers presented at the International Conference on Operations Research (OR 2023), which was held at Universität Hamburg, Germany, August 29–September 1, 2023. More than 700 scientists, practitioners, and students from mathematics, computer science, business/economics, and related fields attended the conference and presented more than 500 papers in plenary presentations, parallel topic streams, as well as special award sessions.

The book discusses classical mathematical optimization, statistics, and simulation techniques. These are complemented by computer science methods and by tools for processing data, designing and implementing information systems. The book also examines recent advances in information technology, which allow big data volumes to be processed and enable real-time predictive and prescriptive business analytics to drive decisions and actions. Lastly, it encompasses addressing problems modeled and treated while considering factors such as uncertainty, risk management, and behavioral issues.

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Kai Hoberg, Joern Meissner

About the editors

Guido Voigt is Chair of Logistics and Supply Chain Management at the University Hamburg, Germany. His research focuses on behavioral operations, supply chain coordination, and closed-loop supply chain management.

Malte Fliedner is Chair of Operations Management at Hamburg University, Germany. His research focuses on resource allocation and scheduling in manufacturing and logistics and comprises the structural analysis of complex planning problems and the development of algorithmic solution approaches for efficient decision support.

Knut Haase is Chair of Transport Economics at the Faculty of Business Administration at the University of Hamburg. His research focuses on quantitative problems in logistics, production, and transport planning.

Wolfgang Brüggemann is Professor of Business Administration, in particular, Operations Research at Hamburg Business School of University Hamburg, Germany. OR applications in various areas, such as production planning, strategic port planning and energy markets, are at the core of his and his team's research interests.

Kai Hoberg is Professor of Supply Chain and Operations Strategy at Kühne Logistics University, Germany. His research focuses on using new technologies and leveraging data to advance supply chain management and is typically combining mixed methods such as data analysis, modeling, or experiments.

Joern Meissner is Full Professor of Supply Chain Management & Pricing Strategy at Kühne Logistics University in Hamburg, Germany. His main research focus is the area of stochastic and dynamic decision-making and in particular applications of machine learning to logistics, manufacturing, supply chain management, and pricing strategy.

Bibliographic Information

Book Title : Operations Research Proceedings 2023

Book Subtitle : Selected Papers of the Annual International Conference of the German Operations Research Society (GOR), Germany, August 29 – September 1, 2023

Editors : Guido Voigt, Malte Fliedner, Knut Haase, Wolfgang Brüggemann, Kai Hoberg, Joern Meissner

Series Title : Lecture Notes in Operations Research

Publisher : Springer Cham

eBook Packages : Business and Management , Business and Management (R0)

Copyright Information : The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2024

Hardcover ISBN : 978-3-031-58404-6 Due: 09 August 2024

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Operation Research

Previous year question paper with solutions for operation research from 2022 to 2022.

Our website provides solved previous year question paper for Operation Research from 2022 to 2022. Doing preparation from the previous year question paper helps you to get good marks in exams. From our OR question paper bank, students can download solved previous year question paper. The solutions to these previous year question paper are very easy to understand.

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IGNOU MMPO-001 Previous Year Exam Question Papers

Download previous year ignou exam question papers for mmpo-001: operations research. question papers are available for june and december term end examinations starting from 2010 to 2023, download mmpo-001 ignou previous year term end exam question paper entitled operations research which is applicable for master of business administration (mba), master of business administration (operations management), post graduate diploma in operations management (mba, mbaom, pgdiom) course(s) in ignou absolutely free click on the appropriate button to get the examination question paper of the year you are looking for. you will get the pdf downloaded instantly checking and comparing previous year question papers for this subject will help you understand which question have been repeated many times. this can help you prepare for your exams effectively..

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