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Do U want to see MBA Project -Operations Research & Project Management
Posted Date: 24 May 2008 Resource Type: Articles/Knowledge Sharing Category: E-Books
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Posted By: sendoorpandi Member Level: Bronze
Rating:  Points: 1
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Operations Research & Project Management
Student name : Santhosh.k
Reg. No : 207231545
Course name : MBA in Engineering
College Name : M S Ramaiah Institute of Technology
Date of project Finished: 25-04-08
Acknowledgment
I express my sincere and grateful appreciation to Asst Prof. Bharat.K for his valuable effort to teach the concept of Operations Research & Project Management and sharing his experience and knowledge.
I would like to thanks Prof. Devakumar, who provided thoughtful review and useful suggestion for understanding and completing the assignment.
I would like to Thanks God and my Parents who has given me the opportunity to pursue my P.G (MBA in Engineering Operation) without any hassles.
I would also like to thank the MSRSAS for providing library & Internet facility to learn the concepts of the subject.
- Midhun Chakravarthi. Ravi
Contents
Title Page i
Declaration ii
Acknowledgement iii
List of Tables v
List of Figures vi
Bibliography 37
Appendix-1 39
Appendix-2 40
Abstract
Operations research uses analyses and techniques from a variety of branches of Mathematics, statistics and other scientific disciplines. Certain analytical results arise repeatedly in applications of operations research to industrial and service operations.
The objective of this assignment is to apply wide range of operations research models and tools in analyzing the different situations and optimizing the results in a concise format to take the critical decisions
The section A explains about the various Operations Research techniques that are used in the business scenario. It gives information about various models like LPP, EOQ, Transportation problems, Assignment cases, Job sequencing and Simulation techniques.
The section B reveals the various types of Project Management Techniques used in the Business like the PERT, CPM, Shortest path, Various Network problems. A clear idea of estimating the project completion time w.r.t money spent on the project has been explained.
And Section C gives the information of X. bar and R charts for checking of homogeneity and relevance. The other part is the Vender rating concept which is very useful in choosing of our suppliers (Venders) based on Quality, Cost Time of delivery etc.
Section A Page no
Q No: 2 Consider the Aditya Birla Retail Company ltd. Determine the optimal distribution for this company to minimize total shipping cost. Evaluate your output.
1-7
Q No: 3Calculate the EOQ and the total costs associated with stocking the item. If the supplier of the item will only deliver batches of 250 units, how are the stocks holding costs affected. Determine an ordering policy for the raw material of Aditya Birla Minacs Worldwide Limited. What is your comment on holding cost?
8-10
Q No: 5 Formulate and compute an assignment that maximizes the profit and minimizes cost of Birla Sun Life Distribution Company Limited (BSDL)
11-15
Q No: 6 Suppose you are the manager of PSI Data Systems and want to minimize the total time taken by the three operators, as this will minimize the electricity consumption. How will you assign the tasks to the operators? Comment on total elapsed time.
15-17
Q. No: 7 Pan Century Surfactants Inc acquired a machine worth $6100 in the year 2000 and its yearly maintenance cost is given in the case. According to you when they should replace the machine and why?
17-19
Q. No: 8 Using Monte-Carlo Simulation techniques estimate the profits and loss for the next ten days of Aditya Birla Nova Company? If the company decides to produces 540 insulator per day, calculate average profit or loss of the company. Comment on efficiency of the technique and suggest improvement over the technique.
20-23
Section B Page no
Q No: 9 Bangalore Mahanagar Palike estimated to construct a fly over in Ring Road; the Construction network is as follows:
23-27
Q No: 10 A Salesman located on a City A (Bangalore) decided to travel to City B (Delhi). He knew the distances of alternative routes from City A to City B. He then drew the highway network map as shown below. The salesman’s problem is to find the shortest route that covers all of the selected cities from city A to city B.
27-30
Section C
Q No:1 Construct X bar R chart of the following data:
Check the values of Homogeneity. Assume values of constants A2 0.58, D3=0 and D4=2.11 and test the homogeneity and state the relevance in any Industry.
30-33
List of figures Page numbers
Fig 1 Network Diagram 23
Fig 2 Network Diagram of project with Critical Path 24
Fig 3 Network Diagram of project with critical path 25
Fig 4 when a salesman has to pass through city 2 & 9. 27
Fig 5 Shortest Route 30
Fig 7 X-Bar Chart 32
Fig 8 R Chart 33
SECTION-A
Question 2: Consider the Aditya Birla Retail Company Ltd. Determine the optimal distribution for this company to maximize total shipping cost. Evaluate your output.
Transportation concerns the movement of products from a source—such as a plant, factory, or work-shop—to a destination—such as a warehouse, customer, or retail store. The goal for any business owner is to minimize transportation costs while also meeting demand for products. Transportation costs generally depend upon the distance between the source and the destination, the means of transportation chosen, and the size and quantity of the product to be shipped.
A programming problem that is concerned with the optimal pattern of the distribution of goods from several points of origin to several different destinations, with the specified requirements at each destination is called as a Transportation problem.
In a transportation problem, the initial feasible solution can be generated by a number of methods. Three of the most commonly methods are
a. The North-West Corner Rule (NWC Rule)
b. Least Cost Method (LCM)
c. Vogel’s Approximation Method (VAM)
After developing the initial feasible solution by any one of the above three methods, the solution thus obtained has to be tested for optimality. Basically 2 methods are widely used:-
a. Stepping Stone Method
b. Modified Distribution (MODI) Method
Applications
The transportation problem was often discussed as an application that would benefit from computerization, because this type of problem can be formulated quantitatively and because such problems are often complex enough to benefit from using a model. Also, the allocation of transportation resources among competing uses is of interest to business decision-makers in a number of different industries. In general, real-world transportation problems are often important!
Airlines make decisions about adding flights. Continental Airlines bases its route and schedule decisions on daily ticket data. Continental uses a computer program developed by American Airlines' Sabre unit. Field quoted Robert Merz, director of network operations at United, "You schedule to maximize profit”.
Recently, Southwest Airlines implemented CALEB(TM) Technologies' Crew Solver DSS to reduce the cost from traffic control delays and mechanical and weather-related disruptions
So using Model-Driven DSS to solve transportation problems can improve profitability.
Mathematical programming provides quantitative bases for management decisions -- bases with which management manipulates and controls various activities to achieve the optimal outcomes of business problems. Management can make better and more effective judgment by use of mathematical programming. However, it is no substitute for the decision maker's ultimate judgment. Source:http://books.google.co.in/books?id=HFY8ROnNSOEC&pg=PA223&lpg=PA223&dq=transportation+problems
Ware houses
W1 W2 W3 Supply
F1 16 20 12 200
F2 14 8 18 160
Factory F3 26 24 16 90
F4 22 16 15 70
F5 11 4 2 90
Demand 180 120 150 450
Here, Demand = 450 & Supply = 610.
Demand /= Supply, supply > Demand,
It is an un balanced Transportation Problem.
Hence in order to Balance this problem, create a dummy column having a Demand of 160.
Ware houses
W1 W2 W3 W4 Supply
F1 16 20 12 0 200
F2 14 8 18 0 160
Factory F3 26 24 16 0 90
F4 22 16 15 0 70
F5 11 4 2 0 90
Demand 180 120 150 160 610
Hence it is a balanced Transportation problem.
Computation of Transportation Cost under North West Corner Rule:
Ware houses
W1 W2 W3 W4 Supply
F1 16
20
12 0 200, (20) 0
F2 14 8
18
0
160, (60) 0
Factories F3 26 24 16
0
90, 0
F4 22 16 15 0
70, 0
F5 11 4 2 0
90, 0
Demand 180, 0 120, (100) 0 150, (90) 0 160, (90) 0 610
Calculation of total price of NWCR:
Factory Warehouse Quantity* Unit price Total price
F1 W1 180 *16 2880
F1 W2 20 *20 400
F2 W2 100 *8 800
F2 W3 60 *18 1080
F3 W3 90 *16 1440
F3 W4 0 *0 0
F4 W4 70 *0 0
F5 W4 90 *0 0
Total 610 6600
Hence the Total cost by NWCR is Rs 6600.
Least Common Method:
Ware House
Factory W1 W2 W3 W4(Dummy) Supply
F1 16
20
(40)12
(160)0
0 (40) 200
F2 (40)14 (120)8 18 0 0 (40) 160
F3 (90)26 24 16 0 (0) 90
F4 (50)22 16 (20)15 0 0 (50)70
F5 11 4 (90)2 0 0 90
Demand 0 (90) (140) 180 0 120 0 (20)(60)150 0 160 610
Calculation of Total Price of LCM:
Factory Warehouse Quantity* Unit price Total price
F1 W3 40*12 480
F1 W4 160 *0 0
F2 W1 40 *14 560
F2 W2 120 *8 960
F3 W1 90 *26 2340
F4 W1 50 *22 1100
F4 W3 20 *15 300
F5 W3 90 *2 180
Total 5920
The total cost of transportation by least cost method is Rs 5920
Vogel's Approximation Method:
W1 W2 W3 W4 Dummy SUPPLY P1 P2 P3 P4 P5
F1
16
20
12
0
200 140 0 12 12 4 4 4
F2
14
8
18 0 160 40 0 8 8 6 6 4
F3
26
24 16 0
90 0
16 - - - -
F4
22
16 15 0
70 0
15 15 - - -
F5
11
4 2
0 90 0
2 2 2 - -
DEMAND 180 140
0 120 0
150 60
0 160 70
0 610
P1 3 4 10 0
P2 3 4 10 0
P3 3 4 10 -
P4 2 12 6 -
P5 2 - 6
P6 2 - - -
Ware houses
W1 W2 W3 W4 Supply
F1 16
20 12
0 200
F2 14
8
18 0
160
Factories F3 26 24 16 0 90
F4 22 16 15 0
70
F5 11 4 2
0 90
Demand 180 120 150 160 610
Calculation of Total Price of VAM:
From To Amount Unit Cost Route Cost
W1 F1 140 16 2240
W1 F3 60 12 720
W2 F1 40 14 560
W2 F2 120 8 960
W3 F5 90 2 180
TOTAL 4660
Conclusion: Compared to NWCR & Least cost method, the VAM method is giving the low cost transportation; hence we can recommend this path
Test for optimality by MODI Method (U-V Method)
Assigning value of u & v with respect to allocated cells cost and allocating delta to cell (1,4)
WAREHOUSE U i
FACTORY W1 W2 W3 DUMMY
F1 (16) (12) ? (0) 0
F2 (14) (8) -2
F3 (0) 0
F4 (0) 0
F5 (2) -10
V j 16 10 12 0
Assigning cost to the empty cell based on the values of u & v.
WAREHOUSE U i
FACTORY W1 W2 W3 DUMMY
F1 • 10 • • 0
F2 • • 10 -2 -2
F3 16 10 12 • 0
F4 16 10 12 • 0
F5 6 0 • 10 -10
V j 16 10 12 0
Actual cost of empty cell
WAREHOUSE
FACTORY W1 W2 W3 DUMMY
F1 • 20 • •
F2 • • 18 0
F3 26 24 16 •
F4 22 16 15 •
F5 11 4 • 0
Reduction of cost from actual to allocated cost for empty cells
WAREHOUSE
FACTORY W1 W2 W3 DUMMY
F1 • 10 • •
F2 • • 8 2
F3 10 14 4 •
F4 6 6 3 •
F5 5 4 • 0
Since all the cost obtained are positive therefore the solution will be considered as a optimal solution
Result of Optimal solution
Factory
Warehouse
Quantity
Unit price Total price
F1
W1 140 16 2240
F1
W3 60 12 720
F2
W1 40 14 560
F2
W2 120 8 960
F5 W3 90 2 180
Total 450 4660
Hence we can say that VAM method can be applied for solving this transportation problem and the minimized cost of transportation is Rs 4660
Question 3: Calculate the EOQ and the total costs associated with stocking the item, If the supplier of the item will only deliver batches of 250 units, how are the stocks holding costs affected. Determine an ordinary policy for the raw materials of Aditya Birla Sun life Distribution Company Ltd.
Definition – EOQ
“EOQ, or Economic Order Quantity, is defined as the optimal quantity of orders that minimizes total variable costs required to order and hold inventory”
Introduction to inventory model: The inventory models are primarily concerned with the optimal stock or inventory policies of the organization. Inventory problems deal with the determination of optimum levels of different inventory items and ordering policies, optimizing a pre-specified standard of effectiveness. It is concerned with the factors such as:
1. Demand per unit time,
2. Cost of placing orders,
3. Costs incurred while keeping the goods in inventory,
4. stock-out costs and
5. Costs of lost sales etc.
If a customer demands a certain quantity of a product, which is not available, then it results in a lost sale. On the other hand, excess inventories mean blocked working capital which is the life blood of modern business. Similarly, in the case of raw materials, shol1age of even a very small item may cause bottlenecks in the production and the entire assemb1y line may came to a halt. Inventory models are also useful in dealing with quantity discounts and multiple products. These models can be of two types –
1. Deterministic and
2. Probabilistic
Applications of inventory model
Inventory models are used in each and every purchase department of the manufacturing companies. All the companies’ keeps inventory for smooth production, meet the unexpected sudden demand, and reduce the lead time, cycle time and operational & sales losses. In purchasing department it is used to calculate various important decision variables such as:-
1. Re-order quantity of material of all the categories A, B, C.
2. Lead-time i.e. order formation to receipt of the materials,
3. Economic order quantity (optimum quantity at which cost of production is minimized)
4. The pessimistic, Optimistic & the most likely level of stock keeping.
5. Duration of Ordering
Companies using EOQ model
1. Padmaja Enterprises, Rice producers uses the EOQ method for the Raw material.
2. Tata Motors Limited, Manufacturer of Four wheeler automobiles, Jamshedpur.
3. Mahindra Tractors, dealers uses this method for the Spare parts inventory control.
CASE PROBLEM
Order policy when inventory carrying cost is expressed as a % age of unit cost
Annual Demand (D) = 1000 units
Cost per order (Co) = Rs. 100
Cost per unit (Cu) = Rs. 40
Interest =18 % of unit cost = 40 x 0.18 = Rs. 7.2
Insurance = 1 % of unit cost = 40 x 0.01 = Rs. 0.4
Allowances for obsolescence = 2 % of unit cost = 40 x 0.02 = Rs. 0.8
Building over head = Rs. 2
Damage & loss = Rs. 1.50
Miscellaneous cost = Rs. 4
Total annual holding per unit (Ch) = 7.2 + 0.4 + 0.8 + 2 + 1.50 + 4 = Rs. 15.90/unit/annum
Calculation for order policy
Economic Order quantity Q*= = = 112.15 or 112 units per order
No of orders per year (No) = = = 8.91 or 9 orders per year
Time between order = = = .0111 years
= 0.0111x12 x 30 = 40 days.
Minimum annual inventory cost =
=
= Rs. 1783.25
If the supplier of the item will deliver only a batch of 250 units then,
Q* = 250 putting in equation Q*=
250 =
250 x 250 =
Ch =
Ch = Rs. 3.2 / unit / annum
Thus the cost of holding will reduce by 79.87 % i.e. from Rs. 15.9 to Rs. 3
Inventory carrying cost is expressed as percentage of annual average inventory investment
Annual Demand (D) = 3600 Kg
Cost per order (Co) = Rs. 36
Cost per unit (Cu) = Rs. 10
Inventory carrying cost (I) = 25 % of the investment in inventories
= 0.25 / unit / annum
Calculation for order policy
If the inventory carrying cost is expressed as a percentage of annual average inventory investment, then
Economic Order quantity Q*= = = 322 Kg per order
No of orders per year (No) = = = 11.18 or 12 orders per year
Time between order = = = .008333 years
= 0.00833 x 12 x 30 = 30 days
Minimum annual inventory cost =
=
= Rs. 804.98 or 805
If the supplier of the item will deliver only a batch of 250 units then,
Q* = 250 putting in equation Q*=
250 =
250 x 250 =
I =
I = 0.4147 or 41.47 %
In this case, the cost of holding will increase by 65.88 % i.e. from 25% to 41.47%
Question 5: Formulate and compute an Assignment that maximizes the profit and minimizes cost of Birla Sun Life Distribution Company Limited (BSDL)
The assignment problem is one of the fundamental combinatorial optimization problems in the branch of optimization or operations research in mathematics. It consists of finding a maximum weight matching in a weighted bipartite graph.
In its most general form, the problem is as follows:
There are a number of agents and a number of tasks. Any agent can be assigned to perform any task, incurring some cost that may vary depending on the agent-task assignment. It is required to perform all tasks by assigning exactly one agent to each task in such a way that the total cost of the assignment is minimized.
If the numbers of agents and tasks are equal and the total cost of the assignment for all tasks is equal
to the sum of the costs for each agent (or the sum of the costs for each task, which is the same thing in this case), then the problem is called the Linear assignment problem. Commonly, when speaking of the Assignment problem without any additional qualification, then the Linear assignment problem is meant.
The Hungarian algorithm is one of many algorithms that have been devised that solve the linear assignment problem within time bounded by a polynomial expression of the number of agents. This method is used to solve the given case.
Application of Assignment Model
Applications of assignment problems are varied in the real world. Certainly it can be useful for the in the task of assigning employees to tasks or machines to production jobs, but its uses are more widespread. It could be used to assign fleets of aircrafts to particular trips, or assigning Buses to routes, or networking computers. The assignment model is also used by government firms and many organizations for assigning the work to the applicant based on the tender values.
It can also be used to assign the work to people properly and reduce the cost, and optimize the profit. Source: http://en.wikipedia.org/wiki/Assignment_problem
Territories
Salesman T1 T2 T3 T4
S1 25 27 28 37
S2 28 34 29 40
S3 35 24 32 33
S4 24 32 25 28
Minimization of cost
Step 1: Row Reduction
Territories
Salesman T1 T2 T3 T4 Min
S1 25 27 28 37 25
S2 28 34 29 40 28
S3 35 24 32 33 24
S4 24 32 25 28 24
Step 2: Column Reduction
Territories
Salesman T1 T2 T3 T4
S1 0 2 3 12
S2 0 6 1 12
S3 11 0 8 9
S4 0 8 1 4
Minimum 0 0 1 4
Step 3: Assignment
Territories
Salesman T1 T2 T3 T4
S1
0 2
2 8
S2 0
6 0 8
S3 11 0 7
5
S4 0
8 0 0
Cost counting
From To Minimized Cost
S1 T1 25
S2 T3 29
S3 T2 24
S4 T4 28
TOTAL 106
In the case of minimization of the cost, the total cost of the Assignment problem is = 106.
Maximization of Profit:
Territories
Salesman T1 T2 T3 T4 Maximum
S1 25 27 28 37 37
S2 28 34 29 40 40
S3 35 24 32 33 35
S4 24 32 25 28 32
Step1: Row Reduction
Territories
Salesman T1 T2 T3 T4
S1 12 10 9 0
S2 12 6 11 0
S3 0 11 3 2
S4 8 0 7 4
Minimum 0 0 3 0
Step 2: Column Reduction
Territories
Salesman T1 T2 T3 T4
S1 12 10 6 0
S2 12 6 8 0
S3 0 11 0 2
S4 8 0 4 4
Step 3: Assignment
T1 T2 T3 T4
S1 12 10 6 0
S2 12
6 8 0
S3 0 11
0 2
S4 8 0 4 4
Step4: Improvement stage:
Since number of allocation is less than 4 so further iteration is required
T1 T2 T3 T4
S1 12 10 6 0 1
S2 12 6 8 0 1
S3
0
11
0 2 2
S4
8 0 4 4 1
1 1 1 2
Number of lines crossing the Zero’s are 3 which is less than 4, so the current assignment can be considered as a first feasible solution but not optimal and. Hence further iterations is required
Step 5: Matrix reduction Minimum Value after the lines passed by is = 6
No line-> Deduct the Minimum
One line-> No change, keep as it is
Two lines-> Add the minimum
Creation of New matrix
T1 T2 T3 T4
S1 6 4 0 0
S2 6 0 2 0
S3 0 11 0 8
S4 8 0 4 10
Assignment:
T1 T2 T3
T4
S1 6 4 0 0
S2 6
0 2 0
S3 0 11
0 8
S4 8 0 4 10
Check for optimality
Number of lines crossing the Zero’s will be 4 which is equal to 4, so the current assignment can be considered as optimal, hence the assignment for the job can be confirmed
Optimal Allocation result
Sales man Territory Cost
S1 T3 28
S2 T4 40
S3 T1 35
S4 T2 32
Maximum Profit = 135
Question 6: Suppose you are the manager of PSI Data Systems and want to maximize the total time taken by the three operators, as this will minimize the electricity consumption. How will you assign the tasks to the operators? Comment on the total elapsed time.
The selection of the appropriate order in which waiting jobs may be served is called "Job sequencing". Although theoretically, it is possible to find the optimum sequence by testing each one, in practice, it is impossible to check each sequence because of the large number of computations involved. For example, if there are 4 jobs to be processed at each of the 5 machines the total number of theoretically possible different sequences will be (4!)5=7,962,62. Obviously, any technique to arrive at minimal elapsed time sequence at least approximately will be quite valuable.
Poor scheduling can leave most of the expensive machine sitting idle while one bottleneck task is performed. Assigning people to jobs, meetings to rooms, or courses to final exam periods are all different examples of scheduling problems.
Assumptions:
1. No operator can process more than one task at a time.
2. Each task once started by a operator should be continued until it is complete.
3. Time between completion of one task and start of second task is negligible
Companies Using Job sequencing are:
The ITC BPL, Bhadradri Paper Plant, which uses the job sequencing technique for assigning jobs to the machines in the production of Paper, is an example of Job sequencing case.
In the production of cars in the large manufacturing plants like Maruti, Tata, it is used.
Manufacturing plants using the CAD/CAM technology uses the job sequencing in operations.
Operators Tasks
1 2 3
A 15 10 9
B 9 15 10
C 8 12 10
Matrix Conversion time for each task for each operators:
Jobs Machine
A B C A+B B+C
1 15 9 8 24 17
2 10 15 12 25 27
3 9 10 10 19 20
Task ordering from 1 to 3 operators
? The minimum time form the above table is 17 i.e., for task 1st, operator II so this task will be scheduled last.
? After eliminating task 1st from table, the minimum time is 19, for 3rd task, operator I thus will be sequenced first.
? Remaining task 2nd will be processed second.
The process sequence should be:
3 2 1
Job Order Sequence
A+B
J3 J2 J1
B+C
Calculation of total elapsed time:
Tasks
Jobs 1 2 3
J3 9 10 10
J2 10
15
12 Min 10+15=25 Max 19+15=34
Total 19 34 46 Min 10+12=24 Max 34+12=46
J1 15 9 8
Total 34 43 54
Note: Here we have to take maximum number.
Conclusion:
Hence the idle time of the machine cannot be avoided in this case but the organization can be made such that the operator 2 & 3 should be called for their work after 9 & 19 hour respectively to avoid the idle time the operator and can used their skill for any other tasks. Also machine 2 & 3, should be started after 9 and 19 hours respectively to reduce the electricity Cost.
Question 7: Pan Century Surfactants Inc acquired a machine worth $ 6100 in the year 2000 and its yearly maintenance cost is given in the case. According to you when they should replace the machine and why?
Replacement Theory:
Replacement theory is concerned with the problem of replacement of machines, electricity bulbs, men, parts etc. The need for replacement arises due to:
? Their deteriorating efficiency.
? Failure or Breakdown
Replacement >
?
Replacement is usually carried out under the following situations:
When existing items have outlived their effective lives and it may not be economical to continue with them anymore.
ii. Items which might have been destroyed either by accident or otherwise.
Items such as machines, equipment etc. follows gradual failure mechanism and they deteriorate with time. Such type of failures has the following effect:-
? Increased expenditure in the form of operating costs
? Decrease in the productivity of the equipment and
? Decreases in the value of the equipment i.e. the resale or salvage value.
Items which follow sudden failure mechanism may fail anytime, thus precipitating cost of failure. The cost of failure in some cases may be quite high as compared to the value of the item itself. Sometime sudden failure of an item may cause loss of production and may also account for damaged or faulty products. In some cases failures may involve safety risks to personnel as well. To avoid the cost of sudden failure, the concern should try to predict when such failures are likely to occur and try to replace the item before it actually fails.
Replacement models are thus divided into four categories
A. Replacement of items that deteriorates
• Replacement of items whose maintenance & repair cost increases with time, Ignoring time value of money
• Replacement of items whose maintenance & repair cost increases with time, and time value of money increases with time.
B. Replacement of items that fails suddenly
• Individual replacement
• Group replacement
When machine should replaced:
The following two main situations the machine should be replaced
? When the machine is not working properly.
? Maintenance or operating cost day by day increasing.
Given Case:
Year 1 2 3 4 5 6 7 8
Maintenance cost in 100 250 400 600 900 1250 1600 2000
Year Maintenance cost in Depreciation Total Cost Avg. Total Cost
1 100 6000 6100 6100
2 250 6000 6350 3175
3 400 6000 6750 2250
4 600 6000 7350 11837.5
5 900 6000 8250 1650
6 1250 6000 9500 1583
7 1600 6000 11100 1585.71
8 2000 6000 13100 1637.5
At 6 th year, 1600>1585.71 => Maintenance cost > average cost at the end of 6th year. So,
At the end of the 6th year (or) at the start of the 7th year the Machine should be replaced.
Conclusion:
Unless the machine is being replaced, its maintenance cost further increases such that the return on Investment for that machine will further be reduced. Its maintenance cost exceeds the original cost of the machine one day. Hence it is better to replace the machine by the 7 th year.
8. Using Monte- Carlo Simulation technique estimate the profits and loss for the next ten days of Aditya Birla Nova Company? If the company decides to produces 540 insulator per day, calculate average profit or loss of the company. Comment on efficiency of the technique and suggest improvement over the technique.
Simulation: ‘Simulation' in general terms can be defined as the representation or imitation of a system in its realistic form A technique for reducing or eliminating analytical errors resulting from interferences, using a reference solution sufficiently similar in quantitative composition to the sample solutions to be analyzed that the interferences in the reference and sample solution are equivalent..
Monte Carlo method
Monte Carlo methods are a class of computational algorithms that rely on repeated random sampling to compute their results. Monte Carlo methods are often used when simulating physical and mathematical systems. Because of their reliance on repeated computation and random or pseudo-random numbers, Monte Carlo methods are most suited to calculation by a computer.
A Monte Carlo method is a technique that involves using random numbers and probability to solve problems.
Applications
Monte Carlo simulation methods are especially useful in studying systems with a large number of coupled degrees of freedom, such as liquids, disordered materials, strongly coupled solids, and cellular structures (see cellular Potts model). More broadly, Monte Carlo methods are useful for modeling phenomena with significant uncertainty in inputs, such as the calculation of risk in business (for its use in the insurance industry, see stochastic modeling). A classic use is for the evaluation of definite integrals, particularly multidimensional integrals with complicated boundary conditions.
Monte Carlo methods in finance are often used to calculate the value of companies, to evaluate investments in projects at corporate level or to evaluate financial derivatives. The Monte Carlo method is intended for financial analysts who want to construct stochastic or probabilistic financial models as opposed to the traditional static and deterministic models.
Monte Carlo methods are very important in computational physics, physical chemistry, and related applied fields, and have diverse applications from complicated quantum chromo dynamics calculations to designing heat shields and aerodynamic forms.
Monte Carlo methods have also proven efficient in solving coupled integral differential equations of radiation fields and energy transport, and thus these methods have been used in global illumination computations which produce photorealistic images of virtual 3D models, with applications in video games, architecture, design, computer generated films, special effects in cinema, business, economics and other fields.
Source: http://www.hindu.com/seta/2004/01/22/stories/2004012200311700.htm
http://en.wikipedia.org/wiki/Monte_Carlo_method
Application areas
Areas of application include:
Graphics, particularly for ray tracing; a version of the Metropolis-Hastings algorithm is also used for ray tracing where it is known as Metropolis light transport
Modeling light transport in biological tissue
Monte Carlo methods in finance
Reliability engineering
In simulated annealing for protein structure prediction
In semiconductor device research, to model the transport of current carriers
Environmental science, dealing with contaminant behavior
Monte Carlo method is used in statistical physics; in particular, Monte Carlo molecular modeling as an alternative for computational molecular dynamics.
Search And Rescue and Counter-Pollution. Models used to predict the drift of a life raft or movement of an oil slick at sea.
In Probabilistic design for simulating and understanding the effects of variability
In Physical chemistry, particularly for simulations involving atomic clusters
In computer science
Las Vegas algorithm
LURCH
Computer Go
General Game Playing
Modeling the movement of impurity atoms (or ions) in plasmas in existing and tokomaks (e.g.: DIVIMP).
In experimental particle physics, for designing detectors, understanding their behavior and comparing experimental data to theory
1 2 3 4
Sales Probability Cumulative of 2 Class Intervals
500 0.1 0.1 0-9
510 0.15 0.25 10-24
530 0.2 0.45 25-44
550 0.35 0.8 44-79
570 0.05 0.85 80-84
600 0.15 1 85-99
Random No Simulation of sales Unsold/ Deficit
10 510 40 Unsold
99 600 -50 Deficit
65 550 0
99 600 -50 Deficit
95 600 -50 Deficit
1 500 50
79 550 0
11 510 40 Un sold
16 510 40 Un sold
20 510 40 Un sold
Average no of insulators not sold per day = 40+50+40+40+40 = 210
10 10
= 21 per Day. Total penalty of UN sold= 21*10*1500 = 31500
Average size of unfilled demand = 50+50+50 = 15 per day
10
Total penalty of Deficit = 150* 500 = 7500
No of units produced = (550*10) = 5500
Total revenue – Expenses = (5500* 50000) – (5500*4000)
= 10000*5500 = 5,50,00,000
Profit = Revenue – Expenses- Penalties
=55000000 – (210*1500) – (150*500)
= 55000000-31500-7500
= 55000000-390000
= 54610000.
Total Profit = 54610000.
SECTION-B
Question 1: Bangalore Mahanagar palike estimated to construct a fly over in Ring Road. The construction network is as follows:
Name of the activity 1-2 1-3 1-4 2-5 3-5 4-6 5-6
Optimistic time (in weeks) 1 1 2 1 2 2 3
Most likely time (in weeks) 1 41 2 1 5 5 6
Pessimistic time (in weeks) 7 7 8 1 14 8 15
(a) Draw a network diagram for the project
(b) Find the minimum project completion time
(c) Find total float, free float and independent float for the various activities
(d) Calculate variance and standard deviation of the project length
(e) What is the probability of completion of project?
i. At least 4 weeks earlier than expected time?
ii. No more than 4 week later than expected time?
Introduction to Networking model: Networking models are extensively used in planning, scheduling and controlling complex projects which can be represented in the form of a net-work of various activities & sub-activities. Two of the most important and commonly used networking models are –
1. Critical Path Method (CPM) and
2. Programme Evaluation & Review Technique (PERT).
The basic difference in PERT and CPM is in how the diagrams are drawn. In PERT, events are placed in circles (or rectangles) to emphasize a point in time. Tasks are indicated by the lines connecting the network of events. In CPM the emphasis is on the tasks, which are placed in circles. The circles are then connected with lines to indicate the relationship between the tasks.
PERT is an integrated project management system designed to manage the complexities of major projects by employing a network of interrelated activities, coordinating optimum cost and time criteria. PERT emphasizes the relationship between the time each activity takes, the costs associated with each phase, and the resulting time and cost for the anticipated completion of the entire project.
PERT is the better known and more extensively applied of the two and it involves, finding the time requirements of a given project, & the allocation of scarce resources to complete the project as scheduled i.e.; within the planned, stipulated time and with minimum cost. The pert has following advantages
• One advantage is the three time estimate process, again useful in identifying difficulties as well as more effective interrelated processes.
• Another advantage is the use of what is termed the management-by-exception principle, whereby data accumulated and analyzed by various means can be applied to the planning and execution of a major project
http://www.referenceforbusiness.com/encyclopedia/Per-Pro/Program-Evaluation-and-Review-Technique-PERT.html
Application of Networking model
PERT originally was designed to plan the manufacturing project. But now a days, in many instances, managers have attempted to apply PERT principles to other types of projects, including hospital planning for such issues as costs and social security, educational planning and development, various accounting functions, and even real estate development.
PERT proved to be an ideal technique for one-of-a-kind projects, using a time network analysis to manage personnel, material resources, and financial requirements. The growth of PERT paralleled the rapid expansion in the defense industry and meteoric developments in the space race also.
(a) Draw a network diagram for the project
Network Diagram:
(b) Find minimum project completion time
Name of activity Optimistic time( In Weeks) Most Likely time( In Weeks) Pessimistic time( In Weeks)
1-2 1 1 7
1-3 1 4 7
1-4 2 2 8
2-5 1 1 1
3-5 2 5 14
4-6 2 5 8
5-6 3 6 15
3 5
4
6 7
2 1
Critical path is: 1 -> 3 -> 5 -> 6
4+6+7 = 17
(c) Find total float, free float and independent float for the various activities
Calculation of Total Float, Free Float & Independent Float:
A B C D E F=E-C G=F-Head Event Slank H=G-Tail Slank
Name of activity Early start Early Finish Late Start Late Finish Total Float Free Float Independent Float
1-2 0 2 7 9 9-2=7 7-7=0 0
1-3 0 4 0 4 4-4=0 0-0=0 0
1-4 2 3 9 12 12-3=9 9-9=0 0
2-5 2 3 9 10 10-3=7 7-0=7 0
3-5 4 10 4 10 10-10=0 0-0=0 0
4-6 5 8 12 17 17-8=9 9-0=9 0
5-6 10 17 10 17 17-17=0 0-0=0 0
(d) Calculate variance and standard deviation of the project length
Name of activity Optimistic time Pessimistic time =(P-O/6)2
1-3 1 7 =7-1/6 = 1*1 = 1
3-5 2 14 =14-2/6 = 2*2 = 4
5-6 3 15 =15-3/6 = 2*2 = 4
= 9
Standard Deviation = = 3
(e) What is the probability of completion of project? (i) At least 4 weeks earlier than expected time? (ii) No more than 4 week later than expected time?
Computation of Projection completion:
Formula Used: Z= (X-
a. 17-4 = 13 weeks
b. 17+4= 21 weeks
X=13 =17 =9
a. In 13 Weeks
= (13-17)/9 = 0.44 (Z Table value for 0.44 = 0.3106)
= -0.5-0.3106*100
= 18.94%
X=21 =17 =9
b. In 21 Weeks
= (21-17)/9 = 0.44(Z Table value for 0.44 = 0.3106)
= 0.5+0.3106*100
= 81.6%
Question 2: A salesman located on a city A (Bangalore), decided to travel to city B (Delhi). He knows the distances of alternative routes from city A to city B. He then draws the highway network map as shown below. The city of origin, city A, is city 1. Destination city, city B, is city 10. Other cites through which the salesman will have to pass are given the numbers 2 to 9. The arrows represent routes between cites and distances in km are indicated for each route. The salesman’s problem is to find the shortest route that covers all of the selected cotes from city A to city B.
10 4
9 9
4 9 8 8
2 7 9
3 4
3 10 4
4
8
The traveling salesman problem (TSP) is a problem in discrete or combinatorial optimization. It is a prominent illustration of a class of problems in computational complexity theory which are classified as NP-hard. Mathematical problems related to the traveling salesman problem were treated in the 1800s by the Irish mathematician Sir William Rowan Hamilton and by the British mathematician Thomas Kirkman. The Traveling Salesman Problem is one of the most intensively studied problems in computational mathematics an example optimal solution for visiting all 24,978 cities in Sweden. Tour has length approximately 72,500 kilometers.
Other Applications of Shortest route problems:
2. Route Planning for Unmanned Aerial Vehicles
The CSP methodology to planning a medium-altitude surveillance mission for a UAV. At the planning stage, and perhaps even during a mission, minimum-risk routes must be determined that are feasible with respect to maneuverability and fuel consumption. We imagine a UAV with capabilities similar to the current Northrop Grumman Hunter MQ-5B, but with better communications capabilities and hence longer range. Cruising speed is 120 kilometers per hour (km/hr), climb and dive rate is 200 meters per minute (m/min), and the aircraft’s mission radius, which will be varied, is at least 500 km.
3. Graph-Coloring Problem which is the problem of finding a coloring of a graph so that the number of colors used is minimal and Quadratic Assignment Problem, the problem of assigning n facilities to n locations so that the costs of the assignment are minimized.
http://www.me.utexas.edu/~jensen/ORMM/computation/unit/combin/routing_multiple.html
http://www.ifi.unizh.ch/ailab/teaching/AL00/chap4.pdf
Salesmen should travel from City A to City B through
1->3->7->9->10
Total Distance traveled = 2+3+4+4 = 13 KM
SECTION-C
1) Construct X bar – R chart of following data.
Introduction - X bar- R chart
Statistical process control is an effective method for improving a firm’s quality and productivity. There has been an increased interest in their effective implementation in American industry, brought about by increased competition and improvements in quality in foreign-made products. Many tools may be utilized to gain the desired information on a firm’s quality and productivity. Some of the more commonly used tools are control charts, which are useful in determining any changes in process performance. These include a variety of charts such as p charts, c charts and x bar and R charts.
Definition - X bar Charts: An x bar chart is used to monitor the average value, or mean, of a process over time. For each subgroup, the x bar value is plotted. The upper and lower control limits define the range of inherent variation in the subgroup means when the process is in control.
Definition - R Chart: An R Chart is a control chart that is used to monitor process variation when the variable of interest is a quantitative measure. Now, what does all this mean? These charts will allow us to see any deviations from desired limits within the quality process and, in effect, allow the firm to make necessary adjustments to improve quality.
Source: http://www.isixsigma.com/dictionary/X-Bar_and_R_Charts-151.htm
http://en.wikipedia.org/wiki/XbarR_chart
http://www.qualityamerica.com/knowledgecente/knowctrWhen_to_Use_an__Xbar__R_Chart.htm
Sl No Mean X-bar Range(R)
1 21.50 2.10
2 22.20 1.10
3 22.70 0.40
4 22.50 1.30
5 23.50 1.60
6 21.60 2.50
7 22.10 3.50
8 22.90 3.80
9 22.10 3.70
10 22.90 2.10
11 23.20 1.80
12 22.50 2.30
13 21.80 1.70
14 22.30 2.10
15 21.90 0.50
16 22.10 1.20
17 21.60 1.80
18 21.80 2.40
19 22.20 4.10
20 21.80 3.70
Sum 445.20 43.70
Formula: Control Limit
X Chart R Chart
X double bar=1/k X-Bar (i)
Control Limit=445.20/20=22.26
UCL=X+A(2)R-Bar
LCL= X-A(2)R-Bar
Upper CL= 22.26+(0.58x2.19)
UCL= 23.53
Lower CL= 22.26-(0.58x2.19)
LCL= 20.99 R-Bar=1/k R(i)
Control Limit =43.70/20=2.19
Upper Control Limit =D(4)R-Bar
Lower Control Limit =D(3)R-Bar
UCL=2.11x2.19=4.61035
LCL=0x2.19=0
X-Bar Chart
R Chart
Bibliography:
Books References:
1. Prem kumar Gupta; D.S. Hira, Operations research, S. Chand & company ltd, Fourth edition, 2007.
2. Martand Telsang, Industrial Engineering And Production Management, S. Chand & company ltd, Second edition, 2004.
3. John M. Nicholas, Project Management for Business and Technology: Principles & Practice, Eastern Economy Edition, Second Edition, 2006.
4. Barath K.A., Opertations Research & Project Management: Module Course Book , MSRSAS, Bangalore.
5. Operations research Frederic S. Hiller, Gerald J. Lieberman
6. The people side of project management, Kliem, R. and Ludin, I. Gower, 1992.
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| Author: Vidya 25 May 2008 | Member Level: Diamond Points : 2 | helpful for mba atudents
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