45
For Slideshare Case: http://books.google.co.in/books?id=Dh19GaFCzrIC&printsec=frontc over&source=gbs_summary_r&cad=0#PPA63,M1

Case Study for Plant Layout :: A modern analysis

Embed Size (px)

DESCRIPTION

A studied analysis into a case calling for optimized plant layouts.

Citation preview

Page 1: Case Study for Plant Layout :: A modern analysis

For Slideshare

Case: http://books.google.co.in/books?id=Dh19GaFCzrIC&printsec=frontcover&source=gbs_summary_r&cad=0#PPA63,M1

Page 2: Case Study for Plant Layout :: A modern analysis

Established in 1949 Core business being manufacturing of

rotating machinery Acquired general engineering plant in 1989 New products to be manufactured Mr. Lamba appointed to develop a new

layout

Page 3: Case Study for Plant Layout :: A modern analysis

Mr.Lamba found out that 6000 components were required in final products out of which 5000 components were to be fabricated on Indus facilities

The approach followed was Gathering reasonably accurate forecast figures Using stratified random sample (4%) for the purpose of

evaluation Using Schnieder’s method for developing alternate layout Getting the proposal checked by production department

Used data to create routing table, a load matrix and adistance matrix chart

Page 4: Case Study for Plant Layout :: A modern analysis
Page 5: Case Study for Plant Layout :: A modern analysis
Page 6: Case Study for Plant Layout :: A modern analysis

Using Table 2 and Table 3, a load distance matrix was developed and

Schnieder’s method was applied to it.

Page 7: Case Study for Plant Layout :: A modern analysis

All the primary operation centers (handling maximum load) are placed as close to raw material stores as possible

Similarly, the secondary operation departments are placed close to primary operation departments

Using this method, the existing layout was transformed into a new layout

Page 8: Case Study for Plant Layout :: A modern analysis
Page 9: Case Study for Plant Layout :: A modern analysis
Page 10: Case Study for Plant Layout :: A modern analysis

The plant’s efficiency improved from 43% to 85.4% The moves distance per time period for path

reduced from 2803.8 to 1403.09 Material handling efficiency reduced from 1023005

unit-load move feet to 517210 unit-load move feet Cost of material handling came down by 50%

The factory plant and production personnel pointed out

that the location of the mechanical assembly andassembly erection departments didn’t allow smoothmovement of painted products.

Page 11: Case Study for Plant Layout :: A modern analysis
Page 12: Case Study for Plant Layout :: A modern analysis

Time constraint: Only six weeks for making the decision

Information constraint: Lack of information from Product division and Process planning departments

Sampling: Stratified data was assumed to be correct

Page 13: Case Study for Plant Layout :: A modern analysis

Factors Considered:• Load and Cubic space utilization• Distance

Factors not Considered:• Safety factors• Working condition for workers  • Cost• Flexibility • Concern of other departments • Customized orders• Optimum usage of floor space

Page 14: Case Study for Plant Layout :: A modern analysis
Page 15: Case Study for Plant Layout :: A modern analysis

Schnieder’s Method has both pros and cons

Pros: Ensures smooth flow of materials through different

operations The high preference given to proximity between primary

operations and raw materials reduced the to-fro movement

Cons: Approach focused at minimizing material handling alone Does not account the costs for shifting from the existing

layout

Page 16: Case Study for Plant Layout :: A modern analysis
Page 17: Case Study for Plant Layout :: A modern analysis

Total material handling effort (TMHE) - single most important and frequently used criterion for layout planning

We look at an alternate layout that is based on a heuristic method that could reduce the time required to bring out a product

Tried to create a “hybrid” layout which looks to introduce a hint of assembly line in a process-based layout

Reason: Continuous lines are usually preferred when time is a factor, since they act to reduce the time spent on a particular WIP when compared to process layouts

Page 18: Case Study for Plant Layout :: A modern analysis

Data regarding sub-divisions of factors (like product mix, market accessibility, etc) is not available, hence ignored

Main factory has not been included in calculations

Assumed portability of units, in line with Lamba’s approach

Page 19: Case Study for Plant Layout :: A modern analysis

Looked at the most frequent and repetitive sequences of departments in the material flow

Arrived at an arrangement that makes the WIP flow between most of them akin to an assembly line, without much back-and-forth between departments

Page 20: Case Study for Plant Layout :: A modern analysis

To →

From ↓

RMS M/c HT Process Shop

Project Stores

Welding Total

RMS 0 7 0 0 0 1 8

M/c 0 0 2 2 3 0 7

HT 0 0 0 7 0 0 7

Process Shop

0 0 0 0 12 0 12

Project Stores

0 0 0 0 0 0 0

Welding 0 0 0 1 0 0 1

Total 0 7 2 10 15 1

Page 21: Case Study for Plant Layout :: A modern analysis

Preceding table based on Table 5.1 in given sample data

A pattern is seen along the diagonal (in red), that enables us to make an arrangement that is in a continuous chain

Important to keep in mind not only the frequencies, but also the loads that are transferred between the departments

Keeping this in mind, arrived at an alternate layout that is continuous and yet would not involve carrying heavy loads too far

Had to resort to some trial-and-error to get there

Page 22: Case Study for Plant Layout :: A modern analysis

The pseudo “line”

Page 23: Case Study for Plant Layout :: A modern analysis

Pro:◦ Objection raised by production personnel about

Mechanical Assembly and Assembly Erection departments may be mitigated to a large extent

Con:◦ Load-distance efficiency is less (64%) as

compared to Lamba’s method (85%)

Page 24: Case Study for Plant Layout :: A modern analysis

Minimization of total space occupied, by area (square footage)

If area is considered, problem is analogous to sheet-cutting problem (pieces of different shapes and sizes need to be cut from a single sheet while minimizing the sheet area used)

Page 25: Case Study for Plant Layout :: A modern analysis

Qualitative method that considers five key factors:

◦ Product (P): What is to be produced?

◦ Quantity (Q): How much of each item will be made?

◦ Routing (R): How will each item be produced?

◦ Supporting services (S): What support will be required for production?

◦ Time (T): When will each item be produced?

Page 26: Case Study for Plant Layout :: A modern analysis

SLP procedures

Page 27: Case Study for Plant Layout :: A modern analysis

Some well known computer programs that use algorithms based on heuristic/qualitative data to optimize layouts are ALDEP (Automated Layout Design Program) and CORELAP (Computerized Relationship Layout Planning).

Page 28: Case Study for Plant Layout :: A modern analysis

Gives a ‘relative importance’ of parameters with each other

Based on the premise that not all factors are equally important, while designing a layout

Factors can be divided into sub-factors and relative weights can be calculated, however not performed here due to lack of data

Page 29: Case Study for Plant Layout :: A modern analysis

Step 1: Formation of AHP initial matrix, based on investigator’s judgment

Step 2: Calculation of proportionate worth of criteria. This is the result

Step 3: Verify result. Multiply initial matrix by average worth

Step 4: Divide final matrix of step 4 by average worth to get l

Consistency Index (CI) = (lavg – n) / (n – 1), where n is the order of the matrix.

Consistency Ratio (CR) = CI/RI (Random Index) If CR << 0.1, then results can be held valid

Page 30: Case Study for Plant Layout :: A modern analysis

Initial Matrix:

  No. of items Volume Distance Weight

No. if items 1 0.5 0.5 0.5

Volume 5 1 0.5 0.8

Distance 7 5 1 1

Weight 7 3 1 1

Page 31: Case Study for Plant Layout :: A modern analysis

  No. of items Volume Distance Weight Avg Worth

No. if items 0.05 0.05 0.15 0.15 0.10

Volume 0.25 0.10 0.15 0.25 0.19

Distance 0.35 0.50 0.35 0.30 0.37

Weight 0.35 0.35 0.35 0.30 0.33

Page 32: Case Study for Plant Layout :: A modern analysis

  No. of items Volume Distance Weight Avg Worth

No. if items 0.05 0.05 0.15 0.15 0.10  0.1195

Volume 0.25 0.10 0.15 0.25 x 0.19

= 0.1820

Distance 0.35 0.50 0.35 0.30 0.37  0.3585

Weight 0.35 0.35 0.35 0.30 0.33  0.3300

Page 33: Case Study for Plant Layout :: A modern analysis

lavg = (2.295+0.95+0.96+1) / 4 = 1.026

Consistency Index (CI) = (lavg – n) / (n – 1), where n is the order of the matrix.CI = (1.026 – 4)/3 = -0.9913

Consistency Ratio CR = CI / RI = -0.9913 / 0.90 = -1.101

Since -1.101 << + 0.10, the results are acceptable.

Calculation of l

Page 34: Case Study for Plant Layout :: A modern analysis
Page 35: Case Study for Plant Layout :: A modern analysis
Page 36: Case Study for Plant Layout :: A modern analysis

Possible Implications

Since buildings cannot be altered, no extra construction or demolishing costs involved

Size of the buildings is immaterial for location of the departments

The process of developing a new layout can still be complicated if we have length and breadth constraints for the blocks

TMHE is the only criteria under consideration in the current method while other factors like minimization of moves,etc. can also be considered

Page 37: Case Study for Plant Layout :: A modern analysis
Page 38: Case Study for Plant Layout :: A modern analysis

‘Not properly located to allow painted products to move out smoothly’ can signify that movement is obstructed due to space constraint or due to narrow exit entrances

The Mechanical and Assembly erection departments which are farther away from the road and aisle can be located much closer

Also check if the material handling equipment is proper and allows smooth flow

Page 39: Case Study for Plant Layout :: A modern analysis
Page 40: Case Study for Plant Layout :: A modern analysis

Assumption

1. No volume restriction, only weight restriction (2000 kg per truck)2. Trucks are considered as the only material movement equipment3. Load is considered in tons4. Main Factory is situated as per layout given below5. The jth department has Nj machines which work simultaneously and

take Tj time to finish one product6. Speed of a truck is considered to be constant at “v” feet / minute7. Dedicated trucks are provided between the nodes and the number of

trucks moving between node i and node j are represented by Xi

8. The main criteria for deciding the number of trucks is the total cost of acquiring the trucks, which is calculated by multiplying the total number of trucks in to the cost of one truck (K). The total cost is to be minimized

Page 41: Case Study for Plant Layout :: A modern analysis

Machine Assembly

Machine Shop

Welding Shop

Process Shop

Project Store

Assembly Errection

HT Plant

RMS

Blade

Plastic

Main Factory

Page 42: Case Study for Plant Layout :: A modern analysis

Fig 2 Load – Hop Distance Map

Page 43: Case Study for Plant Layout :: A modern analysis

Objective Function:

Minimize Z = (X12+X13+X14+X15+X16+X17+X24+X26+X27+X29+X32+X36+X37+X39+X46+X47

+X49+X48+X52+X59+X63+X67+X78+X79+X73+X76+X98+X9,10+X10,8) * K, it should be minimized  

K = Cost of a power vehicle

Constraints Number of vehicles running from ith node to jth node >= (( Time of round trip from ith node to jth node / Time of operation of machine at jth node ) * number of machines at jth node) Example: 

X12 >= ((4d/v)/T2)*N2

  Where X12 = Number of vehicles running from 1st node to 2nd

noded = 1 hop distancev = speed of power trucksT2= Time of operation of a machine at 2nd nodeN2 = Number of machines at 2nd node

Page 44: Case Study for Plant Layout :: A modern analysis

NOTE:

Also we need to consider the recurring cost of truck operations over a long period of time. X12 trucks are moving in between node 1 and 2 which will cost us P= X12*K for acquiring the trucks. Now for example say cost of moving a truck for 1 hop costs C. So total cost for X12 trucks will be (2*(73/X12)*C). So if we consider M period as the life time of the truck, then total recurring cost for X12 will be R= (2*(73/X12)*C)*M. So after M period 

R <= P

Page 45: Case Study for Plant Layout :: A modern analysis

Thank you! Questions at [email protected] please