Timesheet Analysis: OEE Calculation and Pareto Analysis Solution

This assignment provides a detailed solution for calculating Overall Equipment Effectiveness (OEE) and conducting a Pareto analysis based on timesheet data from a manufacturing plant specializing in radiators and heat exchangers. The solution calculates daily availability, performance, and quality metrics to determine the OEE for each day of the week, making assumptions where necessary. It then performs a Pareto analysis to identify the most significant losses affecting the plant's efficiency, highlighting areas such as reel changing intervals, breaks between production runs, and material/cutter setups. Finally, it offers practical recommendations for improvement, focusing on minimizing reel changing intervals, covering breaks to ensure continuous operation, and reducing setup times to enhance production speed and overall efficiency. Desklib offers this solution as part of its collection of solved assignments and past papers, providing students with valuable resources for their studies.

Coursework 1 – Timesheet Analysis
Background
The company manufactures a range of specialist radiator and heat exchangers. A Key
component of all products is the metallic core tubing. Every product range contains these
items. The company produces these tubes on a single machine. Due to the criticality of this
equipment it is important to increase its effectiveness. The process itself entails two raw
material types, depending on product (brass or tin). Reels of raw material are loaded to the
rear, and fed through the machine. The raw material is then folded, welded and cut to the
required length, so producing the finished component.
Current situation
The company has designed a data collection chart and instructed operators to collect data.
Currently there is one week’s worth of data collected. No corrective action has been taken.
All breaks are covered (the machine is planned to run continuously)
The shift starts at 8am and finishes at 4pm
A set-up is defined as a change of material or cutters (or both)
Speed is recorded digitally and reported as an average over the shift.
Tasks
1. Calculate the daily availability, performance, quality and OEE figures (state any
assumptions made).
Solution;
Daily availability
Availability is the percentage of the scheduled time that the machine is available to operate.
It is given by operating time divided by the scheduled time.
availability= Operatingtime
scheduled time
The following are the availabilities for the entire week;
Scheduled time = 8 hours from 8.00am to 4.00pm daily hence;
Monday;
availability= 7.58
8

= 0.947
Tuesday;
availability= 7.33
8
= 0.916
Wednesday;
availability= 6.916
8
= 0.865
Thursday;
availability= 5.06
8
= 0.7075
Friday;
availability= 2.25
8
= 0.2813
Performance
This is also known as the processing rate and it is the speed at which the process runs as a
percentage of its expected speed. It is given by;
performance= materials produced × expected materials ( per hour )
operating time
For the entire week;
Assume that materials expected per day are 50 materials, hence the rate per hour is 6.25
materials
Monday;
performance= 25 ×6.25
7.58
= 20.613 %

Tuesday;
performance= 21 ×6.25
7.33
= 17.906 %
Wednesday;
performance= 2 9× 6.25
6.916
= 26.207 %
Thursday;
performance= 16 ×6.25
5.06
= 19.763 %
Friday;
performance= 19× 6.25
2.25
= 52.778 %
Quality
Quality is the number of good materials produced as a percentage of the total materials input.
It is given by Quality= materials produced−faulty materials
materials produced
We assume that the faulty materials are 5 for every single day;
Monday;
Quality= 25−5
25
= 0.8
Tuesday;

Quality= 21−5
21
= 0.762
Wednesday;
Quality= 2 9−5
29
= 0.828
Thursday;
Quality= 16−5
16
= 0.688
Friday;
Quality= 19−5
19
= 0.875
Overall Equipment Effectiveness (OEE)
This is the multiple of availability, performance and quality of an equipment or machine.
It is given by;
OEE= Availability × Performance ×Quality
The OEE for the entire week is;
Monday;
OEE=0.947 ×20.613 × 0.8
= 15.616 %
Tuesday;
OEE=0.9 16 ×17.906 ×0. 762
= 12.498 %
Wednesday;
OEE=0. 865 ×2 6.207 ×0.8 28

= 18.770%
Thursday;
OEE=0. 705× 19.763× 0. 68 8
= 9.585 %
Friday;
OEE=0. 281× 52.778× 0.8 75
= 12.977 %
2. Pareto analyse the losses and make brief recommendations for improvements.
Pareto analysis is a method used to identify possible solutions where many sources of
comparison are available. It is used to estimate the importance of every possible action and
select the most efficient. It also helps to identify the major causes that need to be considered
and addressed to resolve major problems.
A pareto analysis is done by plotting a graph of possible causes of problems against their
cumulative frequency and the percentage frequency. The causes to the left of the graph, with
the highest frequencies are considered more urgent and important to address.
The pareto analysis graph is as shown below;

A line drawn horizontally from the 80% point meets the line at a point which determines the
most urgent causes to the left of that point.
Recommendations
From the pareto analysis above, it is evident that the following need to be addressed;
i. Reel changing intervals should be reduced to minimum possible levels to maximize
on the time of production which will in turn maximize the efficiency of the system.
ii. The breaks between production should be covered to reduce losses and to increase
the performance of the system. The machine is designed to work continuously and
hence its efficiency is reduced by providing breaks in the operation.
iii. Material and cutter setups should be done at reduced intervals. This will help
improve the seed of production and increase efficiency.