Professional Practices and Business for Engineers
Operations Planning
Assignment 1
Student ID: 2044668
1-18-2021
Contents
PART 1: UNDERSTANDING CUSTOMER SEGMENTS ................................................................ 2
A) Market Characteristics of Segments ........................................................................... 2
Segment A..................................................................................................................... 2
Segment B: .................................................................................................................... 3
Discussion ..................................................................................................................... 4
Summary....................................................................................................................... 5
Likely Household Characteristics: ................................................................................. 5
Feasibility of proposed sizes.......................................................................................... 5
For Segment A............................................................................................................... 6
For Segment B ............................................................................................................... 7
B) Feasible Capacities ..................................................................................................... 8
Part 2: Innovation and Product Planning ............................................................................. 10
a) Introduction to product / system .............................................................................. 10
iii. Given that the development costs of the MVP are £60,000 - what other features do
you propose be developed? Examine the financial implications of including some of
the features that you identified in your product “concept”. What do you propose? . 11
ATAR Model Forecasting ............................................................................................. 12
iv. Is your project proposal worth doing if you take into account the ‘Time Effect of
Money’ (10% discount factor)? ................................................................................... 12
C) Tracking project progress ......................................................................................... 12
Part 3: Production Planning ................................................................................................. 14
A) Workshop operating profitability .............................................................................. 14
i) Contribution Statement for workshop production .................................................. 14
ii) Conduct a Break-Even Analysis for the workshop - show your workings. ............... 14
iii) If expected sales are 6,200 units is there a sufficient Margin of Safety? ................ 15
Part b) ............................................................................................................................. 15
Appendix A .......................................................................................................................... 16
Appendix B ............................................................................................................................ 0
1
PART 1: UNDERSTANDING CUSTOMER SEGMENTS
The market research results on average time spent online per day by different types of UK
households of the segments under consideration are as follows.
Time online per
day (Hours)
Segment A
8.064
11.634
10.192
11.942
9.310
51.142
Sum
Segment B
7.896
8.834
7.644
9.884
9.114
43.372
A) Market Characteristics of Segments
Part (i)
Segment A:
Sum of data points = 51.142 hours
No. of Data Points = N = 5
Mean =
51.142
5
Mean = 10.228 hours
The average online time per day is 10.228 hours for segment A.
Let us calculate the average deviation of the data points in segment A to estimate the
fluctuation in data.
Average Deviation can be calculated as,
Average Deviation =
|10.228−8.064|+|10.228−11.634|+|10.228−10.192|+|10.228−11.942|+|10.228−9.310|
5
2.164+1.406+0.036+1.714+0.918
Average Deviation =
Average Deviation =
5
6.238
5
Average Deviation = 1.248 hours
Data Sets (x)
8.064
11.634
10.192
11.942
9.310
Deviation from Mean
̅ − 𝒙)
(𝒙
2.164
- 1.406
0.036
- 1.714
0.918
̅ − 𝒙)2
(𝒙
4.683
1.977
0.0013
2.937
0.842
10.4403
̅ − 𝒙)𝟐
∑(𝒙
𝒏−𝟏
∑(𝒙
̅ − 𝒙)𝟐
𝒏−𝟏
= 𝟏𝟎.𝟒𝟒𝟎𝟑
𝟒
= 2.610
2
Now, the square root will be taken to find the standard deviation of the data points.
Standard Deviation = √2.610
Standard Deviation = 1.615 hours per day
The range of this segment can be calculated by subtracting the lowest value from the highest.
Range = 11.942 – 8.064
Range = 3.878 hours
For Graphical analysis of the data of segment A, Scatter plot is used.
Segment A
14
12
10
8
6
4
2
0
0
1
2
3
4
5
6
Segment B:
Sum of data points = 43.372 hours
No. of Data Points = N = 5
Mean =
43.372
5
Mean = 8.674 hours
The average online time per day is 8.674 hours for segment B.
Let us calculate the average deviation of the data points in segment A to estimate the
fluctuation in data.
Average Deviation can be calculated as,
Average Deviation =
|8.674−7.896|+|8.674−8.834|+|8.674−7.644|+|8.674−9.884|+|8.674−9.114|
5
0.778+0.160+1.030+1.21+0.44
Average Deviation =
Average Deviation =
5
3.618
5
3
Average Deviation = 0.724 hours
Data Sets (x)
7.896
8.834
7.644
9.884
9.114
̅ − 𝒙)𝟐
∑(𝒙
𝒏−𝟏
̅ − 𝒙)2
(𝒙
Deviation from Mean
̅ − 𝒙)
(𝒙
0.778
0.160
1.030
1.21
0.44
0.605
0.025
1.061
1.464
0.194
3.349
∑(𝒙
̅ − 𝒙)𝟐
𝒏−𝟏
= 𝟑.𝟑𝟒𝟗
𝟒
= 0.837
Now, the square root will be taken to find the standard deviation of the data points.
Standard Deviation = √0.837
Standard Deviation = 0.915 hours per day
The range of this segment can be calculated by subtracting the lowest value from the highest.
Range = 9.884 – 7.644
Range = 2.24 hours
For Graphical analysis of the data of segment B, Scatter plot is used.
Segment B
12
10
8
6
4
2
0
0
1
2
3
4
5
6
Discussion:
The trend of the data points of segment A show a high average usage as compared to segment
B and a higher variability. The higher values of standard deviation and variance show that the
monthly usage of a household in segment A is less predictable as compared to segment B.
By looking at the graphical data from scatter plot, it can be concluded that in segment A, the
data points are more far spread while in that segment B, the data point are closer to each
other. (Devore, 1982) The data of segment B is more reliable and predictable as compared
to that of segment A. While the usage of segment A is more than the usage of segment B.
4
Summary:
Segment A
10.228
1.248
2.610
1.615
Segment B
8.674
0.724
0.837
0.915
Mean
Average Deviation
Variance
Standard Deviation
Likely Household Characteristics:
The possible characteristics of the two market segments are as under.
Segment A:
Segment A has high variability and high usage. If this data is used in a household, the possible
characteristics can be,
1.
2.
3.
4.
5.
6.
The household makes use of smart devices.
The average number of people living in the houses in segment A is large.
Usage of media or television streaming.
The households use video chatting applications.
Usage of peer-to-peer software.
There is high variability so this could be a result of visiting members in the households
of market segment A.
Segment B:
Segment B has low variability and lower usage as compared to A. If this data is used in a
household, the possible characteristics can be,
1. The households in Segment B have internet usage time limits which are followed
making the time spent online per day limited and more predictable.
2. The households use data for smart devices which use same amount of data.
3. The online time spent is still higher which can be a result of online work (Work from
home) by the members of the households of segment B.
Part (ii)
Capacity of a network router = 7.5 hours per day
Occasional peaks = 40% above normal usage = 10.5 hours per day
Probability that online usage for a new household will be below the product capacity (of
normal usage) (Devore, 1982)
Feasibility of proposed sizes
1- High-capacity router
Probability that usage will be greater than high-capacity router limit = 1 - P(Usage<13.4)
Z=
𝑋−𝜇
𝜎
=
13.4−9.451
1.484
=
3.991
1.484
= 2.69
5
Using Z table (Appendix A),
P(Usage<13.4) = 0.99643 = 99.64%
Probability that usage will be greater than high-capacity router limit = 1-0.996 = 0.004 = 0.4%
This shows that the maximum limit of high-capacity router determined by us can fulfill the
need of more than 99% of the users. But it is better to use it only for houses in which medium
capacity router cannot be used. (Ulrich, 2001)
Probability that usage will be greater than the capacity of medium capacity router and less
than high-capacity router = P(Usage<13.4) – P(Usage<8.7)
P(Usage<13.4) – P(Usage<8.7) = 0.99643 - 0.28774 = 0.708 = 70.8%
This shows that 70.8% users will use high-capacity router.
2- Medium Capacity Router
Probability that usage will be greater than the capacity of medium capacity router and less
than high-capacity router = P(Usage<8.7) – P(Usage<6.4)
Z=
𝑋−𝜇
𝜎
=
8.7−9.451
1.484
= −0.751 = - 0.506
1.484
Using Z table (Appendix A),
P(Usage<8.7) = 0.28774 = 28.77%
P(Usage<8.7) – P(Usage<6.4) = 0.28774 - 0.02018 = 0.2675= 26.75%
This shows that 26.75% users will use medium-capacity router.
3- Small Capacity Router
Probability that usage will be less than small-capacity router = P(Usage<6.4)
Z=
𝑋−𝜇
𝜎
=
6.4−9.451
1.484
=
−3.051
1.484
= - 2.05
Using Z table (Appendix A),
P(Usage<6.4) = 0.02018 = 2.02%
This shows that 2.02% users will use small-capacity router.
For Segment A
Data:
Mean = 𝜇 = 10.228
Standard deviation = 𝜎 = 1.615
X = 7.5
P1 = P(usage<7.5) =?
Solution:
𝑋−𝜇
−2.728
Z=
= 7.5−10.228 =
= -1.689
𝜎
1.615
1.615
6
Using Z table (Appendix A)
P1 = P(usage<7.5) = 0.04648 = 4.6%
Now let us calculate the probability that the usage is less than 10.5 hours per day (i.e.,
occasional peak of 40% above normal)
X = 10.5
P2 = P(usage<10.5) =?
Solution:
𝑋−𝜇
0.272
Z=
= 10.5−10.228 =
= 0.168
𝜎
1.615
1.615
Using Z table (Appendix A)
P2 = P(usage<10.5) = 0.56356 = 56.35%
P3 = P(usage>10.5) = 1 – 0.563 = 0.437 = 43.7%
Now, let us find the probability that the usage will be less than 7.5 hours per day and less than
10.5 hours per day, as the router can only handle occasional peaks of 10.5 hours per day. Let
us assume that the router can handle peaks 40% above the normal capacity 5% of the times.
Then let’s see what the probability of usage is between 7.5 and 10.5 hours.
P4 = 0.563 – 0.046 = 0.517 = 51.7%
Conclusion:
The above results show that one router with a capacity of 7.5 hours per day will only be
feasible 4.6% of the times. The other 95.4% of times, the usage will be more than this.
56.35% of times, the usage will be less than 10.5 hours that is the highest usage the router
can handle, and the remaining 43.7% of times, the usage will even cross the occasional peak
value.
We have assumed that the occasional peaks mean that usage should not exceed the peak
value more than 5% of the times, but the probability for this in segment is 51.7%.
The above probability calculations clearly show that this network router or a single network
router with provided specifications will not be feasible for segment A.
For Segment B
Data:
Mean = 𝜇 = 8.674
Standard deviation = 𝜎 = 0.915
X = 7.5
P1 = P(usage<7.5) =?
Solution:
𝑋−𝜇 7.5−8.674 −1.174
Z=
=
=
= -1.283
𝜎
0.915
0.915
7
Using Z table (Appendix A)
P1 = P(usage<7.5) = 0.10027 = 10.03%
Now let us calculate the probability that the usage is less than 10.5 hours per day (i.e.,
occasional peak of 40% above normal)
X = 10.5
P2 = P(usage<10.5) =?
Solution:
𝑋−𝜇
1.826
Z=
= 10.5−8.674 =
= 1.996
𝜎
0.915
0.915
Using Z table (Appendix A)
P2 = P(usage<10.5) = 0.97500 = 97.5%
P3 = P(usage>10.5) = 1 – 0.97500 = 0.025 = 2.5%
Now, let us find the probability that the usage will be less than 7.5 hours per day and less than
10.5 hours per day, as the router can only handle occasional peaks of 10.5 hours per day. Let
us assume that the router can handle peaks 40% above the normal capacity 5% of the times.
Then let’s see what the probability of usage is between 7.5 and 10.5 hours.
P4 = 0.97500 – 0.10027 = 0.875 = 87.5%
Conclusion:
The above results show that one router with a capacity of 7.5 hours per day will only be
feasible 10.03% of the times. The other 89.97% of times, the usage will be more than this.
97.5% of times, the usage will be less than 10.5 hours that is the highest usage the router can
handle, and the remaining 2.5% of times, the usage will even cross the occasional peak value.
We have assumed that the occasional peaks mean that usage should not exceed the peak
value more than 5% of the times, but the probability for this in segment is 87.5%.
The above probability calculations clearly show that this network router or a single network
router with provided specifications will not be feasible for segment B.
B) Feasible Capacities
The firm plans to offer 3 routers for the residential (home) market of Small / Medium / High
sizes - what (normal) capacities (hours/day) do you propose? Explain your rationale.
Usage
7.644
7.896
8.064
8.834
9.114
9.31
9.884
8
Sum
Avg
SD
Range
10.192
11.634
11.942
94.514
9.4514
1.484
4.298
Let us combine the usage of both segments to decide the capacities of router sizes that the
company should offer.
The range of values in both segments is 4.298. let us calculate the normal capacities for a
range of values + 0.85 hours per day from the given range so that the router is still usable if
the usage range of households increase by 10%.
New Range = 4.298 + 0.43 = 4.728 hours per day
The difference in normal capacities = 4.728 = 1.576 ≈ 1.5 hours
Maximum Normal Value for different sizes can be calculated as follows,
Normal Capacity of high-capacity router
11.942 + 1.5 =
13.4 hours per day
Normal Capacity of medium-capacity router
13.442 – 4.728 =
8.7 hours per day
Normal Capacity of small-capacity router
8.714 – 4.728 (1⁄2) =
6.4 hours per day
9
Part 2: Innovation and Product Planning
a) Introduction to product / system
An AI Based device, “Smart Home Device” is proposed for the mentioned needs. This will be
an AI software-based device which will be installed in the house of customer with major
appliances. It will come with sensors and optional manual setting for timer for different
devices. It will also be used to control the intensity of light, speed of appliances and
temperature of air conditioning and central heating systems. This device will be useful in
optimizing the energy usage and remove the wastage.
i. Key features and benefits for 3 market segments
S.No
Market Segment
I.
Early Adopters
Key Features
Artificial intelligence based selflearning device
Benefits
Adjusting controls
according to each home
Voice commands
Easy accessibility
Compatible with google nest
Energy usage optimization
Environment friendly
II.
Greens
CO2 Tracker
Lighting control
HVAC Control
III.
Convenient Seekers
Access control
Control of home appliances
Most widely used by early
adopters of smart home
technology
Lower carbon emissions
and higher savings with
less wastage
The device does not cause
any harm to environment
and helps in reducing
power wastage
The homeowners can
know how much they
have saved on carbon
emissions and can set
their goals with alerts.
Optimal use of day light
and power saving without
compromising on
illumination levels
No need to control HVAC
systems manually they will
be controlled either by
voice commands or
automatic AI controls
Biometric and nonbiometric access control
personalized person to
person according to their
preferences
Automatic control of
home appliances will help
10
Multiple language options
in managing daily tasks
more conveniently
Accessible to people who
speak multiple languages
Security control
Higher security of home
Smart locks
Mistake proofing and
higher security level
ii.
For the MVP, two kinds of features will be removed from the product,
1. Features that provide diversity to the product such as multiple language options etc.
these options will be provided when the product passes the market testing phase and
is set to launch in the market for a diverse set of users.
2. The features that do not require testing and are similar to one another. For example
only management of light intensity will be introduced initially while in actual product,
power management of all major home devices will be introduced to optimize power
usage.
The new set of features for MVP is described as under,
S.No
Market Segment
I.
Early Adopters
II.
Greens
III.
Convenience
Seekers
Key Features
Artificial intelligence based selflearning device
Lighting control
Control of home appliances
Smart locks
Benefits
Adjusting controls
according to each home
Optimal use of day light
and power saving without
compromising on
illumination levels
Automatic control of
home appliances will help
in managing daily tasks
more conveniently
Mistake proofing and
higher security level
iii.
The major cost identified in the smart home device is the integration of the devices. The
proposed product is different from the available smart home devices because it controls the
appliances that are not operated by a software such as lighting control and power usage
control in other devices such as fans. For this, sensors will be mounted, and relays will manage
the output and operating levels of devices that are not already smart.
Another major cost is the development of AI software. This software will remember the
choices of user and manage the home using machine learning.
It is proposed to develop an integration solution and the software first with the features
selected for MVP and then add the remaining features after testing.
11
There are 4 features out of which let us assume one feature i.e., integration of home
appliances will take 4 developers working 5 days a week.
Development of software will take 3 developers working 4 days a week.
Lighting and locks control will be handled by one team of 3 developers working 3 days a week.
The work will be completed in 10 weeks.
4 x 5 x 10 x 150 = £30,000
3 x 4 x 10 x 150 = £18,000
3 x 3 x 10 x 150 = £13,500
Total Development Cost of MVP = 30,000 + 18000 + 13,500 = £61,500
If an MVP is developed with the selected features, then the product will be feasible to launch
in the market for testing. More features cannot be added due budget limitations and those
features are easier to manage in product and already tested in other products.
ATAR Model Forecasting
ATAR Model Forecasting (How the atar model forecasting works, 2019) is done for the
proposed concept of product. This is used to find the feasibility of product. The workings of
model are attached in appendix B.
The ATAR Model shows that the product will be profitable after 3 years. The NPV without
discount factor is £294,971
iv.
The NPV calculations if we take the discount factor into consideration show that the overall
profit after 3 years will be greater than NPV with 10% discount factor. Therefore, the project
proposal is feasible.
C) Tracking project progress
During the development phase, following steps will be taken to track the progress of project.
(Institute, 1996)
1) Project Outline
As the first Step a project outline will be formed in order to roughly know all the steps
included in the project. For this step, the whole project team including technical
experts will be involved.
2) Estimating Resources
The resources required will be estimated and allocated to each department. The
resources will be released for each milestone separately unless there is a special
requirement.
3) Product quality assurance goals
The project team will set the key performance indicators for the project and then
quality will be ensured. The standards for assurance of product quality will also be
determined.
4) Milestones
Project will be divided into milestones.
5) Estimation of activity durations and Deadlines
The activities included in each milestone will be determined by individual department
heads and their durations and deadlines will be set after coordinating with all the
departments according to the project duration.
12
6) Project Tracking Visualization
A project management software such as MS project or Primavera will be used to
manage the project progress and get a better and easier understanding by
visualization tools.
7) Weekly Meetings and follow ups
Weekly follow up and progress meetings will be conducted in which progress and any
difficulties will be discussed. An online platform will also be used for the project team
where any issues faced in an activity will be communicated immediately with the
project team and project manager to be resolved as soon as possible.
13
Part 3: Production Planning
A) Workshop operating profitability:
Price = £400
Variable costs = 0.5 (Revenues)
Annual Capacity = 20,000 Units
Annual Overheads = £680,000
Overheads per unit = £680,000/20,00 = £34
i) Contribution Statement for workshop production
Smart Home Systems
Contribution income statement
For 1000 units
Revenue for 1000 units = 1000 x 400 =
Variable costs = 0.5 (Revenue) =
Contribution margin =
Overheads = 1000 x £34 =
Operating income
Smart Home Systems
Contribution income statement
For 5000 units
Revenue for 5000 units = 5000 x 400 =
Variable costs = 0.5 (Revenue) =
Contribution margin =
Overheads = 5000 x £34 =
Operating income
£400,000
(£200,000)
£200,000
(£34,000)
£166,000
£2,000,000
(£1,000,000)
£1,000,000
(170,000)
£830,000
Smart Home Systems
Contribution income statement
For Multiples of 5000 units
Revenue for 5000X units = 5000X x 400 =
£2,000,000X
Variable costs = 0.5 (Revenue) =
(£1,000,000)X
Contribution margin =
£1,000,000X
Overheads = 5000X x £34 =
(£170,000)X
Operating income
£830,000X
*X is the multiple of 5000 units.
(Kim, 2012)
ii) Conduct a Break-Even Analysis for the workshop - show your workings.
Break Even Analysis for Production
𝐹𝑖𝑥𝑒𝑑 𝐶𝑜𝑠𝑡𝑠
Break Even Quantity = 𝑆𝑎𝑙𝑒𝑠 𝑃𝑟𝑖𝑐𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡−𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
14
£680,000
Break even quantity = £400− £200 = 3400 Units
The break even point for the workshop is 3400 units. The revenue and break even analysis is
shown in graph below.
iii) If expected sales are 6,200 units is there a sufficient Margin of Safety?
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑆𝑎𝑙𝑒𝑠−𝐵𝑟𝑒𝑎𝑘 𝐸𝑣𝑒𝑛 𝑃𝑜𝑖𝑛𝑡
Margin of safety =
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑠𝑎𝑙𝑒𝑠
Margin of Safety =
6200−3400
6200
x100
𝑥 100
Margin of Safety = 45.16%
The margin of safety is 45.16%. this is a sufficient margin of safety because, generally, a
margin of safety above 10% is considered to be good.
Part b)
Contribution Margin = Fixed + Net Income
The current contribution margin ratio is 50% and net annual profit is £3,320,000. If it is
reduced to one third i.e., 16.6%, then the net annual profit will be £648,000. (Kim, 2012)
The percent decrease in profit is 80.48%. This is reduction in profit will not liquidate the
company, but it is great enough to hamper the expansion plans of the company. This
reduction in profit might also affect the future price of product as well as the future or
planned investments of the company.
15
Appendix A
16
17
Appendix B
ATAR Model
ATAR Forecasting Model for Smart Home
Device
Smart Home Device
Yr 0
Size of the target
market (buying
units)
Annual Growth Rate
Yr 1
Yr 2
Yr 3
50
65
85
30%
ATAR VARIABLES
Awareness
10%
15%
25%
Trial rate
30%
45%
30%
Availability
10%
15%
20%
Repeat
10%
10%
10%
50
35
15
5
4
2
80%
75%
5
8
7
10.00%
11.54%
8.38
%
45
32
14
5
8
7
50
39
21
1
1
1
5000.00
5000.00
5500.00
4000.00
3800.00
3000.00
SALES + CUSTOMERS
Trial Customers
New 'Repeat'
Customers
Loyalty rate
Total 'Repeat'
Customers (customer
base)
Brand Penetration
(%
of market)
Sales to 'one-off'
customers
Sales to repeat
customers
TOTAL SALES - units
PRICE + VOLUME
ESTIMATES
Avg purchase
quantity (yr)
Price per unit, after
discounts(to
direct customer,
could be the
retailer)
Cost per unit
Margin per unit
1000.0
1200.0
2500.00
0
Gross Margin %
0
20.0%
24.0%
45.5
%
TOTAL
SALES
REVENUE
Gross margin
Initial investment
(R+D, pre-launch,
MR)
250,00
0
195,000
113,438
50,000
46,800
51,563
100,00
0
75,000
75,000
-193,333
-50,000
-28,200
-23,438
-193,333
-243,333
-271,533
-294,971
193,333
Promotional Spend
Profit contribution after
expenses
Cumulative profit
contribution
CANNIBALIZATION (if
applicable)
% Sales of New Product
from Existing Product
Estimated Margin per
Unit of Existing Product
20%
750.00
Lost Unit Sales of
Existing Product
Lost Gross Margin on
Existing Product
Profit Contribution
(after cannibalization)
Cumulative Profit
Contribution
0
0
0
0
0
0
-193,333
-50,000
-28,200
-23,438
-193,333
-243,333
-271,533
-294,971
KEY FINANCIAL
METRICS
Net Present Value
(NPV)
Internal Rate of Return
Return on Marketing
Investment
(IRR)
#NUM!
(ROMI)
#REF!
Overall profit
(In the 3 years)
Years to payback
Discount Rate (if
applicable, or set to 0%)
$
$
(279,702)
(294,971)
0
10%
1