Datathon 2020 Kaufland – Optimize Retail Supply Chain
Team Name: Raven Delivery
- Borislav Aymaliev
- Gabriela Vasileva
- Irina Naskinova
- Zainab Lawal
- Python, Pandas, MySQL
- Kaufland dataset
Client – Kaufland
In the innovative era we find ourselves in today,we have the ability to not only optimise cost by predicting consumer behaviour from a reflection from data analytics but also implement machine learning processes to avoid unnecessary costs. Here, to manage storage costs, we will reflect on other factors to develop key functions. These key functions will encourage a reward, in this case, as a cut in cost on the amount spent on storage cost.
Two datasets were provided in the form of two CSV files. The first file is named the master data and the second named sales. The initial file conscious data regarding storage costs, expiration periods, and transportation parameters for the items. Whereas,the latter file contains sales information from over the last year for 100 individual items.
The CSV files were initially imported into Excel to get a full feel of the actual possibilities.
From these files, it is key to solve the costs attributed to the following: handling, transportation, storage and wastage. The manipulation of this data is further discussed in data preparation.
Dataset 1 → Data Type:Master Sheet → DataFormat:CSV
Dataset 2 → Data Type:Sales Sheet → DataFormat:CSV
To develop a better understanding ,we created two new tables namely, stock per day and order per day based on information derived from the master sheet and sales sheet provided by our client, Kauflaud.
Stock per Day
The stock per day table examines the amount of stock available daily.
- On 01/01/2019 the items were at maximum quantity
- No sales were made by the end of first day
Current_date states the actual date the stock check occurs on. Item _id and item_name match up to the uniqueidentifier of the items available, this information has been provided by our client.
The next columns have been created by us based on assumptions stated and calculations developed. The following methodology has been executed in accordance to the case details:
[Insert snapshot of table]
Yesterday_qty describes the stock item available on the shop and in the warehouse from the day before. For simplification, yesterday_qty on the first day has been given to be equal to the maximum stock available on the stock shelves. For example, on 01/01/2019, the quantity for dried basil has been given as 40. Therefore, on the first day, Yesterday_qty is given as
Yesterday_qty = maximum stock
On consequent dates, Yesterday_qty is given by the following equation :
Yesterday_qty = Quantity of items in storage on the previous day+The quantity of items on the shop floor – the number of items sold on the previous day.
End of day quantity
it is crucial to take into consideration the items that are currently in stock on a daily basis. Similarly to yesterday’s quantity, we take take the quantity on the shop floor, quantity in storage and the total sold for each item
Yesterday_qty = Quantity of items in storage on the current day+The quantity of items on the shop floor – the number of items sold on the current day. end_day_item_qty=yesterday_qty – (ordered_qty – sold_qty)
It is necessary to identify the items which have expired as this contributes to the wastage cost and this is incurred on the company. Items that have expired need to be checked and identified on a daily basis. We have been given the time taken for an item to expire as mhd.
Determination of expired items
When an item has been delivered, the date is registered as the delivered date. On a daily basis, the number of days which has passed from the delivery (delivered date) and the current date are compared to the mhd. Therefore, we can deduce, if the difference between these dates is greater than the mhd for each item, the item would be disposed of and contribute to waste cost, otherwise, the item will remain in stock.
As an expression:
If delivery date – current date > mhd , consider item as wastage
otherwise keep to item in stock
Quantity of Items Stored
The quantity of items kept in stock
The order per day shows calculations to determine the number of orders made daily. The calculations are broken down further below:
the order ID, Item ID, and item order quantity were obtained from the xx sheet. When an order is made, it is crucial to fill the pallets to the maximum height before this goes on the truck for transportation. Each pallet has been advised to be filled to the maximum height to incur less transportation charges. Each item on a pallet contributes to the maximum pallet height which has been defined as pallet_fraction_per_item_per_order. pallet_fraction_per_item_per_order can be calculated as follows:
actual_order_qty has been defined as the difference between the number sales per item in a day and the maximum stock available to be kept for each item (max_qty), this can be shown as :
actual_order_qty= max_qty – sales_qty
When an order is made, it is crucial to fill the pallets to the maximum height before this goes on the truck for transportation. Each pallet has been advised to be filled to the maximum height to incur less transportation charges
pallet_fraction_per_item_per_order = actual_order_qty/transport_qty( = z +(w/10^n))) /33
df_sale = pd.read_sql(‘SELECT * FROM datathon_sale limit 10’, con=db_connection)
df_masterdata = pd.read_sql(‘SELECT * FROM datathon_masterdata limit 10’, con=db_connection)
|3||192100||Ice Cream stick||ST||24||5304||40||100||0.002||0.000358||480|
item_id order_qty transport_qty min_stock max_stock
count 10.000000 10.00000 10.000000 10.000000 10.00000
mean 219902.000000 55.20000 4729.200000 15.400000 72.80000
std 57165.041337 107.78662 5570.190517 11.117554 37.57304
min 101900.000000 6.00000 264.000000 6.000000 40.00000
25% 189505.000000 12.00000 1392.000000 10.000000 48.50000
50% 223800.000000 22.00000 2976.000000 11.000000 60.00000
75% 260142.500000 29.50000 5646.000000 16.500000 90.00000
max 293220.000000 360.00000 19440.000000 40.000000 150.00000
item_prio_ storage_cost mhd
count 10.000000 10.000000 10.000000
mean 0.002700 0.000592 376.800000
std 0.001567 0.000158 225.000642
min 0.000000 0.000358 98.000000
25% 0.002000 0.000553 243.000000
50% 0.002000 0.000562 326.500000
75% 0.003750 0.000580 530.250000
max 0.005000 0.000944 821.000000
count 10.0 10.000000
mean 40001260.0 11.200000
std 0.0 5.711587
min 40001260.0 4.000000
25% 40001260.0 7.250000
50% 40001260.0 9.500000
75% 40001260.0 15.000000
max 40001260.0 23.000000
item_id the_date sold_qty
count 10.0 10 10.000000
unique NaN 10 NaN
top NaN 2019-01-03 NaN
freq NaN 1 NaN
mean 40001260.0 NaN 11.200000
std 0.0 NaN 5.711587
min 40001260.0 NaN 4.000000
25% 40001260.0 NaN 7.250000
50% 40001260.0 NaN 9.500000
75% 40001260.0 NaN 15.000000
max 40001260.0 NaN 23.000000
df_order = pd.read_sql(‘SELECT * FROM datathon_order limit 10’, con=db_connection)
order_id item_id ordered_units_qty
count 10.0 1.000000e+01 10.000000
mean 12.0 5.500268e+06 1.400000
std 0.0 1.247951e+07 2.221111
min 12.0 3.462000e+05 0.000000
25% 12.0 6.034950e+05 0.000000
50% 12.0 6.582700e+05 0.500000
75% 12.0 7.840875e+05 1.750000
max 12.0 4.000126e+07 7.000000
group_by_date = df_sale.groupby(by=[‘the_date’])
sold_data_avg = group_by_date.mean()
sold_data_count = group_by_date.count()
2019-01-02 40001260 7
2019-01-03 40001260 7
2019-01-04 40001260 16
2019-01-05 40001260 16
2019-01-06 40001260 8
2019-01-07 40001260 23
2019-01-08 40001260 12
2019-01-09 40001260 8
2019-01-10 40001260 4
40001260 2019-01-07 23
Name: sold_qty, dtype: int64
Given the limited time frame, we decided to go ahead with a simplified Q-learning approach. The learning process consists of the following steps:
- Agent action
This is a binary decision to order an item or not to order an item today. It should be decided for all possible items.
The number of items is 103. Therefore our agent can take 2^103 possible actions. To address this problem, we are considering each item independently on its own and the decision is buy or do not buy.
- “Transportation” and “handling” costs calculation
In any case our agent’s actions incur transportation and handling costs. The function to calculate it should be on a per-item basis, so we can attribute these costs to our agent’s individual actions. Very much like in the case description.
- Update of the current state, given the daily sales
This step is essentially an adjustment to the current state of the stock at the shop. Even though 37025 historical sales records are available in the datathon_sales.csv file, they may be insufficient to complete the training of our agent.
Therefore, our approach is to sample data points for each item from its historical distribution, roughly preserving the frequency of its sales during the initial adjustment of the algorithm.
- “Storage” and approximate “waste” costs calculation
At the end of the day, storage and waste costs need to be calculated. Storage costs are easier, given the fact that we already know what was sold during the day.
Waste cost, on the other hand, needs to be attributed to the actions taken the current day, even though items may expire many days into the future. Best results we expect if we were to perform a time series forecast, but for simplicity we are going to use a simple calculation for our expectation.
- Reward formula, given the calculated waste costs.
The outcome of this formula should provide our agent with information on how they performed for this given day. It is the last step of our feedback loop, from which the agent is to learn the impact of their actions.
In essence, based on this result our agent adjusts the probability of ordering each item in the future. And when the process continues on the next day, they will base their decision on the old information, augmented with the new feedback.
This is a simple illustrative model-free Q-learning demo: 002_Model-free_Qlearning_Demo
Next steps are to apply it to our case data.
Reinforcement learning has been selected as the modelling framework.
The main idea we used was to
We had the following major improvements:
Without any of the improvements we had accuracy of around …%, which was very high to begin with.
Although this produced relatively good results, there are cases in which
In terms of time complexity our solution is O(N * M), where N is the number of , and M is .
We used the training data for evaluation of our model (but not training per se).
The results there proved to be .
We noticed that most errors we get are caused by .
We decided to write a custom logic for
We ran the algorithm on all the test data and created a
**Used Libraries and Technologies**
Python for main logic (Open source, https://www.python.org/)
Sources should be available in GitLab. We decided to use separate scripts for separate tasks, since this allowed us to work on specific tasks (e.g. logic) while other are being executed in the background. The sources are in :
This is a simple illustrative Q-learning demo: 002_Model-free_Qlearning_Demo
I find it really nice that you have tried to follow the case suggestion and implement a Reinforcement Learning algorithm to solve it.
Please, don’t hesitate to continue working on this even after the Datatone and finish the article and actually apply RL to the data set.
Great approach regarding the data understanding and good explanation of variables and derived variables. Sound understanding of business process and caveats. This paper would benefit from general explanation of Reinforcement learning framework such as (1).
Although the team explains how they think the environment and the agent is to act I would like to see the complete solution and how it really does in this task.
Congratulations for the hard work!
1 – https://arxiv.org/pdf/2005.01643.pdf
Thank you for your response and providing this extra read.
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I think it’s great that you tried to solve the case using a Reinforcement Learning algorithm in accordance with the case’s suggestion.
Don’t be afraid to keep working on this even after the Datatone is finished, to complete the article, and to actually apply RL to the data set. iscribble
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