Office Rental: Selecting the Best Strategy
The completion from studies in IT and business, two friends prefer on starting a consultancy firm. However, just like any business, they need an office space but as it is very expensive and this may affect their business. They plan defining three strategies that can give them an idea to start up a business.
First Strategy – Rent an office at a prime location, where many customers are prevalent.
|
Favourable Market |
Unfavourable Market |
|
Good profit for the business; despite the costly office space Net profit of $20,000 (2 years) |
Less business, expecting less profit Lose $16,000. |
Second Strategy – Rent an office at a suburb, which is less expensive
|
Favourable Market |
Unfavourable Market |
|
Comparatively good profit for the business Net profit of $15,000 |
Less business, expecting less profit Lose $6,000. |
Third Strategy – Not set up the business
|
Favourable Market |
Unfavourable Market |
|
No business at all Profit/ Loss $0 |
No business at all Profit/ Loss $0. |
Lastly, both friends have different outlook as Ilya is risk taker and Gregor is risk averser.
Being a risk taker, she would prefer strategy 1 and is optimistic towards taking this business to a new level. She would want the profit to maximize out of the three strategies.
|
LIlya Decision |
|||
|
Favourable |
Unfavourable |
Optimum Profit / Loss |
|
|
S1 |
$20,000 |
$16,000 |
$20,000 |
|
S2 |
$15,000 |
$6,000 |
$15,000 |
|
S3 |
$0 |
$0 |
$0 |
Being a risk averse, she would prefer strategy 3 as she is not interested in setting up the business. She would want the profit to minimize out of the three strategies.
|
Gregor Decision |
|||
|
Favourable |
Unfavourable |
Optimum Profit / Loss |
|
|
S1 |
$20,000 |
$16,000 |
$16,000 |
|
S2 |
$15,000 |
$6,000 |
$6,000 |
|
S3 |
$0 |
$0 |
$0 |
As per the question, the chance of favourable market is 55% and that of unfavourable market is 45%. However, the greater expected values can be achieved from strategy 2
|
Expected Values |
|||
|
Favourable |
Unfavourable |
Expected Value |
|
|
S1 |
è20000*0.55 = $11,000 |
è16000*0.45 = $7,200 |
$3,800 |
|
S2 |
è15000*0.55 = $8,250 |
è6000*0.45 = $2,700 |
$5,550 |
|
S3 |
$0 |
$0 |
$0 |
(For expected values: please refer Problem 1 of excel sheet)
With the value of P (for 0 ≤ P ≤ 1), the returns for strategy 1 and strategy 2 is depicted through the plot below.
- The expected return of Strategy 1 is greater than any strategy. Hence, this is when the probability is higher than 0.67 in the favourable market.
èStrategy 1 (0.67 ≤ P ≤ 1)
- The expected return of Strategy 2 is greater than others. Hence, this is when the probability is between 0.29 and 0.66 in the favourable market.
èStrategy 2 (0.29 ≤ P ≤ 0.66)
- The expected return of Strategy 3 is low such that other strategy loses. Hence, this is when the probability is below 0.28 in the favourable market.
èStrategy 3 (0≤ P ≤ 0.28)
Here, TV ads = x1; Radio ads= x2; Billboard ads =x3; Newspaper ads =x4.
Objective function èMinimize Z = 960 x1 + 480 x2 + 600 x3 + 120 x4
Subject to the constraints:
x1 ≤ 10, x2 ≤ 10, x3 ≤ 10, x4 ≤ 10
x1 ≥ 10, x2 ≥ 10
x1 + x2 ≥ 6
960 x1 – 600 x3 – 120 x4 ≥ 0
Also, non-negativity x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0
- Max no. reached by Jim every week
- 6 TV ads and 6 Radio ads are a must.
Part a
Based on the result of simulation, total inventory cost for best case is $ 53,460.00
Worst case is $ 71,000.00, which is the highest cost and the average cost is $ 59,552.00 on the inventory policy.
Part b (i): Re-order point 3; re-order quantity 3
Best case: $ 33,620.00
Worst case: $ 40,580.00
Average case scenario: $ 37,520.00
Part b (ii): Re-order point 7; re-order quantity 7
Best case: $ 83,100.00
Worst case: $ 99,460.00
Average case scenario: $ 92,516.00
Part c
Advertising Budget Allocation for Big Bet Sports Betting Agency
As per the results, the preferred policy to Max would be “Re-order point 3; re-order quantity 3”, this is because average cost incurred is less (in terms of inventory policy) as compared to “Re-order point 7; re-order quantity 7.”
Part 1
Model 1
Predicting the selling price by using area of house
SP = – 34301.5987 + 62.96*Area_of_house
“Leave one out cross validation error = 18153.6193”
Coefficient of determination = 0.7952 ^ 2 = 0.6323. This suggests that 63.23% variation in the model can be explained by area of the house.
The 10 fold cross validation, the errors like training error (relative absolute+ root relative) squared are higher than test mode of 15-fold cross-validation
Figure: “10-fold cross-validation”
Figure: “15-fold cross-validation”
Hence, Area = 2000 ft2, “SP = – 34301.5987 + (62.96*2000) = 91618.4”
Model 2
Predicting the selling price by using no. of bedrooms
SP= 648.6487 + 35168.9189*No._of_Bedrooms
“Leave one out cross validation error = 24313.0086”
Coefficient of determination = 0.5047 ^ 2 = 0.2547. This suggests that 25.47% variation in the model can be explained by no. of bedrooms.
Figure: “10-fold cross-validation”
Figure: “15-fold cross-validation”
Hence, no. of bedrooms = 3, “SP = 648.6487 + (35168.9189 * 3) = 106155”
Model 3
Predicting the selling price by using age as an input
SP= 141448.2518 – 2256.7296*Age
“Leave one out cross validation error = 24313.0086”
Coefficient of determination = 0.8629 ^ 2 = 0.7446. This suggests that 74.46% variation in the model can be explained by age of the house.
Figure: “10-fold cross-validation”
Figure: “15-fold cross-validation”
Hence, age of the house = 24, “SP = 141448.2518 – 2256.7296*24 = 87286.7”
Thus, out of the models, the most valid prediction is given by the third model with independent variable as age of the house.
Part 2
Area and Bedrooms:
Predicting the selling price by using area and no. of bedrooms
SP= -26129.5 + 76.1268*Area_of_house -12403.1*no._of_bedrooms
Coefficient of determination = 0.8616 ^ 2 = 0.7423. This suggests that 74.23% variation in the model can be explained by area and no. of bedrooms.
Figure: “10-fold cross-validation”
Figure: “15-fold cross-validation”
Area and Age of house:
Predicting the selling price by using area and age of house
SP= 69793.9387 + 27.0743*Area -1554.9387*Age_of_house
Coefficient of determination = 0.8798^ 2 = 0.774. This suggests that 77.4% variation in the model can be explained by area and age of house.
Figure: “10-fold cross-validation”
Figure: “15-fold cross-validation”
Age of house and Bedrooms:
Predicting the selling price by using age of house and bedrooms
SP= 99495.77 + 12389*no._of_bedrooms -1985.53*age_of_house
Coefficient of determination = 0.9309^ 2 = 0.8665. This suggests that 86.65% variation in the model can be explained by no. of bedrooms and age of house.
Figure: “10-fold cross-validation”
Figure: “15-fold cross-validation”
Multiple regressions= Age, Area and Bedrooms:
Predicting the selling price by using age of house, area and bedrooms
SP= 70181.01 + 25.1505*Area_of_house -1574.39*age_of_house + 1389.257*no._of_bedrooms
Coefficient of determination = 0.9407^ 2 = 0.8848. This suggests that 88.48% variation in the model can be explained by no. of bedrooms, are of house and age of house.
Figure: “10-fold cross-validation”
Figure: “15-fold cross-validation”
The multiple regression models make clear the variability in the model. It constitutes to be the best model as compared in the first part.
Predicting the price of houses using MLP’s
r = 0.9399.
Hence, coefficient of variation is 0.9396. The variation explained in this model is much higher than any earlier models considered. Thus, the explanation of the variability in the selling prices is a lot higher than all the models developed previously. Also, the less of training error is opt to be 26.81%. Hoever, MLP’s is a better forecast model that is raised than the regular regression model.
Table: “MLP with 1-hidden layer”
As there is increase in hidden layers, it constitutes to a dcrease in the the accuracy of the model.
Table: “MLP with 2-hidden layers”
Preliminary questions
- Dataset contains 17 attributes or features.
- 7 numeric attributes are present.
- 10 categorical (binary) variables are included.
- Examples that data contain is 4521.
- Class variable in the model = y
- As per the model, only 2 possible values can be taken.
- Two values – “yes and no”. “Yes” for account will be opened and “No” for no account to be opened.
Task 1
I. Logistic Regression
Part a: Confusion Matrix
Part b: ROC Curve
Part c: Area under ROC Curve
Part a: Confusion Matrix
Part b: ROC Curve
Part c: Area under ROC Curve
Lift Chart for Logistic Regression Model
Lift Chart for Naïve Bayes Model
Conclusion
The estimation on ROC on two is slightly different such as for Naïve Bayes Model (0.845) and h the logistic regression model (0.888). In this scenario, ROC curve is higher for the Logistic Classifier. However, Bayes Classifier is not superior in comparison to Logistic Classifier.
Even when compared with lift chart Bayes Classifier is not superior in comparison to Logistic Classifier. As a result, “Logistic classifier is a better classifier than Naïve Bayes classifier.”
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