Product Optimization Exercise - IT Computer Science Assignment help

Assignment Task

Questions

1. Using the compensatory rule with logit adjustment: Compute and report our candidate's share, price, margin and expected profit per person under the "proposed market scenario" given above. This question is a strict subset of the next question. If you have done the next question successfully, then all you need to do for this question is report the numbers for Product Candidate 45 which corresponds to our proposed candidate in the "proposed market scenario" given above. Even though this question is a strict subset of the next question, I am asking for it separately. This is because not all students may be able to do the next question successfully as it is a larger and more general computation. Hint to check your answer: The number you should get for expected profit per person is between 4 and 4.5.

2. Discrete Optimization: Consider each of the three levels for each of the five attributes and enumerate all the possibilities in lexical order with Price as the leftmost attribute changing slowest and taking levels sequentially $30, $10, $5, then Time Insulated as the left-second most attribute changing second-slowest and taking values sequentially 0.5 hrs, 1 hrs, 3 hrs and so on. You will have 243 product candidates. (The lexical order produces indices as shown in this file. Please make sure you list your products in that exact same order. The lexical order of products is obtained by looping as shown in the following R code, which has close similarity in python: mugs-products-lexical-order-loop.R.) Again using the compensatory rule, compute and report each candidate's share, price, margin and expected profit per person under the current market scenario. The listing that you submit needs to have all 243 rows. Next, plot all 243 candidates with share on the x-axis and expected profit per person on the y-axis. Hint to check your answer: product candidate 230 has a negative expected profit per customer of between -1.75 and -1.85, and product candidate 106 has an expected profit per customer of between 0.7 and 0.8

3. Identify the product with the highest expected profit per customer. For this optimal product, list the values of the five attributes and its share, price, margin and expected profit per person.

Wherever you use compensatory analysis, use the logit adjustment and use a scaling constant value of c=0.0139. This applies to Q1, Q2, Q3 above and also Optional Task 2 and Optional Task 3.

The Optional Tasks for Extra Credit toward Class Participation

Optional Task 1: Suppose you want to determine the elimination-by-aspects choice of a person choosing from among P products, each having A attributes. You are given the following data structures: a P-by-A matrix containing the rating of each product on each attribute, a vector of length A containing the importance of each attribute, a vector of length A giving the cutoff for each attribute (we are considering the general case where it is possible for the consumer to have different cutoffs for different attributes). Write a function that takes these three data structures as input arguments and produces the elimination-by-aspects choice without using explicit loops like "for", "while", "repeat" or recursion. You can however use implicit loops like in the "apply" family of functions in R or "map" function of python. For extra Credit: Write the function without using any "if" statement and also without using explicit loops. Your solution needs to use the same policy for ties and null sets as given in Q4. It may surprise you to know that this entire EBA function (for an arbitrary number of products and attributes) can be written in less than 10 lines of python or R code, using these languages' array operations,

FYI: My solution program to this task has only 5 lines. Of these 5 lines, 4 lines are operations that occur even with the compensatory model with the broadcast-and-accumulation structure. If you extend those ideas to the EBA context, then you get my 5-line-solution to Optional Task 1.

Optional Task 2: Instead of considering just the three discrete levels for each of the five attributes, consider all possible intermediate levels. You do not need to consider levels outside of the range of levels given (eg you do not need to consider prices greater the $30 or less than $5). Treat "Containment" as a continuous attribute with Slosh Resistant corresponding to level 0, Spill Resistant corresponding to level 0.5 and Leak Resistant corresponding to level 1. With this setup, now you will have what is essentially an infinite number of candidate products. For each such intermediate level not matching the discrete levels provided, compute the preference level by interpolating between the known levels. Similarly, compute the cost by interpolating between the known levels. For example, for Time Insulated = 0.75 hours, take the preference level for Consumer 1 to be 2 (interpolating halfway between 1 and 3 because 0.75 is halfway between 0.5 and 1). By similar logic, for Time Insulated = 0.75 hours, the cost would be $0.75. Your task: Use the compensatory model (interpolation makes programming complicated in the non-compensatory case, and so we will not consider the non-compensatory model for this question). Identify the product with the highest value of expected profit per person. For this optimal product, list the values of the five attributes and its share, price, margin and expected profit per person. Note that this is a continuous optimization problem and you will have to use an optimization routine. In R you will need to use something like the "optim" function which handles bounds and does not need a gradient function to be provided. Warning: you have to be concerned about local optima and these are to be addressed using multiple starting values or a stochastic search method. If this task is too complicated for you, you can consider a simpler version of this task where you are considering a large number (example d=19) of discrete intermediate values and handle the optimization by discrete enumeration. The number of candidates will then be (2*d+3)^5 = 115856201 for even a small number like d=19. A smart way to proceed is to do sequential grid search, starting with a low value of "d" like d=4 and then examining further discrete values in narrower range of discrete values, focusing sequentially only near the grid values that are optimal. In any case, this will involve a lot of computation. In R you can speed up computation by parallel computing using the "multicore" package or the "parallel" package.

Optional Task 3: Instead of optimizing for just one product, one can look for the best PAIR of products to launch simultaneously. We will allow for the possibility that the second product in the pair is the empty product or NULL product, so effectively the firm is offering just a single product rather than two distinct products. For simplicity we will consider the case where we are choosing from just among the original 243 products. The total number of pairs of products where both products are non-NULL is 243*(243-1)/2 = 29403. The number of pairs where one of the products is non-NULL is of course just 243 and these you have already evaluated in Mandatory Question 2 above. Your goal in this Optional Task is to identify the candidate pair of products X and Y with the highest total expected profit per person (adding the expected profit per person for product X and for product Y) using the compensatory rule with logit adjustment. For this optimal pair of products, list for each product in the pair its values of the five attributes and its share, price, margin and expected profit per person. Important note: The best pair is not necessarily the two products in Task 2 above with the two highest values of expected profit per person. To identify the best pair you need to (i) enumerate each of the 29403 pairs, (ii) for each such pair, labeling the products as 3 and 4, compute the probabilities of buying products 1, 2, 3 and 4 , where product 1 and 2 are from Brands A and B as before but now there are TWO products, 3 and 4, from Brand C, (iii) for each pair compute the total of expected profit per person over both 3 and 4 in the pair, (iv) pick the pair with the highest total expected profit per person.

A clever trick to accelerate the process us by skipping steps (ii) and (iii) above for a large fraction of pairs based on the results from Mandatory Question 2. This is because the profit contribution of a candidate in the three product case is an upper bound for the four product case. So we can drop from consideration any pair of candidates whose sum in Mandatory Question 2 is less than the value for the best product of Mandatory Question 2. In fact, we can improve upon this trick by increasing the upper bound progressively as we iterate through the pairs.

Optional Task 4: The table on Product-Price Optimization Slide 18 reported four metrics for each candidate product: Share, Cost, Margin, and Expected Profit per Person. According to the pure algorithmic analytical approach, the best product is in the one with the highest Expected Profit per Person. However, depending on the objectives of the company, your product manager boss may not agree that this is the best product. He/she may argue you need to consider one or more of the remaining three metrics: Share or Cost or Margin. Reflect on the table on Slide 18 or the scatterplot on Slide 19 (and the equivalent outputs from Question 2 and Question 4 above), and propose alternative criteria to identify the best product. For each criterion that you propose, identify the best product for that criterion.

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