Yesterday in class, we started exploring mathematical programming. We started by doing a problem that has a concrete model, and used Lego pieces. We did this to be able to visualize the real world model.

The problem was about a furniture company that makes only two products, tables and chairs. The parts could be used for both tables or chairs. Tables required two large pieces and two small pieces, while the chairs required one large piece and two small pieces. The resources from the company were unlimited, but they would only manufacture as many as they thought it would sell. In the first example, the company could only obtain six large and eight small pieces per day. The profit from each table was $16 and $10 for each chair. In our first question, we had to figure out what the production rates should be. There were four possibilities for production rates and and after making a table and looking at the total profits for each of these options, it was decided the optimal solution was made with 2 tables and 2 chairs. Although all the decision variables are feasible or possible, 2 tables and 2 chairs was the best option.

We then went on to answer a few questions about different amounts of lego pieces available. We were asked if there would be a difference with 7 large pieces instead of 6, and we decided it would. The optimal solution was 3 tables and 1 chair, which would generate $58 instead of $52 in the previous question. But when asked if 9 pieces instead of 8 would make a difference, we concluded it would not and there would not be a new optimal solution. This shows that more pieces is not always necessarily better.

In the last example, we were given 27 small pieces and 18 large pieces per day, and we had to maximize profit. The best way to do this was to have 6 tables and 6 chairs to make $156. We were then given 19 large pieces instead of 18 and we concluded the optimal solution would be 7 tables and 5 chairs. But when given 28 small pieces instead of 27, it would not make a difference and would not change the optimal solution.

In the end, we concluded it is more difficult to solve the second problem than the first because the numbers were bigger. The more pieces available, the higher the numbers get and the more complicated the problem becomes.