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  • Module 2 - Optimization models

    Introduction

    This is a hand-in written by Group 60 (FIRE) for the weekly module #2 in the mathematical modelling course (DAT026) at Chalmers University of Technology.

    Group 60 consists of:

    • Mazdak Farrokhzad

      • 901011-0279

      • twingoow@gmail.com

      • Program: IT

      • Time spent: x hours, y minutes

    • Niclas Alexandersson

      • 920203-0111

      • nicale@student.chalmers.se

      • Program: IT

      • Time spent: x hours, y minutes

    We hereby declare that we have both actively participated in solving every exercise, and all solutions are entirely our own work.

    Problem 1 - TVs and Profit

    Description of the problem

    The problem is about manufacturing TV devices and the profits of selling them.

    We will briefly provide a summary of the problem using variables.

    In this problem, all units manufactured are assumed to be sold. We also assume that the fixed cost of each device are separate so that they are separate in our expressions and functions, therefore we split the cost so that each TV model has half of the given fixed cost. From the provided description of the problem it is not obvious that this choice is wrong or right, it is a choice.

    \(i\) \(i = 1\) \(i = 2\)
    TV size, \(t_i\) \(t_0 = 19"\) \(t_1 = 21"\)
    Retail selling price , \(r_i\) \(r_0 = 3390\) \(r_1 = 3990\)
    Unit cost, \(u_i\) \(u_0 = 1950\) \(u_1 = 2250\)
    Fixed cost, \(f_i\) \(f_0 = 2 \cdot 10^6\) \(f_1 = 2 \cdot 10^6\)
    Price drop per unit sold \(d_i\) \(d_0 = 0.1\) \(d_1 = 0.1\)
    Price drop per unit sold of other TV \(c_i\) \(c_0 = 0.03\) \(c_1 = 0.04\)
    Units sold, \(s_i\) \(s_1\) \(s_2\)
    Average selling price \(a_i\) \(a_1 = r_1 - d_1s_1 - c_1s_2\) \(a_2 = r_2 - d_2s_2 - c_2s_1\)

    Profit and units sold

    The profit of selling TVs of a model can be expressed as: \[p_i = a_is_i - u_is_i - f_i = (a_i - s_i)s_i - f_i\]

    The total profit can be expressed as: \[p = \sum_{i=1}^n p_i = p_1 + p_2\]

    \[\begin{split} p_1 &= a_1s_1 - u_1s_1 - f_1\\ &= (r_1 - d_1s_1 - c_1s_2)s_1 - u_1s_1 - f_1\\ &= r_1s_1 - d_1s_1^2 - c_1s_1s_2 - u_1s_2 - f_1\\ p_2 &= a_2s_2 - u_2s_2 - f_2\\ &= (r_2 - d_2s_2 - c_2s_1)s_2 - u_2s_2 - f_2\\ &= r_2s_2 - d_2s_2^2 - c_2s_1s_2 - u_2s_2 - f_2 \end{split}\]

    So therefore: \[p = (r_1 - d_1s_1 - c_1s_2 - u_1)s_1 - f_1 + (r_2 - d_2s_2 - c_2s_1 - u_2)s_2 - f_2\]

    a) Solving the maximum

    Given the above values, we use the NMaximize function in mathematica to find the optimal value. The parameters will be the function, the constraints for the two unknowns, and the list of which unknown values we wish to find. NMaximize[{(3390 - 0.1 s1 - 0.03 s2 - 1950) s1 - 2*10^6 + (3990 - 0.1 s2 - 0.04 s1 - 2250) s2 - 2*10^6, s1 >= 1, s2 >= 1}, {s1, s2}] This gives us the result: {5.53641*10^6, {s1 -> 4735.04, s2 -> 7042.74}}

    Thus the maximum achievable profit with the provided constants is: \(p \approx 5.53641 \cdot 10^6 SEK\). And to achieve this profit, ca. \(4735\) units of t