The paper “ Applying Management to Science” is a worthy variant of the math problem on mathematics. Based on the charges levied by common carrier tracks for the transportation of each motor from the three different harbors to the four assembling plants, a decision is made on how many motors from which harbor are taken to which assembling plan while incurring the least cost. From the table 1.3 given in the question showing the costs of transportation, the following is deduced from which a decision of how many motors are to be sent from each harbor to each plant is made: The cheapest way to transport a motor to Leipzig is when it comes from Antwerp harbor (£ 61) therefore a maximum number of motors t Leipzig should come from here.
(400 motors) The cheapest way to transport a motor to Nancy is when it comes from Antwerp (£ 40) followed by Le Havre (£ 90) therefore 300 motors and 600 motors are taken to Nancy from Antwerp and Le Havre harbors respectively. The cheapest way t transport a motor to Liege is when it comes from Amsterdam (£ 41) so all the 200 required motors at Liege are taken from here. The cheapest way to transport a motor to Tilburg is when it comes from Le Havre (£ 42) followed by Amsterdam (£ 59.
50) therefore 200 motors and 300 motors respectively transported to Tilburg. This information is then arranged in a spreadsheet model that solves it to give the cheapest cost of transportation and distribution of motors from the harbors to assembly plants is shown. (See 358311 Spread Solutions for Question 1.xls). As indicated above in the deductions made from the table of costs of transportation, it is a representation of the cheapest possible combination that can be set from the conditions given.
It is observable that transportation of these motors from Antwerp harbor to the four plants is the cheapest and therefore the recommendation would be to increase the number of motors at Antwerp from where they are to be collected to the plants to minimize cost. This will meet management’ s requirement for a distribution plan while at the same time being the most economic arrangement involving the distribution of the motors. Report for Question 1 Given that the Auto Power Company wants to distribute these motors to the plants indicated so that relevant companies in need of this service (or cannot stand power interruptions), it must do so in a way that minimizes any possible risks while maximizing returns.
From the information given, the only overt thing of these two that can be determined and adequately controlled to realize the management’ s goal of minimized cost and efficient and timely distribution of the motors to their destinations is the decision of how many motors from which harbors are to be taken to which plant.
This is realized in the spreadsheet model that is designed (and herein attached) to handle this problem. Minimization of costs and maximization of benefits is the one goal to be sought after in the hope of realizing comparative advantage in the market. Question three analysis and results This time series that relates to the sales of this company is analyzed and discussed below. Identification, estimation, and forecasting based on autoregressive order one is obtained from this time series following the general formula: X(t +1) = F0 + F1X(t) + et, where et is a White-Noise series, then Stationary Condition: | F1| < 1, is expressed as a null hypothesis H0 and being tested herein. For a complete analysis and description of this time series, further tests are performed by computation of statistics and checking for both stationary in mean and variance as well testing seasonality (Gershenfeld 78). These findings are given in the tables and graph shown below: Table 3.1 table showing the time-series entries entered for calculation Top of Form t 1 2 3 4 X(t) t 15 16 17 18 X(t) t 29 30 31 32 X(t) t 43 44 45 46 X(t) t 57 58 59 60 X(t) Table 3.2 table showing results from the calculation Mean X(t) 80.7631579 Variance X(t) 80.8978532 Auto relation 0.5270426 Its Standard error 0.2003089 Parameter F1 0.5254951 Its Standard error 0.205511 Parameter F0 38.1909341 Its Standard error 16.6952101 H0: Stationary Condition F1‹ 1 Little or no evidence against the stationary condition Table 3.3 Table showing Diagnostic results for analysis White Noise Analysis and Diagnostic Mean 0 Variance 64.917333 Mean: The First half -1.3426414 Mean: The second half 1.2083772 Variance: The First-half 60.5670286 Variance: The second half 81.0999415 First-order serial-correlation 0.1901538 Second-order serial-correlation -0.5102268 First partial serial-correlation 0.1901538 Second partial serial-correlation -0.5668828282427032 Durbin-Watson statistic 1.49112 Mean absolute noise 6.50349 Normality Condition Evidence against normality 1st White Noise i).
-12.5489 ii). -4.5184 iii). 5.5492 iv). 8874 v). 9.8714 vi). -1.8155 vii). 5.7649 viii). 6.2982 1st Step Ahead Forecast 1).
80.8874 2). 80.6969 3). 80.5967 4). 80.5441 One method of Detrending a time series is by calculation. A sample of how this is done is shown below: Given a bounded time series xt, integration first converts this into an unbounded process Xt: Next, Xt is divide into time windows of length L samples, and a local least squares straight line fit (the local trend) is calculated by minimizing the squared error E2 with respect to the slope and intercept parameters a, b:Then, the root-mean-square deviation from the trend, the fluctuation, is calculated over every window at every time scale: This finalizes the detrending of the time series by calculation. The forecast for the next four quarters; Q1, Q2, Q3, and Q4 of 2009 are calculated to be as indicated in Table 3.3 above as: Q1 Q@ Q3 Q4 The year 2009 80.8874 80.6969 80.5967 80.5441 ConclusionThe method used for the analysis of this time series is simplified by the use of the named software designed to do the calculations with high accuracy and effectiveness.
These calculations are done using a Time Series Analysis and Forecasting software obtained from http://home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/Autoreg.htm
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