Question

At the Boise Cascade Corporation, lumber mill logs arrive by truck and are scaled (measured to determine the number of board feet) before they are dumped into a log pond. The figure below shows the basic flow. Scaling Stations Trucks Enter Log Pond The mill manager must determine how many scale stations to have open during various times of the day. If too many stations are open, the scalers will have excessive idle time and the cost of scaling will be unnecessarily high. On the other hand, if too few scale stations are open, some log trucks will have to wait. The manager has studied the truck arrival patterns and has determined that throughout the day, trucks randomly arrive at 11 per hour on average. Each scale station can scale 4 trucks per hour (15 minutes each). If the manager knew how many trucks would arrive during the hour, she would know how many scale stations to have open. For example, 0-4 trucks, open 1 scale station; 5-8 trucks, open 2 scale stations, etc. However, the number of trucks is a random variable and is uncertain. Your task is to provide guidance for the decision. 1) Based on the mean rate of 11 trucks per hour, what probability distribution should be used? 2) Use the probabilities fromthe first probability distribution above to create a probability distribution for the number of scale stations to open. For example, 0-4 trucks, open 1 scale station; 5-8 trucks, open 2 scale stations, etc. What is the expected number of scale stations needed? Are there additional factors that should be considered? What are your recommendations for handling those factors? 3)

Answer 1

The number of trucks arriving per hour is a discrete random variable, with a given mean rate. It can be best modelled as Poisson distribution, and can be written as

$P(x,\mu )=\frac{e^{-\mu }\mu ^{x}}{x!}\rightarrow \textup{ Given, }\mu =11$

Now we are more interested to obtain the number of scale stations to open. Here, we note that a particular range of the parameter 'x' in above distribution corresponds to the number of stations required. Hence, we need to evaluate the sum of those probabilities, and map them to #stations (N). Let us map them as per below

$P(N=1)\equiv P(0,\mu )+P(1,\mu )+P(2,\mu )+P(3,\mu )+P(4,\mu )$

$=e^{-11 }[\frac{11 ^{0}}{0!}+\frac{11 ^{1}}{1!}+\frac{11 ^{2}}{2!}+\frac{11 ^{3}}{3!}+\frac{11 ^{4}}{4!}]\approx 0.0151$

$P(N=2)\equiv P(5,\mu )+P(6,\mu )+P(7,\mu )+P(8,\mu )$

  $=e^{-11 }[\frac{11 ^{5}}{5!}+\frac{11 ^{6}}{6!}+\frac{11 ^{7}}{7!}+\frac{11 ^{8}}{8!}]\approx0.2169$

$P(N=3)\equiv P(9,\mu )+P(10,\mu )+P(11,\mu )+P(12,\mu )$

  $=e^{-11 }[\frac{11 ^{9}}{9!}+\frac{11 ^{10}}{10!}+\frac{11 ^{11}}{11!}+\frac{11 ^{12}}{12!}]\approx0.4567$

$P(N=4)\equiv P(13,\mu )+P(14,\mu )+P(15,\mu )+P(16,\mu )$

  $=e^{-11 }[\frac{11 ^{13}}{13!}+\frac{11 ^{14}}{14!}+\frac{11 ^{15}}{15!}+\frac{11 ^{16}}{16!}]\approx0.25538$

$P(N=5)\equiv P(17,\mu )+P(18,\mu )+P(19,\mu )+P(20,\mu )$

$=e^{-11 }[\frac{11 ^{17}}{17!}+\frac{11 ^{18}}{18!}+\frac{11 ^{19}}{19!}+\frac{11 ^{20}}{20!}]\approx0.05125$

$P(N=5)\equiv P(21,\mu )+P(22,\mu )+P(23,\mu )+P(24,\mu )$

  $=e^{-11 }[\frac{11 ^{21}}{21!}+\frac{11 ^{22}}{22!}+\frac{11 ^{23}}{23!}+\frac{11 ^{24}}{24!}]\approx0.00447$

Similarly, probabilities for higher number of scaling stations required can be evaluated. But since the sum of above probabilities is 0.9998, so for practical purposes we can stop at this point. The average number of scaling stations required can now be evaluated as

$E(N)=\sum n\times P(N=n)$

$=1\times 0.0151+2\times 0.2169+3\times 0.4567+4\times0.05125+5\times 0.00447\approx 2.04$

The expected no of scale stations needed is very close to the value 2.

There could be many additional factors. Traffic conditions may sometimes create some bunch of trucks to be stranded, which will then arrive altogether, thus creating a sudden burst in demand. The scaling station is also a machine in itself, which is subject to failure. So allowance must be made for such a n eventuality. For ex, at least 6 scaling stations should be operationsla to create redundancy.