5.4: The Exponential Distribution
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The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length of time, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.
Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, marketing studies have shown that the amount of money customers spend in one trip to the supermarket follows an exponential distribution.
Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts.
The random variable for the exponential distribution is continuous and often measures a passage of time, although it can be used in other applications. Typical questions may be, “what is the probability that some event will occur within the next \(x\) hours or days, or what is the probability that some event will occur between \(x_1\) hours and \(x_2\) hours, or what is the probability that the event will take more than \(x_1\) hours to perform?” In short, the random variable \(X\) equals (a) the time between events or (b) the passage of time to complete an action, e.g. wait on a customer. The probability density function is given by:
\[f(x)=\frac{1}{\mu} e^{-\frac{x}{\mu}}\nonumber\]
where \(\mu\) is the historical average waiting time.
An alternative form of the exponential distribution formula recognizes what is often called the decay factor. The decay factor simply measures how rapidly the probability of an event declines as the random variable \(X\) increases. When the notation using the decay parameter \(\lambda\) is used, the probability density function is presented as:
\[f(x)=\lambda e^{-\lambda x}\nonumber\]
where \(\lambda=\frac{1}{\mu}\)
In order to calculate probabilities for specific probability density functions, the cumulative density function is used. The cumulative density function (cdf) is simply the integral of the pdf and is:
\[F(x)=\int_{0}^x\left[\frac{1}{\mu} e^{-\frac{t}{\mu}}\right]dt=1-e^{-\frac{x}{\mu}}\nonumber\]
Example \(\PageIndex{1}\)
Let \(X\) = amount of time (in minutes) a postal clerk spends with a customer. The time is known from historical data to have an average amount of time equal to four minutes.
It is given that \(\mu = 4\) minutes, that is, the average time the clerk spends with a customer is 4 minutes. Remember that we are still doing probability and thus we have to be told the population parameters such as the mean. To do any calculations, we need to know the mean of the distribution: the historical time to provide a service, for example. Knowing the historical mean allows the calculation of the decay parameter, \(\lambda\).
\(\lambda=\frac{1}{\mu}\). Therefore, \(\lambda=\frac{1}{4}=0.25\).
When the notation used the decay parameter, \(\lambda\), the probability density function is presented as \(f(x)=\lambda e^{-\lambda x}\), which is simply the original formula with \(\lambda\) substituted for \(\frac{1}{\mu}\), or \(f(x)=\frac{1}{\mu} e^{-\frac{1}{\mu} x}\).
To calculate probabilities for an exponential probability density function, we need to use the cumulative density function. As shown below, the curve for the cumulative density function is:
\(f(x) = 0.25e^{–0.25x}\) where \(x\) is at least zero and \(\lambda = 0.25\).
For example, \(f(5) = 0.25e^{(-0.25)(5)} = 0.072\). In other words, the function has a value of .072 when \(x = 5\).
The graph is as follows:
Notice the graph is a declining curve. When \(x = 0\),
\(f(x) = 0.25e^{(−0.25)(0)} = (0.25)(1) = 0.25 = m\). The maximum value on the y-axis is always \(m\), one divided by the mean.
Exercise \(\PageIndex{1}\)
The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.
Example \(\PageIndex{2}\)
a. Using the information in Example \(\PageIndex{1}\), find the probability that a clerk spends four to five minutes with a randomly selected customer.
- Answer
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a. Find \(P (4 < X < 5)\).
The cumulative distribution function (CDF) gives the area to the left.
\(P(X < x) = 1 – e^{–\lambda x}\)
\(P(X < 5) = 1 – e^{(–0.25)(5)} = 0.7135\) and \(P(x < 4) = 1 – e^{(–0.25)(4)} = 0.6321\)
\(P(4 < X < 5)= 0.7135 – 0.6321 = 0.0814\)
Exercise \(\PageIndex{2}\)
The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?
Example \(\PageIndex{3}\)
On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.
a. What is the probability that a computer part lasts more than 7 years?
- Answer
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a. Let \(x =\) the amount of time (in years) a computer part lasts.
\(\mu = 10\) so \(\lambda=\frac{1}{\mu}=\frac{1}{10}=0.1\)
Find \(P(X > 7)\). Draw the graph.
\(P(X > 7) = 1 – P(X < 7)\).
Since \(P(X < x) = 1 – e^{–\lambda x}\) then \(P(X > x) = 1 – ( 1 –e^{–\lambda x}) = e^{–\lambda x}\)
\(P(X > 7) = e^{(–0.1)(7)} = 0.4966\). The probability that a computer part lasts more than seven years is \(0.4966\).
b. On the average, how long would five computer parts last if they are used one after another?
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b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.
d. What is the probability that a computer part lasts between 9 and 11 years?
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d. Find \(P (9 < X < 11)\). Draw the graph.
\(P(9 < X < 11) = P(X < 11) – P(X < 9) = (1 – e^{(–0.1)(11)}) – (1 – e^{(–0.1)(9)}) = 0.6671 – 0.5934 = 0.0737\). The probability that a computer part lasts between 9 and 11 years is \(0.0737\).
Exercise \(\PageIndex{3}\)
On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?
Example \(\PageIndex{4}\)
Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter \(\frac{1}{12}\). If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes.
What is \(lambda, \mu\), and \(\sigma\)? The probability that you must wait more than 5 minutes is _______ .
- Answer
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\(\lambda = \frac{1}{12}\)
\(\mu = 12\)
\(\sigma = 12\)
\(P(X > 5) = e^{-5/12} = 0.6592\)
Example \(\PageIndex{5}\)
The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.
- On average, how many minutes elapse between two successive arrivals?
- When the store first opens, how long on average does it take for three customers to arrive?
- After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.
- After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.
- Is an exponential distribution reasonable for this situation?
- Answer
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a.Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.
b.Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.
c.Let \(X =\) the time between arrivals, in minutes. By part a, \(\mu = 2\), so \(\lambda = \frac{1}{2}= 0.5\).
d.\(P(X > 5) = 1 – P(X < 5) = 1 – (1 – e^{(-0.5)(5)}) = e^{–2.5} \approx 0.0821\). This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.
The cumulative distribution function is \(P(X < x) = 1 – e^{(-0.5)(x)}\)
Therefore \(P(X < 1) = 1 – e^{(–0.5)(1)} = 0.3935\).
Memoryless Property of the Exponential Distribution
Recall that the amount of time between customers for the postal clerk discussed earlier is exponentially distributed with a mean of two minutes. Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property . The exponential and geometric probability density functions are the only probability functions that have the memoryless property. Specifically, the memoryless property says that
\(P(X > r + t | X > r) = P (X > t)\) for all \(r \geq 0\) and \(t \geq 0\)
For example, if five minutes have elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation.
\(P(X > 5 + 1 | X > 5) = P(X > 1) = e^{(-0.5)(1)} = 0.6065\).
This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival.
The exponential distribution is often used to model the longevity of an electrical or mechanical device. In Example \(\PageIndex{3}\), the lifetime of a certain computer part has the exponential distribution with a mean of ten years. The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is \(P(X > 17|X > 10) = P(X > 7) = 0.4966\), where the vertical line is read as "given".
Example \(\PageIndex{6}\)
Refer back to the postal clerk again where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?
The decay parameter of \(X\) is \(\lambda = \frac{1}{4} = 0.25\).
The cumulative distribution function is \(P(X < x) = 1 – e^{–0.25x}\).
We want to find \(P (X > 7|X > 4)\). The memoryless property says that \(P (X > 7|X > 4) = P (X > 3)\), so we just need to find the probability that a customer spends more than three minutes with a postal clerk.
This is \(P(X > 3) = 1 – P(X < 3) = 1 – (1 – e^{–0.25⋅3}) = e^{–0.75} \approx 0.4724\).
Relationship between the Poisson and the Exponential Distribution
There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of \(\mu\) units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean \(\mu\). Recall that if \(X\) has the Poisson distribution with mean \(\mu\), then \(P(X=x)=\frac{\mu^{x_{e}-\mu}}{x !}\).
The formula for the exponential distribution: \(P(X=x)=m e^{-m x}=\frac{1}{\mu} e^{-\frac{1}{\mu} x}\) Where \(m =\) the rate parameter, or \(\mu =\) average time between occurrences.
We see that the exponential is the cousin of the Poisson distribution and they are linked through this formula. There are important differences that make each distribution relevant for different types of probability problems.
First, the Poisson has a discrete random variable, \(X\), where time; a continuous variable is artificially broken into discrete pieces. We saw that the number of occurrences of an event in a given time interval, \(X\), follows the Poisson distribution.
For example, the number of times the telephone rings per hour. By contrast, the time between occurrences follows the exponential distribution. For example. The telephone just rang, how long will it be until it rings again? We are measuring length of time of the interval, a continuous random variable, exponential, not events during an interval, Poisson.
The Exponential Distribution v. the Poisson Distribution
A visual way to show both the similarities and differences between these two distributions is with a time line.
The random variable for the Poisson distribution is discrete and thus counts events during a given time period, \(t_1\) to \(t_2\) on Figure \(\PageIndex{8}\), and calculates the probability of that number occurring. The number of events, four in the graph, is measured in counting numbers; therefore, the random variable of the Poisson is a discrete random variable.
The exponential probability distribution calculates probabilities of the passage of time, a continuous random variable. In Figure \(\PageIndex{8}\) this is shown as the bracket from t 1 to the next occurrence of the event marked with a triangle.
Classic Poisson distribution questions are "how many people will arrive at my checkout window in the next hour?".
Classic exponential distribution questions are "how long it will be until the next person arrives," or a variant, "how long will the person remain here once they have arrived?".
Again, the formula for the exponential distribution is:
\[f(x)=\lambda e^{-\lambda x} \operatorname{or} f(x)=\frac{1}{\mu} e^{-\frac{1}{\mu} x}\nonumber\]
We see immediately the similarity between the exponential formula and the Poisson formula.
\[P(x)=\frac{\mu^{x} e^{-\mu}}{x !}\nonumber\]
Both probability density functions are based upon the relationship between time and exponential growth or decay. The “e” in the formula is a constant with the approximate value of 2.71828 and is the base of the natural logarithmic exponential growth formula. When people say that something has grown exponentially this is what they are talking about.
An example of the exponential and the Poisson will make clear the differences been the two. It will also show the interesting applications they have.
Poisson Distribution
Suppose that historically 10 customers arrive at the checkout lines each hour. Remember that this is still probability so we have to be told these historical values. We see this is a Poisson probability problem.
We can put this information into the Poisson probability density function and get a general formula that will calculate the probability of any specific number of customers arriving in the next hour.
The formula is for any value of the random variable we chose, and so the x is put into the formula. This is the formula:
\[f(x)=\frac{10^{x} e^{-10}}{x !}\nonumber\]
As an example, the probability of 15 people arriving at the checkout counter in the next hour would be
\[P(x=15)=\frac{10^{15} e^{-10}}{15 !}=0.0611\nonumber\]
Here we have inserted x = 15 and calculated the probability that in the next hour 15 people will arrive is .061.
Exponential Distribution
If we keep the same historical facts that 10 customers arrive each hour, but we now are interested in the service time a person spends at the counter, then we would use the exponential distribution. The exponential probability function for any value of x, the random variable, for this particular checkout counter historical data is:
\[f(x)=\frac{1}{.1} e^{-x / 1}=10 e^{-10 x}\nonumber\]
To calculate \(\mu\), the historical average service time, we simply divide the number of people that arrive per hour, 10 , into the time period, one hour, and have \(\mu = 0.1\). Historically, people spend 0.1 of an hour at the checkout counter, or 6 minutes. This explains the .1 in the formula.
There is a natural confusion with \(\mu\) in both the Poisson and exponential formulas. They have different meanings, although they have the same symbol. The mean of the exponential is one divided by the mean of the Poisson. If you are given the historical number of arrivals you have the mean of the Poisson. If you are given an historical length of time between events you have the mean of an exponential.
Continuing with our example at the checkout clerk; if we wanted to know the probability that a person would spend 9 minutes or less checking out, then we use this formula. First, we convert to the same time units which are parts of one hour. Nine minutes is 0.15 of one hour. Next we note that we are asking for a range of values. This is always the case for a continuous random variable. We write the probability question as:
\[p(x \leq 9)=1-10 e^{-10 x}\nonumber\]
We can now put the numbers into the formula and we have our result.
\[p(x=.15)=1-10 e^{-10(.15)}=0.7769\nonumber\]
The probability that a customer will spend 9 minutes or less checking out is \(0.7769\).
We see that we have a high probability of getting out in less than nine minutes and a tiny probability of having 15 customers arriving in the next hour.


2.2: Minterms and MATLAB Calculations
The concepts and procedures in this unit play a significant role in many aspects of the analysis of probability topics and in the use of MATLAB throughout this work.
Minterm vectors and MATLAB
The systematic formulation in the previous module Minterms shows that each Boolean combination, as a union of minterms, can be designated by a vector of zero-one coefficients. A coefficient one in the \(i\)th position (numbering from zero) indicates the inclusion of minterm \(M_i\) in the union. We formulate this pattern carefully below and show how MATLAB logical operations may be utilized in problem setup and solution.
Suppose \(E\) is a Boolean combination of \(A, B, C\). Then, by the minterm expansion,
\(E = \bigvee_{J_E} M_i\)
where \(M_i\) is the \(i\)th minterm and \(J_E\) is the set of indices for those \(M_i\) included in \(E\). For example, consider
\(E = A(B \cup C^c) \cup A^c (B \cup C^c)^c = M_1 \bigvee M_4 \bigvee M_6 \bigvee M_7 = M(1, 4, 6, 7)\)
\(F = A^c B^c \cup AC = M_0 \bigvee M_1 \bigvee M_5 \bigvee M_7 = M(0, 1, 5, 7)\)
We may designate each set by a pattern of zeros and ones (\(e_0, e_1, \cdot\cdot\cdot, e_7\)). The ones indicate which minterms are present in the set. In the pattern for set \(E\), minterm \(M_i\) is included in \(E\) iff \(e_i = 1\). This is, in effect, another arrangement of the minterm map. In this form, it is convenient to view the pattern as a minterm vector , which may be represented by a row matrix or row vector [\(e_0, e_1, \cdot\cdot\cdot, e_7\)]. We find it convenient to use the same symbol for the name of the event and for the minterm vector or matrix representing it . Thus, for the examples above,
\(E \sim \) [0 1 0 0 1 0 1 1] and \(F \sim\) [1 1 0 0 0 1 0 1]
It should be apparent that this formalization can be extended to sets generated by any finite class.
Minterm vectors for Boolean combinations
If \(E\) and \(F\) are combinations of \(n\) generating sets, then each is represented by a unique minterm vector of length \(2^n\). In the treatment in the module Minterms , we determine the minterm vector with the aid of a minterm map. We wish to develop a systematic way to determine these vectors.
As a first step, we suppose we have minterm vectors for \(E\) and \(F\) and want to obtain the minterm vector of Boolean combinations of these.
- The minterm expansion for \(E \cup F\) has all the minterms in either set. This means the \(j\)th element of the vector for \(E \cup F\) is the maximum of the \(j\)th elements for the two vectors.
- The minterm expansion for \(E \cap F\) has only those minterms in both sets. This means the \(j\)th element of the vector for \(E \cap F\) is the minimum of the \(j\)th elements for the two vectors.
- The minterm expansion for \(E^c\) has only those minterms not in the expansion for \(E\). This means the vector for \(E^c\) has zeros and ones interchanged. The \(j\)th element of \(E^c\) is one iff the corresponding element of \(E\) is zero.
We illustrate for the case of the two combinations \(E\) and \(F\) of three generating sets, considered above
\(E = A(B \cup C^c) \cup A^c (B \cup C^c)^c \cup A^c (B \cup C^c)^c \sim\) [0 1 0 0 1 0 1 1] and \(F = A^c B^c \cup AC \sim\) [1 1 0 0 0 1 0 1]
Then
\(E \cup F \sim\) [1 1 0 0 1 1 1 1], \(E \cap F \sim\) [0 1 0 0 0 0 0 1], and \(E^c \sim\) [1 0 1 1 0 1 0 0]
MATLAB logical operations
MATLAB logical operations on zero-one matrices provide a convenient way of handling Boolean combinations of minterm vectors represented as matrices. For two zero-one matrices \(E, F\) of the same size
- \(E|F\) is the matrix obtained by taking the maximum element in each place.
- \(E\text{&}F\) is the matrix obtained by taking the minimum element in each place.
- \(E^c\) is the matrix obtained by interchanging one and zero in each place in \(E\).
Thus, if \(E, F\) are minterm vectors for sets by the same name, then \(E|F\) is the minterm vector for \(E \cup F\), \(E\text{&}F\) is the minterm vector for \(E \cap F\), and \(E = 1 - E\) is the minterm vector for \(E^c\).
This suggests a general approach to determining minterm vectors for Boolean combinations.
Start with minterm vectors for the generating sets.
Use MATLAB logical operations to obtain the minterm vector for any Boolean combination.
Suppose, for example, the class of generating sets is \(\{A, B, C\}\). Then the minterm vectors for \(A\), \(B\), and \(C\), respectively, are
\(A = \) [0 0 0 0 1 1 1 1] \(B = \) [0 0 1 1 0 0 1 1] \(C = \) [0 1 0 1 0 1 0 1]
If \(E = AB \cup C^c\), then the logical combination \(E = (A \text{&} B)\ |\ C\) of the matrices yields \(E =\) [1 0 1 0 1 0 1 1].
MATLAB implementation
A key step in the procedure just outlined is to obtain the minterm vectors for the generating elements \(\{A, B, C\}\). We have an m-function to provide such fundamental vectors. For example to produce the second minterm vector for the family (i.e., the minterm vector for \(B\)), the basic zero-one pattern is replicated twice to give
0 0 1 1 0 0 1 1
The function minterm(n,k) generates the k th minterm vector for a class of n generating sets.
minterms for the class {a, b, c}.
>> A = minterm(3,1) A = 0 0 0 0 1 1 1 1 >> B = minterm(3,2) B = 0 0 1 1 0 0 1 1 >> C = minterm(3,3) C = 0 1 0 1 0 1 0 1
minterm patterns for the boolean combinations
\(F = AB \cup B^c C\) \(G = A \cup A^c C\)
F = (A&B)|(~B&C) F = 0 1 0 0 0 1 1 1 >> G = A|(~A&C) G = 0 1 0 1 1 1 1 1 >> JF = find(F)-1 % Use of find to determine index set for F JF = 1 5 6 7 % Shows F = M(1, 5, 6, 7)
These basic minterm patterns are useful not only for Boolean combinations of events but also for many aspects of the analysis of those random variables which take on only a finite number of values.
Zero-one arrays in MATLAB
The treatment above hides the fact that a rectangular array of zeros and ones can have two quite different meanings and functions in MATLAB.
A
numerical matrix
(or vector) subject to the usual operations on matrices..
A
logical array
whose elements are combined by a. Logical operators to give new logical arrays; b. Array operations (element by element) to give numerical matrices; c. Array operations with numerical matrices to give numerical results.
Some simple examples will illustrate the principal properties.
>>> A = minterm(3,1); >> B = minterm(3,2); >> C = minterm(3,3); >> F = (A&B)|(~B&C) F = 0 1 0 0 0 1 1 1 >> G = A|(~A&C) G = 0 1 0 1 1 1 1 1 >> islogical(A) % Test for logical array ans = 0 >> islogical(F) ans = 1 >> m = max(A,B) % A matrix operation m = 0 0 1 1 1 1 1 1 >> islogical(m) ans = 0 >> m1 = A|B % A logical operation m1 = 0 0 1 1 1 1 1 1 >> islogical(m1) ans = 1 >> a = logical(A) % Converts 0-1 matrix into logical array a = 0 0 0 0 1 1 1 1 >> b = logical(B) >> m2 = a|b m2 = 0 0 1 1 1 1 1 1 >> p = dot(A,B) % Equivalently, p = A*B' p = 2 >> p1 = total(A.*b) p1 = 2 >> p3 = total(A.*B) p3 = 2 >> p4 = a*b' % Cannot use matrix operations on logical arrays ??? Error using ==> mtimes % MATLAB error signal Logical inputs must be scalar.
Often it is desirable to have a table of the generating minterm vectors. Use of the function minterm in a simple “for loop” yields the following m-function.
The function mintable(n) Generates a table of minterm vectors for \(n\) generating sets.
mintable for three variables
>> M3 = mintable(3) M3 = 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1
As an application of mintable, consider the problem of determining the probability of \(k\) of \(n\) events. If \(\{A_i: 1 \le i \le n\}\) is any finite class of events, the event that exactly \(k\) of these events occur on a trial can be characterized simply in terms of the minterm expansion. The event \(A_{kn}\) that exactly \(k\) occur is given by
\(A_{kn} = the\ disjoint\ union\ of\ those\ minterms\ with\ exactly\ k\ positions\ uncomplemented\)
In the matrix these are the minterms corresponding to columns with exactly \(k\) ones. The event BknBkn that \(k\) or more occur is given by
\(B_{kn} = \bigvee_{r = k}^{n} A_{rn}\)
If we have the minterm probabilities, it is easy to pick out the appropriate minterms and combine the probabilities. The following example in the case of three variables illustrates the procedure.
the software survey (continued)
In the software survey problem, the minterm probabilities are
\(pm =\) [0 0.05 0.10 0.05 0.20 0.10 0.40 0.10]
where \(A =\) event has word processor, \(B =\) event has spread sheet, \(C =\) event has a data base program. It is desired to get the probability an individual selected has \(k\) of these, \(k = 0, 1, 2, 3\).
Solution
We form a mintable for three variables. We count the number of “successes” corresponding to each minterm by using the MATLAB function sum, which gives the sum of each column. In this case, it would be easy to determine each distinct value and add the probabilities on the minterms which yield this value. For more complicated cases, we have an m-function called csort (for sort and consolidate) to perform this operation.
>> pm = 0.01*[0 5 10 5 20 10 40 10]; >> M = mintable(3) M = 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 >> T = sum(M) % Column sums give number T = 0 1 1 2 1 2 2 3 % of successes on each >> [k,pk] = csort(T,pm); % minterm, determines % distinct values in T and >> disp([k;pk]') % consolidates probabilities 0 0 1.0000 0.3500 2.0000 0.5500 3.0000 0.1000
For three variables, it is easy enough to identify the various combinations “by eye” and make the combinations. For a larger number of variables, however, this may become tedious. The approach is much more useful in the case of Independent Events , because of the ease of determining the minterm probabilities.
Minvec procedures
Use of the tilde \(\sim\) to indicate the complement of an event is often awkward. It is customary to indicate the complement of an event \(E\) by \(E^c\). In MATLAB, we cannot indicate the superscript, so we indicate the complement by \(E^c\) instead of \(\sim E\). To facilitate writing combinations, we have a family of minvec procedures (minvec3, minvec4, ..., minvec10) to expedite expressing Boolean combinations of \(n = 3, 4, 5, \cdot\cdot\cdot, 10\) sets. These generate and name the minterm vector for each generating set and its complement.
boolean combinations using minvec3
We wish to generate a matrix whose rows are the minterm vectors for \(\Omega = A \cup A^c, A, AB, ABC, C,\) and \(A^c C^c\), respectively.
>> minvec3 % Call for the setup procedure Variables are A, B, C, Ac, Bc, Cc They may be renamed, if desired >> V = [A|Ac; A; A&B; A&B&C; C; Ac&Cc]; % Logical combinations (one per % row) yield logical vectors >> disp(V) 1 1 1 1 1 1 1 1 % Mixed logical and 0 0 0 0 1 1 1 1 % numerical vectors 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 1 0 1 0 0 0 0 0
Minterm probabilities and Boolean combination
If we have the probability of every minterm generated by a finite class, we can determine the probability of any Boolean combination of the members of the class. When we know the minterm expansion or, equivalently, the minterm vector, we simply pick out the probabilities corresponding to the minterms in the expansion and add them. In the following example, we do this “by hand” then show how to do it with MATLAB .
Consider \(E = A (B \cup C^c) \cup A^c (B \cup C^c)^c\) and \(F = A^c B^c \cup AC\) of the example above, and suppose the respective minterm probabilites are
\(p_0 = 0.21\), \(p_1 = 0.06\), \(p_2 = 0.29\), \(p_3 = 0.11\), \(p_4 = 0.09\), \(p_5 = 0.03\), \(p_6 = 0.14\), \(p_7 = 0.07\)
Use of a minterm map shows \(E = M(1, 4, 6, 7)\) and \(F = M(0, 1, 5, 7)\). so that
\(P(E) = p_1 + p_4 + p_6 + p_7 = p(1, 4, 6, 7) = 0.36\) and \(P(F) = p(0, 1, 5, 7) = 0.37\)
This is easily handled in MATLAB.
- Use minvec3 to set the generating minterm vectors.
-
Use
logical
matrix operations
\(E = (A \text{&} (B|Cc))|(Ac\text{&}((B|Cc)))\) and \(F = (Ac \text{&} Bc)|(A\text{&}C)\)
to obtain the (logical) minterm vectors for \(E\) and \(F\) - If \(pm\) is the matrix of minterm probabilities, perform the algebraic dot product or scalar product of the pmpm matrix and the minterm vector for the combination. This can be called for by the MATLAB commands PE = E*pm' and PF = F*pm'.
The following is a transcript of the MATLAB operations.
>> minvec3 % Call for the setup procedure Variables are A, B, C, Ac, Bc, Cc They may be renamed, if desired. >> E = (A&(B|Cc))|(Ac&~(B|Cc)); >> F = (Ac&Bc)|(A&C); >> pm = 0.01*[21 6 29 11 9 3 14 7]; >> PE = E*pm' % Picks out and adds the minterm probabilities PE = 0.3600 >> PF = F*pm' PF = 0.3700
solution of the software survey problem
We set up the matrix equations with the use of MATLAB and solve for the minterm probabilities. From these, we may solve for the desired “target” probabilities.
>> minvec3 Variables are A, B, C, Ac, Bc, Cc They may be renamed, if desired. Data vector combinations are: >> DV = [A|Ac; A; B; C; A&B&C; Ac&Bc; (A&B)|(A&C)|(B&C); (A&Bc&C) - 2*(Ac&B&C)] DV = 1 1 1 1 1 1 1 1 % Data mixed numerical 0 0 0 0 1 1 1 1 % and logical vectors 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 0 0 -2 0 1 0 0 >> DP = [1 0.8 0.65 0.3 0.1 0.05 0.65 0]; % Corresponding data probabilities >> pm = DV\DP' % Solution for minterm probabilities pm = -0.0000 % Roundoff -3.5 x 10-17 0.0500 0.1000 0.0500 0.2000 0.1000 0.4000 0.1000 >> TV = [(A&B&Cc)|(A&Bc&C)|(Ac&B&C); Ac&Bc&C] % Target combinations TV = 0 0 0 1 0 1 1 0 % Target vectors 0 1 0 0 0 0 0 0 >> PV = TV*pm % Solution for target probabilities PV = 0.5500 % Target probabilities 0.0500
An alternate approach
The previous procedure first obtained all minterm probabilities, then used these to determine probabilities for the target combinations. The following procedure does not require calculation of the minterm probabilities. Sometimes the data are not sufficient to calculate all minterm probabilities, yet are sufficient to allow determination of the target probabilities.
Suppose the data minterm vectors are linearly independent, and the target minterm vectors are linearly dependent upon the data vectors (i.e., the target vectors can be expressed as linear combinations of the data vectors). Now each target probability is the same linear combination of the data probabilities. To determine the linear combinations, solve the matrix equation
\(TV = CT * DV\ which\ has\ the\ MATLAB\ solution\ CT = TV/DV\)
Then the matrix \(tp\) of target probabilites is given by \(tp = CT * DP'\). Continuing the MATLAB procedures above, we have:
>> CT = TV/DV;
>> tp = CT*DP'
tp = 0.5500 0.0500
The procedure mincalc
The procedure mincalc performs calculations as in the preceding examples. The refinements consist of determining consistency and computability of various individual minterm probabilities and target probilities. The consistency check is principally for negative minterm probabilities. The computability tests are tests for linear independence by means of calculation of ranks of various matrices. The procedure picks out the computable minterm probabilities and the computable target probabilities and calculates them.
To utilize the procedure, the problem must be formulated appropriately and precisely, as follows:
Use the MATLAB program
minvecq
to set minterm vectors for each of
q
basic events.
Data
consist of Boolean combinations of the basic events and the respective probabilities of these combinations. These are organized into two matrices:
- The data vector matrix \(DV\) has the data Boolean combinations– one on each row. MATLAB translates each row into the minterm vector for the corresponding Boolean combination. The first entry (on the first row) is A | Ac (for \(A \bigvee A^c\)), which is the whole space. Its minterm vector consists of a row of ones.
- The data probability matrix \(DP\) is a row matrix of the data probabilities. The first entry is one, the probability of the whole space.
The objective is to determine the probability of various target Boolean combinations. These are put into the target vector matrix \(TV\), one on each row. MATLAB produces the minterm vector for each corresponding target Boolean combination.
Computational note . In mincalc, it is necessary to turn the arrays DV and TV consisting of zero-one patterns into zero-one matrices. This is accomplished for DV by the operation . and similarly for TV. Both the original and the transformed matrices have the same zero-one pattern, but MATLAB interprets them differently.
Usual case
Ṡuppose the data minterm vectors are linearly independent and the target vectors are each linearly dependent on the data minterm vectors. Then each target minterm vector is expressible as a linear combination of data minterm vectors. Thus, there is a matrix \(CT\) such that \(TV = CT * DV\). MATLAB solves this with the command \(CT = TV/DV\) The target probabilities are the same linear combinations of the data probabilities . These are obtained by the MATLAB operation \(tp = DP * CT'\).
Cautionary notes
The program mincalc depends upon the provision in MATLAB for solving equations when less than full data are available (based on the singular value decomposition). There are several situations which should be dealt with as special cases. It is usually a good idea to check results by hand to determine whether they are consistent with data. The checking by hand is usually much easier than obtaining the solution unaided, so that use of MATLAB is advantageous even in questionable cases.
The Zero Problem
. If the total probability of a group of minterms is zero, then it follows that the probability of each minterm in the group is zero. However, if mincalc does not have enough information to calculate the separate minterm probabilities in the case they are not zero, it will not pick up in the zero case the fact that the separate minterm probabilities are zero. It simply considers these minterm probabilities not computable.
Linear dependence
. In the case of linear dependence, the operation called for by the command CT = TV/DV may not be able to solve the equations. The matrix may be singular, or it may not be able to decide which of the redundant data equations to use. Should it provide a solution, the result should be checked with the aid of a minterm map.
Consistency check
. Since the consistency check is for negative minterms, if there are not enough data to calculate the minterm probabilities, there is no simple check on the consistency. Sometimes the probability of a target vector included in another vector will actually exceed what should be the larger probability. Without considerable checking, it may be difficult to determine consistency.
In a few unusual cases, the command CT = TV/DV does not operate appropriately, even though the data should be adequate for the problem at hand. Apparently the approximation process does not converge.
MATLAB Solutions for examples using mincalc
software survey
% file mcalc01 Data for software survey minvec3; DV = [A|Ac; A; B; C; A&B&C; Ac&Bc; (A&B)|(A&C)|(B&C); (A&Bc&C) - 2*(Ac&B&C)]; DP = [1 0.8 0.65 0.3 0.1 0.05 0.65 0]; TV = [(A&B&Cc)|(A&Bc&C)|(Ac&B&C); Ac&Bc&C]; disp('Call for mincalc') >> mcalc01 % Call for data Call for mincalc % Prompt supplied in the data file >> mincalc Data vectors are linearly independent Computable target probabilities 1.0000 0.5500 2.0000 0.0500 The number of minterms is 8 The number of available minterms is 8 Available minterm probabilities are in vector pma To view available minterm probabilities, call for PMA >> disp(PMA) % Optional call for minterm probabilities 0 0 1.0000 0.0500 2.0000 0.1000 3.0000 0.0500 4.0000 0.2000 5.0000 0.1000 6.0000 0.4000 7.0000 0.1000
computer survey
% file mcalc02.m Data for computer survey minvec3 DV = [A|Ac; A; B; C; A&B&C; A&C; (A&B)|(A&C)|(B&C); ... 2*(B&C) - (A&C)]; DP = 0.001*[1000 565 515 151 51 124 212 0]; TV = [A|B|C; Ac&Bc&C]; disp('Call for mincalc') >> mcalc02 Call for mincalc >> mincalc Data vectors are linearly independent Computable target probabilities 1.0000 0.9680 2.0000 0.0160 The number of minterms is 8 The number of available minterms is 8 Available minterm probabilities are in vector pma To view available minterm probabilities, call for PMA >> disp(PMA) 0 0.0320 1.0000 0.0160 2.0000 0.3760 3.0000 0.0110 4.0000 0.3640 5.0000 0.0730 6.0000 0.0770 7.0000 0.0510
% file mcalc03.m Data for opinion survey minvec4 DV = [A|Ac; A; B; C; D; A&(B|Cc)&Dc; A|((B&C)|Dc) ; Ac&B&Cc&D; ... A&B&C&D; A&Bc&C; Ac&Bc&Cc&D; Ac&B&C; Ac&Bc&Dc; A&Cc; A&C&Dc; A&B&Cc&Dc]; DP = 0.001*[1000 200 500 300 700 55 520 200 15 30 195 120 120 ... 140 25 20]; TV = [Ac&((B&Cc)|(Bc&C)); A|(B&Cc)]; disp('Call for mincalc') >> mincalc03 Call for mincalc >> mincalc Data vectors are linearly independent Computable target probabilities 1.0000 0.4000 2.0000 0.4800 The number of minterms is 16 The number of available minterms is 16 Available minterm probabilities are in vector pma To view available minterm probabilities, call for PMA >> disp(minmap(pma)) % Display arranged as on minterm map 0.0850 0.0800 0.0200 0.0200 0.1950 0.2000 0.0500 0.0500 0.0350 0.0350 0.0100 0.0150 0.0850 0.0850 0.0200 0.0150
The procedure mincalct
A useful modification, which we call mincalct , computes the available target probabilities, without checking and computing the minterm probabilities. This procedure assumes a data file similar to that for mincalc, except that it does not need the target matrix \(TV\), since it prompts for target Boolean combination inputs. The procedure mincalct may be used after mincalc has performed its operations to calculate probabilities for additional target combinations.
(continued) Additional target datum for the opinion survey
Suppose mincalc has been applied to the data for the opinion survey and that it is desired to determine \(P(AD \cup BD^c)\). It is not necessary to recalculate all the other quantities. We may simply use the procedure mincalct and input the desired Boolean combination at the prompt.
>> mincalct Enter matrix of target Boolean combinations (A&D)|(B&Dc) Computable target probabilities 1.0000 0.2850
Repeated calls for mcalct may be used to compute other target probabilities.


3.1: Soil Texture and Structure
Learning Objectives
- Differentiate the three soil separates (sand, silt, and clay) based on their particle size diameters.
- Determine the percentages of sand, silt, and clay in selected soil samples using data collected from the hydrometer method of particle size analysis.
- Estimate the textural class using the texture-by-feel method on selected soil samples.
- Use a textural triangle to determine the textural class of a soil.
- Understand the relationship between particle size and specific surface area.
Soil texture and structure are considered “master variables”, meaning that texture and structure directly influence a large number of other soil properties. For example, in comparing a clayey soil and a sandy soil, one would expect the clayey soil to have larger specific surface area, more cation exchange capacity, more total porosity, less macroporosity, and more organic matter than the sandy soil. Thus, by simply knowing the texture of the soil, inferences can be made in regard to many soil properties. Here, soil texture will be determined quantitatively using the hydrometer method, and estimated using the texture by feel method. Different soil structure types will also be observed. What you learn about texture and structure in these activities will be used later during the soil pit field trips to describe soil profiles in the field.
Materials
- Three soils of known textures
- Sodium hexametaphosphate
- Squirt bottles filled with tap water
- Lab balances accurate to the nearest 0.01 g
- Soil test cylinders, 1L volume (Item #200231000, Kimble ™ Kimax™ Soil Test Cylinders)
- Hydrometers calibrated in units of g/L (Item #13-202-133 Fisherbrand™ Soil Analysis ASTM Hydrometer)
- Digital thermometers accurate to the nearest 0.1°C
- Milkshake mixers and stainless steel milkshake mixer cups (Hamilton Beach HMD400 120V Triple Spindle Commercial Drink Mixer, Hamilton Beach, Glen Allen, Virginia, U.S.)
Recommended Reading
- Eye on Agriculture Today: Soil Texture by Feel (KSREVideos, 2010)
- Soil Profiling: Structure (KSREVideos, 2011c)
- Estimating Soil Texture by Feel (Presley and Thien, 2008)
Prelab Assignment
Using the recommended reading and viewing resources and the introduction to this lab, consider the questions listed below. These definitions/questions will provide a concise summary of the major concepts to be addressed in the lab. They will also serve as the basis for the post-lab quiz and are useful study notes for exams.
- Define and explain the difference between soil texture and soil structure.
- Create a table that lists the sizes for sand, silt, and clay particles using USDA criteria.
- What is a textural triangle?
- What is Stoke’s Law and how is it used in soil science?
- Define specific surface area and explain its relationship to size of soil particles.
Introduction
Soil texture refers to the proportions of sand (2.0 – 0.05 mm in diameter), silt (0.05 – 0.002 mm), and clay (less than 0.002 mm). The relative proportions determine the textural class. Soil texture influences nearly every aspect of soil use and management. Many of the physical and chemical properties of the soil depend on how fine (clayey) or coarse (sandy) a soil is. Soil texture is a permanent feature unless soils are subjected to rapid erosion, deposition, or removal.
Moreover, much of the reactivity of soils is related to the amount of surface area available. As the average particle size decreases, the surface area per unit weight increases (see Table 7.1).
| Seperate | Diameter (mm) | Particles/g | Specific surface area (cm^2/g) |
|---|---|---|---|
| Very course sand | 2.00 - 1.00 | 90 | 11 |
| Coarse sand | 1.00 - 0.50 | 720 | 23 |
| Medium sand | 0.50-0.25 | 5,700 | 45 |
| Fine sand | 0.25-0.10 | 46,000 | 91 |
| Very fine sand | 0.10-0.05 | 722,000 | 277 |
| Silt | 0.05-0.002 | 5,776,000 | 454 |
| Clay | < 0.002 | 90,260,000,000 | 8,000,000 |
Nearly any type of land management will be influenced by texture. Table 7.2 provides a summary of soil management factors related to texture.
| Water relations | Sandy soils | Loamy soils | Clayey soils |
|---|---|---|---|
|
Infiltration—entry of surface water into the
soil. Opposite of run-off potential |
Rapid | Medium to slow |
Very rapid if
cracks present; slow if no cracks |
|
Percolation—internal water drainage and
leaching |
Excessive | Good | Fair-Poor |
| Water Storage—available for plant use | Very low | Medium | High |
|
Aeration—movement of oxygen into root
zone |
Very good | Moderate | Poor |
|
Tillage and Erosion
Tillage power required |
Low | Medium | High |
| Tillability—ease of seed-bed preparation | Easy | Medium | Difficult |
|
Erodibility
Wind erosion hazard |
High | Low | Medium |
| Water erosion hazard | Low | High | Low to medium |
|
Chemical Relations
Fertility potential (nutrient storage) |
Low | Medium | High |
|
Chemical Recommendations—rates per
acre |
Low | Medium | High |
Activity 1: Textural Triangle
Soils with similar distributions of sand, silt, and clay have similar properties and are therefore grouped into the same soil textural class. Twelve textural classes are recognized, and their compositions are designated on a textural triangle (Figure 7.1). Study the arrangement of the triangle. Each corner represents 100% of either sand, silt, or clay, and each represents 0 to 100% of a given fraction. The proportions of sand, silt, and clay define the twelve classes. Figure 7.1 depicts a soil with 20% sand, 25% silt, and 55% clay. Those three lines intersect within the boundaries of the “Clay” textural class, so the soil is a clay soil.
Using that same procedure, determine the textural class names for the soils described in Table 7.3, and plot each result on the textural triangle in Figure 7.2.
| % Sand | % Silt | % Clay | Textural Class | |
|---|---|---|---|---|
| 1. | 33 | 33 | 34 | |
| 2. | 55 | 30 | 15 | |
| 3. | 80 | 5 | 15 | |
| 4. | 25 | 60 | 15 | |
| 5. | - | 20 | 50 | |
| 6. | 60 | - | 30 | |
| 7. | 40 | 40 | - |
Activity 2: Estimating Soil Texture by Feel
A soil scientist often needs to estimate soil texture while in the field or when laboratory data on the amounts of sand, silt, and clay are not available. With practice, you can learn to estimate texture by simply feeling or manipulating a moist sample.
To learn this technique, consider a simplified and generalized version of the textural triangle (Fig. 5.3). This modified triangle consists of three tiers based on approximate clay content. Clays are very cohesive, plastic, and can be easily molded. Clay loams are intermediate in clay content, cohesiveness, and ease of molding. Loams are soils low enough in clay content to possess little cohesiveness and are more difficult to mold. Sands do not form stable forms when molded.
Using the procedure outlined in Figure 7.4, determine the texture on the samples provided in the laboratory. Working the sample at the proper moisture content is very important. The sample must be moistened throughout. Achieving the correct moisture condition may take several minutes. After moistening and mixing soil to the proper consistency, perform the ribbon test, the grittiness test, and the smoothness test as described in the diagram. Try to estimate the texture of the samples, then check the answers provided. After you have calibrated your fingers on the practice samples, determine the texture for the unidentified samples provided, and enter your estimate in Table 7.3.
For the unidentified samples, record the ribbon length, predominant wet feel (smooth, gritty, neither), and textural class names determined by the feel method shown in Table 7.3.
| Does is make a ball? | Does it make a ribbon? | Ribbon Length | Predominate Wet Feel | Textural Class | |
|---|---|---|---|---|---|
| 1. | |||||
| 2. | |||||
| 3. |
Activity 3: Particle Size Analysis by Hydrometer
Particle size analysis is based on the principle that different size particles fall through a fluid at different rates.
A particle falling in a fluid is subjected to 3 forces: gravity, buoyancy, and friction (Figure 7.5). The gravity and buoyancy forces are constant, but the frictional force increases as the velocity increases (like the drag force on an airplane increases as it goes faster). Because of this increasing frictional force, the particle eventually reaches a constant velocity (terminal velocity). Constant velocity occurs when the sum of the forces acting on the particle is zero, or acceleration is zero.
Stoke’s Law is derived by setting up an equation containing the three forces acting on the particle when acceleration is zero:
\[\text{Force of gravity }=\text{ force of buoyancy }+\text{ force of friction}\]These forces are determined from the following relationships:
Force of Gravity
\[\text{Force of gravity }=\text{ mass of particle }\times\text{ acceleration of gravity}\] \[\text{Mass of particle }=\text{ Volume of particle }\times\text{ density of particle}\]Therefore,
\[\text{Force of gravity }=\text{ volume of particle }\times\text{ density of particle}\times\text{ acceleration of gravity}\]Force of Buoyancy
\[\text{Force of buoyancy }=\text{ mass of water displaced by particle }\times\text{ acceleration of gravity}\] \[\text{Mass of displaced water }=\text{ volume of particle }\times\text{ density of water}\]Therefore,
\[\text{Force of buoyancy }=\text{ volume of particle }\times\text{ density of water}\times\text{ acceleration of gravity}\]Friction Force
Friction force is a function of the size of the particle, the velocity of the particle, and the viscosity of the water.
\[\text{Friction force }=6π\times\text{ viscosity }\times\text{ radius of particles}\times\text{ velocity}\]After substituting the appropriate components of the forces into the equation, it can be solved for the terminal velocity:
\[\text{Velocity }=\frac{\text{ particle diameter}^2\times\text{ acceleration due to gravity}\times(\text{ particle density}-\text{ liquid density})}{18\times\text{viscosity of liquid}}\]Note that the larger the diameter of the particle, the faster it settles (sand grains will settle faster than silt particles, which will settle faster than clay particles). Also, the density and viscosity of water vary with temperature, so the velocity of settling will be influenced by the temperature of the water (Figure 7.6).
Stoke’s Law can be simplified by specifying the temperature of the water:
\[\text{Velocity }(\text{cm s}^2)=\text{K}\times(\text{diameter of particle in cm})^2\]Where K (cm -1 s -1 ) is a constant incorporating water density and viscosity and acceleration due to gravity. Because velocity is distance/time, this equation can be solved for time required for a particle of a specified diameter to fall a given distance.
\[\text{Velocity }=\frac{\text{distance}}{\text{time}}=\text{K}\times(\text{diameter})^2\]Therefore,
\[\text{Time}=\frac{\text{distance}}{\text{K}\times(\text{diameter})^2}\]Thus, for a particle of a given diameter, the time required for the particle to fall a specified distance can be calculated.
Consider this example: How long will it take a 0.05-mm particle to fall 10 cm in water at 25°C?
At 25°C, K = 10,000 cm -1 s -1 . Substituting this value and the diameter (0.005 cm) into the above equation yields:
\[\text{Time}=\frac{10\text{ cm}}{10000\text{ cm}^{-1}\text{ s}^{-1}\times(0.005\text{ cm})^2}=\frac{10}{0.25\text{ cm s}^{-1}}=40\text{s}\]Thus, after 40 seconds, the upper 10 cm of a soil-water suspension is completely free of all particles 0.05 mm or larger, so it is free of sand and contains only silt and clay.
Let’s do the same calculation for a 0.002-mm particle, which is the upper limit of the clay range.
\[\text{Time}=\frac{10\text{ cm}}{10000\text{ cm}^{-1}\text{ s}^{-1}\times(0.002\text{ cm})^2}=\frac{10}{0.0004\text{ cm s}^{-1}}=6.94\text{hr}\]Thus, after 6 hours and 56 minutes, the upper 10 cm of a soil-water suspension is free of all particles 0.002 mm or larger (sand and silt), so it contains only clay particles.
As we can see, Stokes Law can be used to determine when a volume of a soil-water suspension will be devoid of soil particles larger than a given size. Then we can measure the concentration of soil remaining in that volume. For example, after 40 seconds, we can measure the concentration of soil in suspension in the upper 10 cm of a suspension and thus determine how much clay + silt are present.
Under natural conditions, sand, silt, and clay particles are bound together in aggregates. These aggregates must be broken down so soil particles act independently of each other. For example, an aggregate of clay particles would behave as a silt particle, a phenomenon we want to avoid.
Dispersion is a 2-step, chemical/mechanical process. First, sodium hexametaphosphate (like the dishwashing detergent “Calgon®”) is added to a soil-water suspension to increase electronegativity of soil clays; it causes a repulsive force between clay particles. Then the suspension is stirred vigorously (milkshake mixer or blender) to assure complete dispersion. The repulsive forces generated by the chemical treatment tend to stabilize the dispersed condition. Dispersion thus assures that aggregated clay particles do not behave like silt-sized or sand-sized particles.
Soil organic matter is an important binding agent, so it first must be removed by oxidation (using hydrogen peroxide, for example). In soils with very low in organic matter, this step is often omitted. It will not be used in this exercise.
After proper dispersion, sand, silt, and clay can be separated and quantified by allowing the particles to settle in water. (NOTE: a sieve is commonly used to quantify sand content of a soil sample, and then a hydrometer to quantify silt and clay content of the remaining particles. We will use a hydrometer for all three particle sizes in this lab activity.)
One method of determining the concentration of soil in suspension is using a hydrometer to measure the density of the suspension. The hydrometer is commonly used in field labs. In this exercise, we’ll use a hydrometer calibrated to read directly in g/L of suspension.
Simplified Hydrometer Procedure
Chemical dispersion (performed in the previous lab)
- Weigh out 30.0 g of dry soil (assume oven-dry) into a 250-ml Erlenmeyer flask.
- Wash sides of flask with distilled water from a wash bottle.
- Add 100 ml of distilled water using a graduated cylinder, and add 10 ml of sodium hexametaphosphate solution (500 g/L) from the dispenser on the sodium hexametaphosphate bottle.
- Swirl to mix.
- Wash sides of flask with distilled water from a wash bottle.
- Cover the flask with Parafilm and label the flask with your lab section and table number as well as soil type. Store the flasks in the location specified by your instructor until next laboratory period.
After chemical dispersion
- Quantitatively transfer the dispersed sample into a metal dispersion cup (a milkshake mixer cup), fill to half full with distilled water, and mix for 5 minutes as directed by your instructor. (“Quantitatively transfer” means to transfer all the sample.)
- Quantitatively transfer the sample into a 1-liter sedimentation cylinder.
- Fill cylinder to the 1000-ml mark with distilled water.
-
Suspend all soil particles in the sedimentation cylinder using one of the following methods:
- Plunger method: Carefully insert stirring plunger and move up and down the full length of the cylinder for 30 seconds, ensuring that all particles are thoroughly mixed. Hold base of cylinder firmly with other hand.
- Stopper method: Place a rubber stopper (one properly sized for the sedimentation cylinders) into the top of the cylinder, and holding both the bottom of the cylinder and the top of the stopper, mix the solution vigorously by inverting the cylinder and turning it right side up repeatedly for 30 seconds.
- Record time when stirring is stopped and plunger removed, or when the stoppered cylinder is returned (quickly but gently) right-side-up to the laboratory bench.
- Immediately insert the hydrometer slowly and carefully. Read the hydrometer (top of the meniscus) at exactly 40 seconds after stirring was stopped or the cylinder was returned to the bench. (Note: if a soil is high in organic matter that was not removed before beginning this experiment, bubbles will form at the surface. To disperse the bubbles, add three to five drops of alcohol to the surface of the solution immediately after inserting the hydrometer.
- Repeat steps 4 through 6 until readings are within 0.5 units of each other. Record the reading on the data sheet.
-
Determine the temperature of the suspension and make corrections for temperature and for the blank as described in Table 7.4:
- For each degree above 20°C, add 0.36 to the hydrometer reading.
- For each degree below 20°C, subtract 0.36 from the hydrometer reading.
- Because the soil suspension also contains sodium hexametaphosphate, each hydrometer reading must be corrected to account for the effect of sodium hexametaphosphate on density. A “blank” cylinder has been set up containing only water and sodium hexametaphosphate. Record the hydrometer reading from this cylinder and subtract that value from all hydrometer readings.
- Your lab instructor has conducted measurements for the 7-hour reading and will provide the data to you to complete your soil texture determination.
- Calculations:
Remember that these hydrometer readings are in g/L. Because the volume in the cylinder is one liter, the readings gives the amount of soil in the cylinder. For example, the 7-hour reading indicates the amount of clay (g) in the cylinder. Therefore, the percent clay is
\[\text{Percent clay}=\frac{\text{corrected 7 hr reading}}{\text{mass of dry sample}}\times100%\]The 40-second reading is used to calculate percent silt + clay:
\[\text{Percent silt }+\text{ clay}=\frac{\text{corrected 40 s reading}}{\text{mass of dry sample}}\times100%\]Then the silt and sand percentages can be determined.
\[\text{Percent silt }=(\text{percent silt}+\text{percent clay})-\text{percent clay}\] \[\text{Percent sand }=100-(\text{percent silt}+\text{percent clay})\]Use Table 7.4 to help determine the textural class of your samples. Label all results in the proper units.
Table 7.5. Soil Texture by Hydrometer Method Data.
| Row | Formula | Soil A | Soil B | Soil C | |
|---|---|---|---|---|---|
| a | Dry weight of soil, g | ||||
| b | Average of 40-second hydrometer readings | ||||
| c | Hydrometer reading from blank | ||||
| d | b-c | Hydrometer reading corrected for blank | |||
| e | Temperature of suspension, first readings | ||||
| f | d + [(e - 20) x 0.36] |
40-second hydrometer reading corrected for
temperature |
|||
| g | 7-hour hydrometer reading | ||||
| h | Hydrometer reading from blank | ||||
| i | g-h | Hydrometer reading corrected for blank | |||
| j | Temperature of suspension, second reading | ||||
| k | i + [(j - 20) x 0.36] |
7-hr hydrometer reading corrected for
temperature |
|||
| l | f ÷ a x 100 | Percent silt + clay | |||
| m | k ÷ a x 100 | Percent clay | |||
| n | l-m | Percent silt | |||
| o | 100 - m - n | Percent sand | |||
| Textural class (from triangle) |
Table adapted from King et al. (2003)
Soil Texture by Hydrometer Method Calculations
Activity 4: Soil Structure
Table 7.6. Summary of soil structure types described in mineral soils.
| Structure Type | Aggregate Description | Usual Location |
|---|---|---|
| Granular |
Relatively nonporous, small and
spheroidal peds; not fitted to adjoining aggregates |
A horizon |
| Platy |
Aggregates are plate-like. Plates
often overlap and impair permeability. |
E horizon |
| Angular blocky |
Block-like peds bounded by
other aggregates whose sharp angular faces form the cast for the ped. The aggregates often break into smaller blocky peds. |
B horizon |
| Subangular blocky |
Block-like peds bounded by
other aggregates whose rounded subangular faces form the cast for the ped. |
B horizon |
| Prismatic |
Column-like peds without
rounded caps. Other prismatic aggregates form the cast for the ped. Some prismatic aggregates break into smaller blocky peds. |
B horizon |
| Columnar |
Column-like peds with rounded
caps bounded laterally by other columnar aggregates that form the cast for the peds. |
B horizon in alkali soils |
Carefully examine the different structure types listed in Table 7.6 and shown in Figure 7.7, then answer the following questions.
How does blocky structure affect permeability of clayey soils?
How would a structureless massive condition affect permeability?
How would manipulation (such as tillage) of a wet clayey soil affect structure?
Assignment: Online Quiz
A quiz will be available online. Please access it as directed by your instructor.


4.5: Projections
Orthogonality
When the angle between two vectors is \(\frac π 2\) or \(90^0\), we say that the vectors are orthogonal. A quick look at the definition of angle (Equation 12 from "Linear Algebra: Direction Cosines") leads to this equivalent definition for orthogonality:
\[(x,y)=0⇔x\; \mathrm{and}\;y\;\mathrm{are\;orthogonal}. \nonumber \]
For example, in Figure 1(a), the vectors \(x=\begin{bmatrix}3\\1\end{bmatrix}\) and \(y=\begin{bmatrix}-2\\6\end{bmatrix}\) are clearly orthogonal, and their inner product is zero:
\[(x,y)=3(−2)+1(6)=0 \nonumber \]
In Figure 1(b), the vectors \(x=\begin{bmatrix}3\\1\\0\end{bmatrix}\), \(y=\begin{bmatrix}−2\\6\\0\end{bmatrix}\), and \(z=\begin{bmatrix}0\\0\\4\end{bmatrix}\) are mutually orthogonal, and the inner product between each pair is zero:
\[(x,y)=3(−2)+1(6)+0(0)=0 \nonumber \]
\[(x,z)=3(0)+1(0)+0(4)=0 \nonumber \]
\[(y,z)=−2(0)+6(0)+0(4)=0 \nonumber \]
Exercise \(\PageIndex{1}\)
Which of the following pairs of vectors is orthogonal:
- \(x=\begin{bmatrix}1\\0\\1\end{bmatrix},y=\begin{bmatrix}0\\1\\0\end{bmatrix}\)
- \(x=\begin{bmatrix}1\\1\\0\end{bmatrix},y=\begin{bmatrix}1\\1\\1\end{bmatrix}\)
- \(x=e_1,y=e_3\)
- \(x=\begin{bmatrix}a\\b\end{bmatrix} , y=\begin{bmatrix}−b\\a\end{bmatrix}\)
We can use the inner product to find the projection of one vector onto another as illustrated in Figure 2. Geometrically we find the projection of \(x\) onto \(y\) by dropping a perpendicular from the head of \(x\) onto the line containing \(y\). The perpendicular is the dashed line in the figure. The point where the perpendicular intersects \(y\) (or an extension of \(y\)) is the projection of \(x\) onto \(y\), or the component of \(x\) along \(y\). Let's call it \(z\).
Exercise \(\PageIndex{1}\)
Draw a figure like Figure 2 showing the projection of y onto x.
The vector \(z\) lies along \(y\), so we may write it as the product of its norm \(||z||\) and its direction vector \(u_y\) :
\[z=||z||u_y=||z||\frac {y} {||y||} \nonumber \]
But what is norm \(||z||\)? From Figure 2 we see that the vector \(x\) is just \(z\), plus a vector \(v\) that is orthogonal to \(y\):
\[x=z+v,(v,y)=0 \nonumber \]
Therefore we may write the inner product between \(x\) and \(y\) as
\[(x,y)=(z+v,y)=(z,y)+(v,y)=(z,y) \nonumber \]
But because \(z\) and \(y\) both lie along \(y\), we may write the inner product \((x,y)\) as
\[(x,y)=(z,y)=(||z||u_y,||y||u_y)=||z||||y||(u_y,u_y)=||z||||y||||u_y||^2=||z||||y|| \nonumber \]
From this equation we may solve for \(||z||=(x,y)||y||\) and substitute \(||z||\) into Equation \(\PageIndex{6}\) to write \(z\) as
\[z=||z||\frac y {||y||}=\frac {(x,y)}{||y||} \frac y {||y||}= \frac {(x,y)}{(y,y)}y \nonumber \]
Equation \(\PageIndex{10}\) is what we wanted–an expression for the projection of \(x\) onto \(y\) in terms of \(x\) and \(y\).
Exercise \(\PageIndex{1}\)
Show that \(||z||\) and \(z\) may be written in terms of \(\cosθ\) for \(θ\) as illustrated in Figure 2:
\(||z||=||x||\cosθ\)
\(z=\frac {||x||\cosθ}{||y||}y\)
Orthogonal Decomposition
You already know how to decompose a vector in terms of the unit coordinate vectors,
\[\mathrm{x}=\left(\mathrm{x}, \mathrm{e}_{1}\right) \mathrm{e}_{1}+\left(\mathrm{x}, \mathrm{e}_{2}\right) \mathrm{e}_{2}+\cdots+\left(\mathrm{x}, \mathrm{e}_{n}\right) \mathrm{e}_{n} \nonumber \]
In this equation, (x, \(e_k\))\(e_k\)(x,ek)ek is the component of \(x\) along \(e_k\), or the projection of \(x\) onto \(e_k\), but the set of unit coordinate vectors is not the only possible basis for decomposing a vector. Let's consider an arbitrary pair of orthogonal vectors \(x\) and \(y\):
\((\mathrm{x}, \mathrm{y})=0\).
The sum of \(x\) and \(y\) produces a new vector \(w\), illustrated in Figure 3, where we have used a two-dimensional drawing to represent \(n\) dimensions. The norm squared of \(w\) is
\[\begin{aligned}
\|\mathrm{w}\|^{2} &=(\mathrm{w}, \mathrm{w})=[(\mathrm{x}+\mathrm{y}),(\mathrm{x}+\mathrm{y})]=(\mathrm{x}, \mathrm{x})+(\mathrm{x}, \mathrm{y})+(\mathrm{y}, \mathrm{x})+(\mathrm{y}, \mathrm{y}) \\
&=\quad\|\mathrm{x}\|^{2}+\|\mathrm{y}\|^{2}
\end{aligned} \nonumber \]
This is the Pythagorean theorem in \(n\) dimensions! The length squared of \(w\) is just the sum of the squares of the lengths of its two orthogonal components.
The projection of \(w\) onto \(x\) is \(x\), and the projection of \(w\) onto \(y\) is \(y\):
\(w=(1) x+(1) y\)
If we scale \(w\) by \(a\) to produce the vector z=\(a\)w, the orthogonal decomposition of \(z\) is
\(\mathrm{z}=a \mathrm{w}=(\mathrm{a}) \mathrm{x}+(\mathrm{a}) \mathrm{y}\).
Let's turn this argument around. Instead of building \(w\) from orthogonal vectors \(x\) and \(y\), let's begin with arbitrary \(w\) and \(x\) and see whether we can compute an orthogonal decomposition. The projection of \(w\) onto \(x\) is found from \(\mathrm{z}=\frac{\|\mathrm{x}\| \cos \theta}{\|\mathrm{y}\|} \mathrm{y}\).
\[w_{x}=\frac{(w, x)}{(x, x)} x \nonumber \]
But there must be another component of \(w\) such that \(w\) is equal to the sum of the components. Let's call the unknown component \(w_y\). Then
\[\mathrm{w}=\mathrm{w}_{\mathrm{x}}+\mathrm{w}_{\mathrm{y}} \nonumber \]
Now, since we know \(w\) and \(w_x\) already, we find \(w_y\) to be
\[\mathrm{w}_{\mathrm{y}}=\mathrm{w}-\mathrm{w}_{\mathrm{x}}=\mathrm{w}-\frac{(\mathrm{w}, \mathrm{x})}{(\mathrm{x}, \mathrm{x})} \mathrm{x} \nonumber \]
Interestingly, the way we have decomposed \(w\) will always produce \(w_x\) and \(w_{>y}\) orthogonal to each other. Let's check this:
\[\begin{array}{l}
\begin{aligned}
\left(\mathrm{w}_{\mathrm{x}}, \mathrm{w}_{\mathrm{y}}\right) &=\left(\frac{(\mathrm{w}, \mathrm{x})}{(\mathrm{x}, \mathrm{x})} \mathrm{x}, \mathrm{w}-\frac{(\mathrm{w}, \mathrm{x})}{(\mathrm{x}, \mathrm{x})} \mathrm{x}\right) \\
&=\frac{(\mathrm{w}, \mathrm{x})}{(\mathrm{x}, \mathrm{x})}(\mathrm{x}, \mathrm{w})-\frac{(\mathrm{w}, \mathrm{x})^{2}}{(\mathrm{x}, \mathrm{x})^{2}}(\mathrm{x}, \mathrm{x}) \\
&=\quad \frac{(\mathrm{w}, \mathrm{x})^{2}}{(\mathrm{x}, \mathrm{x})}-\frac{(\mathrm{w}, \mathrm{x})^{2}}{(\mathrm{x}, \mathrm{x})} \\
&= \qquad \qquad 0
\end{aligned}\\
\end{array} \nonumber \]
To summarize, we have taken two arbitrary vectors, \(w\) and \(x\), and decomposed \(w\) into a component in the direction of \(x\) and a component orthogonal to \(x\).


7.2: Market Ratios
PE Ratio = Market Price per share ÷ Earnings per share
Note: Earnings per share ( “EPS” ) = Net Income divided by the number of shares the company has (“Number of shares outstanding”) . For example, if a company earned $1 million and has 500,000 shares outstanding, its EPS is $2 per share.
This, rather than the dollar price, may be used as a measure of “cheapness” or “expensiveness” of a stock. “Value” stocks are generally characterized by low PE ratios. A V alue Stock may be thought of as such because its price – relative to its earnings – represents a “good buy” for the money, i.e., cheaper.The analyst must be cognizant of the possibility that a low PE may also mean that its future earnings prospects are generally deemed unfavorable.
In contrast, Growth Stocks will reflect (very) high PEs on the basis of generally optimistic views of future earnings growth. Does this mean that earnings will necessarily catch up with price? Let’s say that a company’s PE = $50/$1 = 50 times earnings. That is a very high ratio. If the market anticipates next year’s earnings to double, the stock may be said to be trading at just $50/$2 = 25 times next year’s earnings , a more reasonable ratio.
High PEs are the effect of the high expectations the market has for future growth in a company’s earnings. As a company’s earnings grow in the future, the multiple that one paid for his shares goes down, making the purchase, with the benefit of hindsight, a good choice.
For example, if a company earns $1 per share and sells for $50, its PE will equal 50x earnings, quite high, quite “expensive” for what you get. If , however , the earnings next year do indeed grow to say, $5 per share, the buyer at last year’s price w ho purchased the stock for a meager $50, or only 10x next year’s earnings , got a good deal . Therefore, companies with h igh growth expectations will manifest higher PE ratios than the boring companies with low growth prospects. A stock analyst might have said that the PE ratio is now 50x, but only 10x next year’s expected earnings – if s/he so prophesied.
At the time this is being written, average PE ratios for the S& P 500 Index of stocks are approximately “2 0 times earnings.” Why would someone pay so much for each dollar of earnings?
 It is generally very easy to buy and sell a stock. All one needs to do is place an order with a stockbroker (and pay for it within the required three-day settlement period for stock). There is an added or premium value in this liquidity. That is why “Private Equity” investments can be so profitable. A private company will sell at a much lower PE ratio due to the lack of liquidity.If the company is later sold in an IPO (“initial public offering” – the point at which time a stock is sold to the public for the first time), there will be much profit to go around!
There are certain demerits to the PE Ratio. For instance, Earnings are subject to accrual accounting manipulation s . Moreover, companies with very low or negative earnings yield a meaningless PE ratio. A company with low earnings m a y command a ridiculously high PE. A company with negative earnings will not yield a PE at all – it is not computable!
Dividend Yield = D per share ÷ P per share
One may view this as the Cash-on-cash Return that a stock may provide. If I pay $10 for Yawr Co. stock and it pays an annual dividend of $1, my “ cash-on- cash return ” or dividend yield is 10%. Some industries will reflect higher dividend yields than others . Dividends are usually paid from current net income, but if there are losses , dividends would have to b e paid from retained earnings. Typically, mature, rather than growth companies , pay high dividends. This yield ignores price changes.
It is eminently noteworthy that the dividend yield excludes any capital growth, which is integral to Yawr Co. stock’s Total Return . In other words, the Total Return equals the dividend yield plus the capital gain (or loss) expressed in percentage terms. If you purchased this stock for $10 and sold it at $15, your capital gain would be 50%. The Total Return is: 10% + 50% = 60%.
It is often considered a positive Signal that a company maintains its dividend in the face of losses. This Signaling indicates management’s optimism about future earnings prospects. The dividend itself thus boils down to a “human decision” by the firm’s board of directors, rather than a measure of business performance per se. If the board feels that the future is good, it may choose to pay dividends even if the company is presently losing money.
Dividend Payout Ratio = PR = D Total ÷ NI = DPS ÷ EPS
This ratio indicates the percentage of net income, which is paid out to common shareholders as dividends. If a firm earns $1,000,000 and pays $100,000 in dividends, its payout ratio is 10%. If there is preferred stock, the preferred dividends would first have to be deducted from the net income. The PR formula would be adjusted: PR = (Common Stock Dividends) ÷ (Net Income – Preferred Stock Dividends).
Retention Rate = RR = 1 – PR = (NI – D Total ) ÷ NI = ARE ÷ NI
This ratio indicates how much of the company’s earnings are retained internally for future growth and, as such, together with the firm’s ROE impact the firm’s growth potential. On the next page, we discuss the relationship between earnings retention and firm’s growth. (This will be explicated further below in the “EFN Model.”) Naturally, PR + RR = 1.0 = 100%.


13.1.3: Why Early Failure Can Lead to Success Later
Learning Objectives
By the end of this section, students will be able to:
- Determine several reasons for business failure and explore strategies to overcome them
- Understand the root of fear of failure
- Learn to spot signs of fear of failure and take steps to overcome it
Your goal in your venture is to achieve success—ideally, fairly quickly—but most successful entrepreneurs experience some failures along the way. While these setbacks can be discouraging, they provide lessons and experiences that can lead to eventual success. Whether a new business is a retail store, restaurant, hair salon, consulting company, tech firm, or manufacturing plant, the truth is that many businesses fail within the first couple of years. According to Bloomberg, only 20 percent of new businesses find success within their first eighteen months. Data from entrepreneur.com and the Small Business Administration suggest that only 30 percent of new ventures succeed to the ten-year mark. For women and minorities, the percentages are even lower. Success and longevity are possible, of course, but it takes a lot of work, and sometimes it takes a second or third try.
Entrepreneurs recognize that failure is part of the success of being a business owner. Business failure is the ending of a business due to the lack of goal attainment, which can mean low levels of revenue and profits, or not meeting investors’ expectations. Business failure can result in the loss of assets—such as revenue, equipment, and capital—and can cause trauma for the business owner. However, these failures often help entrepreneurs improve the outcomes for their next business, as they have now learned valuable lessons that can be applied to new projects.
As Jack Ma, billionaire founder of Alibaba (the biggest and most profitable online retailer in China) said, “No matter what one does, regardless of failure or success, the experience is a form of success in itself.” Ma experienced many failures—including being rejected from job opportunities and universities he wanted to attend, and having his pitch rejected by all but one person in a room full of friends. But he persevered, founded Alibaba, learned as he went, and today runs China’s largest online retailer with almost $24 billion in revenue.
Another company, Quirky, launched by Ben Kaufman in 2009, was an initial failure. Quirky was a platform that allowed inventors to submit their ideas to a panel of Quirky experts who would then manufacture the product at a low price and sell it to different markets. Initially, Quirky gained $185 million in funding and had support partnerships from Amazon, Bed Bath & Beyond, and Best Buy. Thousands of people flocked to the site and submitted their inventions to get seen and voted on. Unfortunately, Quirky struggled to sell many of the products at a sustainable profit margin and filed for bankruptcy in 2015 after investors stopped funding the venture. Fortunately, one of its businesses, the smart-home business section, was spun off and sold as Wink in September 2017. The company was rebooted by a new set of entrepreneurs with a better business model.
Common Contributors to Failure
There are some common reasons for failure that often combine to end a business. However, there are often ways to prevent these failures from happening. CB Insights, a company that mines and analyzes data for companies and uses machine learning to help them answer complicated strategic questions, researched the factors that contributed to the failures of 101 startups. Figure <IP_ADDRESS> shows some of the main factors they identified in their research.
In the following discussion, we expand on some of these reasons for failure and offer advice on how to avoid them. Considering the following failures will give you an idea of what may go wrong in a business but will also help you spot and correct them before you launch.
Marketing Failure
Some of the most common failures result from marketing missteps.
- Lack of differentiation of product or service . Creating or running a company with a me-too mentality and without a clear unique selling proposition or difference from other established businesses can be detrimental. Answering questions such as “Why should customers leave other companies to buy my product?” “What is a benefit I provide that no one else does?” and “If I open a yogurt parlor in a commercial area, what will help my business differentiate itself from all the other yogurt, ice cream, dessert, and fast food places?” can help create ways to be special rather than a me-too business. This also needs to be defined in your mission, within your products and their benefits, and a clearly defined competitive advantage that can be translated into a superior value proposition.
- Missing the right customer . Not reaching the target market can lead to failure. Perhaps for your travel tour business, you targeted single people ages 18–34 who like to experience travel in a different way, but maybe your true target—the people who want to book your tours—are couples with children and those 50 and over. You would have sent out marketing communications to the wrong age group. Or, the right message is not being conveyed and your positioning is off, preventing you from connecting with your true consumer. Conducting market research and testing your communication with various targets can help better define your customer.
Management Failure
These are some typical management challenges that can prove detrimental.
- Passion wanes or not the right business . Sometimes the reason a business fails is due to a simple reason: The founder has lost excitement over the business and is no longer interested in making it succeed. Other times, the problem may stem from not having the right focus or idea, which can come from a lack of awareness about trends in the market. Joel Delgado, a young entrepreneur who started El Paso’s first escape room called El Paso Disaster Room 915 in 2015, developed this business with three of his friends right after college. Although the business was very well received by young people who wanted this new form of entertainment, a few years later, he decided to close down the business and pursue a career in teaching. His passion had waned. Sometimes, entrepreneurs get bored with their startup, but there are ways to keep the business exciting by changing the product offerings.
- Fighting founders . Not agreeing over the strategies and direction of a business can deter companies from moving forward. If the founders are not working together and disagree too much, this can help dissolve the business quickly. Differing points of view can be healthy for a business, as these can bring other complementing perspectives, but when conflict doesn’t get resolved, it can be difficult to continue moving forward. Some business partners have found themselves ousted by their business partners because the relationships have suffered a deep strain. This happened to Steve Jobs in 1985 when he was let go by his own company. Often, it is difficult to assess when conflict will arise, so it is recommended that business partners work with people who have the same love for the business and its core values, even if they bring different strengths and temperaments to the table. This may reduce fights and sustain relationships. Having a contract in place that defines who is in charge of what part of the business, their compensation, and what happens if anyone departs, is a good way to ensure that the investment won’t be lost if partners decide to part ways.
- Lack of planning . One of the surest roads to failure is to not have a business plan or marketing plan. Even using the lean startup method requires some sort of planning before embarking on the process. Planning helps business owners flesh out ideas and understand the business better. Following a business plan and a marketing plan can help solidify where the business is headed, whether its viable, and what resources are required to make it work. For example, imagine a business owner who opens a restaurant in an already-crowded market only to realize how much competition existed in the area. If the owner had researched and planned ahead, the venture could have been differentiated through the type of food offered, the location, or different promotional tactics to build clientele. Failure to plan in a situation like this can result in struggling to keep afloat and can lead to successive challenges. For example, the owner might raise prices to compensate for lower-than-expected revenue, but higher prices may drive away an already fragile customer base, and the venture may not be sustainable.
Financial Failure
Cash flow, debt, and capital are just some of the many financial factors that play a large role in startup success or failure.
- Lack of cash . Lacking cash to operate the business can trample its operations and lead to a quick decline. If there is no cash flow, it can also indicate that perhaps the business model is not working. For example, consider our fictional restaurant owner. A lack of customers means a lack of cash flowing in as revenue. Loans can provide some assistance while working to build the clientele, but loans can go only so far. If the business doesn’t grow and develop a sustainable cash flow, its ability to operate will come to an end.
- Too much debt . Significant debt can hurt a business because it needs to pay back its lenders. In the previous example, if the restaurant owner takes on too much debt, it may struggle to pay it back due to its lack of cash flow.
- Lack of capital . Not having sufficient capital can deter the business from expanding or even from meeting customer demand. Having a clear business plan can help determine the financial requirements, which are the estimated sales and profits required for the successful attainment of goals. Financing for the business can come from an entrepreneur’s own savings, bank loans, investors, and even friends and family. Having the right funding to start or grow a business can make or break the business. For instance, our restaurant owner would need enough capital to invest in kitchen equipment, appliances, and furniture to start the business. Undercapitalization in a startup could shut the business down before it even gets up and running. Once the business is running, ensuring that financial goals are met through metrics such as sales and profits can help prevent shortages of cash flow, which is essential to keeping a business alive.
Innovation Failure
Innovation can be tricky because it requires creativity, risk, and often some subjectivity, taking into consideration feelings and intuition in decision making. Lack of innovation and missteps in innovation can be obstacles to success.
- Lack of innovation or failing to change effectively . Companies that don’t change their strategies, technology, or products run the risk of jeopardizing the business. Similarly, those that do change but not with the right adjustments risk failure. The build-measure-learn loop can help avoid these pitfalls by discerning what consumers really want and need by asking for their feedback.
Blockbuster, an instant hit out of the gates in the 1980s, failed to innovate, or pivot its business model, and went out of business. The video store revolutionized the media and movie industry by displaying empty boxes of its titles on shelves categorized by genre (the VHS tapes were kept behind the counter) and keeping late hours for night owls and last-minute weekenders wanting to catch a flick. Using a computer system to track videos as they were rented, most of its profit was made by the fees charged for late movie returns. In 1987, the company was bought by Wayne Huizenga, an American entrepreneur who owned several businesses in different industries and who developed Blockbuster into a successful business by modeling McDonald’s approach. The enterprise grew from twenty locations to over 9,000 locations. Blockbuster was then acquired in 1993 by media giant Viacom in an intricate transaction. As streaming and related technology entered the market, Blockbuster failed to innovate by not making necessary pivots to its delivery of entertainment and so could not compete with new technology. Over the ensuing decade, it became more common to see Blockbuster stores with “store closing” banners, as shown in Figure <IP_ADDRESS>, than it did to see a thriving store. There were other reasons why Blockbuster failed. Embedded with the lack of technology was the lack of fostering a culture of creativity from all of its employees, which hindered innovation. It lacked visionary leadership. It needed a culture of collaboration and teamwork to facilitate communication with stakeholders, such as employees, customers, and upper administration. Finally, it closed all but one of its stores in 2013. By then, Netflix had completely changed the landscape of digital streaming services and left Blockbuster without a ticket to the show.

Proactively Considering Failure
Consider the list of reasons for failure presented in Figure <IP_ADDRESS>. Come up with solutions to each of those reasons, focusing on a business of your choice.
- How would you avoid the failure?
- What strategies or goals can you have in place to avoid the pitfalls of failure?
Fear of Failure
Even the fear of failure can be enough to lead a business to fail. Fear can freeze entrepreneurs and force them into a corner instead of advancing their businesses; it can freeze them from reaching potential clients and being profitable. When companies struggle, the owners may experience many emotions, such as pain, grief, shame, humiliation, self-blame, anger, and hopelessness. Business failure is hard to separate from personal failure, as the business is often associated with the identity of the entrepreneur. Managing these emotions can help business owners heal and continue moving forward to their next business. Reaching out for help to a trusted mentor or therapist can help provide guidance in dealing with these feelings.
There are likewise many stories of entrepreneurs who, despite their fears, continued working toward their goal and were successful. They kept going, determined that fear would not run their lives. Tony Robbins, author and serial entrepreneur, says that people are afraid of failure because it is painful, and people try to avoid pain and suffering at all costs. “Failure,” he says, “is the ultimate loss.” But overcoming fear doesn’t have to be painful if you understand it as a program or detrimental thinking running in your mind. People are hardwired to believe they are not good enough or can’t do it. These patterns can be embedded in people’s minds and can deter creativity and success. Some signs of fear of failure are listed in Table <IP_ADDRESS>.
|
Fear can freeze up entrepreneurs. The important thing is to let go of old thinking patterns and adopt a positive outlook. It also helps to move forward despite fear and have mentors who can help deal with negative thoughts and emotions.
Fear of failure often stems from thoughts of inadequacy or a belief that you don’t have the experience and skills necessary to succeed, that you’re not smart enough, and so on. Thoughts like these can prevent a person from starting or advancing their business. Individuals with a high level of stress about failure tend to lack self-esteem, have anxiety, be perfectionists, and tend to avoid new or unfamiliar things at all costs. Fortunately, these traits and behaviors can be controlled and conquered.
Most entrepreneurs will tell you that, at some point, they had to fight their fears before they had any success.
Most entrepreneurs who succeed will tell you that behind every success there were many failures. Many people in general think that those who are great and successful were born that way. But that is simply not true.
Take Michael Jordan, for example. One of the most successful basketball players of all time, with five championship rings and thousands of points and assists, Jordan often said that he owed his success to failure. “I’ve missed more than 9,000 shots in my career. I’ve lost almost 300 games. Twenty-six times I’ve been trusted to take the game-winning shot and missed.  And that is why I succeed.” Michael Jordan, a kid who didn’t make the varsity team in his high school, got his act together and propelled himself to stardom because he didn’t give up when he failed. The same goes for entrepreneurs: Just because it didn’t work the first time, it doesn’t mean that they can’t try again.
James Dyson, founder and inventor of the Dyson vacuum cleaner, had failed 5,126 times before he came up with his Dual Cyclone vacuum cleaner in 1993, fifteen years after he created the first version. When asked about how failure helped him, he said, “Failure is interesting—it’s part of making progress. You never learn from success, but you do learn from failure. When I created the Dual Cyclone vacuum, I started out with a simple idea, and by the end, it got more audacious and interesting. I got to a place I never could have imagined because I learned what worked and didn’t work.” Dyson also says that he continuously embraces risk and failure, and allows his employees to explore that: “Nothing beats the thrill of invention. Letting people go out and try their ideas, getting them totally involved, and unleashing new thinking. They’re not bound to any methodology—in fact, the stranger and riskier, the better.”
Fear of Failure around the World
According to research, fear of failure is influenced by people’s upbringing and cultural backgrounds. The Global Entrepreneurship Monitor (GEM) , an organization that researches entrepreneurship around the world, has studied this topic. According to its 2018–2019 report, Americans are not as afraid to fail in business as people from many other countries are. This report includes those who would like to start a business but feel stuck because of fear of failure. Fear of failure is highest in Greece, Italy, Russia, and Cyprus, and lowest in Latin America and Africa. Usually, the fear is lowest in countries where there are few jobs and where people have to become entrepreneurial to survive.

GEM Data
Go to the Global Entrepreneurship Monitor at http://www.gemconsortium.org/ and look up the report on entrepreneurship for the United States. Compare it with two other countries.
- What are the differences and similarities between the countries?
- What can you do to address some of the factors in the US that hinder entrepreneurial development? What can be done to improve entrepreneurial conditions in the two other countries you examined?
GEM also reports that women are usually more afraid of failure than men and show less confidence in their abilities (Figure <IP_ADDRESS>). The research also shows that women tend to open ventures in consumer industries, whereas male-driven startups are often in manufacturing and tech sectors, and men receive more capital and incentives to open those businesses. This can be attributed to the lack of women in science and tech industries. Other studies corroborate GEM’s findings, showing that women are more afraid to start a business, they don’t receive as much funding, and feel they have to prove themselves more than men to be taken seriously.
What Failure Teaches You
Some would argue that, in reality, failure does not exist. There is only perceived failure, or obstacles that can become steps up the ladder to a better outcome. The importance in failing lies in learning how to get back up again. As Eric Ries dove into failure with his first company and felt the deep pain and disappointment of having to let go of an unsuccessful idea, he applied his lessons to create his new virtual reality company IMVU. IMVU then became a very successful experiment from which he derived the lean startup method. He was able to pinpoint his shortcomings and find ways to better himself and his performance. With his new company, he was able to get the product to market faster without it being perfect and ask for much-needed feedback from customers that he hadn’t received before.
Converting Lessons Learned to Success
Most successful business owners will tell you that success didn’t come to them easily. They had to persevere in pursuit of their goals to attain them. Converting their failures to lessons often led them to higher success than they had imagined. Entrepreneurs who can turn their obstacles into positive lessons can resurrect from failure.
Take Evan Williams, a visionary who launched a software tool to help users easily publish blogs. That venture, Blogger.com, was launched in 1999 and purchased by Google in 2003. The next year, Williams decided to establish Odeo, a platform for creating and sharing podcasts. Apple had been working on a similar platform and shared it earlier on iTunes to the demise of Odeo. Williams didn’t want to go head to head with iTunes, as this could potentially destroy his company, so he found a way to go in a different direction and establish a new way of sharing status updates and other data. Williams took that failure, and with a few friends, co-founded Twitter. We all know how that worked out: Twitter is one of the most popular microblogging tech platforms.
LINK TO LEARNING
Check out Evan Williams’s TedTalk where he takes you through iterations from actual Twitter users.
Kathryn Minshew, shown in Figure <IP_ADDRESS>, also experienced failure but has learned from it. In 2010, she quit her job to create Pretty Young Professionals (PYP). She invested $25,000 of her own money to develop a networking platform for women who were smart and passionate about their careers. After a few months, she and the three other co-founders started to disagree on some issues, including their approach to advertising. As they hadn’t completely formalized the ownership of PYP, two of the founders lost their investments, including Minshew. After experiencing the loss of her investment, her friendships, and the original idea, Minshew persevered with a new opportunity and established The Muse. This platform is similar to PYP but added job listings, workshops, and advice. Currently, the site has over 4 million users and is a major competitor of LinkedIn. Minshew didn’t let her first disappointment prevent her from starting a new venture. She applied the lessons learned in building the original platform to the creation of a new one—with the right contracts and right people in place. Minshew now speaks at business conferences and events, sharing her mistakes and how she used them to lift herself up to launch a new venture properly. She hopes that her experience helps other entrepreneurs avoid the pitfalls she went through so they can be as successful.
Being able to learn from different types of failure is as important as experiencing success because you get to know your strengths and how to leverage them. Failures can be seen as stepping stones rather than embarrassments. Feeling comfortable takes practice. Table <IP_ADDRESS> shares several tips on dealing with these dynamics.
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Coping with Fear
When coping with fear of failure, there are two strategies that entrepreneurs can use. The first way to reduce fear of failure is to modify your business strategy by changing the target outcomes for your business. Instead of aiming to make $50,000 in the first six months, you can be open to the idea of a leaner start, learning about your customers and how best to serve them. Learning/growth goals can be just as valuable as hitting a revenue target in terms of overall success and longevity. Also, if a goal is not met, don’t consider it a total failure: Refocus and ask yourself what you learned from the experience and how to apply that knowledge to a revised goal. More than likely you will find this experience has enhanced your toolbox to help you make changes within the business or start a new one.
A second way to reduce high stress is to practice some sort of meditation or breathing exercise that can help lower the anxiety created by fearful thinking. Reaching out to a mentor, support group, therapist, or counselor can also help alleviate fearful thinking.


14.12: Ratio Analysis of Financial Information
Learning Outcomes
- Perform a ratio analysis on a financial statement
Ratio analysis can be used in a variety of ways to glean information about the financial health of a business.
Ok, so ratios are another amazing way to notice variances in assets, liabilities, income and expenses. There are tons of different ratios we could look at but let’s take a couple and examine them for Simply Yoga. Take a look at their balance sheet.
| Current Assets : | |
| Cash | 9550 |
| Accounts Receivable | 900 |
| Prepaid Expenses | 1100 |
| Total Current Assets | 11550 |
| Property and Equipment : | |
| Yoga Props (less accum depr) | 1500 |
| Total property and Equip. | 1500 |
| Total Assets | 13050 |
| Current Liabilities: | |
| Accounts Payable | 710 |
| Payroll Taes Payable | 672 |
| Payroll Taxes Payable | 1382 |
| Total Current Liabilities | |
| Long Term Liabilities | |
| Loan Payable | 6500 |
| Stockholder’s Equity | |
| Common Stock | 1000 |
| Retained Earnings | 4186 |
| Total Equity | 5168 |
| Total Liabilities | 13050 |
Let’s talk first about the working capital ratio. The formula is:
Working capital= current assets−current liabilities
$10,168=$11,550−$1382
So, this shows that Simply Yoga has plenty of funds to pay current liabilities, which is a good thing! But , it also shows that they are holding more funds in a very liquid account, which may be better used to pay off any higher interest debt, such as their loan payable. This is an area for review, right?
The current ratio is another way to look at the ability of a company to cover short term debt.
Current Ratio = Current assets/Current Liabilities
8.36=$11,550/$1,382
What this tells us is that Simply Yoga has enough current assets to cover their current liabilities 8.36 times. Again , this is a good thing, unless they are paying a crazy amount of interest somewhere else. Might that cash be better used to pay off that loan they have sitting on the books?
Learn More
There are tons of ratios out there, that depending on the company and what you are looking for, may be helpful! Check out this great resource for ratio analysis . More ratio formulas can be found here . These ratios may come in handy, so go ahead and print out the list!
What might be some other ratios you would consider for Simply Yoga from the list?
Practice Questions