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Solution for Let a Be a Square Matrix Such that a 2 − a + I = O , Then Write a − 1 Interms of A. - CBSE (Commerce) Class 12 - Mathematics

Question

Let A be a square matrix such that $A^2 - A + I = O$, then write $A^{- 1}$  interms of A.

Solution

$\text{ Given: } A^2 - A + I = O$
$A^{- 1} \left( A^2 - A + I \right) = A^{- 1} O (\text{ Pre - multiplying both sides because }A^{- 1}\text{ exists})$
$\left( A^{- 1} A^2 \right) - \left( A^{- 1} A \right) + A^{- 1} I = O ( A^{- 1} O = O)$
$\Rightarrow A - I + A^{- 1} = O ( A^{- 1} I = A^{- 1} )$
$\Rightarrow A^{- 1} = I - A$

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Solution Let a Be a Square Matrix Such that a 2 − a + I = O , Then Write a − 1 Interms of A. Concept: Determinant of a Square Matrix.
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Mathematics » Functions III » Exponential Functions

# Laws of Logarithms

## Laws of Logarithms

In earlier grades, we used the following exponential laws for working with exponents:

• $${a}^{m} \times {a}^{n}={a}^{m+n}$$
• $$\cfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}$$
• $${(ab)}^{n}={a}^{n}{b}^{n}$$
• $${\left(\cfrac{a}{b}\right)}^{n}=\cfrac{{a}^{n}}{{b}^{n}}$$
• $${({a}^{m})}^{n}={a}^{mn}$$

where $$a > 0$$, $$b > 0$$ and $$m, n \in ℤ$$.

The logarithmic laws are based on the exponential laws and make working with logarithms much easier.

### Logarithmic laws:

• $$\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0)$$
• $$\log_{a}{x} = \cfrac{\log_{b}{x}}{\log_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)$$
• $$\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$$
• $$\log_{a}{\cfrac{x}{y}} = \log_{a}{x} – \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)$$

The last two logarithmic laws in the list above are not covered in this section. They are discussed at the end of the tutorial and are included for enrichment only.

### Logarithmic law:

$\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0 )$

\begin{align*} \text{Let } {\log}_{a}{x} &= m \ldots (1)\qquad (x > 0 ) \\ \therefore x &= a^{m} \\ \therefore (x)^{b} &= ( a^{m} )^{b} \\ \therefore x^{b} &= a^{bm} \\ \text{Change to logarithmic form: }\log_{a}(x^{b}) &= bm \\ \text{And subst}: \quad m &= {\log}_{a}{x} \\ \therefore {\log}_{a}{x^{b}} &= b{\log}_{a}{x} \end{align*}

In words: the logarithm of a number which is raised to a power is equal to the value of the power multiplied by the logarithm of the number.

## Example

### Question

Determine the value of $$\log_{3}{{27}^{4}}$$.

### Use the logarithmic law to simplify the expression

\begin{align*} \log_{3}{{27}^{4}} &= 4 \log_{3}{{27}}\\ &= 4\log_{3}{{3}^{3}} \\ &= (4 \times 3) \log_{3}{3}\\ &= 12 (1) \\ &= 12 \end{align*}

$$\log_{3}{{27}^{4}} = 12$$

### Special case:

$\log_{a}{\sqrt[b]{x}} = \cfrac{\log_{a}{x}}{b} \qquad (x > 0 \text{ and } b > 0 )$

The following is a special case of the logarithmic law $$\log_{a}{x^{b}} = b \log_{a}{x}$$:

\begin{align*} \log_{a}{\sqrt[b]{x}} &= \log_{a}{{x}^{\frac{1}{b}}} \\ &= \cfrac{1}{b} \log_{a}{x} \\ &= \cfrac{\log_{a}{x}}{b} \end{align*}

### Logarithmic law:

$${\log}_{a}x=\cfrac{{\log}_{b}{x}}{{\log}_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)$$

It is often necessary or convenient to convert a logarithm from one base to another base. This is referred to as a change of base.

\begin{align*} \text{Let } \quad \log_{a}{x} &= m \\ \therefore x &= a^{m} \\ \text{Consider the fraction: } \quad & \cfrac{\log_{b}{x}}{\log_{b}{a}} \\ \text{Substitute } x = a^{m}: \quad \cfrac{\log_{b}{x}}{\log_{b}{a}} &= \cfrac{\log_{b}{a^{m}}}{\log_{b}{a}} \\ &= m ( \cfrac{\log_{b}{a}}{\log_{b}{a}} ) \\ &= m (1) \\ \therefore \cfrac{\log_{b}{x}}{\log_{b}{a}}&= \log_{a}{x} \end{align*}

### Special applications:

\begin{align*} (1) \qquad \log_{a}{x} &= \cfrac{\log_{x}{x}}{\log_{x}{a}} \\ \therefore \log_{a}{x} &= \cfrac{1}{\log_{x}{a}} \\ & \\ & \\ (2) \qquad \log_{a}{\cfrac{1}{x}} &= \log_{a}{x^{-1}} \\ \therefore \log_{a}{\cfrac{1}{x}} &= – \log_{a}{x} \end{align*}

## Example

### Question

Show: $$\log_{2}{8} = \cfrac{\log{8}}{\log{2}}$$

### Simplify the right-hand side of the equation

\begin{align*} \text{RHS } &= \cfrac{\log{8}}{\log{2}} \\ &= \cfrac{\log{2^{3}}}{\log{2}} \\ &= 3 ( \cfrac{\log{2}}{\log{2}} ) \\ &= 3(1) \\ &= 3 \end{align*}

### Simplify the left-hand side of the equation

\begin{align*} \text{LHS } &= \log_{2}{8} \\ &= \log_{2}{2^{3}} \\ &= 3 \log_{2}{2} \\ &= 3(1) \\ &= 3 \end{align*}

We have shown that $$\log_{2}{8} = \dfrac{\log{8}}{\log{2}} = 3$$.

## Example

### Question

If $$a = \log{2}$$ and $$b = \log{3}$$, express the following in terms of $$a$$ and $$b$$:

1. $$\log_{3}{2}$$
2. $$\log_{2}{\cfrac{10}{3}}$$

### Use a change of base to simplify the expressions

1. \begin{align*} \log_{3}{2} &= \cfrac{\log{2}}{\log{3}} \\ &= \cfrac{a}{b} \end{align*}
2. \begin{align*} \log_{2}{\cfrac{10}{3}} &= \cfrac{\log{\cfrac{10}{3}}}{\log{2}} \\ &= \cfrac{\log{10} – \log{3}}{\log{2}} \\ &= \cfrac{1- b}{a} \end{align*}

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# 6.2: Principles of Thermal Processing of Packaged Foods

Ricardo Simpson

Departamento de Ingeniería Química y Ambiental, Universidad Técnica Federico Santa María, Valparaíso, Chile

Centro Regional de Estudios en Alimentos y Salud (CREAS) Conicyt-Regional GORE Valparaíso Project R17A10001, Curauma, Valparaíso, Chile

Helena Nuñez

Departamento de Ingeniería Química y Ambiental, Universidad Técnica Federico Santa Maria, Valparaíso, Chile

Cristian Ramírez

Departamento de Ingeniería Química y Ambiental, Universidad Técnica Federico Santa María, Valparaíso, Chile

Centro Regional de Estudios en Alimentos y Salud (CREAS) Conicyt-Regional GORE Valparaíso Project R17A10001, Curauma, Valparaíso, Chile

 Key Terms
 Heat transfer Bacterial inactivation Food sterilization Microorganism heat resistance Decimal reduction time Commercial sterilization

### Introduction

Thermal processing of foods, like cooking, involves heat and food. However, thermal processing is applied to ensure food safety and not necessarily to cook the food. Thermal processing as a means of preservation of uncooked food was invented in France in 1795 by Nicholas Appert, a chef who was determined to win the prize of 12,000 francs offered by Napoleon for a way to prevent military food supplies from spoiling. Appert worked with Peter Durand to preserve meats and vegetables encased in jars or tin cans under vacuum and sealed with pitch and, by 1804, opened his first vacuum-packing plant. This French military secret soon leaked out, but it took more than 50 years for Louis Pasteur to provide the explanation for the effectiveness of Appert’s method, when Pasteur was able to demonstrate that the growth of microorganisms was the cause of food spoilage.

The preservation for storage by thermal treatment and removal of atmosphere is known generically as canning, regardless of what container is used to store the food. The basic principles of canning have not changed dramatically since Appert and Durand developed the process: apply enough heat to food to destroy or inactivate microorganisms, then pack the food into sealed or “airtight” containers, ideally under vacuum. Canned foods have a shelf life of one to four years at ordinary temperatures, making them convenient, affordable, and easy to transport.

Outcomes

After reading this chapter, you should be able to:

• • Identify the role of heat transfer concepts in thermal processing of packaged foods
• • Describe the principles of commercial sterilization of foods
• • Describe the inactivation conditions needed for some example microorganisms important for food safety
• • Define some sterilization criteria for specific foods
• • Apply, in simple form, the main thermal food processing evaluation techniques

### Concepts

The main concepts used in thermal processing of foods include: (a) heat transfer; (b) heat resistance of microorganisms of concern; and (c) bacterial inactivation.

#### Heat Transfer

The main heat transfer mechanisms involved in the thermal processing of packaged foods are convection and conduction. Heat transfer by convection occurs due to the motion and mixing of flows. The term natural convection refers to the case when motion and mixing of flow is caused by density differences in different locations due to temperature gradients. The term forced convection refers to the case when motion and mixing of flow is produced by an outside force, e.g., a fan. Heat transfer by conduction occurs when atoms and molecules collide, transferring kinetic energy. Conceptually, atoms are bonded to their neighbors, and if energy is supplied to one part of the solid, atoms will vibrate and transfer their energy to their neighbors and so on.

The main heat transfer mechanisms involved in the thermal processing of packaged foods are shown in Figure 6.2.1. Although the figure shows a cylindrical can (a cylinder of finite diameter and height), a similar situation will arise when processing other types of packaging such as glass containers, retortable pouches, and rigid and semi-rigid plastic containers. In general, independent of shape, food package sizes range from 0.1 L to 5 L (Holdsworth and Simpson, 2016).

The main mechanism of heat transfer from the heating medium (e.g., steam or hot water) to the container or packaging is convection. Then heat transfers by conduction through the wall of the container or package. Once inside the container, heat transfer through the covering liquid occurs by convection, and in solid foods mainly by conduction. In case of liquid foods, the main mechanism is convection.

The rate of heat transfer in packaged foods depends on process factors, product factors, and package types. Process factors include retort temperature profile, process time, heat transfer medium, and container agitation. Product factors include food composition, consistency, initial temperature, initial spore load, thermal diffusivity, and pH. Factors related to package type are container material, because the rate of heat transfer depends on thermal conductivity and thickness of the material, and container shape, because the surface area per unit volume plays a role in the heat penetration rate.

For liquid foods, the heating rate is determined not only by the thermal diffusivity α, but also by the viscosity. The thermal diffusivity is a material property that represents how fast the heat moves through the food and is determined as:

$a=K_{t}/(\rho\ C_{p})$

where α = thermal diffusivity (m2/s)

• Kt = thermal conductivity (W/m-K)
• ρ = density (kg/m3)
• Cp = specific heat (W/s-kg-K)

It is extremely difficult to develop a theoretical model for the prediction of a time-temperature history within the packaging material. Therefore, from a practical point of view, a satisfactory thermal process (i.e., time-temperature relationship) is usually determined using the slowest heating point, the cold spot, inside the container.

#### Heat Resistance of Microorganisms of Concern

The main objective in the design of a sterilization process for foods is the inactivation of the microorganisms that cause food poisoning and spoilage. In order to design a safe sterilization process, the appropriate operating conditions (time and temperature) must be determined to meet the pre-established sterilization criterion. To establish this criterion, it is necessary to know the heat resistance of the microorganisms (some examples are given in Table 6.2.1), the thermal properties of the food and packaging, and the shape and dimensions of the packaged food. From these, it is possible to determine the retort temperature and holding time (that is, the conditions for inactivation), how long it will take to reach that temperature (the come-up time), and how long it will take to cool to about 40°C (the cooling time) (Holdsworth and Simpson, 2016).

Table $$\PageIndex{1}$$: Some typical microorganisms heat resistance data (Holdsworth and Simpson, 2016).
Organism Conditions for Inactivation

Vegetative cells

10 min at 80°C

Yeast ascospores

5 min at 60°C

Fungi

30–60 min at 88°C

Thermophilic organisms:

Bacillus stearothermophilus

4 min at 121.1°C

Clostridium thermosaccharolyticum

3–4 min at 121.1°C

Mesophilic organisms:

Clostridium botulinum spores

3 min at 121.1°C

Clostridium botulinum toxins Types A & B

0.1–1 min at 121.1°C

Clostridium sporogenes

1.5 min at 121.1°C

Bacillus subtilis

0.6 min at 121.1°C

The pH of the food is extremely relevant to the selection of the sterilization process parameters, i.e., retort temperature and holding time, because microorganisms grow better in a less acid environment. That is why the standard commercial sterilization process is based on the most resistant microorganism (Clostridium botulinum) at the worst-case scenario conditions (higher pH) (Teixeira et al., 2006). The microorganism heat resistance is greater in low-acid products (pH ≥ 4.5–4.6). On the other hand, medium-acid to acidic foods require a much gentler heat treatment (lower temperature) to meet the sterilization criterion. Based on that, foods are classified into three groups:

• low-acid products: pH > 4.5–4.6 (e.g., seafood, meat, vegetables, dairy products);
• medium-acid products: 3.7 < pH < 4.6 (e.g., tomato paste);
• acidic products: pH < 3.7 (e.g., most fruits).

#### Bacterial Inactivation

Abundant scientific literature supports the application of first-order kinetics to quantify bacterial (spores) inactivation as (Esty and Meyer, 1922; Ball and Olson, 1957; Stumbo, 1973, Holdsworth and Simpson, 2016):

$(\dfrac{dN}{dt})_{I} = -kN$

where $$N$$ = viable bacterial (microbial) concentration (microorganisms/g) after process time t

• t = time
• I = inactivation
• k = bacterial inactivation rate constant (1/time)

Instead of k, food technologists have utilized the concept of decimal reduction time, D, defined as the time to reduce bacterial concentration by ten times. In other words, D is the required time at a specified temperature to inactivate 90% of the microorganism’s population. A mathematical expression that relates the rate constant, k, from Equation 6.2.2 to D is developed by separating variables and integrating the bacterial concentration from the initial concentration, N0, to N0/10 and from time 0 to D, therefore obtaining:

$k=\dfrac{ln\ 10}{D} = \dfrac{2.303}{D}$

or

$D = \dfrac{ln\ 10}{k} = \dfrac{2.303}{k}$

where k = reaction rate constant (1/min)

D = decimal reduction time (min)

A plot of the log of the survivors (log N) against D is called a survivor curve (Figure 6.2.2). The slope of the line through one log cycle (decimal reduction) is −1/D and

$log\ N = log\ N_{0} - \dfrac{t}{D}$

where N = number of survivors

N0 = N at time zero, the start of the process

#### Temperature Dependence of the Decimal Reduction Time, D

Every thermal process of a food product is a function of the thermal resistance of the microorganism in question. When the logarithm of the decimal reduction time, D, is plotted against temperature, a straight line results. This plot is called the thermal death time (TDT) curve (Figure 6.2.3). From such a plot, the thermal sensitivity of a microorganism, z, can be determined as the temperature change necessary to vary TDT by one log cycle.

Bigelow and co-workers (Bigelow and Esty, 1920; Bigelow, 1921) were the first to coin the term thermal death rate to relate the temperature dependence of D. Mathematically, the following expression has been used:

$log\ D = log\ D_{ref}-\dfrac{T-T_{ref}}{z}$

or

$D = D_{ref}10^{\dfrac{T-T_{ref}}{z}}$

where D = decimal reduction time at temperature T (min)

Dref = decimal reduction time at reference temperature Tref (min)

z = temperature change necessary to vary TDT by one log cycle (°C), e.g., normally z = 10°C for Clostridium botulinum

T = temperature (°C)

Tref = reference temperature (normally 121.1°C for sterilization)

The D value is directly related to the thermal resistance of a given microorganism. The more resistant the microorganism to the heat treatment, the higher the D value. On the other hand, the z value represents the temperature dependency but has no relation to the thermal resistance of the target microorganism. Then, the larger the z value the less sensitive the given microorganism is to temperature changes. D values are expressed as DT. For example, D140 means the time required to reduce the microbial population by one log cycle when the food is heated at 140°C.

#### Food Sterilization Criterion and Calculation

Sterilization means the complete destruction or inactivation of microorganisms. The food science and engineering community has accepted the utilization of a first-order kinetic for Clostridium botulinum inactivation (Equation 6.2.2). Again, this pathogen is the target microorganism in processes that use heat to sterilize foods. Theoretically, the inactivation time needed to fully inactivate Clostridium botulinum is infinite. According to Equations 6.2.2 and 6.2.3 and assuming a constant process temperature and that k is constant, the following expression is obtained:

$N_{f} = N_{0}e^{-kt} = N_{0}e^{-\dfrac{ln10}{D}t}$

This equation shows that the final concentration of Clostridium botulinum (Nf) tends to zero when time (t) tends to infinity; therefore, it is not possible to reach a final concentration equal to zero for the target microorganism. Thus, it is necessary to define a sterilization criterion (or commercial sterilization criterion) to design a process that guarantees a safe product within a finite time.

The level of microbial inactivation, defined by the microbial lethality value or cumulative lethality, is the way in which the sterilization process is quantified. Specifically, the sterilizing value, denoted by F0, is the required time at 121.1°C to achieve 12 decimal reductions (12D). In other words, F0 is the time required to reduce the initial microorganism concentration from N0 to N0/1012 at the process temperature of 121.1°C.

The 12D sterilization criterion is an extreme process (i.e., overkill) designed to ensure no cells of C. botulinum remain in the food and, therefore, prevent illness or death. According to the FDA (1972), the minimum thermal treatment for a low-acid food should reach a minimum F0 value of 3 min (that is larger than 12D; D for C. botulinum at 121.1°C is 0.21 min, then 12 × 0.21 = 2.52 min, which is lower than 3 min). Thus, a thermal process for commercial sterilization of a food product should have an F0 value greater than 3 minutes.

The F0 attained for a food can be calculated easily when the temperature at the center of the food during the thermal processing is known by:

$F_{0}=\int^{t}_{0}10^{\dfrac{T-T_{ref}}{z}}dt$

where F0 = cumulative lethality of the process from time 0 to the end of the process (t)

T = temperature measured at the food cold spot, which is the place in the food that heats last

Tref = temperature of microorganism reference; for sterilization of low-acid foods, Tref = 121.1°C for C. botulinum

z = temperature change necessary to reduce D value by ten times; in the case of sterilization of low-acid foods, z = 10°C for C. botulinum

t = process time to reach F0

Equation 6.2.9 can be calculated according to the general method proposed by Bigelow and co-workers 100 years ago (Bigelow et al., 1920; Simpson et al., 2003).

If the food is heated instantaneously to 121.1°C and maintained at this temperature for 3 min, then the F0 value for this process will be 3 min. From Equation 6.2.9,

$$F_{0}=\int^{t}_{0}10^{\dfrac{121.1-121.1}{10}}dt=\int^{t}_{0}10^{0}dt=\int^{3}_{0}1dt$$

Since the time interval is between 0 to 3 min, then the integral solution is 3 min or $$\int^{3}_{0}1dt$$= 3 – 0 = 3. However, in practice, due to the resistance of the food to the transfer of heat, the thermal sterilization process requires a longer time in order to get a F0 ≥ 3 min, because a significant part of the processing time is needed to raise the cold-spot temperature of the food and later to cool the food.

### Applications

#### Commercial Sterilization Process

A general, simplified flow diagram for a typical commercial canning factory is presented in Figure 6.2.4.

Stage 1: Selecting and preparing the food as cleanly, rapidly, and precisely as possible. Foods that maintain their desirable color, flavor, and texture through commercial sterilization include broccoli, corn, spinach, peas, green beans, peaches, cherries, berries, sauces, purees, jams and jellies, fruit and vegetable juices, and some meats (Featherstone, 2015). The preparation must be performed with great care and with the least amount of damage and loss to minimize the monetary cost of the operation. If foods are not properly handled, the effectiveness of the sterilization treatment is compromised.

Stage 2: Packing the product in hermetically sealable containers (jars, cans, or pouches) and sealing under a vacuum to eliminate residual air. A less common approach is to sterilize the food first and then aseptically package it (aseptic processing and packaging of foods).

Stage 3: Stabilizing the food by sterilizing through rigorous thermal processing (i.e., high temperature to achieve the correct degree of sterilization or the target destruction of the microorganisms present in the food), followed by cooling of the product to a low temperature (about 40°C), at which enzymatic and chemical reactions begin to slow down.

Stage 4: Storing at a temperature below 35°C, the temperature below which food-spoilage organisms cannot grow.

Stage 5: Labeling, secondary packaging, distribution, marketing, and consumption. Although not part of the thermal process per se, this stage addresses the steps required for commercialization of the treated foods.

Stage 3, thermal processing, is the focus of this chapter. The aim of the thermal process is to inactivate, by the effect of heat, spores and microorganisms present in the unprocessed product. The thermal process is performed in vessels known as retorts or autoclaves to achieve the required high temperatures (usually above 100°C).

As depicted in Figure 6.2.5, a typical sterilization process has three main steps: come-up time, operator process time, and cooling. The first step, the come-up time (CUT), is the time required to reach the specified retort temperature (TRT), i.e., the target temperature in the retort. The second step is the holding time (Pt), also called operator process time, which is the amount of time that the retort temperature must be maintained to ensure the desired degree of lethality. This depends on the target microorganism or the expected microbiological contamination. The final step is the cooling, when the temperature of the product is decreased by introducing cold water into the retort. The purpose of cooling the food is to minimize the excessive (heat) processing of the food, and avoid the risk of thermophilic microorganism development. During the cooling cycle, it may be necessary to inject sterile air into the food packaging to avoid sudden internal pressure drops and prevent package deformation.

The concepts described in this chapter describe the key principles for applying a thermal process to packaged food to achieve the required lethality for food safety. These concepts can be used to design a thermal process to ensure adequate processing time and food safety while avoiding over processing the packaged food. This should ensure safe, tasty, and nutritious packaged foods.

### Examples

Example $$\PageIndex{1}$$

Example 1: Calculation of microbial count after a given thermal process

Problem:

The D120°C value for a microorganism is 3 minutes. If the initial microbial contamination is 1012 cells per gram of product, how many microorganisms will remain in the sample after heat treatment at 120°C for 18 minutes?

Solution

Calculate the number of remaining cells using Equation 6.2.5 with N0 = 1012 cells/g, t = 18 minutes, and D120°C = 3 minutes.

From Equation 6.2.5,

$$log\ N_{(t)} =log\ N_{0}-\dfrac{t}{D}$$

$$log\ N_{(18)} =log\ 10^{12}\dfrac{\text{cells}}{g}-\dfrac{18\text{ min}}{3\text{ min}}$$

Solving for N(18) yields:

$$N_{(18)} =10^{6}\text{ cells/g}$$

Discussion:

Starting with a known microbial concentration (N0), the final concentration of a specific microorganism for a given thermal process at constant temperature can be calculated if the thermal resistance of the microorganism at a given temperature is known. In this case, D120°C = 3 min.

Example $$\PageIndex{2}$$

Example 2: Calculation of z value for a particular microorganism

Problem:

D of a given bacterium in milk at 65°C is 15 minutes. When a food sample that has 1010 cells of the bacterium per gram of food is heated for 10 minutes at 75°C, the number of survivors is 2.15 × 103 cells. Calculate z for this bacterium.

Solution

First, calculate D at the process temperature of 75°C, D75°C, using Equation 6.2.5. Then calculate z using Equation 6.2.6 with D65°C = 15 minutes, N0 = 1010 cells/g, and t = 10 minutes at T = 75°C.

$$log\ N_{(t)} = log\ N_{0} - \dfrac{t}{D}$$ (Equation $$\PageIndex{5}$$)

$$log\ 2.15\times10^{3} \text{ cells/g} = log\ 10^{10} \text{ cells/g}-\dfrac{10\text{ min}}{D_{75^\circ C}}$$

and D75°C = 1.5 min.

To calculate z, recall Equation 6.2.6:

$$log\ D = log\ D_{ref} - \dfrac{T-T_{ref}}{z}$$

Solving for z, Equation 6.2.6 can be expressed as:

$$z = \dfrac{\Delta T}{log(\dfrac{D_{1}}{D_{2}})}$$

with ∆T = (75 – 65)°C, D1 = D65°C and D2 = D75°C,

$$z = \dfrac{75-65}{log(\dfrac{15}{1.5})}=10 ^\circ C$$

Discussion:

As previously explained, the z value represents the change in process temperature required to reduce the D value of the target microorganism by ten times. In this case, the z value is 10°C and accordingly the D value was reduced 10 times, from 15 minutes to 1.5 minutes.

Example $$\PageIndex{3}$$

Example 3: Lethality of thermal processing of a can of tuna fish

Problem:

Table 6.2.2 presents the values of temperature measured in the retort (TRT) and the temperature measured at the cold spot of a can of tuna fish (Tcold spot) during a thermal process. The total process time was 63 min until the product was cold enough to be withdrawn from the retort.

1. (a) Determine CUT (the time required to come up to TRT), operator process time Pt, and cooling time.
2. (b) Determine the lethality value (F0) attained for the can of tuna fish.
Table $$\PageIndex{2}$$: Retort temperature (TRT) and cold spot (Tcold spot) during thermal processing of tuna fish in a can.
Time (min) TRT (°C) Tcold spot (°C)

0.97

29.7

45.0

1.97

39.7

45.0

2.97

49.7

45.0

3.97

59.7

45.0

4.97

69.7

44.9

5.97

79.7

44.9

6.97

89.7

44.8

7.97

99.7

44.7

8.97

109.7

44.7

9.97

119.7

44.8

10.97

120.0

45.0

11.97

120.0

45.4

12.97

120.0

46.0

13.97

120.0

46.9

14.97

120.0

48.0

15.97

120.0

49.3

16.97

120.0

50.8

17.97

120.0

52.6

18.97

120.0

54.4

19.97

120.0

56.4

20.97

120.0

58.5

21.97

120.0

60.6

22.97

120.0

62.8

23.97

120.0

65.0

24.97

120.0

67.1

25.97

120.0

69.3

26.97

120.0

71.4

27.97

120.0

73.5

28.97

120.0

75.5

29.97

120.0

77.5

30.97

120.0

79.4

31.97

120.0

81.2

32.97

120.0

83.0

33.97

120.0

84.7

34.97

120.0

86.3

35.97

120.0

87.9

36.97

120.0

89.4

37.97

120.0

90.8

38.97

120.0

92.2

39.97

120.0

93.6

40.97

120.0

94.8

41.97

120.0

96.0

42.97

120.0

97.2

43.97

120.0

98.3

44.97

120.0

99.3

45.97

120.0

100.3

46.97

120.0

101.3

47.97

120.0

102.2

48.97

120.0

103.0

49.97

120.0

103.9

50.97

120.0

104.7

51.97

120.0

105.4

52.97

120.0

106.1

53.97

120.0

106.8

54.97

120.0

107.4

55.97

120.0

108.0

56.97

120.0

108.6

57.97

120.0

109.2

58.97

120.0

109.7

59.97

120.0

110.2

60.97

120.0

110.7

61.97

120.0

111.1

62.97

120.0

111.6

63.97

120.0

112.0

64.97

120.0

112.4

65.97

120.0

112.8

66.97

120.0

113.1

67.97

120.0

113.4

68.97

120.0

113.8

69.97

120.0

114.1

70.97

120.0

114.4

71.97

120.0

114.6

72.97

120.0

114.9

73.97

120.0

115.2

74.97

120.0

115.4

76

25.0

115.6

77

25.0

115.8

78

25.0

116.0

79

25.0

116.2

80

25.0

116.2

81

25.0

116.0

82

25.0

115.5

83

25.0

114.6

84

25.0

113.4

85

25.0

111.8

86

25.0

110.0

87

25.0

107.9

88

25.0

105.6

89

25.0

103.1

90

25.0

100.6

91

25.0

97.9

92

25.0

95.3

93

25.0

92.6

94

25.0

89.9

95

25.0

87.3

96

25.0

84.7

97

25.0

82.1

98

25.0

79.6

99

25.0

77.2

100

25.0

74.9

101

25.0

72.6

102

25.0

70.5

103

25.0

68.4

104

25.0

66.3

105

25.0

64.4

106

25.0

62.6

107

25.0

60.8

108

25

59.08

109

25

57.46

110

25

55.91

111

25

54.43

112

25

53.01

113

25

51.67

114

25

50.38

115

25

49.15

116

25

47.99

117

25

46.87

118

25

45.81

119

25

44.8

120

25

43.84

121

25

42.92

122

25

42.05

123

25

41.22

124

25

40.43

Solution

1. (a) To determine CUT and Pt, plot TRT and Tcold spot against time, which produces the thermal profiles in Figure 6.2.6.

Figure 6.2.6 shows that the CUT is approximately 10 min and Pt, during which process temperature is maintained constant at 120°C, is approximately 64 min.

1. (b) The lethality value, F0, can be obtained through numerical integration of Equation 6.2.9 using the trapezoidal rule (Patashnik, 1953). The calculations can be completed as follows or using software such as Excel.

As presented in Table 6.2.3, for each time, we can evaluate Equation 6.2.9:

2. $$F_{0}=\int^{t}_{0}10^{\dfrac{T-121.1}{10}}dt$$ (Equation $$\PageIndex{9}$$)

where T = Tcold spot and Tref and z-value for Clostridium botulinum are 121.1°C and 10°C, respectively.

1. Given that F0 corresponds to the integral of 10[(Tcold spot −Tref)/z], this can be solved numerically by the trapezoidal rule method, i.e., by determining the area under the curve by dividing the area into trapezoids, computing the area of each trapezoid, and summing all trapezoidal areas to yield F0. (More details about the trapezoidal rule are included in the appendix.) The calculations are summarized in Table 6.2.3. In this particular case, F0 was about 6.07 min. The change of F0 along the thermal process is shown as the blue line in Figure 6.2.7.
Table $$\PageIndex{3}$$: Numerical integration of Equation 6.2.9 for the estimation of F0.
Time (min) TRT (°C) Tcold spot (°C) (Tcold spotTref)/z 10[(Tcold spot −Tref)/z] Trapezoidal Area Sum of Areas

0.97

29.67

45

−7.6

0.000

0.000

0.000

1.97

39.67

45

−7.6

0.000

0.000

0.000

2.97

49.67

44.99

−7.6

0.000

0.000

0.000

3.97

59.67

44.97

−7.6

0.000

0.000

0.000

4.97

69.67

44.93

−7.6

0.000

0.000

0.000

5.97

79.67

44.85

−7.6

0.000

0.000

0.000

6.97

89.67

44.76

−7.6

0.000

0.000

0.000

7.97

99.67

44.69

−7.6

0.000

0.000

0.000

8.97

109.67

44.68

−7.6

0.000

0.000

0.000

9.97

119.67

44.77

−7.6

0.000

0.000

0.000

10.97

120

45

−7.6

0.000

0.000

0.000

11.97

120

45.41

−7.6

0.000

0.000

0.000

12.97

120

46.03

−7.5

0.000

0.000

0.000

13.97

120

46.88

−7.4

0.000

0.000

0.000

14.97

120

47.97

−7.3

0.000

0.000

0.000

15.97

120

49.29

−7.2

0.000

0.000

0.000

16.97

120

50.83

−7.0

0.000

0.000

0.000

17.97

120

52.55

−6.9

0.000

0.000

0.000

18.97

120

54.42

−6.7

0.000

0.000

0.000

19.97

120

56.41

−6.5

0.000

0.000

0.000

20.97

120

58.49

−6.3

0.000

0.000

0.000

21.97

120

60.63

−6.0

0.000

0.000

0.000

22.97

120

62.79

−5.8

0.000

0.000

0.000

23.97

120

64.97

−5.6

0.000

0.000

0.000

24.97

120

67.14

−5.4

0.000

0.000

0.000

25.97

120

69.29

−5.2

0.000

0.000

0.000

26.97

120

71.41

−5.0

0.000

0.000

0.000

27.97

120

73.48

−4.8

0.000

0.000

0.000

28.97

120

75.5

−4.6

0.000

0.000

0.000

29.97

120

77.46

−4.4

0.000

0.000

0.000

30.97

120

79.36

−4.2

0.000

0.000

0.000

31.97

120

81.2

−4.0

0.000

0.000

0.000

32.97

120

82.97

−3.8

0.000

0.000

0.001

33.97

120

84.67

−3.6

0.000

0.000

0.001

34.97

120

86.31

−3.5

0.000

0.000

0.001

35.97

120

87.89

−3.3

0.000

0.001

0.002

36.97

120

89.4

−3.2

0.001

0.001

0.003

37.97

120

90.84

−3.0

0.001

0.001

0.004

38.97

120

92.22

−2.9

0.001

0.002

0.005

39.97

120

93.55

−2.8

0.002

0.002

0.007

40.97

120

94.81

−2.6

0.002

0.003

0.010

41.97

120

96.02

−2.5

0.003

0.004

0.014

42.97

120

97.17

−2.4

0.004

0.005

0.018

43.97

120

98.27

−2.3

0.005

0.006

0.024

44.97

120

99.31

−2.2

0.007

0.007

0.032

45.97

120

100.31

−2.1

0.008

0.009

0.041

46.97

120

101.27

−2.0

0.010

0.012

0.053

47.97

120

102.17

−1.9

0.013

0.014

0.067

48.97

120

103.04

−1.8

0.016

0.017

0.084

49.97

120

103.86

−1.7

0.019

0.021

0.105

50.97

120

104.65

−1.6

0.023

0.025

0.130

51.97

120

105.39

−1.6

0.027

0.029

0.159

52.97

120

106.1

−1.5

0.032

0.034

0.193

53.97

120

106.78

−1.4

0.037

0.040

0.233

54.97

120

107.42

−1.4

0.043

0.046

0.279

55.97

120

108.04

−1.3

0.049

0.053

0.332

56.97

120

108.62

−1.2

0.056

0.060

0.393

57.97

120

109.18

−1.2

0.064

0.068

0.461

58.97

120

109.7

−1.1

0.072

0.077

0.538

59.97

120

110.21

−1.1

0.081

0.086

0.624

60.97

120

110.69

−1.0

0.091

0.096

0.720

61.97

120

111.14

−1.0

0.101

0.106

0.826

62.97

120

111.57

−1.0

0.111

0.117

0.943

63.97

120

111.99

−0.9

0.123

0.129

1.072

64.97

120

112.38

−0.9

0.134

0.140

1.212

65.97

120

112.75

−0.8

0.146

0.153

1.365

66.97

120

113.11

−0.8

0.159

0.165

1.530

67.97

120

113.44

−0.8

0.171

0.178

1.708

68.97

120

113.76

−0.7

0.185

0.191

1.899

69.97

120

114.07

−0.7

0.198

0.205

2.104

70.97

120

114.36

−0.7

0.212

0.219

2.323

71.97

120

114.63

−0.6

0.225

0.233

2.555

72.97

120

114.9

−0.6

0.240

0.247

2.802

73.97

120

115.15

−0.6

0.254

0.261

3.063

74.97

120

115.38

−0.6

0.268

0.275

3.338

76

25

115.61

−0.5

0.282

0.290

3.628

77

25

115.83

−0.5

0.297

0.304

3.932

78

25

116.02

−0.5

0.310

0.316

4.248

79

25

116.17

−0.5

0.321

0.322

4.570

80

25

116.19

−0.5

0.323

0.315

4.885

81

25

115.97

−0.5

0.307

0.290

5.175

82

25

115.45

−0.6

0.272

0.248

5.422

83

25

114.58

−0.7

0.223

0.196

5.618

84

25

113.36

−0.8

0.168

0.143

5.761

85

25

111.81

−0.9

0.118

0.097

5.858

86

25

109.96

−1.1

0.077

0.062

5.920

87

25

107.87

−1.3

0.048

0.038

5.958

88

25

105.57

−1.6

0.028

0.022

5.980

89

25

103.12

−1.8

0.016

0.012

5.992

90

25

100.57

−2.1

0.009

0.007

5.999

91

25

97.94

−2.3

0.005

0.004

6.003

92

25

95.26

−2.6

0.003

0.002

6.005

93

25

92.58

−2.9

0.001

0.001

6.006

94

25

89.91

−3.1

0.001

0.001

6.007

95

25

87.26

−3.4

0.000

0.000

6.007

96

25

84.66

−3.6

0.000

0.000

6.007

97

25

82.11

−3.9

0.000

0.000

6.007

98

25

79.63

−4.1

0.000

0.000

6.007

99

25

77.22

−4.4

0.000

0.000

6.007

100

25

74.88

−4.6

0.000

0.000

6.007

101

25

72.62

−4.8

0.000

0.000

6.007

102

25

70.45

−5.1

0.000

0.000

6.007

103

25

68.35

−5.3

0.000

0.000

6.007

104

25

66.34

−5.5

0.000

0.000

6.007

105

25

64.41

−5.7

0.000

0.000

6.007

106

25

62.55

−5.9

0.000

0.000

6.007

107

25

60.78

−6.0

0.000

0.000

6.007

108

25

59.08

−6.2

0.000

0.000

6.007

109

25

57.46

−6.4

0.000

0.000

6.007

110

25

55.91

−6.5

0.000

0.000

6.007

111

25

54.43

−6.7

0.000

0.000

6.007

112

25

53.01

−6.8

0.000

0.000

6.007

113

25

51.67

−6.9

0.000

0.000

6.007

114

25

50.38

−7.1

0.000

0.000

6.007

115

25

49.15

−7.2

0.000

0.000

6.007

116

25

47.99

−7.3

0.000

0.000

6.007

117

25

46.87

−7.4

0.000

0.000

6.007

118

25

45.81

−7.5

0.000

0.000

6.007

119

25

44.8

−7.6

0.000

0.000

6.007

120

25

43.84

−7.7

0.000

0.000

6.007

121

25

42.92

−7.8

0.000

0.000

6.007

122

25

42.05

−7.9

0.000

0.000

6.007

123

25

41.22

−8.0

0.000

0.000

6.007

124

25

40.43

−8.1

0.000

0.000

6.007

Discussion:

The cumulative lethality, F0, was about 6.01 min, meaning that the process is safe according to FDA requirements, i.e., F0 ≥ 3 min (see the Food Sterilization Criterion and Calculation section above).

Example $$\PageIndex{4}$$

Example 4: Lethality of thermal processing of a can of mussels

Problem:

Temperatures measured in the retort and the temperature measured at the cold spot of a can of mussels during a thermal process performed at 120°C were recorded. The total process time was 113 min until the product was cold enough to be withdrawn from the retort. The measured thermal profiles (TRT and Tcold spot) were plotted, as was done in Example 6.2.3. The resulting plot (Figure 6.2.8) shows that CUT was approximately 10 min and Pt was approximately 53 min. The lethality value, F0, was obtained through numerical integration of Equation 6.2.9. In this case, F0 attained in the mussels can with a processing temperature of 120 °C was 2.508 min. The evolution of F0 along the thermal process is shown in Figure 6.2.9 as the blue line.

Discussion:

The cumulative lethality, F0, attained along the thermal process was 2.5 min, meaning that the process is not safe according to FDA requirements (F0 ≥ 3 min). Thus, the thermal processing time of canning process of mussels must be extended in order to reach the safety value recommended by the FDA.

Example $$\PageIndex{5}$$

Example 5: Processing time at different retort temperatures

Problem:

Determine the required processing time to get a lethality of 6 min (F0 = 6 min) when the retort temperature is (a) 120°C and considered equal to the cold spot temperature, (b) 110°C, and (c) 130°C.

Solution

The F0 is typically set for the 12D value to give a 12 log reduction of heat-resistant species of mesophilic spores (typically taken as C. botulinum). The Tref = 121.1°C and z = 10°C. Therefore, Equation 6.2.9 can be used directly by replacing T by the retort temperature, given that cold spot temperature can be assumed equal to retort temperature:

$$F_{0} = \int^{t}_{0}10^{\dfrac{T-121.1}{z}}dt$$ (Equation $$\PageIndex{9}$$)

$$6 = \int^{t}_{0}10^{\dfrac{120-121.1}{10}}dt$$

$$t = \dfrac{6}{10^{\dfrac{120-121.1}{10}}}$$

Solving the integral yields a processing time, t, of 7.7 min.

1. (b) When the temperature of the retort is reduced to 110°C, the lethality must be maintained at 6 min. Solving Equation 6.2.9:

$$6 = \int^{t}_{0} 10^{\dfrac{110-121.1}{10}} dt$$

1. gives the required processing time t of 77.2 min.
2. (c) When the temperature of the retort is increased to 130°C, and maintaining the F0 = 6 min, the processing time is reduced to 0.77 min

$$6 = \int^{t}_{0} 10^{\dfrac{130-121.1}{10}} dt$$

Discussion:

The results showed that as the temperature in the food increased in 10°C increments, the processing time was reduced by one decimal reduction. This variation is due to a z value of 10°C.

### Appendix: The Trapezoidal Rule

A trapezoid is a four-sided region with two opposite sides parallel (Figure 6.2.10). The area of a trapezoid is the average length of the two parallel sides multiplied by the distance between the two sides. In Figure 6.2.11, the area (A) under function f(x) between points x0 and xn is given by:

$A=\int^{b}_{a} f(x)dx$

An approximation of the area A is the sum of the areas of the individual trapezoids (T), where T can be calculated using Equation 6.2.11:

$T=\dfrac{1}{2}\Delta x_{1}[f(x_{0})+f(x_{1})]+\dfrac{1}{2}\Delta x_{2}[f(x_{1})+f(x_{2})]+…+\dfrac{1}{2}\Delta x_{n}[f(x_{n-1})+f(x_{n})]$

where $$\Delta x_{i}=x_{i}-x_{i-1}$$, for i = 1, 2, 3, . . . , n

In the particular case where ∆x1 = ∆x2 = ∆x3 = . . . = ∆xn = ∆x, Equation 6.2.11 can be expressed as:

$T=\Delta x[\dfrac{f(x_{0})}{2} + f(x_{1})+ f(x_{2})+ f(x_{3})+…+\dfrac{f(x_{n})}{2}$

or, in the following reduced form:

$T=\Delta x[\dfrac{f(x_{0})}{2} +\sum^{n-1}_{i=1}f(x_{i})+\dfrac{f(x_{n})}{2}$

Finally, to estimate area A under the trapezoidal rule,

$A=\int^{x_{n}}_{x_{0}}f(x)dx\cong\dfrac{1}{2}\Delta x_{1}[f(x_{0})+f(x_{1})]+\dfrac{1}{2}\Delta x_{2}[f(x_{1})+f(x_{2})]+…+\dfrac{1}{2}\Delta x_{n}[f(x_{n-1})+f(x_{n})]$

When all intervals are of the same size (∆x1 = ∆x2 = ∆x3 = . . . = ∆xn = ∆x), the following expression can be applied:

$A=\int^{x_{n}}_{x_{0}}f(x)dx\cong\Delta x(\dfrac{f(x_{0})}{2} +\sum^{n-1}_{i=1}f(x_{i})+\dfrac{f(x_{n})}{2}) =\dfrac{1}{2}\Delta x(f(x_{0}) +2\sum^{n-1}_{i=1}f(x_{i})+f(x_{n}))$

Example $$\PageIndex{6}$$

Example

Problem:

Using the heat penetration data at the cold spot of a canned food in Table 6.2.4, calculate the cumulative lethality, F0, in the range of 23 to 27 min using the trapezoidal rule.

Table $$\PageIndex{4}$$: Heat penetration data at the slowest heating point.

Time (min) Temperature (C)

. . .

. . .

23

118.5

24

118.7

25

118.9

26

119.1

27

119.3

. . .

. . .

Solution

From Equation 6.2.9,

$$F_{0}=\int^{27}_{23}10^{\dfrac{T-121.1}{10}}dt$$

Applying the trapezoidal rule and considering that all time steps are equal (∆t = 1 min), calculate F0 using Equation 6.2.15,

$$F_{0}=\int^{27}_{23}10^{\dfrac{T-121.1}{10}}dt \cong \dfrac{1}{2}[f(23)+2f(24)+2f(25)+2f(26)+f(27)]$$

where ∆t = 1 (1 min interval), and:

$$f(23)=10^{\dfrac{118.5-121.1}{10}} = 0.549541$$

$$f(24)=10^{\dfrac{118.7-121.1}{10}} = 0.57544$$

$$f(25)=10^{\dfrac{118.9-121.1}{10}} = 0.6025596$$

$$f(26)=10^{\dfrac{119.1-121.1}{10}} = 0.63095734$$

$$f(27)=10^{\dfrac{119.3-121.1}{10}} = 0.66069345$$

Replacing into equation (6.2.15):

$$F_{0}=\int^{27}_{23}10^{\dfrac{T-121.1}{10}}dt \cong \dfrac{1}{2}(0.549541+2\times0.57544+2\times0.6025596+2\times 0.63095734+0.66069345)$$

Therefore, F0 ~ 2.41407394 ~ 2.41 min.

Discussion:

The applied process to sterilize the target food is not safe since F0 < 3 minutes.

### Image Credits

Figure 1. Simpson, R. (CC By 4.0). (2020). Main heat transfer mechanisms involved in the thermal processing of packaged foods. Retrieved from onlinelibrary.wiley.com

Figure 2. Holdsworth, S. Donald-Simpson, R. (CC By 4.0). (2020). Semilogarithmic survivor curve. Retrieved from https://www.springer.com/la/book/9783319249025

Figure 3. Holdsworth, S. Donald-Simpson, R. (CC By 4.0). (2020). Thermal death time (TDT) curve. Retrieved from https://www.springer.com/la/book/9783319249025.

Figure 4. Simpson, R. (CC By 4.0). (2020). Stages of a typical food commercial canning factory.

Figure 5. Ramírez, C. (CC By 4.0). (2020). Temperature profiles for a typical thermal process, where CUT is come-up time and Pt is operator time.

Figure 6. Ramírez, C. (CC By 4.0). (2020). Temperature profile of thermal processing data in table 2.

Figure 7. Ramírez, C. (CC By 4.0). (2020). Thermal process temperature profiles including the cumulative lethality value (F at any time t).

Figure 8. Ramírez, C. (CC By 4.0). (2020). Temperature profile of thermal processing data (Table 4).

Figure 9. Ramírez, C. (CC By 4.0). (2020). Thermal process temperature profiles including the cumulative lethality value (F at any time t).

Figure 10. Simpson, R. (CC By 4.0). (2020). Example of a trapezoid.

Figure 11. Simpson, R. (CC By 4.0). (2020). Curve divided into n equal parts each of length ΔX.

### References

Ball, C. O., & Olson, F. C. (1957). Sterilization in food technology—Theory, practice and calculations. New York, NY: McGraw-Hill.

Bigelow, W. D. (1921). The logarithmic nature of thermal death time curves. J. Infectious Dis., 29(5), 528-536. https://doi.org/10.1093/infdis/29.5.528.

Bigelow, W. D., & Esty, J. R. (1920). The thermal death point in relation to time of typical thermophilic organisms. J. Infectious Dis., 27(6), 602-617. doi.org/10.1093/infdis/27.6.602.

Bigelow, W. D., Bohart, G. S., Richardson, A. C., & Ball, C. O. (1920). Heat penetration in processing canned foods. Bull. No. 16. Washington, DC: Research Laboratory, National Canners Association.

Esty, J. R., & Meyer, K. F. (1922). The heat resistance of the spores of B. botulinus and allied anaerobes. J. Infectious Dis., 31(6), 650-663. https://doi.org/10.1093/infdis/31.6.650.

FDA. (1972). Sterilizing symbols. Low acid canned foods. Inspection technical guide. Ch. 7. ORO/ETSB (HFC-133). Washington, DC: FDA.

Featherstone, S. (2015). 7: Retortable flexible containers for food packaging. In A complete course in canning and related processes (14th ed.). Vol. 2: Microbiology, packaging, HACCP and ingredients (pp. 137-146). Sawston, Cambridge, U.K.: Woodhead Publ. https://doi.org/10.1016/B978-0-85709-678-4.00007-5.

Holdsworth, S. D., & Simpson, R. (2016). Thermal processing of packaged foods (3rd ed.). Springer. doi.org/10.1007/978-3-319-24904-9.

Patashnik, M. (1953). A simplified procedure for thermal process evaluation. Food Technol., 7(1), 1-6.

Simpson, R., Almonacid, S., & Teixeira, A. (2003). Bigelow’s general method revisited: Development of a new calculation technique. J. Food Sci., 68(4), 1324-1333. doi.org/10.1111/j.1365-2621.2003.tb09646.x.

Stumbo, C. R. (1973). Thermobacteriology in food processing (2nd. ed.). New York, NY: Academic Press.

Teixeira, A., Almonacid, S., & Simpson, R. (2006). Keeping botulism out of canned foods. Food Technol., 60(2), Back page.

This page titled 6.2: Principles of Thermal Processing of Packaged Foods is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Nicholas M. Holden, Mary Leigh Wolfe, Jactone Arogo Ogejo, & Enda J. Cummins (Virginia Tech Libraries' Open Education Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


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The Notebook
İstanbul, Turkey

Shannon Entropy & Data Compression

In this article, I'll try to explain Shannon Entropy with one of its use case: data-compression. First, I'll provide a small introduction about data-representation and a need for compression with very brief information about its requirements. Consequently, I'll derive the proof of well-known entropy formula in the equation below that estimates the average minimum number of bits needed to encode information with priorly known probabilities:

H(X) = -\sum_{i}p_{i}\log_{b}p_{i}

Data Representation and Compression

We all know that the way that our digital devices store data as ones and zeros, aka in binary format. So, our understanding of the smallest chunk of information is: "whether there is a voltage or not". We represent letters of a text, pixels of an image, etc. in binary format.

Below, I provide an ASCII (American Standard Code for Information Interchange) encoded "hello" below. In ASCII, each letter -and other control characters, such as \texttt{DEL} and numbers like 1, 2 etc.- consists 8 bits. So, the text that we see is "hello", but the information that a digital device stores are:

So, there are 26 letters in the English alphabet, and we can say that $5$ bits would be enough to encode all characters, since :

2^{5} = 32 > 26

Nevertheless, ASCII has 127 different values --See Reference 1, which at least requires 7 bits per value. The rest of the document tries to answer the following question:

"What would be the minimum number of bits that we have to use if we knew the distribution of letters for a particular text that we have?"

So, as it could be seen in Figure above --and we all well-know, some of the letters are more often used than the others. Here, letter "e" is the most commonly used one in English. So, would it be better to represent "e" with fewer bits than the others, if the letter distribution of our text would be exactly the same with the distribution above. If so, how less could it be?

Uniquely Decodable Codes

A code is a mapping of source messages into codewords. As an example, ASCII code above maps letters (as a source message) to codewords --such as letter "h" to number 01101000.

Here, a code is distinct if each codeword is distinguishable from every other (i.e., the mapping from source messages to codewords is one-to-one). A distinct code is uniquely decodable if every codeword is identifiable in a sequence of codewords.

Prefix Codes
A uniquely decodable code is called a prefix code (or prefix-free code) if it has the prefix property, which requires that no codeword is a proper prefix of any other codeword.

The code below is decodable but it is not a prefix code because the codeword for "a" is a prefix for the codeword for "b". This means that we can not instantaneously decode "a" without waiting for the next bit of data (to determine whether it is actually "a" or just the first half of "b".)

Alternatively, the code below is a prefix code, since no codeword is a prefix of another codeword.

Kraft's Inequality

Kraft's inequality states that given a list of positive integers (n_{1}, n_{2}, \dots, n_{r}) there exists a prefix code with a set of codewords (\sigma_{1}, \sigma_{2}, \dots, \sigma_{r}) where the length of each codeword |\sigma_{i}| = n_{i}, \forall_{i} if and only if:

\sum_{i=1}^{r} s^{-n_{i}} \leq 1

where s is the size of the alphabet S.

In our previous example code, the alphabet is simply the binary digits S = \{0, 1\}, and therefore the size of the alphabet s = 2. We can easily check that indeed the inequality holds:

2^{-1} + 2^{-3} + 2^{-4} + 2^{-4} \leq 1

Proof
As it could be observed in the following Code Tree, there are s^{n_{j}} combination of codewords where we could chose for j_{th} level if there was no prefix constraints, but due to the codeword at a prior level k < j, s^{n_{j} - n_{k}} amount of codewords are forbidden, since they contain \sigma_{k} as a prefix. In other words, to satisfy prefix-code requirements, each codeword must be represented by a leaf node, since it has to eliminate its descendants as codewords.

So, a codeword at level j has s^{n_{r}-n_{j}} descendants (where n_{r} is the length of longest codeword) at the highest level. For instance 1 has 2^{3-1} descendants which are "111", "110", "101" and "100".

Besides, descendant sets of codewords must be disjoint to satisfy prefix-free requirement. As an example, "1" and "10" has joint descendants, which are "101" and "100", hence a prefix-free codeword can not contain both "1" and "10".

So, we know that the total number of forbidden codewords at step j is:

\sum_{i=1}^{j-1} s^{n_j - n_i}

As an example, if our level j=3, we would have 6 forbidden codewords which can also be observed from the Code Tree above:

\sum_{i=1}^{j-1} s^{n_j - n_i} = 2^{(3-1)} + 2^{(3-2)} = 6

For each level j, we possibly have s^{n_j} codewords; and this amount should exceed the number of forbidden codewords for level j to have a prefix codeword set. Here we apply this lemma for highest level r below, which implies the Kraft Inequality:

\sum_{i=1}^r s^{n_r - n_i} \leq s^{n_r} \\ \Rightarrow \sum_{i=1}^{r} s^{-n_{i}} \leq 1

Shannon Entropy

If the codeword \sigma_{i} occurs with probability p_{i} in the particular set, the expected length of code is:

\sum_{i=1}^r p_{i}n_{i}

and Shannon Entropy estimates the average minimum number of bits needed to encode is:

\sum_{i=1}^r p_{i}\log_{s}\frac{1}{p_{i}}

So, here we calculate the difference between the Shannon entropy and the expected length of the code is:

\sum_{i=1}^r p_{i}\log_{s}\frac{1}{p_{i}} - \sum_{i=1}^r p_{i}n_{i} = \sum_{i=1}^r p_{i}(\log_{s}\frac{1}{p_{i}} - \log_{s}s^{n_{i}})

=\sum_{i=1}^r p_{i}(\log_{s}\frac{1}{p_{i}} + \log_{s}s^{-n_{i}})

=\sum_{i=1}^r p_{i}\log_{s} \frac{s^{-n_{i}}}{p_{i}}

\leq \log_{s}(\sum_{i=1}^r p_{i} \frac{s^{-n_{i}}}{p_{i}})

= \log_{s} (\sum_{i=1}^r s^{-n_{i}})

\leq \log_{s} 1 = 0

where we simply use the fact that \log_{s} is a concave function and apply Kraft's inequality. So, here we prove that Shannon entropy converges the expected length of the prefix-free code.

References:


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## Articles published in Uncategorized

So, you’ve taken a practice test!  Maybe you’re closing in on the score you want, or maybe you still have some distance left to travel.  Regardless of which scenario applies to you, “reading” your practice test data is an incredibly crucial element to GRE progress!

I write this assuming I don’t need to discuss looking at your score, comparing overall quantitative to overall verbal, etc.  Everyone looks at the ‘big’ numbers.  The question is, what eureka moments can we gain from a deeper analysis?

There are three components to analyzing a practice test: analyzing timing, analyzing accuracy by question type, and analyzing accuracy by topic tested.

Analyzing Timing

You can’t analyze your timing until you know what your timing should look like.

Does anything stand out to you in image above?  Why do some questions take you less than one minute, while some take more than three?  We expect some variation across different questions – Reading Comprehension should take longer than Text Completion or Sentence Equivalence, and Data Interpretation questions (especially the first DI question) will usually take longer than Quantitative Comparison.  But why are there such wide swings in question time within the QC question?  And I can’t help but notice that the two Discrete Quantity questions both took less time than the vast majority of the QC questions.  Perhaps this is someone who is skilled in math, but doesn’t yet truly grasp the logic underlying the QC questions.

A review of QC as a question type is probably called for from this practice test.

Another element of timing is more fundamental (and the above image captures this concept also).  Do you know when to let a question go, guess, and move on to the next question?  Any time you spent three minutes on a question, you had a problem letting go.  Right or wrong, that question hurt you.

Bottom line, when you analyze timing in a practice test, you want to see two things: 1) question by question timing – were you able to let a question go when needed, and 2) question type timing – do you have the desired timing for each question type?

Now it’s time to Generate an Assessment Report!!!  (It’s exciting because it’s got three exclamation points 🙂  )

Analyzing Accuracy by Question Type

Take a few moments and see what you can find in the image above.  Don’t worry, I’ll wait  🙂

Seriously, there’s a lot you should consider here.  If you haven’t been looking for at least 5 minutes, you haven’t spent enough time.  And although I *said* this is the Accuracy analysis portion of this post, we’re not done with timing.

First, let’s talk good decisions vs poor decisions.  Good decisions – on TC, you know when to get out of a question.  Look at the Average Time Wrong vs Average Time right for the Harder and Devilish TC questions.  That’s what we want to see!  This indicates you recognize when more time will/won’t pay off.  (Maybe… more on this in a moment.)

So why aren’t you making the same decisions in SE?

Finally, why, why, why are you spending five minutes – on AVERAGE across three questions – in RC?  What’s going on here?  There’s some leeway in RC, because of the time needed to read and process a longer passage, but not five minutes leeway.

On the Easier RC question that you missed, you missed it in one minute.  This indicates you were confident in your answer.  Confident in the wrong answer – somewhere in this question is a trap that you fell for, and you need to figure out what that trap was!

Back to the TC timing: one possibility is you know when to get out of TC, and that’s why your wrong answers take less time than the right answers.  Another, more disturbing possibility, is you’re cheating yourself on TC time.  How do I know this?  Look at the variation between TC and SE accuracy – it’s not huge, but the discrepancy is there.  Why is TC accuracy lower?

Finally, the most obvious element of this analysis is that RC is your lowest accuracy.  Time to go back and study!!!

Analyzing Accuracy by Topic Tested

This issue cannot be addressed by looking at one image – you will generate an assessment report, and view the Analysis by Content Area and Topic.  There are a few things you’re looking for here.

First, and foremost, are you seeing accuracy and speed in topics you’ve studied?  If you haven’t studied Geometry yet, who cares if your Geometry accuracy is 20%!  But you’ve spent two full weeks reviewing algebra, so why are you missing 2 out of 3 function/formula questions?  Bright side though, your accuracy in Quadratics is through the roof!!

Obviously that paragraph is a hypothetical, but notice two things: first, you need to decide which area(s) deserve your analysis; second, you need to look not just at the overall topic, but also at the subtopics.

You’re looking for improvements and discrepancies.  Which areas are strong?  Which are weak?  Do you have a mix of strong and weak areas in one major topic?  These are all question you need to ask yourself.

BUT you need to take this with a grain of salt – don’t neglect to consider the difficulty of the individual questions!  Yes, maybe you missed 2 function questions.  But they were both Devilish difficulty!  You’re not weak in this area, you just got hit by some of the worst questions.

Finally, don’t neglect to examine timing in this area of analysis.  Yes, you were accurate in Rates questions.  But you spent 4 minutes on them.  Time to study!!

Final Thoughts

I hope you’ve found this helpful.  If you go back and look at my previous GRE blog posts, I think you’ll notice that this post contains many, many more rhetorical questions.  That’s the point of practice test analysis.  In the test, and when you’re studying, the computer, or the book, or whatever study source you’re using is asking you questions.

Analyzing your practice tests is the time for you to ask the questions.  What are the weak areas?  Strong areas?  Why am I performing differently in Word Problems vs Geometry?

And there’s one question you must ask, which I haven’t addresses, simply because of how much space it would require – Are you seeing improvement???

Every time you take a practice test, from the second practice test on to the last, look at the most recent test, do all this analysis.  Then look at the test prior – what’s changed?  What has stayed the same?  Have you improved in your weaknesses, and have strengths remained strong?

A practice test doesn’t teach you anything in and of itself – but it tells you where you are, and where you’re moving, and what you *should* be teaching yourself.

Good Luck!!!

### Manhattan Prep Instructor Wows With Record-Setting Jeopardy Win

Yesterday, the whole wide internet was shocked by the Jeopardy dominance of “This dude Michael”, but here at Manhattan Prep we weren’t surprised at all. We already knew that dude was smart and we knew that dude had the mathematical wherewithal to bet $7,000 without batting an eyelash. Here at MPrep we know that dude as Michael Bilow (one of those people who command such respect that he must always be talked about using his last name lest anyone in earshot mistakenly attribute an anecdote or joke to some less deserving Michael). On Jeopardy, he lived up to his legend taking home the fourth highest single-day winnings in Jeopardy history:$57,198.

Michael Bilow joined the Manhattan Prep family in 2011 using his perfect GRE score and spectacular teaching chops to secure a role as an LA-based GRE instructor. A few years later we realized we needed more Bilow in our business so we asked him to join the Marketing Department. He took a position as our Business Data Analyst, while continuing to teach GRE classes and pursue his PhD. After seeing him flawlessly juggle those responsibilities, we never had any doubt that he would take the Jeopardy world by fire.

By now the whole country knows of Bilow’s intellectual prowess, but we know so much more. Michael is a dedicated practitioner of improv, a delightful presence in Google Hangout meetings, and a stylish dresser. We can’t wait for his next trip to the New York City headquarters so he can buy us a drink with his winnings after he takes a quick a nap in a tutoring pod.

Congrats, Michael Bilow! Keep it up!

### How to Switch from the GRE to the GMAT

Lately, we’ve been talking about how to decide which test to take. What if you decide to switch from the GRE to the GMAT? That’s what we’ll tackle today! (Next time, we’ll talk about what to do if you want to switch from the GMAT to the GRE.)

### How do I study?

The overall way that you want to study doesn’t actually change that much; rather, you’ll just need to change what you are studying, as discussed later in this article.

First, you’ll need to determine whether the way that you’ve already been studying is actually the optimal way. If not, then you’ll need to make some changes, regardless of whether you stick with the GRE or switch to the GMAT.

The GRE and the GMAT are both executive reasoning tests; that is, the test makers want to know how you think and make decisions. You of course need to know content (certain facts, rules, formulas) in order to do well on either test, but that level of study is not enough; you also need to lift yourself to a second level of understanding that allows you to think your way through these sometimes bizarrely-worded problems as effectively and efficiently as possible.

Follow the two links I put in the last paragraph. Take some time to just think about the concepts presented there. Has this been your approach to studying so far? If so, great. Keep thinking and working in that way.

If not, however, recognize that you’re going to need to start studying with this new mindset, regardless of whether you take the GRE or the GMAT.

### What are my strengths and weaknesses?

Any time you’re developing or revising a study plan, you’ll want to put together a solid analysis of your strengths and weaknesses.

If you have been studying for the GRE for a while, then you should have some practice CAT data. (If not, or if it has been more than 6 weeks since you last took a CAT, then you’ll need to take one to get the data. Make sure to take the test under official conditions, including the essays, length of breaks, and so on.)

Analyze your most recent two CATs (this link tells you how to analyze Manhattan Prep CATs). If you haven’t taken MPrep CATs, you can still read through that link to get an idea of how you want to analyze your data from another test. Your goal is to split all question types and content into one of three buckets:

Bucket 1: Strengths. I’ll still study and practice these but not as heavily as other areas.

Bucket 2: Low-hanging Fruit: These are my easiest opportunities for improvement. Careless mistakes. Things that I get wrong fast. Things that I get right but just a little too slowly.

Bucket 3: Weaknesses. These are areas that I’ll ignore until I’ve worked out my Bucket 2 issues. Things that I’m likely to get wrong even if I give myself unlimited time. Things that I get right but way too slowly. Things that use up way too much mental energy, even if I get them right.

Your primary focus until your next practice test will be working a lot to improve Bucket 2, while maintaining Bucket 1 skills and getting Bucket 3 questions wrong fast (yes, seriously!).

[Aside: there are certain things that will stay in Bucket 3 forever. I’m terrible at combinatorics and I’m pretty bad at 3D geometry. That’s been true since my very first practice GRE, more than 10 years ago! When I see these, I’ll give it a look in case the problem is very similar to one that I do remember how to do, but otherwise, I pick my favorite letter and move on.]

Okay, now that you know what your strengths and weaknesses are, you need to familiarize yourself with the differences between the GRE and the GMAT.

### What new things do I have to learn?

The Essays and Integrated Reasoning

You won’t care as much about one difference, so let’s get it out of the way. At the beginning of the GRE, you write two essays. The GMAT also asks you to write an essay but in place of the second essay you’ll have to do the Integrated Reasoning section, a multiple-choice section that mixes quant and verbal skills.

This section is different enough from the others that you will have to study how to answer these questions and how to manage your time during the section. At the time of this publication (in March 2015), schools aren’t using IR scores much, so this section is less important, though this could change in the future.

Quant

Next, for the quant section of the test, you’re going to need to learn about one different question type contained on the GMAT: Data Sufficiency (DS).

The GMAT dives more deeply into number properties, story problems, and some algebra concepts, so you may need to get GMAT books for these topics versus continuing to use your GRE books.

The timing on the two tests is also quite different, so you’ll have to learn how to handle 37 questions in 75 minutes on the GMAT, or about 2 minutes per question on average.

Verbal

Most of your new efforts on verbal will be geared towards the grammar question type, Sentence Correction (SC). You’ll definitely need to get some materials that teach you the grammar and meaning issues that are tested on SC.

Again, if you are already using Manhattan Prep materials, you can use what you already have for Reading Comprehension (RC), but you will need to get new materials for Critical Reasoning (CR). The CR question types on the GRE are also tested on the GMAT, but the GMAT contains additional CR question types that don’t appear on the GRE.

Again, the timing will be different on the GMAT. You’ll need to answer 41 verbal questions in 75 minutes, spending about 1 minute 20 seconds on SC, 2 minutes on CR, and about 6 to 8 minutes total for RC passages and questions.

### How do I make a study plan?

We’ve already talked about part of the process (analyzing your strengths and weaknesses). You may decide to take a class or work with a tutor, in which case your teacher will give you specific assignments . If not, you’ll need to develop your own study plan.

### Takeaways for switching from GRE to GMAT

(1) Make sure that you’re going into your studies with the right overall mindset (executive reasoning!) and that you know how to lift yourself to the “second level” of study.

(2) Begin your studies by concentrating on the aspects that are new to you: the different question types and topics that are tested on the GMAT. Once you build those skills up to a competent level, you’ll review all aspects and question types.


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## Convex functions, partial orderings, and statistical applications.(English)Zbl 0749.26004

Mathematics in Science and Engineering. 187. Boston, MA etc.: Academic Press. xiii, 467 p. (1992).
There are many books on inequalities and on convex functions but then there are so many inequalities around — more or less connected to convex functions — that no book of reasonable size can contain all. The present book takes a pretty good stab at it — as long as the inequalities are satisfied by convex functions in general and, except in Chapters 13-17, not by particular ones (thus showing a similar trait to some complex function theory courses where the only complex function a student sees is “$$f$$”). So, for instance, when following the subject index the reader looks up “Minkowski’s inequality” he or she may find something different from what could be expected.
Speaking about subject indices, this reviewer having stood on both sides of the fence, can appreciate that the typical author finds it tiresome to put them together. But the typical reader, in particular of such encyclopedical works (not meant, we hope, to be read cover to cover in linear order) relies heavily on them. So, for example, “monotonic $$n$$- tuple”, “isotonic functional” might be sorely missed (only “inequalities” are under I) as well as properties $$A1,A2,\dots,L1,L2,K_ 1,K_ 2$$ in the index or in the list of notations.
The book is not self-contained: Not only are results mentioned only by reference, but so are even definitions, for instance in Theorem 8.3: “…If $$F_ n$$ is a quasi-arithmetic mean (for definition see Bullen,…) then …”. Since there is no proof, the reader can understand this theorem only if she or he looks it up elsewhere. This is quite strange because either a direct definition or specialization from the “generalized mean” defined on p. 107 would take only a couple of lines. — Also, not all definitions are marked as such. — In Remark 15.22, counterexamples would be appreciated which show why the two conjectures there formulated are false.
Chapters 1-10 are solidly filled wall to wall with inequalities; even the word “application” refers there to other inequalities. Chapter 11 contains “geometric” inequalities, Chapter 12 majorization and Schur convexity, while Chapters 13-17 give applications to probability and statistics (random variables, their moments, density functions, distributions, arrangement ordering).
All over there are many interesting historical remarks, including corrections to conventional wisdom about who has priority to what definition or result. {L. Fejér is on the receiving end of one of these remarks. Another quotation from Fejér may be of interest in this context: “The history of mathematics serves to prove that nobody discovered anything: there was always somebody who knew it before.”} Some of these remarks also correct published statements and point out simpler proofs (usually by the authors). All in all, a very useful book, but there is room for improvement.

### MSC:

 26A51 Convexity of real functions in one variable, generalizations 26-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to real functions 26-02 Research exposition (monographs, survey articles) pertaining to real functions 26D15 Inequalities for sums, series and integrals 26D10 Inequalities involving derivatives and differential and integral operators 26D20 Other analytical inequalities 60E15 Inequalities; stochastic orderings


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# Hoare logic

Hoare logic (also known as Floyd–Hoare logic or Hoare rules) is a formal system with a set of logical rules for reasoning rigorously about the correctness of computer programs. It was proposed in 1969 by the British computer scientist and logician Tony Hoare, and subsequently refined by Hoare and other researchers.[1] The original ideas were seeded by the work of Robert W. Floyd, who had published a similar system[2] for flowcharts.

## Hoare triple

The central feature of Hoare logic is the Hoare triple. A triple describes how the execution of a piece of code changes the state of the computation. A Hoare triple is of the form

${\displaystyle \{P\}C\{Q\}}$

where ${\displaystyle P}$  and ${\displaystyle Q}$  are assertions and ${\displaystyle C}$  is a command.[note 1] ${\displaystyle P}$  is named the precondition and ${\displaystyle Q}$  the postcondition: when the precondition is met, executing the command establishes the postcondition. Assertions are formulae in predicate logic.

Hoare logic provides axioms and inference rules for all the constructs of a simple imperative programming language. In addition to the rules for the simple language in Hoare's original paper, rules for other language constructs have been developed since then by Hoare and many other researchers. There are rules for concurrency, procedures, jumps, and pointers.

## Partial and total correctness

Using standard Hoare logic, only partial correctness can be proven, while termination needs to be proved separately. Thus the intuitive reading of a Hoare triple is: Whenever ${\displaystyle P}$  holds of the state before the execution of ${\displaystyle C}$ , then ${\displaystyle Q}$  will hold afterwards, or ${\displaystyle C}$  does not terminate. In the latter case, there is no "after", so ${\displaystyle Q}$  can be any statement at all. Indeed, one can choose ${\displaystyle Q}$  to be false to express that ${\displaystyle C}$  does not terminate.

Total correctness can also be proven with an extended version of the While rule.[citation needed]

In his 1969 paper, Hoare used a narrower notion of termination which also entailed absence of any run-time errors: "Failure to terminate may be due to an infinite loop; or it may be due to violation of an implementation-defined limit, for example, the range of numeric operands, the size of storage, or an operating system time limit."[3]

## Rules

### Empty statement axiom schema

The empty statement rule asserts that the ${\displaystyle {\textbf {skip}}}$  statement does not change the state of the program, thus whatever holds true before ${\displaystyle {\textbf {skip}}}$  also holds true afterwards.[note 2]

${\displaystyle {\dfrac {}{\{P\}{\textbf {skip}}\{P\}}}}$

### Assignment axiom schema

The assignment axiom states that, after the assignment, any predicate that was previously true for the right-hand side of the assignment now holds for the variable. Formally, let ${\displaystyle P}$  be an assertion in which the variable ${\displaystyle x}$  is free. Then:

${\displaystyle {\dfrac {}{\{P[E/x]\}x:=E\{P\}}}}$

where ${\displaystyle P[E/x]}$  denotes the assertion ${\displaystyle P}$  in which each free occurrence of ${\displaystyle x}$  has been replaced by the expression ${\displaystyle E}$ .

The assignment axiom scheme means that the truth of ${\displaystyle P[E/x]}$  is equivalent to the after-assignment truth of ${\displaystyle P}$ . Thus were ${\displaystyle P[E/x]}$  true prior to the assignment, by the assignment axiom, then ${\displaystyle P}$  would be true subsequent to which. Conversely, were ${\displaystyle P[E/x]}$  false (i.e. ${\displaystyle \neg P[E/x]}$  true) prior to the assignment statement, ${\displaystyle P}$  must then be false afterwards.

Examples of valid triples include:

• ${\displaystyle \{x+1=43\}y:=x+1\{y=43\}}$
• ${\displaystyle \{x+1\leq N\}x:=x+1\{x\leq N\}}$

All preconditions that are not modified by the expression can be carried over to the postcondition. In the first example, assigning ${\displaystyle y:=x+1}$  does not change the fact that ${\displaystyle x+1=43}$ , so both statements may appear in the postcondition. Formally, this result is obtained by applying the axiom schema with ${\displaystyle P}$  being (${\displaystyle y=43}$  and ${\displaystyle x+1=43}$ ), which yields ${\displaystyle P[(x+1)/y]}$  being (${\displaystyle x+1=43}$  and ${\displaystyle x+1=43}$ ), which can in turn be simplified to the given precondition ${\displaystyle x+1=43}$ .

The assignment axiom scheme is equivalent to saying that to find the precondition, first take the post-condition and replace all occurrences of the left-hand side of the assignment with the right-hand side of the assignment. Be careful not to try to do this backwards by following this incorrect way of thinking: ${\displaystyle \{P\}x:=E\{P[E/x]\}}$ ; this rule leads to nonsensical examples like:

${\displaystyle \{x=5\}x:=3\{3=5\}}$

Another incorrect rule looking tempting at first glance is ${\displaystyle \{P\}x:=E\{P\wedge x=E\}}$ ; it leads to nonsensical examples like:

${\displaystyle \{x=5\}x:=x+1\{x=5\wedge x=x+1\}}$

While a given postcondition ${\displaystyle P}$  uniquely determines the precondition ${\displaystyle P[E/x]}$ , the converse is not true. For example:

• ${\displaystyle \{0\leq y\cdot y\wedge y\cdot y\leq 9\}x:=y\cdot y\{0\leq x\wedge x\leq 9\}}$ ,
• ${\displaystyle \{0\leq y\cdot y\wedge y\cdot y\leq 9\}x:=y\cdot y\{0\leq x\wedge y\cdot y\leq 9\}}$  ,
• ${\displaystyle \{0\leq y\cdot y\wedge y\cdot y\leq 9\}x:=y\cdot y\{0\leq y\cdot y\wedge x\leq 9\}}$  , and
• ${\displaystyle \{0\leq y\cdot y\wedge y\cdot y\leq 9\}x:=y\cdot y\{0\leq y\cdot y\wedge y\cdot y\leq 9\}}$

are valid instances of the assignment axiom scheme.

The assignment axiom proposed by Hoare does not apply when more than one name may refer to the same stored value. For example,

${\displaystyle \{y=3\}x:=2\{y=3\}}$

is wrong if ${\displaystyle x}$  and ${\displaystyle y}$  refer to the same variable (aliasing), although it is a proper instance of the assignment axiom scheme (with both ${\displaystyle \{P\}}$  and ${\displaystyle \{P[2/x]\}}$  being ${\displaystyle \{y=3\}}$ ).

### Rule of composition

Hoare's rule of composition applies to sequentially executed programs ${\displaystyle S}$  and ${\displaystyle T}$ , where ${\displaystyle S}$  executes prior to ${\displaystyle T}$  and is written ${\displaystyle S;T}$  (${\displaystyle Q}$  is called the midcondition):[4]

${\displaystyle {\dfrac {\{P\}S\{Q\}\quad ,\quad \{Q\}T\{R\}}{\{P\}S;T\{R\}}}}$

For example, consider the following two instances of the assignment axiom:

${\displaystyle \{x+1=43\}y:=x+1\{y=43\}}$

and

${\displaystyle \{y=43\}z:=y\{z=43\}}$

By the sequencing rule, one concludes:

${\displaystyle \{x+1=43\}y:=x+1;z:=y\{z=43\}}$

Another example is shown in the right box.

### Conditional rule

${\displaystyle {\dfrac {\{B\wedge P\}S\{Q\}\quad ,\quad \{\neg B\wedge P\}T\{Q\}}{\{P\}{\textbf {if}}\ B\ {\textbf {then}}\ S\ {\textbf {else}}\ T\ {\textbf {endif}}\{Q\}}}}$

The conditional rule states that a postcondition ${\displaystyle Q}$  common to ${\displaystyle {\textbf {then}}}$  and ${\displaystyle {\textbf {else}}}$  part is also a postcondition of the whole ${\displaystyle {\textbf {if}}\cdots {\textbf {endif}}}$  statement. In the ${\displaystyle {\textbf {then}}}$  and the ${\displaystyle {\textbf {else}}}$  part, the unnegated and negated condition ${\displaystyle B}$  can be added to the precondition ${\displaystyle P}$ , respectively. The condition, ${\displaystyle B}$ , must not have side effects. An example is given in the next section.

This rule was not contained in Hoare's original publication.[1] However, since a statement

${\displaystyle {\textbf {if}}\ B\ {\textbf {then}}\ S\ {\textbf {else}}\ T\ {\textbf {endif}}}$

has the same effect as a one-time loop construct

${\displaystyle {\textbf {bool}}\ b:={\textbf {true}};{\textbf {while}}\ B\wedge b\ {\textbf {do}}\ S;b:={\textbf {false}}\ {\textbf {done}};b:={\textbf {true}};{\textbf {while}}\ \neg B\wedge b\ {\textbf {do}}\ T;b:={\textbf {false}}\ {\textbf {done}}}$

the conditional rule can be derived from the other Hoare rules. In a similar way, rules for other derived program constructs, like ${\displaystyle {\textbf {for}}}$  loop, ${\displaystyle {\textbf {do}}\cdots {\textbf {until}}}$  loop, ${\displaystyle {\textbf {switch}}}$ , ${\displaystyle {\textbf {break}}}$ , ${\displaystyle {\textbf {continue}}}$  can be reduced by program transformation to the rules from Hoare's original paper.

### Consequence rule

${\displaystyle {\dfrac {P_{1}\rightarrow P_{2}\quad ,\quad \{P_{2}\}S\{Q_{2}\}\quad ,\quad Q_{2}\rightarrow Q_{1}}{\{P_{1}\}S\{Q_{1}\}}}}$

This rule allows to strengthen the precondition and/or to weaken the postcondition. It is used e.g. to achieve literally identical postconditions for the ${\displaystyle {\textbf {then}}}$  and the ${\displaystyle {\textbf {else}}}$  part.

For example, a proof of

${\displaystyle \{0\leq x\leq 15\}{\textbf {if}}\ x<15\ {\textbf {then}}\ x:=x+1\ {\textbf {else}}\ x:=0\ {\textbf {endif}}\{0\leq x\leq 15\}}$

needs to apply the conditional rule, which in turn requires to prove

${\displaystyle \{0\leq x\leq 15\wedge x<15\}x:=x+1\{0\leq x\leq 15\}}$ ,   or simplified
${\displaystyle \{0\leq x<15\}x:=x+1\{0\leq x\leq 15\}}$

for the ${\displaystyle {\textbf {then}}}$  part, and

${\displaystyle \{0\leq x\leq 15\wedge x\geq 15\}x:=0\{0\leq x\leq 15\}}$ ,   or simplified
${\displaystyle \{x=15\}x:=0\{0\leq x\leq 15\}}$

for the ${\displaystyle {\textbf {else}}}$  part.

However, the assignment rule for the ${\displaystyle {\textbf {then}}}$  part requires to choose ${\displaystyle P}$  as ${\displaystyle 0\leq x\leq 15}$ ; rule application hence yields

${\displaystyle \{0\leq x+1\leq 15\}x:=x+1\{0\leq x\leq 15\}}$ ,   which is logically equivalent to
${\displaystyle \{-1\leq x<15\}x:=x+1\{0\leq x\leq 15\}}$ .

The consequence rule is needed to strengthen the precondition ${\displaystyle \{-1\leq x<15\}}$  obtained from the assignment rule to ${\displaystyle \{0\leq x<15\}}$  required for the conditional rule.

Similarly, for the ${\displaystyle {\textbf {else}}}$  part, the assignment rule yields

${\displaystyle \{0\leq 0\leq 15\}x:=0\{0\leq x\leq 15\}}$ ,   or equivalently
${\displaystyle \{{\textbf {true}}\}x:=0\{0\leq x\leq 15\}}$ ,

hence the consequence rule has to be applied with ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  being ${\displaystyle \{x=15\}}$  and ${\displaystyle \{{\textbf {true}}\}}$ , respectively, to strengthen again the precondition. Informally, the effect of the consequence rule is to "forget" that \{x=15\} is known at the entry of the ${\displaystyle {\textbf {else}}}$  part, since the assignment rule used for the ${\displaystyle {\textbf {else}}}$  part doesn't need that information.

### While rule

${\displaystyle {\dfrac {\{P\wedge B\}S\{P\}}{\{P\}{\textbf {while}}\ B\ {\textbf {do}}\ S\ {\textbf {done}}\{\neg B\wedge P\}}}}$

Here ${\displaystyle P}$  is the loop invariant, which is to be preserved by the loop body ${\displaystyle S}$ . After the loop is finished, this invariant ${\displaystyle P}$  still holds, and moreover ${\displaystyle \neg B}$  must have caused the loop to end. As in the conditional rule, ${\displaystyle S}$  must not have side effects.

For example, a proof of

${\displaystyle \{x\leq 10\}{\textbf {while}}\ x<10\ {\textbf {do}}\ x:=x+1\ {\textbf {done}}\{\neg x<10\wedge x\leq 10\}}$

by the while rule requires to prove

${\displaystyle \{x\leq 10\wedge x<10\}x:=x+1\{x\leq 10\}}$ ,   or simplified
${\displaystyle \{x<10\}x:=x+1\{x\leq 10\}}$ ,

which is easily obtained by the assignment rule. Finally, the postcondition ${\displaystyle \{\neg x<10\wedge x\leq 10\}}$  can be simplified to ${\displaystyle \{x=10\}}$ .

For another example, the while rule can be used to formally verify the following strange program to compute the exact square root ${\displaystyle x}$  of an arbitrary number ${\displaystyle a}$ —even if ${\displaystyle x}$  is an integer variable and ${\displaystyle a}$  is not a square number:

${\displaystyle \{{\textbf {true}}\}{\textbf {while}}\ x\cdot x\neq a\ {\textbf {do}}\ {\textbf {skip}}\ {\textbf {done}}\{x\cdot x=a\wedge {\textbf {true}}\}}$

After applying the while rule with ${\displaystyle P}$  being ${\displaystyle {\textbf {true}}}$ , it remains to prove

${\displaystyle \{{\textbf {true}}\wedge x\cdot x\neq a\}{\textbf {skip}}\{{\textbf {true}}\}}$ ,

which follows from the skip rule and the consequence rule.

In fact, the strange program is partially correct: if it happened to terminate, it is certain that ${\displaystyle x}$  must have contained (by chance) the value of ${\displaystyle a}$ 's square root. In all other cases, it will not terminate; therefore it is not totally correct.

### While rule for total correctness

If the above ordinary while rule is replaced by the following one, the Hoare calculus can also be used to prove total correctness, i.e. termination[note 3] as well as partial correctness. Commonly, square brackets are used here instead of curly braces to indicate the different notion of program correctness.

${\displaystyle {\dfrac {<\ {\text{is a well-founded ordering on the set}}\ D\quad ,\quad [P\wedge B\wedge t\in D\wedge t=z]S[P\wedge t\in D\wedge t

In this rule, in addition to maintaining the loop invariant, one also proves termination by way of an expression ${\displaystyle t}$ , called the loop variant, whose value strictly decreases with respect to a well-founded relation ${\displaystyle <}$  on some domain set ${\displaystyle D}$  during each iteration. Since ${\displaystyle <}$  is well-founded, a strictly decreasing chain of members of ${\displaystyle D}$  can have only finite length, so ${\displaystyle t}$  cannot keep decreasing forever. (For example, the usual order ${\displaystyle <}$  is well-founded on positive integers ${\displaystyle \mathbb {N} }$ , but neither on the integers ${\displaystyle \mathbb {Z} }$  nor on positive real numbers ${\displaystyle \mathbb {R} ^{+}}$ ; all these sets are meant in the mathematical, not in the computing sense, they are all infinite in particular.)

Given the loop invariant ${\displaystyle P}$ , the condition ${\displaystyle B}$  must imply that ${\displaystyle t}$  is not a minimal element of ${\displaystyle D}$ , for otherwise the body ${\displaystyle S}$  could not decrease ${\displaystyle t}$  any further, i.e. the premise of the rule would be false. (This is one of various notations for total correctness.) [note 4]

Resuming the first example of the previous section, for a total-correctness proof of

${\displaystyle [x\leq 10]{\textbf {while}}\ x<10\ {\textbf {do}}\ x:=x+1\ {\textbf {done}}[\neg x<10\wedge x\leq 10]}$

the while rule for total correctness can be applied with e.g. ${\displaystyle D}$  being the non-negative integers with the usual order, and the expression ${\displaystyle t}$  being ${\displaystyle 10-x}$ , which then in turn requires to prove

${\displaystyle [x\leq 10\wedge x<10\wedge 10-x\geq 0\wedge 10-x=z]x:=x+1[x\leq 10\wedge 10-x\geq 0\wedge 10-x

Informally speaking, we have to prove that the distance ${\displaystyle 10-x}$  decreases in every loop cycle, while it always remains non-negative; this process can go on only for a finite number of cycles.

The previous proof goal can be simplified to

${\displaystyle [x<10\wedge 10-x=z]x:=x+1[x\leq 10\wedge 10-x ,

which can be proven as follows:

${\displaystyle [x+1\leq 10\wedge 10-x-1  is obtained by the assignment rule, and
${\displaystyle [x+1\leq 10\wedge 10-x-1  can be strengthened to ${\displaystyle [x<10\wedge 10-x=z]}$  by the consequence rule.

For the second example of the previous section, of course no expression ${\displaystyle t}$  can be found that is decreased by the empty loop body, hence termination cannot be proved.

## Notes

1. ^ Hoare originally wrote "${\displaystyle P\{C\}Q}$ " rather than "${\displaystyle \{P\}C\{Q\}}$ ".
2. ^ This article uses a natural deduction style notation for rules. For example, ${\displaystyle {\dfrac {\alpha ,\beta }{\phi }}}$  informally means "If both ${\displaystyle \alpha }$  and ${\displaystyle \beta }$  hold, then also ${\displaystyle \phi }$  holds"; ${\displaystyle \alpha }$  and ${\displaystyle \beta }$  are called antecedents of the rule, ${\displaystyle \phi }$  is called its succedent. A rule without antecedents is called an axiom, and written as ${\displaystyle {\dfrac {}{\quad \phi \quad }}}$ .
3. ^ "Termination" here is meant in the broader sense that computation will eventually be finished; it does not imply that no limit violation (e.g. zero divide) can stop the program prematurely.
4. ^ Hoare's 1969 paper didn't provide a total correctness rule; cf. his discussion on p.579 (top left). For example Reynolds' textbook (John C. Reynolds (2009). Theory of Programming Languages. Cambridge University Press.), Sect.3.4, p.64 gives the following version of a total correctness rule: ${\displaystyle {\dfrac {P\wedge B\rightarrow 0\leq t\quad ,\quad [P\wedge B\wedge t=z]S[P\wedge t  when ${\displaystyle z}$  is an integer variable that doesn't occur free in ${\displaystyle P}$ , ${\displaystyle B}$ , ${\displaystyle S}$ , or ${\displaystyle t}$ , and ${\displaystyle t}$  is an integer expression (Reynolds' variables renamed to fit with this article's settings).

## References

1. ^ a b Hoare, C. A. R. (October 1969). "An axiomatic basis for computer programming" (PDF). Communications of the ACM. 12 (10): 576–580. doi:10.1145/363235.363259. Archived from the original on 2016-03-04.CS1 maint: BOT: original-url status unknown (link)
2. ^ R. W. Floyd. "Assigning meanings to programs." Proceedings of the American Mathematical Society Symposia on Applied Mathematics. Vol. 19, pp. 19–31. 1967.
3. ^ p.579 upper left
4. ^ Huth, Michael; Ryan, Mark. Logic in Computer Science (second ed.). CUP. p. 276. ISBN 052154310X.


text:
Acta Chimica Sinica ›› 1999, Vol. 57 ›› Issue (4): 339-344.

Original Articles

### 反应NH~4ClO~4+Mg+K~2Cr~2O~7的非线性化学动力学2: 固相 振荡燃烧的化学模 型

1. 北京理工大学力学工程系国家重点实验室
• 出版日期:1999-04-15 发布日期:1999-04-15

### The nonlinear chemical reaction kinetics of NH~4ClO~4+Mg+K~2Cr~2O~7 2: Chemical models of the solid phase oscillatory combustion

Feng Changgen;Liu Zhaomiao;Zeng Qingxuan;Xu Shiying

• Online:1999-04-15 Published:1999-04-15

The non-isothermic three-variable chemical model of the oscillatory combustion of NH~4ClO~4+Mg+K~2Cr~2O~7 system is established in the present paper. The nonlinear reaction kinetic character of this system is described by the results of numerical simulation. By applying mathematical methods for nonlinear analysis, the mechanism of the nonlinear chemical kinetics of the solid phase combustion is studied. The results show that the system has rich content of the character of the nonlinear chemical kinetics. From the result of numerical simulation it can be seen that with the changing of the value of the nondimensional variable μ of NH~4ClO~4 when μ is between μ=0.00547863 and μ=1.47185, the oscillatory period is different. When μ=0.5678, 0.591, 0.6378, 0.6388 and 0.6478, the period -1, period -2, period -4, pseudo-period and period -3 oscillations are obtained respectively. It can be recognized that the change of NH~4ClO~4 has influence on the oscillatory period. The result of numerical simulation is same to the reaction mechanism analysis of the experimental study. The non-isothermic three-variable chemical model studied in the paper not only considers the influence of concentration on the oscillatory period, but also analyzes the effect of temperature on the oscillatory combustion. The nonlinear numerical analysis method is combined with the chemical kinetics reaction mechanism analysis of the pyrotechnics combustion in the present paper. And the experimental and chemical models study are developed.

CLC Number:


text:
## How to check if a block has alignfull using a conditional?

Question

Are there any conditional queries to check if a block has an alignfull class for example?

I’m using the below which when part of another function adds the color overlay class to my body classes, but ideally this should only happen if the block is set to alignfull.

if ( 'core/cover' === blocks[0]['blockName'] ) { // Would like to check if this block is set to alignfull before adding the below. if ( isset(blocks[0]['attrs']['overlayColor'] ) ) {
$classes[] = 'has-' .$blocks[0]['attrs']['overlayColor'] . '-header';
}

}

in progress 0
1 year 2020-08-10T04:11:24-05:00 1 Answer 53 views 0

1. Hi there,

Yes, there is something you can use to check if a block is alignfull. There is an attribute to see if the block you are checking it’s been aligned full.

In your case I would use (in fact, is  what I’m using these days on my starter Genesis  Theme):

function genesis_first_block_body_classes( $classes ) { if ($blocks[0][‘blockName’] === ‘core/cover’ && blocks[0][‘attrs’][‘align’] === ‘full’ ) {classes[] = ‘first-block-cover-full’;

}

return \$classes;

}

This checks if the first block is a cover block and it’s been aligned with full-width. Then it adds the designated class to the body. Of course, you can whatever class you want to.

I hope that helps.