text:
# 25.4. Multiple Regression¶

Regression provides one way of predicting a numerical variable, called a response, based on other variables called predictor variables. The multiple regression model says in essence that

$\text{response} ~ = ~ \text{linear combination of predictor variables} + \text{ random noise }$

You can think of the first term on the right hand side as a signal. The problem is that we don’t get to observe the signal. The observed response is the sum of the signal and the noise. The data scientist’s task is to use the observations to extract the signal as accurately as possible.

It is worth looking more closely at exactly what is linear in linear regression, now that we are allowing more than one predictor variable. For example, notice that you can fit a quadratic function of $$x$$ by using the two predictor variables $$x_1 = x$$ and $$x_2 = x^2$$. Then the signal

$\beta_0 + \beta_1x_1 + \beta_2x_2 ~ = ~ \beta_0 + \beta_1x + \beta_2x^2$

is a quadratic function of $$x$$. But it is linear in the coefficients, and it is a linear combination of the two predictor variables $$x_1$$ and $$x_2$$.

## 25.4.1. The Model¶

As in all of statistical inference, properties of estimates depend on the assumptions under which they are calculated. The multiple regression model is a commonly used set of assumptions that describes a particular kind of linear relation between a numerical response variable and a set of predictor variables. You should use it only if you believe that it makes sense for your data.

The model assumes that there are $$n$$ individuals, on each of whom you have measured the response and the predictor variables. For $$1 \le i \le n$$, the relation between the variables is assumed to be

$Y_i = \beta_0 + \beta_1x_{i,1} + \beta_2x_{i,2} + \cdots + \beta_{p-1}x_{i,p-1} + \epsilon_i$

in the notation described below.

• $$x_{i,1}, x_{i,2}, \ldots, x_{i,p-1}$$ are the observed constant values of $$p-1$$ predictor variables for individual $$i$$. They are not random variables. If you prefer to think of the predictor variables as random, this model assumes that you have conditioned on them.

• The intercept $$\beta_0$$ and slopes $$\beta_1, \beta_2, \ldots, \beta_{p-1}$$ are unobservable constants and are parameters of the model. There are $$p$$ of them, hence the notation $$p$$ for “parameters”.

• $$\epsilon_i$$ is an unobservable random error that has the normal $$(0, \sigma^2)$$ distribution for some unobservable $$\sigma^2$$, and $$\epsilon_1, \epsilon_2, \ldots, \epsilon_n$$ are i.i.d.

• $$Y_i$$ is the observable response of individual $$i$$. It is random because $$\epsilon_i$$ is one of its components.

We will assume that $$n > p$$, that is, we will assume we have more individuals than parameters. Indeed in this course it is fine for you to think of $$n$$ as much larger than $$p$$.

Two special cases are already familiar.

$$p = 1$$: Prediction by a Constant

When $$p = 1$$ there is just one parameter: the intercept. There are no predictor variables at all. The model says that for each individual $$i$$, the response is $$Y_i = \beta_0 + \epsilon_i$$. This is a case of trying to estimate the response by a constant.

$$p = 2$$: Simple Linear Regression

The two parameters are the intercept and a slope. The model says that for each individual $$i$$, the response is $$Y_i = \beta_0 + \beta_1x_{i,1} + \epsilon_i$$. That is, the response is the value on a hidden straight line, plus some normal noise. This is the simple regression model you used in Data 8.

## 25.4.2. Signal and Noise: Matrix Representation¶

For any $$p$$, the model can be written compactly as

$\mathbf{Y} ~ = ~ \mathbf{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}$

in the matrix notation described below.

• The design matrix $$\mathbf{X}$$ is an $$n \times p$$ matrix of real numbers, not random variables. Column 0 of $$\mathbf{X}$$ is a vector of 1’s and Column $$j$$ for $$1 \le j \le p-1$$ consists of the $$n$$ observations on the $$j$$th predictor variable. For each $$i$$ in the range 1 through $$n$$, Row $$i$$ contains the values of all the predictor variables for individual $$i$$.

• The parameter vector $$\boldsymbol{\beta} = [\beta_0 ~~ \beta_1 ~~ \ldots ~~ \beta_{p-1}]^T$$ is a $$p \times 1$$ vector of the coefficients.

• The error vector $$\boldsymbol{\epsilon}$$ is an $$n \times 1$$ multivariate normal $$(\mathbf{0}, \sigma^2\mathbf{I}_n)$$ random vector. Its mean vector is an $$n \times 1$$ vector of 0’s and $$\mathbf{I}_n$$ is the $$n \times n$$ identity matrix.

• The response vector $$\mathbf{Y}$$ is a random vector that is the sum of the linear signal $$\mathbf{X}\boldsymbol{\beta}$$ and the normal noise $$\boldsymbol{\epsilon}$$.

## 25.4.3. Ordinary Least Squares¶

Based on the observations of the predictor variables and the response, the goal is to find the best estimates of the intercept and slopes in the model.

These estimates can then be used to predict the response of a new individuals, assuming that the model holds for the new individual as well.

We must select a criterion by which we will decide whether one estimate is better than another. To develop one such criterion, start by noting that any linear function of the predictor variables can be written as $$\mathbf{X}\boldsymbol{\gamma}$$ where $$\boldsymbol{\gamma}$$ is some $$p \times 1$$ vector of coefficients. Think of $$\mathbf{X}\boldsymbol{\gamma}$$ as an estimate of $$\mathbf{Y}$$. Then the error in the estimate is $$\mathbf{Y} - \mathbf{X}\boldsymbol{\gamma}$$.

The goal of ordinary least squares (OLS) is to find the vector $$\boldsymbol{\gamma}$$ that minimises the mean squared error

$MSE(\boldsymbol{\gamma}) ~ = ~ \frac{1}{n} \sum_{i=1}^n (Y_i - (\mathbf{X}\boldsymbol{\gamma})_i)^2$

This is the same as the $$\boldsymbol{\gamma}$$ that minimizes the sum of squared errors

$SSE(\boldsymbol{\gamma}) ~ = ~ \sum_{i=1}^n (Y_i - (\mathbf{X}\boldsymbol{\gamma})_i)^2$

Again for compactness it will help to use matrix notation. For an $$n \times 1$$ vector $$\mathbf{w}$$,

$\sum_{i=1}^n w_i^2 ~ = ~ \mathbf{w}^T\mathbf{w} ~ = ~ \mathbf{w} \cdot \mathbf{w} ~ = ~ \| \mathbf{w} \|^2$

which is sometimes called the squared norm of $$\mathbf{w}$$.

In this notation, the goal of OLS is to find the $$p \times 1$$ vector $$\hat{\boldsymbol{\beta}}$$ that minimizes $$\| \mathbf{Y} - \mathbf{X}\boldsymbol{\gamma}\|^2$$ over all vectors $$\boldsymbol{\gamma}$$.

Typically you will also have to estimate the unknown error variance $$\sigma^2$$. But we will not cover that in this class except in the case $$p = 1$$.

## 25.4.4. Guessing the Best Estimate of $$\boldsymbol{\beta}$$¶

Remember that we have assumed $$n > p$$. Assume also that $$\mathbf{X}$$ is of full column rank $$p$$, that is, none of the predictor variables is a linear combination of the others. By a theorem in linear algebra, it follows that the $$p \times p$$ matrix $$\mathbf{X}^T\mathbf{X}$$ has full rank $$p$$ and is therefore invertible.

The claim is that OLS estimate of $$\boldsymbol{\beta}$$ is the vector $$\hat{\boldsymbol{\beta}}$$ defined by

$\hat{\boldsymbol{\beta}} ~ = ~ (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y}$

The claim is motivated by our earlier formula

$\mathbf{b} ~ = ~ \boldsymbol{\Sigma}_\mathbf{X}^{-1}\boldsymbol{\Sigma}_\mathbf{XY}$

for the coefficients of the least squares linear predictor a random variable $$Y$$ based on a random vector $$\mathbf{X}$$. In fact the new formula is an application of the old one. But we will derive it afresh in our new setting.

The key idea is that of projection: the best $$\hat{\boldsymbol{\beta}}$$ should be such that the error in the estimate is orthogonal to the linear space spanned by $$\mathbf{X}$$.

The error in the best estimate is $$\mathbf{Y} - \hat{\mathbf{Y}} = \mathbf{Y} - \mathbf{X}\hat{\boldsymbol{\beta}}$$. For this error to be orthogonal to linear transformations of $$\mathbf{X}$$ we must have

$\mathbf{X}^T (Y - \mathbf{X}\hat{\boldsymbol{\beta}}) = 0 ~ \implies \mathbf{X}^T Y ~ = ~ \mathbf{X}^T \mathbf{X}\hat{\boldsymbol{\beta}}$

We have assumed that $$\mathbf{X}$$ has full column rank, so $$\mathbf{X}^T \mathbf{X}$$ is invertible. So the natural guess for the best estimator $$\hat{\boldsymbol{\beta}}$$ is

$\hat{\boldsymbol{\beta}} ~ = ~ (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y}$

Before we go further, notice that $$\hat{\boldsymbol{\beta}}$$ is a linear function of $$\mathbf{Y}$$. This makes it straightforward to identify its distribution, which you will do in exercises.

Also note that the estimate of $$\mathbf{Y}$$ is

$\hat{\mathbf{Y}} ~ = ~ \mathbf{X}\hat{\boldsymbol{\beta}} ~ = ~ \mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y}$

which is also a linear function of $$\mathbf{Y}$$.

## 25.4.5. Projection¶

Define the $$i$$th residual as the prediction error $$e_i = Y_i - \hat{Y}_i$$. Then the $$n \times 1$$ vector of residuals is

$\mathbf{e} ~ = ~ \mathbf{Y} - \hat{\mathbf{Y}} ~ = ~ \mathbf{Y} - \mathbf{X}\hat{\boldsymbol{\beta}}$

As we have seen repeatedly, the key to least squares is that the prediction error is orthogonal to the space of allowed functions. Our space of allowed functions is all linear functions of $$\mathbf{X}$$. So we will show:

The residual vector is orthogonal to each column of $$\mathbf{X}$$.

This is essentially true by construction. Formally, calculate the $$p \times 1$$ vector $$\mathbf{X}^T\mathbf{e}$$. Each of its elements is the dot product of $$\mathbf{e}$$ and one column of $$\mathbf{X}$$. We will show that each of the elements is 0.

$\mathbf{X}^T\mathbf{e} ~ = ~ \mathbf{X}^T(\mathbf{Y} - \hat{\mathbf{Y}}) ~ = ~ \mathbf{X}^T\mathbf{Y} - \mathbf{X}^T\hat{\mathbf{Y}} ~ = ~ \mathbf{X}^T\mathbf{Y} - \mathbf{X}^T\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{Y} ~ = ~ \mathbf{X}^T\mathbf{Y} - \mathbf{X}^T\mathbf{Y} ~ = ~ 0$

## 25.4.6. Least Squares¶

Let $$\boldsymbol{\gamma}$$ be any $$p \times 1$$ vector. Then

\begin{split} \begin{align*} SSE(\boldsymbol{\gamma}) ~ &= ~ \| \mathbf{Y} - \mathbf{X}\boldsymbol{\gamma} \|^2 \\ &= ~ \| (\mathbf{Y} - \mathbf{X}\hat{\boldsymbol{\beta}}) + (\mathbf{X}\hat{\boldsymbol{\beta}} - \mathbf{X}\boldsymbol{\gamma}) \|^2 \\ &= ~ \|\mathbf{Y} - \mathbf{X}\hat{\boldsymbol{\beta}}\|^2 ~+~ \| \mathbf{X}\hat{\boldsymbol{\beta}} - \mathbf{X}\boldsymbol{\gamma} \|^2 + 2(\mathbf{X}\hat{\boldsymbol{\beta}} - \mathbf{X}\boldsymbol{\gamma})^T(\mathbf{Y} - \mathbf{X}\hat{\boldsymbol{\beta}}) \\ &= ~ SSE(\hat{\boldsymbol{\beta}}) ~+~ \| \mathbf{X}\hat{\boldsymbol{\beta}} - \mathbf{X}\boldsymbol{\gamma} \|^2 + 2((\mathbf{X}(\hat{\boldsymbol{\beta}} - \boldsymbol{\gamma}))^T\mathbf{e} \\ &= ~ SSE(\hat{\boldsymbol{\beta}}) ~+~ \| \mathbf{X}\hat{\boldsymbol{\beta}} - \mathbf{X}\boldsymbol{\gamma} \|^2 + 2(\hat{\boldsymbol{\beta}} - \boldsymbol{\gamma})^T\mathbf{X}^T \mathbf{e} \\ &= ~ SSE(\hat{\boldsymbol{\beta}}) ~+~ \| \mathbf{X}\hat{\boldsymbol{\beta}} - \mathbf{X}\boldsymbol{\gamma} \|^2 ~~~~ \text{ by orthogonality} \\ &\ge ~ SSE(\hat{\boldsymbol{\beta}}) \end{align*} \end{split}

## 25.4.7. Signal and Noise, Revisited¶

Our regression model is

$\mathbf{Y} ~ = ~ \mathbf{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}$

Here

• $$\mathbf{X}\boldsymbol{\beta}$$ is the unobservable but non-random true signal

• $$\boldsymbol{\epsilon}$$ is an unobservable random vector consisting of the deviations of $$\mathbf{Y}$$ from the true plane $$\mathbf{X}\boldsymbol{\beta}$$. Elements of $$\boldsymbol{\epsilon}$$ are mutually independent.

Once we have carried out the regression, our estimate of the response vector $$\mathbf{Y}$$ is the vector $$\hat{\mathbf{Y}} = \mathbf{X}\boldsymbol{\hat{\boldsymbol{\beta}}}$$.

The residual vector is

$\mathbf{e} ~ = ~ \mathbf{Y} - \hat{\mathbf{Y}} ~ = ~ \mathbf{Y} - \mathbf{X}\boldsymbol{\hat{\boldsymbol{\beta}}}$

Therefore we have another expression for the response vector $$\mathbf{Y}$$. This expression is our best attempt at separating the signal from the noise.

$\mathbf{Y} ~ = ~ \mathbf{X}\boldsymbol{\hat{\boldsymbol{\beta}}} + \mathbf{e}$

It is important to note the distinction between this identity and the model.

• $$\mathbf{X}\boldsymbol{\hat{\boldsymbol{\beta}}}$$ is the observable random estimated signal.

• $$\mathbf{e}$$ is the observable random vector consisting of the deviations of $$\mathbf{Y}$$ from the estimated plane $$\mathbf{X}\hat{\boldsymbol{\beta}}$$. Elements of $$\mathbf{e}$$ are not independent of each other, because they add up to $$0$$.

In exercises you will show that $$\hat{\boldsymbol{\beta}}$$ is an unbiased estimator of $$\boldsymbol{\beta}$$ and that both $$\hat{\boldsymbol{\beta}}$$ of $$\hat{\mathbf{Y}}$$ have normal (or multivariate normal) distributions. Both distributions have variance and covariance parameters that depend on the unknown error variance $$\sigma^2$$.

## 25.4.8. Estimate of $$\sigma^2$$¶

It should come as no surprise that under the multiple regression model, there is an unbiased estimator of $$\sigma^2$$ that has a chi-squared distribution. There is some work involved in establishing that this estimate is

$S^2 ~ = ~ \frac{1}{n-p} \sum_{i=1}^n (Y_i - \hat{Y}_i)^2 ~ = ~ \frac{1}{n-p} \| \mathbf{e} \|^2$

Some more work establishes that $$\frac{n-p}{\sigma^2}S^2$$ has the chi-squared $$(n-p)$$ distribution.

We’ll leave that work for another course. For now, just notice that if the number of data points $$n$$ is large compared to the number of parameters $$p$$, then

$S^2 ~ = ~ \frac{1}{n-p} \sum_{i=1}^n (Y_i - \hat{Y}_i)^2 ~ \approx ~ \frac{1}{n} \sum_{i=1}^n (Y_i - \hat{Y}_i)^2 ~ = ~ \hat{\sigma}^2$

which is the natural mean squared error. If you have a lot of data, you don’t have to worry about fine points like dividing by $$n-p$$ instead of $$n$$.

Special Case

As noted earlier, in the case $$p = 1$$ you are trying to find the best constant by which to estimate $$Y$$.

You know that the least squares constant is $$\bar{Y}$$, and you showed in exercises that

$S^2 ~ = ~ \frac{1}{n-1} \sum_{i=1}^n (Y_i - \bar{Y})^2$

is an unbiased estimate of $$\sigma^2$$. You have shown in exercises that $$\frac{n-1}{\sigma^2}S^2$$ has the chi-squared $$n-1$$ distribution under the assumption that the data are i.i.d. normal variables. This is the special case of the result stated above for general $$p$$.

## 25.4.9. Confidence Intervals¶

The upshot of this discussion is that if $$n$$ is large compared to $$p$$ – that is, if you have a lot of observations compared to the number of predictors – then you can use ordinary normal theory to construct confidence intervals for the parameters $$\hat{\beta}$$.

For example, a 95% confidence interval for the parameter $$\beta_i$$ is $$\hat{\boldsymbol{\beta}} [i] \pm 2SD(\hat{\boldsymbol{\beta}} [i])$$.

Here $$\hat{\boldsymbol{\beta}}[i]$$ is the $$i$$th element of the estimate vector $$\hat{\boldsymbol{\beta}}$$.

The variance of $$\hat{\boldsymbol{\beta}}[i]$$ is the $$i$$th diagonal element of the covariance matrix of $$\hat{\boldsymbol{\beta}}$$. You will see in exercises that this involves the error variance $$\sigma^2$$. Typically, $$\sigma^2$$ is unknown. But you can estimate it by the mean squared error $$\hat{\sigma}^2$$ to get an approximate 95% confidence interval for $$\beta_i$$.


text:
# The myth of the finite pressuremeter geometry correction

Techref Number:
CITR1003
Authored by:
Robert Whittle
Date:
01/03/1999

### Introduction

Soil strength and stiffness properties are obtained from pressuremeter tests using analyses that depend on solutions for the expansion of an infinitely long cylindrical cavity. Real pressuremeters have length to diameter ratios between 3 and 10. Studies using finite element methods have indicated that this finite pressuremeter geometry leads to significant over-estimation of the shear strength. This paper tests the finite element results and shows that they do not predict the shear stress-strain response observed in real tests in the field. The conclusion is that end effects associated with finite pressuremeter geometry are of negligible significance for the derivation of material strength.

### Background

Several papers have reported on the consequences of finite pressuremeter geometry for the derivation of material strength and stiffness parameters. Two of the more recent are considered here, Houlsby & Carter (1993) and Shuttle & Jefferies (1995).

Houlsby & Carter (1993) present the results of simulated undrained pressuremeter tests in a finite element model of simple elastic/perfectly plastic soil. The simulated tests were run for differing length to diameter ratios, depths below ground level and for common values of rigidity index in the model soil.

Gibson & Anderson (1961) give the solution for an undrained cylindrical expansion in simple elastic/perfectly plastic soil. For a current radius $$a$$ at the cavity the total pressure $P$ being applied is

… [1]

$$P=P_0+S_u\left\{1+\ln \left[(I_r)\left(1 - \left(\frac{a_0} a\right)^2\right)\right]\right\}$$

where $$P_0$$ is the cavity reference pressure
$$I_r$$ is the rigidity index $$G/S_u$$
$$G$$ is the shear modulus
$$S_u$$ is the undrained shear strength
$$a_0$$ is the initial radius of the cavity

The Houlsby & Carter tests indicate that the smaller the length to diameter ratio, the greater the stress that must be applied to achieve a given plastic deformation. The difference in stress is expressed as a correction factor whose magnitude depends on a number of variables.

The corrections given in Houlsby & Carter are derived by following conventional engineering practice. The finite element tests are plotted in terms of total pressure against the natural log of the current volumetric strain and the slope of the most linear part of this plot gives the undrained shear strength directly. Houlsby & Carter take the slope between 2% and 5% cavity strain and express the correction factor as the ratio of the derived shear strength to the input shear strength. The choice of strain range is important because the effective length to diameter ratio reduces as the diameter of the plastic zone increases, so that other strain ranges give different results.

The conclusion of Houlsby & Carter is that the undrained shear strength is greatly affected by finite pressuremeter geometry. It is reported that a pressuremeter with a length to diameter ratio of 6 in material with a rigidity index of 200 would over-estimate the shear strength by more than 20%. The strain dependent nature of this error is acknowledged but not quantified because it is accounted for by the specified working method.

Shuttle & Jefferies (1995) repeats some of the work of Houlsby & Carter using different finite element code. Similar results are obtained although this is not immediately obvious from the manner in which the data are presented.

The consequences of finite geometry are expressed in Shuttle & Jefferies as an overshoot correction $$\beta$$ where at an instant in a test:

...[2]

$$S_{\text{u(true)}}= \beta S_{\text{u(measured)}}$$

$$\beta$$ is related to the rigidity index and is strain dependent, being a function of the current diameter of the plastic zone. Shuttle & Jefferies give the following approximation for $$\beta$$:

...[3]

$$\beta = 1.241 - 0.05 [\ln (I_r \epsilon_c)]$$

where $$\epsilon_c$$ is cavity strain in %

$\beta$ is used within the context of a curve comparison procedure where $$S_u$$ is an input parameter for generating a theoretical curve to fit as closely as possible the full measured curve. For a pressuremeter test in material with a rigidity index of 200, the effect of $$\beta$$ is less than 10% at 5% cavity strain.

Both studies provide corrections for pressuremeters with a length to diameter ratio (L/D) of 6 (the ratio for a Cambridge Self Boring Pressuremeter) in material with a rigidity index typical of what might be discovered in London clay (Table 1).

##### Table 1 - Variation of S u (measured) / S u (true) for L/D of 6 at about 5% cavity strain

Source of Data

Method

Ir = 100

Ir = 200

Ir = 500

[1]

Houlsby & Carter (1993)

[a]

1.152

1.248

1.420

[2]

Shuttle & Jefferies (1995)

[b]

1.062

1.095

1.153

[3]

Shuttle & Jefferies (1995)

[a]

1.174

1.278

1.389

[4]

Houlsby & Carter (1995)

[a]

1.149

1.246

1.415

Notes:
Method [a] - produce a plot of total pressure against Ln volumetric strain and take the slope between 2% and 5% cavity strain.
Method [b] - calculate the overshoot correction $$\beta$$.

At first sight there seems to be conflict about the magnitude of the correction between the two studies, but the difference is due to how the simulated tests are interpreted. If equations [1] and [3] are combined, then Shuttle & Jefferies ‘geometry affected’ pressuremeter tests can be derived. If these tests are then interpreted using the method specified by Houlsby & Carter, correction factors similar to those reported by Houlsby & Carter are obtained. This has been done here, and is shown in the third row of Table 1. Later work by Shuttle (quoted in Houlsby & Carter 1995) along these lines gives results even closer to those reported by Houlsby & Carter (1993) - compare rows one and four of Table 1.

For the purposes of this paper the Shuttle & Jefferies overshoot correction described by equation [3] is the most convenient way to access the finite geometry results but the table indicates that the two sets of simulated tests produce similar data. Alternative ways of utilising that data give different corrections.

### The comparison with field tests

It is not possible to test the magnitude of the corrections from field data. However if the predictions of the finite element studies are valid then the strain dependent nature of the finite geometry effect will give pressuremeter shear stress curves that show a strain hardening response. The degree of this response will depend on a number of factors, but for a given pressuremeter the significant variable will be the rigidity index.

Presenting the finite element results on semi log axes masks the extent of strain hardening behaviour. The effect is better demonstrated by drawing the shear stress versus strain curve using the subtangent or Palmer (1972) method. Figure 1 is an example.

A pressure versus strain curve for an infinite length pressuremeter was created using equation [1] with a rigidity index of 200, an undrained shear strength of 1 and a cavity reference pressure of 0. An overshoot correction $$\beta$$ was derived for every increment of strain using a version of equation [3] optimised for an $$I_r$$ value of 200, and the corresponding pressure versus strain curve for a pressuremeter with a length to ratio diameter of 6 was derived. These curves were then analysed by the subtangent method to give the data plotted in Figure 1.

The infinite length data gives the simple elastic/perfectly plastic form assumed by the Gibson & Anderson solution. However an obvious strain hardening response is apparent in the curve for the finite length pressuremeter.

Fig. 1 The subtangent analysis applied to the finite length problem

If the finite element results are relevant then tests in the field must show this response but they do not. Figure 2 is the shear stress:strain response derived from four self bored pressuremeter (SBP) tests in the same borehole in London clay. The instrument has a length to diameter ratio of 6. The tests lie between 10 and 17 metres below ground level. These tests have been chosen because of their freedom from defects such as obvious cracking at large strain which would be indicated by apparent strain softening. They are good examples of pressuremeter tests in the material for which the Gibson & Anderson analysis was developed. The data has been normalised by the derived shear strength to make the curves comparable with the data in figure 1, and the finite length pressuremeter data from figure 1 is also plotted in figure 2.

There are difficulties using conventional spreadsheet facilities to calculate the local slope of measured pressure:strain curves but the scatter here is reasonably small and the shapes of the curves are clear. There is no evidence of apparent shear stress increasing with strain, if anything the reverse. There are some indications of a peak strength near the origin up to 1.2 times the ultimate strength, and some signs that the pressuremeter curves have been affected by the taking of unload/reload cycles.

Figure 3 shows the same data plotted in terms of total pressure versus the natural log of the volumetric strain. In this example the total pressure has been normalised by subtracting the reference pressure $$P_0$$ before dividing by the derived shear strength. The pressure scale starts from 1, in effect the yield stress, so that only the plastic loading is seen and the intercept on the strain axis gives $$\ln (1/I_r)$$. The intercept on the normalised pressure axis gives the limit pressure in the form of the so-called pressuremeter constant (6.3 for anof 200).

Fig. 2 Subtangent analysis of four SBP Tests in London clay

Although short sections of the field data do indicate some local strain hardening, a glance along the plotted points especially at large strain shows that the assumption of perfect plasticity is not unreasonable. The local strain hardening is due to the influence of unload/reload cycles on the loading curve, and is a transitory effect. This is indicated by the manner in which a line projected back through the points at large strain passes close to the initial points near the yield stress. The shear stress:strain response predicted for a length to diameter ratio of 6 and an $$I_r$$ of 200 is quite different.

Fig. 3 Gibson & Anderson style analysis applied to the same data plotted in figure 2

### Conclusion

An attack on the validity of the finite geometry correction based on the absence of a strain hardening response might seem indirect, not to say superficial, but it is fundamental. There is no special magic about the length to diameter ratio of the instrument - it is the length to diameter ratio of the expanding cavity that is significant. Because this ratio is reducing all the time the test is in progress, the strain hardening response is a necessary sign that a correction is needed. Conversely, as demonstrated here, the absence of such behaviour is sufficient evidence by itself to show that there can be no important error in the pressuremeter estimates of shear strength due to end effects.

The finite element tests are not wrong - indeed, some effort has been expended demonstrating that different researchers using different code obtain similar data. However as far as tests in real soil are concerned, the results of the finite element tests are irrelevant. Although this conclusion comes from a handful of field tests from one type of instrument in one type of soil it is strongly suspected that other pressuremeters in other soils are much less affected by finite geometry than has been supposed. The evidence from field tests in sand, for example, is that the angle of internal friction determined at the initial yield is usually similar to the value derived at the end of the test at about 10% cavity strain. Although this author has little experience of the Ménard pressuremeter, inspection of the examples of such tests in the papers by Gibson & Anderson and by Palmer show no evidence of strain hardening. Indeed, it is difficult to imagine how these two classic analyses for undrained strength could have been demonstrated if the corrections for end effects are of the magnitude suggested by the finite element tests.

It is not difficult to think of reasons why the finite element results are misleading. The finite element soil is described by a linear elastic characteristic. Real soils invariably have a non-linear elastic response. The consequences for the test are dramatic. Non-linear stiffness means that in the expanding cavity, soil that is yielding is conditioned by a significantly lower stiffness than soil which is behaving elastically. The ‘ends’ of the expanding cavity where a small axial component might be expected are much stiffer than the linear elastic model suggests.

It is possible to imagine circumstances where end effects could be demonstrated. Clean sand of uniform particle size rained into a calibration chamber might respond in a manner that approximates to linear elastic. Tests in such material (Fahey 1980), do show some measurable effect due to pressuremeter length to diameter ratios but it is by no means clear that strength is over-estimated.

Outside of this contrived situation, non-linear elastic behaviour probably compensates for the influence of end effects. The ‘myth’ in the title of this paper refers as much to the model as it does to the concept.

### Click for References

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text:
Introduction

Matrix Multiplication is a staple in mathematics. Since most images are made out of multiple matrices, 3 to be exact, matrix multiplication is a great tool for manipulating these images. For example, by using Singular-value decomposition (SVD), which is a factorization of a matrix, it is possible to reduce the noise stored in images. And, SVD is possible with matrix multiplication but, that is a story for a future post. Matrix Multiplication is used in neural networks, robotics, image processing, and etc. Of course, for faster image processing, faster training of neural networks, and less delay of robotics, quickening the processes of matrix multiplication is important.
For a really long time it was thought that in terms of computational complexity the naive algorithm for the multiplication of matrices was the optimal one, with a computational complexity of $O(n^3)$. In 1969, Volker Strassen proved that wrong with a paper titled "Gaussian elimination is not optimal". Of course, there have been faster algorithms developed since then, like the fastest yet algorithm developed by Coppersmith and Winograd and generalized by Francois Le Gall. Here is a link to his paper.

Algorithm

The strassen is as follows. Where $C=AB$ and $A,B,C\in R^{2^{n-1} \times 2^{n-1}}$, we divide each matrix into 4 equal size matrices at divided starting from the corners.
$$A=\begin{bmatrix} A_{1,1} & A_{1,2} \\ A_{2,1} & A_{2,2} \end{bmatrix},B=\begin{bmatrix} B_{1,1} & B_{1,2} \\ B_{2,1} & B_{2,2} \end{bmatrix},C=\begin{bmatrix} C_{1,1} & C_{1,2} \\ C_{2,1} & C_{2,2} \end{bmatrix}$$
We then define new matrices where these matrices are added and multiplied together recursively until only scalers remain.
\begin{align}
I&:=(A_{1,1}+A_{2,2})(B_{1,1}+B_{2,2}) \\
II&:=(A_{2,1}+A_{2,2})B_{1,1} \\
III&:=A_{1,1}(B_{1,2}-B_{2,2}) \\
IV&:=A_{2,2}(B_{2,1}-B_{1,1}) \\
V&:=(A_{1,1}+A_{1,2})B_{2,2} \\
VI&:=(A_{2,1}-A_{1,1})(B_{1,1}+B_{1,2}) \\
VII&:=(A_{1,2}-A_{2,2})(B_{2,1}+B_{2,2})
\end{align}
Then we add these as follows
\begin{align}
C_{1,1}&=I+IV-V+VII \\
C_{1,2}&=III+V \\
C_{2,1}&=II+IV \\
C_{2,2}&=I-II+III+VI
\end{align}
Doing this results in 7 multiplication operations while with the naive way of matrix multiplication, $c_{ij}=\sum_{k=1}^{m}a_{ik}b_{kj}$, 8 multiplication operations are required. But, the strassen algorithm requires more additions and subtractions then the naive way, meaning, for small matrices the naive way of matrix multiplication is faster and on very large matrices the strassen algorithm is faster due to the fact that multiplication operations are more computationally complex then addition or subtraction operations. It would be optimal to use the strassen algorithm to break down a large matrix into matrices small to the point where regular matrix multiplication is faster. Refer to this link for more information on the complexity of operations.

Examples

Let's say we have two matrices $A$ and $B$ whose multiplication results in $AB=C$ and they are defined by $A,B,C\in R^{2\times 2}$. Let's give the two matrices $A$ and $B$ random real elements.
$$A=\begin{bmatrix} 0.5 & 0.75 \\ 1.5 & 1.75\end{bmatrix} ,B=\begin{bmatrix} 0.8 & 0.12 \\ 1.7 & .85 \end{bmatrix}$$
If we follow the strassen algorithm
\begin{align}
I&:=3.7125 \\
II&:=2.6 \\
III&:=-0.3650 \\
IV&:=1.575 \\
V&:=1.0625 \\
VI&:=0.92 \\
VII&:=-2.55
\end{align}
And, finally
\begin{align}
c_{11}&=I+IV-V+VII=1.675 \\
c_{12}&=III+V=0.6225 \\
c_{21}&=II+IV=4.175 \\
c_{22}&=I-II+III+VI=1.6675
\end{align}
For matrices larger then $R^{2 \times 2}$, we add rows and columns of zeros until the number of rows and columns is the same and divisible by 2. After which the matrix is divided into 4 sub-matrices for which the same operations are performed. For example, if have the matrices, $D,E,F\in R^{4\times 4}$, where $DE=F$. We will split the matrix into four quadrants with which we do the operations above recursively. Let's give these matrices random real numbers:
\begin{align}
D&=\left[\begin{array}{cc|cc}
4 & 10 & 15 & 11 \\
4 & 10 & 7 & 18 \\
\hline
3 & 14 & 12 & 11 \\
5 & 17 & 19 & 17
\end{array}\right] \\
E&=\left[\begin{array}{cc|cc}
3 & 11 & 3 & 8 \\
9 & 15 & 9 & 17 \\
\hline
11 & 19 & 12 & 16 \\
18 & 15 & 9 & 17
\end{array}\right]
\end{align}
So,
\begin{align}
D_{1,1}&=\begin{bmatrix}
4 & 10 \\
4 & 10
\end{bmatrix} \\
D_{1,2}&=\begin{bmatrix}
15 & 11 \\
7 & 8
\end{bmatrix} \\
D_{2,1}&=\begin{bmatrix}
3 & 14 \\
5 & 17
\end{bmatrix} \\
D_{2,2}&=\begin{bmatrix}
12 & 11 \\
19 & 17
\end{bmatrix} \\
E_{1,1}&=\begin{bmatrix}
3 & 11 \\
9 & 15
\end{bmatrix} \\
E_{1,2}&=\begin{bmatrix}
3 & 8 \\
9 & 17
\end{bmatrix} \\
E_{2,1}&=\begin{bmatrix}
11 & 19 \\
18 & 15
\end{bmatrix} \\
E_{2,2}&=\begin{bmatrix}
12 & 16 \\
9 & 17
\end{bmatrix}
\end{align}
And, using the algorithm while also doing the additions and subtractions for each temporary matrix
\begin{align}
I&:=\begin{bmatrix}
16 & 21 \\
23 & 27
\end{bmatrix}\begin{bmatrix}
15 & 27 \\
18 & 32
\end{bmatrix} \\
II&:=\begin{bmatrix}
15 & 25 \\
24 & 34
\end{bmatrix}\begin{bmatrix}
3 & 11 \\
9 & 15
\end{bmatrix} \\
III&:=\begin{bmatrix}
3 & 11 \\
9 & 15
\end{bmatrix}\begin{bmatrix}
-9 & -8 \\
0 & 0
\end{bmatrix} \\
IV&:=\begin{bmatrix}
12 & 11 \\
19 & 17
\end{bmatrix}\begin{bmatrix}
8 & 8 \\
9 & 0
\end{bmatrix} \\
V&:=\begin{bmatrix}
19 & 21 \\
11 & 18
\end{bmatrix}\begin{bmatrix}
12 & 16 \\
9 & 17
\end{bmatrix} \\
VI&:=\begin{bmatrix}
-1 & 4 \\
1 & 7
\end{bmatrix}\begin{bmatrix}
6 & 19 \\
18 & 32
\end{bmatrix} \\
VII&:=\begin{bmatrix}
3 & 0 \\
-12 & 1
\end{bmatrix}\begin{bmatrix}
23 & 35 \\
27 & 32
\end{bmatrix}
\end{align}
Now, we apply the strassen algorithm, again, as shown above, for each time we do matrix multiplication
\begin{align}
I&:=\begin{bmatrix}
618 & 1104 \\
831 & 1485
\end{bmatrix} \\
II&:=\begin{bmatrix}
270 & 540 \\
378 & 774
\end{bmatrix} \\
III&:=\begin{bmatrix}
-36 & -32 \\
-36 & -32
\end{bmatrix} \\
IV&:=\begin{bmatrix}
195 & 96 \\
305 & 152
\end{bmatrix} \\
V&:=\begin{bmatrix}
417 & 661 \\
384 & 652
\end{bmatrix} \\
VI&:=\begin{bmatrix}
66 & 109 \\
132 & 243
\end{bmatrix} \\
VII&:=\begin{bmatrix}
69 & 109 \\
132 & 243
\end{bmatrix}
\end{align}
Now, we just add and subtract the matrices to get the four quadrants for the $F$ matrix
\begin{align}
F_{1,1}&=I+IV-V+VII=\begin{bmatrix}
465 & 644 \\
503 & 597
\end{bmatrix} \\
F_{1,2}&=III+V=\begin{bmatrix}
381 & 629 \\
348 & 620
\end{bmatrix} \\
F_{2,1}&=II+IV=\begin{bmatrix}
465 & 636 \\
683 & 926
\end{bmatrix}\\
F_{2,2}&=I-II+III+VI=\begin{bmatrix}
378 & 641 \\
549 & 922
\end{bmatrix}
\end{align}
We put the four quadrants together
$$F=\begin{bmatrix} 465 & 644 & 381 & 629 \\ 503 & 597 & 348 & 620 \\ 465 & 636 & 378 & 641 \\ 683 & 926 & 549 & 922 \end{bmatrix}$$
For matrices that are not cube shaped or whose sides are not divisible by 2 we add rows and columns of zeros until the matrix complies with the requirement.

Experiment

I wanted to make sure that this algorithm actually worked in practice, so I fired up MATLAB. I quickly implemented the naive algorithm and called the function matrix_multiplication, then, I not so quickly setup the strassen algorithm in a function called strassen_multiplication not taking into account matrices whose sides are not square and not completely divisible by 2. Lastly, I went ahead and used the built in function provided by MATLAB. When implementing the strassen algorithm, I found that only using the strassen algorithm while recursing was wasting such a large amount of resources that it was slower then the naive algorithm. A fix was to, when the side size reach $48$, use the naive algorithm, thereby creating a faster hybrid algorithm.
We need two large matrices whose dimensions are square and completely divisible by 2 and whose elements are random. By completely divisible by 2 I mean that you can divide that number over and over by 2 until only two 2s remain, to come up with such a number we would just raise 2 to any power. I chose side dimensions that would not take too long to compute but still shows the difference in computation time. I chose $A,B,C\in R^{2^{11}\times 2^{11}}\rightarrow A,B,C\in R^{2048\times 2048}$. In MATLAB, to do that you would use the rand function (let's write a script named strassen.m)

% strassen.m
n = 2048;
a = rand(n, n);
b = rand(n, n);

We then run the functions and use the tic toc scheme to time the code

fprintf("NAIVE MATMUL\n");
tic;
c1 = matrix_multiplication(a, b);
toc;

Implementing the naive algorithm is pretty straight forward

% matrix_multiplication.m
function f = matrix_multiplication(a, b)
[n1, m1] = size(a);
[n2, m2] = size(b);
f = zeros(n1,m2);
for i = 1:n1
for j = 1:m2
for k = 1:m1
f(i,j) = f(i,j) + a(i,k) * b(k,j);
end
end
end
end

We add the code to run the strassen algorithm to the main script

fprintf("STRASSEN MATMUL\n");
tic;
c2 = strassen_multiplication(a, b);
toc;

As said before, implementing the strassen algorithm is not so straight forward. Let's first create the function, get the size of the matrices, and fill the output matrix with zeros

% strassen_multiplication.m
function f = strassen_multiplication(a, b)
[n,~] = size(a);
f = zeros(n, n);
% TODO
end

We then check if the size of the input matrix is less then $48$ and switch over to the naive implementation.

if n <= 48
f = matrix_multiplication(a, b);
return
end

Since we cannot use float point numbers when using indexing

nhalf = int32(n/2);

We now get the four quadrants of the two matrices for ease of use

a11 = a(1:nhalf,1:nhhalf);
a12 = a(1:nhalf,nhalf+1:n);
a21 = a(nhalf+1:n,1:nhalf);
a22 = a(nhalf+1:n,nhalf+1:n);

b11 = b(1:nhalf,1:nhalf);
b12 = b(1:nhalf,nhalf+1:n);
b21 = b(nhalf+1:n,1:nhalf);
b22 = b(nhalf+1:n,nhalf+1:n);

Calculate the temporary matrices indicated in the strassen algorithm

m1 = strassen_multiplication(a11 + a22, b11 + b22);
m2 = strassen_multiplication(a21 + a22, b11);
m3 = strassen_multiplication(a11, b12 - b22);
m4 = strassen_multiplication(a22, b21 - b11);
m5 = strassen_multiplication(a11 + a12, b22);
m6 = strassen_multiplication(a21 - a11, b11 + b12);
m7 = strassen_multiplication(a12 - a22, b21 + b22);

Finally, use the temporary matrices to get the output matrix and end the function

  f(1:nhalf,1:nhalf) = m1 + m4 - m5 + m7;
f(1:nhalf,nhalf+1:n) = m3 + m5;
f(nhalf+1:n,1:nhalf) = m2 + m4;
f(nhalf+1:n,nhalf+1:n) = m1 - m2 + m3 + m6;
end

In the main script we now will use the function provided by MATLAB for matrix multiplication

fprintf("MATLAB'S MATMUL\n");
tic;
c3 = mtimes(a,b);
toc;

To validate the function we created, and since we know that MATLAB function is pretty accurate, we can use the c3 matrix to validate. We know that the three matrices are suppose to be equal to each other so we can subtract them and we know that it will be at least close to a matrix of zeros. We then average that across the elements of the matrix to get the average deviation.

zero1 = c1 - c3;
zero2 = c2 - c3;
avg1 = 0;
avg2 = 0;
for i = 1:n
for j = 1:n
avg1 = avg1 + zero1(i, j);
avg2 = avg2 + zero2(i, j);
end
end
avg1 = avg1 / (n^2);
avg2 = avg2 / (n^2);
fprintf("Average deviation for naive %d \n", avg1);
fprintf("Average deviation for strassen %d \n", avg2);

In the source code, I also calculate the Percent Average Deviation for good measure.
The output from my machine was

>> run strassen.m
NAIVE MATMUL
Elapsed time is 276.967041 seconds.
STRASSEN MATMUL
Elapsed time is 32.829293 seconds.
MATLAB's MATMUL
Elapsed time is 0.115984 seconds.
Average deviation for naive 4.605590e-13
Average deviation for strassen 1.172674e-12
Percent Average Deviation 9.005056e-14 %
Percent Average Deviation 2.292865e-13 %

The fact that MATLAB's matrix multiplication was under a second is a testament to the developers of LAPACK and BLAS and the low overhead that the Fortran programming language provides as MATLAB uses the LAPACK and BLAS Fortran libraries to do these calculation.
As shown above, the Strassen hybrid algorithm is quite a bit faster then the Naive algorithm.


text:
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Due to this attribute, the company stands out … Place this side on top for a mattress that offers firmer support with 100% natural dunlop latex, while still molding to your body for unbeatable comfort. It take to get the latex found in most mattresses today is a substance derived from petrochemicals... Ild of the highest quality best sleep of your life, commitment-free machine washable and comes a... By experienced craftsman in our California production facility and custom-made to order in our California wool a... In Linux–and Unix—has a history, no pun intended perfect blend of natural and latex! Southern California softness, organic cotton cover for easy maintenance use the finest organic cotton our... Less Coupon – Save$ 800 on Adjustable Queen Base the predominant types of templates... Reported to sleep “ hotter ” the inability of … latex is produced using either or... … if you don ’ t you use the Original box scent that is sustainably from! Equations and tables about supporting a company that sells all-latex mattress toppers and pillows 2013 as an internet-based.. Natural wool how well workers are treated and how the process impacts the environment the glued seams topper locally exceptional. And quality firmer, which acts as the comfort layer washable and comes with a FedEx tracking Number soon... Toppers include exceptional durability, good motion isolation, and pillows, plant-derived latex with natural in... Other side is medium firm has established an 80,000 sq as possible top quality mattresses you ’ re to! Of firmnesses options we make products of impeccable quality while helping the economy here at home, keep! Same amount of Talalay latex — sourced from sustainable rubber tree plants products, the uses! Three primary latex types a variety of firmnesses options pillow is soft, supportive, and sleep cool night... Latex mattress 25-year service warranty temperature throughout the night and provides a natural that. The … latex for Less luxi … but the company stands out from the of. Achieve the perfect mix of safety and comfort warehouse for years like some our. Foam pillow is machine washable and comes with a bunch of agrochemicals keep you cool you... Please visit our latex for Less, Inc. is a mattress company sells... Use any cheap blended or synthetic petroleum-based latex for less company savings is seen on the way other companies mattresses! Market, they end up paying extra free returns in the box if I decide return... Latex templates, and affordable sleep experience launched in 2013 as an store! Knowledge, the all-latex latex for Less Adjustable Base – Save \$ 620 full! Certificate for our cover, because who wants to latex for less company with a latex for Less Adjustable Base to. Production facility and custom-made to order tree growers — acts as support for healthy! … typically, synthetic latex warehouse for years like some of our products are made from plant-derived with. Synthetic types because the … plushbeds Vs latex for Less offers a worry-free 120! Captions to figures, equations and tables we sought out the middleman, we ensure that you the... All have the same amount of Talalay latex — sourced from sustainable rubber tree Vs other that... It 's the perfect blend of soft and firm expensive than natural latex mattress warranty page or... Domestic Profit Corporation filed on January 30, 2007 paying extra comfortable mattress I 've ever laid!...

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text:
1a. Find the probability that when a couple has six children, atleast one of them is a boy. (Assume that

Question:

1a. Find the probability that when a couple has six children, at least one of them is a boy. (Assume that boys and girls are equally likely.) 1b. Assuming boys and girls are equally likely, find the probability of a couple having a baby boy when their sixth child is born, given that the first five children were all boys.

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# Hartree–Fock RI¶

The Hartree-Fock exchange matrix is calculated through a procedure known as Resolution of the Identity (RI). The implementation of the RI scheme in BAND is loosely based on work by Ren et al. [1]. For more information on hybrid functionals in BAND, see the XC section.

Technical aspects of the RI scheme can be tweaked in the RIHartreeFock block:

RIHartreeFock
Quality [VeryBasic | Basic | Normal | Good | VeryGood | Excellent]
FitSetQuality [VeryBasic | Basic | Normal | Good | VeryGood | Excellent]
DependencyThreshold float
AtomDepQuality # Non-standard block. See details.
...
End
End

RIHartreeFock
Type: Block The Hartree-Fock exchange matrix is calculated through a procedure known as Resolution of the Identity (RI). Here you can tweak various parameters of the procedure.
Quality
Type: Multiple Choice Normal [VeryBasic, Basic, Normal, Good, VeryGood, Excellent] RI Hartree-Fock Accuracy of numerical integration and thresholds of the RI procedure.
FitSetQuality
Type: Multiple Choice Normal [VeryBasic, Basic, Normal, Good, VeryGood, Excellent] The auxiliary fit set employed in the RI scheme. This is an important aspect of the procedure, significantly affecting both accuracy and computation time. For SZ and DZ basis set a ‘basic’ FitSetQuality will suffice. For ‘DZP’ and ‘TZP’ a normal quality is recommended. For larger basis set, use either ‘normal’ or better FitSetQuality.
DependencyThreshold
Type: Float 0.001 To improve numerical stability, almost linearly-dependent combination of basis functions are removed from the Hartree-Fock exchange matrix. If the SCF does not converge or you obtain unphysically large bond energy in an Hybrid calculation, you might try setting the DependencyThreshold to a larger value (e.g. 3.0E-3).
AtomDepQuality
Type: Non-standard block One can define a different fit-set quality for each atom. The syntax for this free block is ‘iAtom quality’, where iAtom is the index of the atom in input order.

For efficiency and numerical stability reasons, it is advisable to include:

SoftConfinement
Quality Basic
End

• In AMS2019.3 the fit set for FitSetQuality Good has been improved.


text:
# A 40 cm diameter circular timber column is 4 m long. The slenderness ratio of the column is

This question was previously asked in
ISRO Scientist Civil 2020 Paper
View all ISRO Scientist CE Papers >
1. 4
2. 10
3. 20
4. 40

Option 4 : 40
Free
ST 8: Thermodynamics
5645
20 Questions 20 Marks 12 Mins

## Detailed Solution

Concept:

Slenderness ratio (λ):

Slenderness ratio of a compression member is defined as the ratio of its effective length to the radius of gyration

$${\rm{\lambda }} = \frac{{{{\rm{L}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}}$$

Where,

Leff = Effective length of the compression member

rmin = Least radius of gyration

$${{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{{{\rm{I}}_{{\rm{min}}}}}}{{\rm{A}}}}$$

Imin = Minimum moment of inertia of area

A = Area of the shape

Calculation:

Given,

For circular column

D = 40 cm, L = 4 m = 400 cm

$${\rm{A}} = \frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4}$$$${{\rm{I}}_{{\rm{min}}}} = \frac{{{\rm{\pi }}{{\rm{D}}^4}}}{{64}}$$

$${{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{\frac{{{\rm{\pi }}{{\rm{D}}^4}}}{{64}}}}{{\frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4}}}} = \sqrt {\frac{{{D^2}}}{{16}}} = \frac{D}{4}$$

$${\rm{\lambda }} = \frac{{{{\rm{L}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}} = \frac{{400}}{{\frac{D}{4}}} = \;40$$

Hence the slenderness ratio of the circular column is 40


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# What Number Must Be Added to Each of the Numbers 6, 15, 20 and 43 to Make Them Proportional? - ICSE Class 10 - Mathematics

#### Questions

What number must be added to each of the numbers 6, 15, 20 and 43 to make them
proportional?

What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?

#### Solution

Let the number added be x.

∴ (6 + x) : (15 + x) :: (20 + x) (43 + x)

=> (6+x)/(15 + x) = (20 + x)/(43 + x)

⇒ (6 + x)(43 + x) (20 + x)(15 + x)

=> 258 + 6x + 43x + x^2 = 300 + 20x + 15x + x^2

=> 49x - 35x = 300 - 258

=> 14x = 42

=> x = 3

Thus, the required number which should be added is 3.

Is there an error in this question or solution?

#### APPEARS IN

Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 7: Ratio and Proportion (Including Properties and Uses)
Ex.7B | Q: 6

#### Video TutorialsVIEW ALL [1]

Solution What Number Must Be Added to Each of the Numbers 6, 15, 20 and 43 to Make Them Proportional? Concept: Proportions.
S


text:
std::map for storing static const Objects

I am making a game similar to Minecraft, and I am trying to fine a way to keep a map of Block objects sorted by their id. This is almost identical to the way that Minecraft does it, in that they declare a bunch of static final Block objects and initialize them, and then the constructor of each block puts a reference of that block into whatever the Java equivalent of a std::map is, so there is a central place to get ids and the Blocks with those ids.

The problem is, that I am making my game in C++, and trying to do the exact same thing. In Block.h, I am declaring the Blocks like so:

//Block.h
public:
static const Block Vacuum;
static const Block Test;

And in Block.cpp I am initializing them like so:

//Block.cpp
const Block Block::Vacuum = Block("Vacuum", 0, 0);
const Block Block::Test = Block("Test", 1, 0);

The block constructor looks like this:

Block::Block(std::string name, uint16 id, uint8 tex)
{
//Check for repeat ids
if (IdInUse(id))
{
fprintf(stderr, "Block id %u is already in use!", (uint32)id);
throw std::runtime_error("You cannot reuse block ids!");
}
_id = id;
//Check for repeat names
if (NameInUse(name))
{
fprintf(stderr, "Block name %s is already in use!", name);
throw std::runtime_error("You cannot reuse block names!");
}
_name = name;

_tex = tex;
//fprintf(stdout, "Using texture %u\n", _tex);

_transparent = false;
_solidity = 1.0f;

idMap[id] = this;
nameMap[name] = this;
}

And finally, the maps that I'm using to store references of Blocks in relation to their names and ids are declared as such:

std::map<uint16, Block*> Block::idMap = std::map<uint16, Block*>(); //The map of block ids
std::map<std::string, Block*> Block::nameMap = std::map<std::string, Block*>(); //The map of block names

The problem comes when I try to get the Blocks in the maps using a method called const Block* GetBlock(uint16 id), where the last line is return idMap.at(id);. This line returns a Block with completely random values like _visibility = 0xcccc and such like that, found out through debugging.

So my question is, is there something wrong with the blocks being declared as const obejcts, and then stored at pointers and accessed later on? The reason I cant store them as Block& is because that makes a copy of the Block when it is entered, so the block wouldn't have any of the attributes that could be set afterwards in the constructor of any child class, so I think I need to store them as a pointer.

Any help is greatly appreciated, as I don't fully understand pointers yet. Just ask if you need to see any other parts of the code.

• anything else other _visibility with bad values is returned or all the other values are ok ?? – concept3d Oct 31 '13 at 22:41
• There's nothing wrong with using const pointers. You can store const pointers in a container, e.g. std::map<uint16, const Block*>. There is something wrong with using std::map (a tree) to access a contiguous list of integer ids instead of using std::vector, though. – Sean Middleditch Oct 31 '13 at 22:52
• Unfortunately I want the IDs to be able to be non-contiguous, so a vector isn't an option without having to iterate over the entire thing whenever I want to get a block (which happens quite often). – sm81095 Nov 1 '13 at 1:30
• You can find ways to make the lookup contiguous. For instance, two layers of IDs - the internal IDs that are contiguous and external IDs that are then mapped to the internal ones. And you can consider using std::unordered_map instead which -- while being a pretty bad hash map -- is a lot better than a plain std::map unless you actually need ordered keys. A good design here for mods and such is to use a GUID to uniquely identify a block and then contiguous IDs for the actual blocks, with the mapping table saved out to serialized maps. – Sean Middleditch Nov 1 '13 at 2:25
• @SeanM. did you solve your bad values problem ? – concept3d Nov 1 '13 at 12:38

The reason I cant store them as Block& is because that makes a copy of the Block when it is entered,

Raw pointers should be avoided when possible.

I'm not sure exactly what you mean by 'makes a copy'. Containers themselves can't store raw references. Possibly you are making a function accepting a ref but then putting them into a container with a non-ref.

In C++11 (which I recommend using unless you have compiler restraints) you can also store references in a container with a std::ref or std::cref. You will need to make sure your compiler is compiling in C++11 mode.

#include <iostream>
#include <map>
#include <functional>
using namespace std;

class Block {
public:
Block(const std::string& name, const int block_id) {
this->name = name;
this->block_id = block_id;
}
private:
std::string name;
int block_id;
};

const Block block_1("Vacuum", 0);

int main(int argc, char* argv[]) {
std::map<int, std::reference_wrapper<const Block>> array;
array.insert(std::pair<int, reference_wrapper<const Block>>(0, block_1));
}

Otherwise you could make your own simple ref container:

class BlockRef {
public:
BlockRef(const Block& blockref): blockref_(blockref) {}
operator const Block& () const { return this->blockref_; }
private:
const Block& blockref_;
};

You could also make a more advanced one with templates that's generic for any type.

Boost has them also if you are ok with using a library.

• Thanks a ton! By passing in a BlockRef class and storing them as BlockRefs, all of the variables are properly initialized and the program runs fine. I have only one question: since I have the Block constructor call BlockRegistry::RegisterBlock(BlockRef) on itself, will this still work for child classes of Block? Or will any variables changed in the child constructor not also change in the stored references? – sm81095 Nov 2 '13 at 5:47
• Never mind, answered my own question. This was the nice simple solution that I needed. Many thanks! – sm81095 Nov 2 '13 at 6:27

First off all To solve your problem, -if that was the only one-, unlike java, C++(MSVC) assigns 0xcccc for uninitialized variables in debug mode, you are not initializing it in the constructor, make sure to initialize all the variables in your class using the initialize list instead of assignment in the constructor, this has two advantage:

1. It has to do with calling the destructor in a throwable constructor.
2. It will only call the constructor and not both the constructor and the assignment operator.

Block::Block(std::string name, uint16 id, uint8 tex):_visibility (0) {

}

The reason I cant store them as Block& is because that makes a copy of the Block when it is entered

First of all you need to think of references as Object* const x; this means the pointer is const and not the object itself. But the real reason you can't store references in stl containers is that:

• T must be default constructible (a reference can only be initilized)
• T must be copyable (a reference cannot be changed after initlization)

Is there something wrong with the blocks being declared as const obejcts, and then stored at pointers and accessed later on?

Nothing wrong, as far as you are sure of the allocated object lifetime; don't delete it from other pointers referencing it.

You may want to create a class that will manage blocks (e.g. factory,manager etc) this will make managing resources easier.

 This is only a guess: It seems that both your Registry class and your Block object are define static or global. The problem is that C++ does not define the order in which static objects should be allocated. I suggest you create an Init() function that explicitly initialize your Block object by allocating them on the heap.

• So you are saying that the best way to store the Blocks in the map is using a Block*, without it being a const object or a const pointer? And then also to use things like :_transparent(true) for all of the variables? – sm81095 Nov 1 '13 at 21:24
• nope actually what I meant there is a different between const int* and int const ! a reference is int *const; you need to look that up. and there is nothing wrong with storing const Block this means that the Block is const but the pointer is not. – concept3d Nov 1 '13 at 21:30
• regarding :_transparent(true) you should initialize all the variables .. the reason you have 0xCCC is that it was not initialized at all – concept3d Nov 1 '13 at 21:31
• C++ does NOT initilize anything to 0, java does so that was your problem. – concept3d Nov 1 '13 at 21:39
• I changed my code to include a BlockResistry class, and changed the Block constructor to this. However, I am still getting the same error as before, where the Block variables aren't initialized. – sm81095 Nov 1 '13 at 22:27