Business Decision AnalysisDr Gary Simpson, Dr Panagiotis Ganotakis

g.p.m.simpson@aston.ac.uk, ganotap1@aston.ac.ukNB270 / NB268

Statistical Techniques & Operational Research

Week1 Decision Analysis2 The Binomial and Poisson Distributions 3 The normal distribution, estimation4 Confidence limits including the use of the t-distribution5 Hypothesis Tests for means and proportions6 Chi-squared hypothesis tests7 Introduction to Differentiation8 Optimisation using Differentiation 9 Transportation Problem10 Linear Programming (formulation & graphical solution).11 LP (interpreting LINDO output)12 Revision

Method of Assessment100% by examination. The examination will be an open-book examination and lasts for 2 hours.It is designed to test overall understanding of the module and will include formulation of models, interpretation of computer output, use of statistical tables and calculations by-hand where appropriate.

Books to buyStandard tables required by all students:Lindley DV & Scott WF (1984), New Cambridge Elementary Statistical Tables, 2nd Edition, Cambridge: Cambridge University Press.

Main Recommended texts:

All students are expected to read the course notes and work through the examples on the tutorial sheets. These two books are recommended for further reading (help for those students that feel they need to read from an additional source):Either these two books

A1. Bradley, T (2008), Essential Mathematics for Economics and Business, John Wiley & Sons.

&A2. Bradley, T (2007), Essential Statistics for Economics and Business, John Wiley &

Sons.

or this single book:

B. Curwin J & Slater R (2008), Quantitative Methods for Business Decisions, 6th Edition, Thomson learning.

Both where the essential reading for BN1105 Quantitative Techniques

Other texts:

Anderson, Sweeney, Williams, Freeman & Shoesmith (2007), Statistics for Business and Economics, Thomson learning.

Dewhurst F (2006), Quantitative Methods for Business Management, 2nd Edition, London: McGraw-Hill Education Europe.

Oakshott L (2001), Essential Quantitative Methods for Business, 2nd Edition, Basingstoke: Palgrave.

Wisnieswki M (2005), Quantitative Methods For Decision Makers, 4th Edition, London: Prentice Hall.

Useful Online Sources:

Lindo, provide a demo version of their linear programming package, which can be downloaded from their website www.lindo.com Hunt &Tyrrell (2000), http://www.coventry.ac.uk/discussThis web-site provides helpful material for some of the statistics parts of this module

Gustman G (2003) StatPrimer (Version 6.4)http://www.sjsu.edu/faculty/gerstman/StatPrimer/

SPSS Video tutorialshttp://www.stat.tamu.edu/spss.php

Statistical Techniques

The purpose of decision analysis is to enable a decision maker to choose between several decision options in a way that is best, or optimal, in some sense. The decision will often be affected by future events that are not under the control of the decision maker; such events are called states of nature.We shall use d1, d2, … to denote decision options and s1, s2, … to denote states of nature.The result of choosing a particular decision option di in conjunction with a particular state of nature sj is called the payoff V(di,sj) for that particular pair di and sj. It is simplest to think of this payoff as a direct monetary payment, but in some contexts it is preferable to use the concept of utility to take account of the relative importance of different outcomes. An extra £1000 would not make much difference to a millionaire but would to an impoverished student. So the utility of the £1000 would greater for the student than for the millionaire.

A firm’s decision problem

A firm has to make a decision on which of three types of equipment to buy (A, B or C) and has worked out the payoff (in £1000) it will receive depending on whether the firm can secure a contract of type I or a contract of type II with its most important customer. (One of the two contracts will be agreed.)

States of natureDecision Option Contract type I: s1 Contract type II: s2

Choose A: d1 250 70Choose B: d2 160 150Choose C: d3 90 170

A decision tree, similar to a probability tree, can be used to show the same information as the table. Within a decision tree we need to distinguish between decision nodes, which we will represent by a square, and chance nodes, represented by a circle.

Maximisation without probabilities

Although it is desirable to have some measure of the probabilities of the various possible states of nature, it is not always possible to get reliable estimates of these probabilities. There are several different approaches to decision making without probabilities and it is important to understand why they often lead to different “optimal” decisions.

1. Optimistic approach (Maximax payoff)

This method simply chooses the decision giving the largest possible payoff in the whole table. It is sometimes called the maximax payoff because it can be found by working down the payoff table, row by row for each decision option di writing down Vmax(di), the maximum payoff for any of the states of

nature. The decision option chosen is the one with the largest Vmax(di), that is the maximum of the maximum payoffs.

States of nature

Decision Option Contract I: s1 Contract II: s2 Vmax(di)Choose A: d1 250 70 250Choose B: d2 160 150 160Choose C: d3 90 170 170

2. Pessimistic approach (Maximin payoff)

Here we consider the worst outcome for each decision option di which we call Vmin(di). Then we have to identify the maximum of these Vmin(di). This maximises the minimum payoff and so is called the maximin payoff.

States of natureDecision Option Contract I: s1 Contract II: s2 Vmin(di)

Choose A: d1 250 70 70Choose B: d2 160 150 150Choose C: d3 90 170 90

3. The Hurwicz criterion

This represents a balance between the maximax and maximin approaches. Define the index of optimism to be a number between 0 and 1, then choose the decision option di which gives the maximum value of H(di) = Vmax(di) +(1)Vmin(di).When =1, this is the same as the maximax approach, and when = 0, this is the same as the maximin approach. For our example, we will consider = 0.5.

States of natureDecision Option s1 s2 Vmax(di) Vmin(di) H(di)

Choose A: d1 250 70 250 70 160Choose B: d2 160 150 160 150 155Choose C: d3 90 170 170 90 130

4. Minimax regret

This approach considers how much we might regret choosing a decision option once the state of nature is revealed. The regret or opportunity loss, R(di,sj), is the difference between the best of all payoffs for a particular state of nature sj and the payoff when option di is chosen and state sj occurs:

R(di,sj) = Vmax(sj) V(di,sj)

So in our example, Vmax(s1) = 250 and Vmax(s2) = 170 and the following table shows the regret values

States of natureDecision Option s1 s2 Rmax (di)

Choose A: d1 250 - 250 = 0 170 – 70 = 100 100Choose B: d2 250 - 160 = 90 170 – 150 = 20 90Choose C: d3 250 - 90 = 160 170 – 170 = 0 160

The option di which minimises the (maximum) regret Rmax(di) is chosen.

Decision analysis using probabilities

When we have reasonable estimates of the probabilities, P(sj), of the various states of nature, the optimal decision option can be chosen on the basis of the expected values of the payoffs.

In the example, if the management thinks that there is a 60% chance of getting contract I, and believes that there is a 40% chance of getting contract II then

States of natureDecision Option Contract I: s1 Contract II: s2 EV(di)

Choose A: d1 250 70 178Choose B: d2 160 150 156Choose C: d3 90 170 122

P(sj) 0.6 0.4

So the best decision is d1 (option A) with an expected return of 178.

Using a process called backward induction, we can also calculate this result using a decision tree.

Backward induction runs backwards through the tree in the following way:For each chance node, the expected value of the node is calculated by multiplying the probabilities at the edges leaving this node by the payoffs at the end of the respective edge and summing up over all edges belonging to this chance node.

E.g. for the chance node following decision d1:P(s1) V(d1,s1)+ P(s2) V(d1,s2)= 0.6250 + 0.470 = 178

For a decision node, the value of the node is calculated as the maximum of the values of the nodes following it.E.g. for the first node, we need to find: max{178, 156, 122} = 178.Hence, the optimal choice is d1 (option A) again, as it leads to the highest expected profit.

Expected value of perfect information

If the decision maker had a better idea of which state of nature will actually occur, this would lead to an improvement in the expected payoff. However, if the cost of obtaining the required information exceeded the improvement in payoff, no benefit would result.We therefore need a method for determining an upper limit for the budget for obtaining more information on the future state of nature.

The payoff table shows that if the company gets contract I, the optimal decision is d1 whereas d3 is the optimal decision if the company gets contract II.If the decision maker was able to gain perfect information about which type of contract the company will get, the expected value of following this combined decision strategy would be

E(Vmax | PI) = V(s1)P(s1) + V(s2)P(s2) = 2500.6 + 1700.4 = 218

If no information was available about the actual state of nature then we have already found the best policy is d1 with an expected value of 178 (which is £178,000).So, the Expected Value of Perfect Information is

EVPI = E(Vmax | PI) - EVmax = 218 - 178 = 40

This £40,000 represents the additional expected payoff that could be obtained if perfect information about the future state of nature was available. It would therefore be uneconomic to pay more than £40k for extra information – no matter how good that information might be. However, each amount up to £40k should be paid for perfect information, as the expected payoff can be increased.

Random Variables In statistics, the events we are concerned about often represent numerical observations (e.g. the daily demand for a product). In such cases we imagine the sample space to be made up of points that correspond to that quantity, so the random variable X may take a particular values x. Note that a capital letter is generally used for the random variable and the corresponding small letter for value.The probability of the event that “X takes the value x” is written P(X=x)

Discrete Probability Distributions If a discrete random variable X takes some value x with probability P(X=x) = f(x)then f(x) is called the probability function of X. The sum of individual f(x) terms for x up to and including some value r is called the cumulative distribution function (c.d.f.) and is denoted by F(r).Note the probability that X is less than or equal to r is P(Xr) = F(r) = f(0) +f(1) +…+f(r)

Mean of a Discrete Random Variable The theoretical mean of a discrete random variable is denoted by ,and is also called the expected value E(X) is calculated in a similar way to the sample mean for repeated data but the relative

frequencies are replaced by the probability f(x). Mean = E Xall x

xf x

Variance of a Discrete Random Variable The theoretical variance of a discrete random variable is denoted by ,or Var(X) is given by,

2

all x= Var X x f x2 2

Binomial Distribution If the random variable X represents the number of successes in n independent trials of a random experiment in which there are only two possible outcomes from each trial; a ‘success’ with probability p and a ‘failure’ with probability q=1-p

Then the probability function is given by:

Expected number of success, mean of the binomial distribution = np. The variance = npq Cambridge Table 1 shows P(Xr)=F(r) for different n and p values.

Example There are 20 computers in a room; the probability that a computer is working is 0.95. Assuming that breakdown of a computer is independent to the breakdown of the other computers, a) what is the probability that all 20 computers are working ? b) What is the probability that exactly 17 computers are working ? c) What is the probability that 17 or more computers are working ?a) n = 20 = number of trials ‘success’ = working so p = 0.95 and q= 1-p = 1-0.95 =0.05

P(all working) = P X = 20 20 20

20 20 200 95 0 05C . . 0 3585. (to 4 d.p. )

b) n = 20 ‘success’ = working so p = 0.95 and q= 1-p = 1-0.95 =0.05

P(17 working) = P X = 17 20 17

17 20 170 95 0 05C . . 0 0596. (to 4 d.p. )

c) n = 20 ‘success’ = not working so p = 0.05 and q= 1-p = 0.95

P( 3 or less not working) = P(X3) = F(3) = 0.9841 (From Table 1 page 22)Note the tables can also be use for parts a) and b) for exampleP(17 working) = P(3 not working) = P(X3) - P(X2) = F(3)-F(2) = 0.9841 -0.9245 = 0.0596

Poisson Distribution If the random variable X represents the number of events observed in a fixed ‘observation frame’ (usually a specified period of time) and the expected number of events is . If the events occur at random, in such a way that they are independent of each other, the mean rate of occurrence is constant and are relatively rare they are said to follow a Poisson process and the probability function is given

by: P X = x f xx

x

!exp , the mean is = andthe variance is =

Cambridge Table 2 shows P(Xr)=F(r) for specified =

Example A telephone help-line receives an average of 2 calls per minute. Assuming that the calls follow a Poisson process calculate the probability that the help-line receives a) no calls in a 3 minute period. b) exactly 8 calls in a 3 minute period. c) 6 calls or less in a 4 minute period. d) at least 6 and no more than 10 calls in a 4 minute period.Mean no of calls per minute = 2 so the mean no. of calls per 3 minutes = 23 = 6

a) P X = 0

x

x!exp

6

06

0

!exp

1

10 002479. 0 0025. (to 4 dp)

b) P X = 8

x

x!exp

6

86

8

!exp

1679616

403200 002479. 0 1033. (to 4 dp)

c) Mean no. of calls per 4 minutes = 24 = 8, P X 6 F( ) .6 0 3134 (From table 2 page 28)d)

P X 106 10 6 10 5 0 8159 0 1912 0 6247 P X P X P X P X . . .

Poisson Approximation to the binomial If for a binomial distribution with n>20 and p<0.1 then set = np and use the Poisson distribution as an approximation.

ExampleIf 1% of the population have a peanut allergy find the probability that in a randomly selected group of 200 people more than 4 have a peanut allergy.n = 200, ‘success’ = peanut allergy so p = 0.01 and q = 0.99 , = np = 2000.01=2Using Poisson Tables on page 25 P X 4 1 4 1 0 9473 0 0527P X . .

Continuous probability distributions.Cumulative distribution function, F(x) = P(X x) , as for discrete X, but is a smooth function of x.Probability density function, f(x) = rate of change of F(x) = dF(x)/dx ; the pdf is not itself a probability. Areas under the pdf curve represent probabilities:F(x0) = area under f(x) to the left of x0.

Mean, = total are under the curve of xf(x) = all x xf(x) dxVariance, = total are under the curve of (x)2f(x) = all x (x)2f(x) dx = all x x2f(x) dx –2

Standard deviation, .

The normal distribution.The Normal distribution is also called the Gaussian distribution and is the most important of all continuous probability distributions. Many naturally occurring quantities have this kind of distribution, and in the correct conditions it can be used to provide a good approximation of many other distributions -including discrete distributions.Note: Just because a random variable is not normally distributed does not make it abnormal !

x

f(x)

It is common practice to refer to the normal distribution which has mean and standard deviation as N(,2). Note that it is the variance not the standard deviation that is given. So N(75,16) stands for a normal distribution with mean 75 and variance 16, hence a standard deviation of 4.To avoid having to integrate the pdf of the normal distribution, tables are available of the standard normal distribution, N(0,1) and problems for any other mean and/or standard deviation are restated as equivalent problems with mean 0 and standard deviation 1.

Standardised Normal Deviate A random variable which has a standard normal distribution is usually denoted by Z and its pdf is denoted by (z) and its cdf by (z) = P(Z<z)If we have a problem involving some random variable X which has the distribution N(,2). Then to find the probabilities we first find the standardised deviate subtracting the mean then dividing by the standard deviation.

So the standardised deviate corresponding to a typical value x is zx

Using tables to find normal probabilities. To use tables of (z) to find the probabilities for X which is distributed as N(,2). Note the following relationships:

1. P X xx

z

2. P X xx

z

1 1

3. z z1

The probability density function pdf for the normal curve depends on two parameters, and which represent its mean and standard deviation.The pdf is a symmetrical bell shaped curve centred on x = About 66% probability area lies between and Nearly all the probability area lies between and

ExampleThe average income in a country is known to be £12,800 with a standard deviation £2400, assuming that incomes are normally distributed, working in units of £1000, calculate the probability that:a) A randomly chosen individual from that country has an income of greater than £20000 ?

1 3P Z

1 3

1 0 99865. 0 00135.

b) A randomly chosen individual from that country has an income of less than £10000?

P Z 1 1667.

1 1667.

1 1 1667 .

1 0 8783. 0 1217.

c) A randomly chosen individual from that country has an income of between £10000 and £14000?

P Z P Z0 5 1 1667. .

0 5 1 1667. .

0 6915 1 1 1667. .

0 6915 1 0 8783. . 0 5698.

Normal Approximation to the BinomialIf X has a binomial distribution with n>10 and p near 0.5 then you can use the normal approximation. Set = np and 2 = npq. As n gets larger the approximation is valid for values of p further from 0.5. Note: Because the normal distribution describes continuous random variables and the binomial distribution describes discrete random variables a continuity correction is applied.

P X

x P Zx 0 5.

P Z

x np

npq

0 5.

Normal Approximation to the PoissonIf X has a Poisson distribution with >20 then you can use the normal approximation.Set and 2 = . Note: Because the normal distribution describes continuous random variables and the Poisson distribution describes discrete random variables a continuity correction is applied.

P X

x P Zx 0 5.

P Z

x 0 5.

Estimation and confidence limits.Appropriate sample statistics may be used to provide Point Estimates of unknown values of parameters in theoretical models. Thus the sample mean and sample standard deviation provide point estimates of and in a normal distribution.

However, for many purposes we also need some measure of the accuracy of this kind of point estimate. The usual way of providing this is to calculate confidence limits for the unknown parameter value. Broadly speaking these are the values between which the unknown parameter is expected to lie with a specified probability. The probability is usually multiplied by 100 and expressed as a percentage, which is called the confidence level.

The difference between the confidence limits is called the confidence interval. Higher confidence levels result in the confidence interval being wider. A sensible balance between a reasonably high confidence level and a narrow confidence interval is often obtained by using a 95% confidence level.

With the 100(1-)% confidence level, we are allowing a probability of /2 that the parameter lies below the lower confidence limit and a probability of /2 that it lies above the upper confidence limit.

For the 95% confidence level, /2 =0.025 (or 2.5%). The confidence limits are found by considering the distribution of some statistic that depends upon the sample data and upon the unknown parameter.

Confidence Limits for Normal The distribution required here is that of the statistic t

x

s n

, found in Table 10.

This is similar to that of the standard normal distribution but varies according to a parameter called the degrees of freedom, “nu”). When calculating confidence limits for , the degrees of freedom are one less than the sample size, so = n 1. The value of t with this which has probability /2 lying beyond it in the right tail of the distribution is t/2.

The 100(1)% confidence limits for are then given by:

ExampleA car manufacturer wants to estimate the average miles per gallon consumed by it new model in typical urban traffic conditions. In random sample of 25 urban traffic trials the new model yields a sample mean of 32.0 miles per gallon and a sample standard deviation of s = 2.6 miles per gallon. Assuming consumption is normally distributed, calculate the 95% confidence limits for the mean.

n = 25, degrees of freedom = 25 1 =24.Since 100(1-)=95, this tells us that 100 = 2.5, we look up the 2.5% point for =24 in Table 10.

xt s

n /2

32 02 064 2 6

25.

. .

32 05 3664

5.

. 32 0 1 073. .

So the 95% confidence limits for the mean are 30.9 to 33.1 miles per gallon (to 1 d.p.)

Confidence Limits for Binomial p.

If a series of n binomial trials yields x successes then the observed proportion of successes is px

n

and the observed proportion of failures is q p 1 .

Approximate 100(1-)% confidence limits for p are:

where z/2 is the value of z which cuts off probability /2 in the right tail of the standard normal distribution (see Table 5). For 95% confidence, the required z value is the 2.5% point in Table 5, which is 1.9600.

ExampleIn a survey of 100 randomly selected students 68 stated that the mess in the kitchen belonged their flatmates. Find the 95% confidence interval for the proportion of students who blamed the mess on their flatmates.

.p 68

1000 680 . .q p 1 1 0 68 0 320

/p zpq

n 2

0 680 1 96

0 680 0 320

100. .

. . 0 680 1 96 0 04665. . .

0 680 0 091. .The 95% confidence limits for the percentage of students who stated that the mess was caused by their flatmates are 59% and 77%.

Choice of Sample Size for Specified AccuracyIn advance of taking a sample, it is not always clear how close the confidence limits for an unknown parameter will be. However the eventual aim may be to be able to express the estimate for the parameter within some acceptable margin of error.

Consider the case of finding 95% confidence limits for binomial p,

which would be given by:

Suppose that we wanted our final estimate to be in the form p e then

so, for a 95% confidence limit npq

e

1 96.

giving ne

pq

1 962

.

ExampleSuppose we wished to know to within ±1% the percentage of shoppers who regularly use a Tesbury Supermarket loyalty card. From a small sample of 200 shoppers, it has been estimated that about 35% do use these cards. Find how many more shoppers should be sampled.

ne

pq

1 962

.

1 96

0 010 35 0 65

2.

.. . 8750 i.e. a further 8550 shoppers.

If we had not taken got the information from the small sample we would have to assume the worst case which would be that . This would give n = 9604. So using the initial sample will save asking approximately 850 shoppers.

Hypothesis TestsWhen we considered estimation problems, we assumed that we had data generated by a known type of model - normal, binomial etc.- but we did not know the value of some parameter(s) in that model, so we had to estimate parameters such as or p by using information from sample data.

We now consider two different types of problem:(1) Assuming that we know the type of model, we wish to test some assumption about parameter

values. This differs from estimation in that some value of the unknown parameter has to be proposed for testing.

(2) We wish to test some assumption about the type of model itself.

In both cases, the basic idea is to check whether it is reasonable to conclude that the sample that has actually been observed was really generated by the assumed model. We cannot be absolutely certain about the conclusion but we try to ensure that it has a high probability of being correct.

To do this, we must first decide upon our basic assumption,which is called the null hypothesis and is denoted by H0.Using the sample data, we calculate the value of some test statisticthat depends upon H0 and upon the sample.If the value obtained for the test statistic would have a high probability if H0 were truethen we accept H0 ; otherwise we reject H0 in favour of some pre-specified alternative hypothesis, usually denoted by H1 or HA.

The procedure can be broken down into five steps:

Step 1: State H0 and H1.

Step 2: Select the test statistic.

Step 3: Specify the significance level, denoted by (“alpha”), which is the lowest probability for accepting H0.Also find the critical value(s) of the test statistic,which will mark the borderline between accepting and rejecting H0.

Step 4: Assuming that H0 is true, use the sample data to calculate the value of the test statistic.

Step 5: Compare calculated and critical values of the test statistic,hence decide whether to accept or reject H0.

To illustrate the steps in more detail, we consider the following example.

Example: Testing for normal when is unknownSuppose that the scheduled time for a particular rail journey is 120 minutes. It is thought that journey times are normally distributed but there is uncertainty about whether the mean really is 120 minutes.

Step 1: Working in minutes, set up the null hypothesis that the mean is 120, which we write as:H0 : = 120 .

The simplest alternative hypothesis is that the mean might differ from this 120 in either direction (so trains might, on average, be either faster or slower), in which case we write:

H1 : 120 ;This is understood to cover the two possibilities: < 120 or > 120 .

Step 2: As test statistic we use t, which depends upon the sample mean , sample standard deviation, sample size and the value of :

tx

s n

/.

We can see that t will have a large positive value if the sample mean is much larger than or a large negative value if it is much smaller than . As for confidence limits on , the number of degrees of freedom for t is = n 1 .

Step 3: We usually set the significance level at 5%, so that if the observed results have a probability of less than 0.05 we shall reject H0 .Note: The critical values will depend upon H1 .

If we have specified that might shift in either direction from 120, then we shall have two critical values, marking out a critical region in two tails of the t-distribution. The total probability associated with both tails is , with /2 lying to the right of t/2 (the upper critical value) and /2 lying to the left of t/2 (the lower critical value).

120xX

H1with < 120 H1with > 120

H0 with = 120

Using the 5% level of significance, the critical values for a two-tailed test are therefore the 2.5% points on the t-distribution, shown in Table 10. If we take a sample of 10 actual journey times, we have = 10 1 = 9 and the critical values are:

tcrit = t0.025 = 2.262 .

Step 4: Only now do we actually look at the observed journey times.Suppose that the sample of 10 journey times gave a mean of 124 minutes and a standard deviation of 8 minutes. Using = 120 as if H0 were true, the calculated value of the test statistic is:

tcalc = (124 -120)/(8/10) = 1.58 .

Step 5: Since tcalc lies between the critical values ( 2.262), we accept H0 and conclude that the mean journey time could be 120 minutes. If we had found that tcalc < 2.262 or > 2.262, we

would have rejected H0 in favour of H1 and concluded that the mean had shifted from 120 minutes.

Note: If we had decided to test the alternative hypothesis that the shift in the mean could only have occurred in one direction, then we would have chosen a critical value that cut off in just one tail of the t-distribution. If the shift could only be to the right then consider the right tail, if it could only be to the left then consider the left tail.

Exactly the same principles apply in carrying out any of the following tests.

A Summary of Hypothesis Tests

1. Single sample test for normal when is unknown , use tx

s n

0 with = n 1

2. Single sample test for binomial p , use z

p p

p p n

0

0 01

3. Comparing normal means from two paired samplesConsider individual differences and carry out the test on the differences using the t statistic as in test 1 above.

4. Comparing normal means from two independent samples .

Calculate a pooled estimate of the common variance

sn s n s

n np2 1 1

22 2

2

1 2

1 1

2

and use with = n1+n2 2

5. Comparing proportions from two independent samples .Calculate a pooled estimate of the common p which is

px x

n n

1 2

1 2and use

Each of these tests is illustrated by a worked example on the following pages.

Example (Single sample test for normal when is unknown – one tail)Mr A Thug, a mugger believes that Aston students attending a nightclub carry on average £15 in cash on them and so are not worth bothering with.In a sample of 25 such students the mean amount of cash carried was found to be £16 with s = £4. You may assume that the amount of cash carried is normally distributed.Test Mr A Thug’s hypothesis at the 5% level of significance.Step 1: State the null and alternative hypothesisThe null hypothesis is the statement that says the claim is true. H0 : = 0 = £15 The alternative hypothesis is the statement that says the claim is False. H1 : > £15 Step 2: Select the test statisticThis is a single sample test for normal when is unknown.

So we use the test statistic: tx

s n

0 with = n 1 = 25 1 = 24

Step 3: Specify the level of significance and find the critical value(s)5% level of significance so = 0.05. This is a one tailed test so tcrit = 1.711

Step 4: Calculate the test statistic

tx

s n

0 16 15

4 25

1

4 51 25.

Step 5: Compare the sample statistic with the critical value and make a decision tcrit 1.25 Accept Ho There is insufficient evidence to reject Mr A Thug’s claim.

Example (Single sample test for binomial p – two tails)A lecturer has claimed that 80% of their students can perform a hypothesis test.A random sample of 36 students where examined and 21 where able to perform such a test.Is there evidence at the 5% level of significance that the claim is wrong? Step 1: State the null and alternative hypothesisThe null hypothesis is the statement that says the claim is true. H0 : p0 = 0.8 The alternative hypothesis is the statement that says the claim is False. H1 : p0 0.0.8Step 2: Select the test statistic

This is a single sample test for binomial p so we use the test statistic: z

p p

p p n

0

0 01

Step 3: Specify the level of significance and find the critical value(s)5% level of significance so = 0.05. This is a two tailed test zcrit = ±1.960Step 4: Calculate the test statistic

z

p p

p p n

0

0 01

0 58333 0 80

0 8 1 0 8 36

. .

. .

0 21667

0 16 36

.

.

0 21667

0 06673 25

.

..

Step 5: Compare the sample statistic with the critical value and make a decision |-3.25| |zcrit | so reject Ho as the claim is sufficiently unlikely. So accept H1

Example (Comparing normal means from two paired samples)Five workers were timed on a particular task, before and after they had received training.

-1.96

2.5%

z0 1.96

2.5%Accept H0

Reject H0 Reject H0

In the following table x denotes the time taken before training and y denotes the time after training,both measured in minutes:

Worker A B C D E x 240 260 270 250 255 y 231 254 272 245 253

Carry out a suitable t-test at the 5% level of significance to determine whether the mean job time after training is significantly less than that before training.

SolutionLet the mean job time before training be 1 and that after training be 2 . Also let the mean of the differences, d = x y be D. From the rules for expectation, it can be shown that D = 1 2 .We can then proceed with the usual five steps for testing.Step 1: State the hypothesesThese may be stated either in the form (1) H0 : 1 = 2 (the means before and after are equal)

versus H1 : 1 > 2 (mean after training is smaller)

or (2) H0 : D = 0 (mean of differences is zero)versus H1 : D > 0 (mean of differences is positive)

Whichever form is used, it is clear that H1 is concerned with a shift in one direction only,so we shall use a one-tail test.Step 2: Select the test statisticWe are going to carry out a version of Test 1, using the individual differences, d = x y , as if they were the x-values. Denoting their sample mean by d and their sample standard deviation by sd we therefore use the test statistic

td

s n

D

d

/

Step 3: Specify the level of significance and find the critical value(s)The question specified the 5% level, so = 0.05 . Since we are using a one-tail test, the critical value is t0.05 for = n 1 = 5 1 = 4 , which is 2.132 .Step 4: Calculate the test statisticFrom the table of data, find the individual differences and then calculate their mean and standard deviation.

Worker A B C D E x 240 260 270 250 255 y 231 254 272 245 253

xy = d 9 6 2 5 2 d = 4.0 , sd = 4.183

tcalc = (4.0 0) / (4.183/5) = 4 5 / 4.183 = 2.138

Step 5: Compare the sample statistic with the critical value and make a decisionOur calculated value of 2.138 is only very slightly greater than the critical value 2.132. In a situation like this we would normally try to obtain more information before making a decision. In this case it would be sensible to observe more workers before and after training, so as to obtain a larger sample.

Note that if we had asked the vaguer question “Is there evidence of a change in the mean job time?” this would have led to a two-tail test, for which the critical values would have been 2.776.Our calculated value would still have been 2.138, which would lie between the critical values, leading us to accept the null hypothesis of no change in the mean.

Example (Comparing normal means from two independent samples)A review in a motoring magazine claimed that cars of Type 1 have a mean fuel consumption that is 5 miles per gallon (mpg) more than that for cars of Type 2. A test was carried out using 8 cars of Type 1 and 10 cars of Type 2. The sample mean was found to be 43mpg for Type 1 and 40 mpg for Type 2; the sample standard deviation was 3.0 for Type 1 and 2.5mpg for Type2. Carry out a suitable t-test at the 5% level of significance to determine whether the difference in the underlying mean consumption for the two types of car really is 5mpg as claimed in the magazine.

SolutionLet the mean consumption be 1 for Type 1 and 2 for Type 2. Similarly, use labels 1 and 2 for sample statistics relating to Type 1 and Type 2 respectively. For this type of problem it is useful to introduce the notation 0 for the difference 1 2 when H0 is true. We then proceed with the usual five steps.

Step 1: State the hypothesesH0 : 1 2 = 0 = 5.0 versus H1 : 1 2 0 = 5.0 (2-tail test).

Step 2: Select the test statisticIn order to do this, we assume that the population variances for the two types are equal, so that1

2 = 22 = some unknown 2 . A formal test can be carried out to check that this assumption is valid

but for small samples it nearly always does prove to be valid, so we omit the test. However, we need to estimate the unknown common variance 2 .; to do this, we pool information about variance from both samples to find the pooled estimate of the common variance, sp

2 .

sn s n s

n np2 1 1

22 2

2

1 2

1 1

2

( ) ( ) .

The test statistic is then

tx x

sn np

( )1 2 0

1 2

1 1

with = n1 + n2 2 .

Step 3: Level of significance and critical value(s)Using the 5% level for a two-tail test with = n1 + n2 2 = 8 + 10 2 = 16 ,the critical values for t are the 2.5% points for 16 degrees of freedom , 2.120 .

Step 4: Calculate the test statisticWe know that n1 = 8 , x1 43 , s1 = 3.0 , n2 = 10 , x2 40 , s2 = 2.5 ,hence sp

2 = [(8-1)(3.0)2 + (10-1)(2.5)2] / (8 + 10 -2) = 119.25 / 16 = 7.4531and so sp = 2.73 .Then tcalc = [(43 - 40) - 5] /[2.73(1/8 +1/10)] = 2 / 1.295 = 1.54 .

Step 5: Compare the sample statistic with the critical value and make a decisionSince 1.54 lies between the critical values , 2.120 , we accept the null hypothesis and conclude that the magazine’s claim is valid.

Example (Comparing proportions from two independent samples)Samples of male and female students were selected as part of a survey of holiday destinations.Out of 120 male students, 50 had been abroad (outside UK), and out of 80 female students, 42 hadbeen abroad. Is it reasonable to conclude that the same proportion of male and female students in general had been abroad?

SolutionLet suffix 1 denote male students and suffix 2 denote female students. Let the proportions ofthe male and female student populations who had been abroad be p1 and p2 respectively and letthe corresponding observed proportions be p1 and p 2 .

Step 1: H0: p1 = p2 = some common unknown p versus H1: p1 p2 (two-tail test).

Step 2: Calculate a pooled estimate of the common p ,

px x

n n

1 2

1 2

and then the test statistic is zp p

p p n n

( )[( / ) ( / )]

1 2

1 21 1 1.

Step 3: Two-tailed test at 5% level so critical values are the 2.5% points of the standard normal distribution, z = 1.96 (from Table 5).

Step 4: To evaluate the test statistic, we first need to calculate the observed proportions for the two samples and the pooled estimate of the common proportion.

Hence

zcalc

0 4167 0 5250

0 46 0 54 1 120 1 80

0 1083

0 07191 505

. .

( . )( . )[( / ) ( / )]

.

.. .

Step 5: Since zcalc lies between the critical values, we accept the null hypothesis and conclude that the underlying proportions of male students and female students who have been abroad is the same.

Chi-squared Tests for Expected and Observed Frequencies

We now consider a family of tests that are based on the following idea. If p is the probability of a particular event happening when some random experiment is performed once, then the expected number of such events when the experiment is performed n times will be np. This expected frequency will usually be different from the frequency that is actually observed when the experiment is performed n times. If the expected and observed frequencies are very different, this may indicate that the wrong probability has been used to calculate the expected frequency.

If we consider an experiment that results in just one type of event (“success”) or its complement (“failure”), hypotheses about the value of p for one or two samples can be tested using an appropriate z-test (Test 2 or Test 5 on page 19 above). However, when there are more than two possible classes of outcome from the experiment, more general methods are needed and one of the most widely used of these is based on the statistic 2 which is called “chi-squared”.[Note that, in English, the “ch” in “chi” is pronounced like a “k”, as in “chaos”, “chorus”, “echo” and other words derived from Greek.]

Suppose that there are k classes of possible outcome from the experiment and that class i has probability pi for i = 1 to k. If we perform the experiment n times, the expected frequency of a result in class i will be npi , which we shall write as Ei . Writing Oi for the corresponding observed frequency, we then define the chi-squared statistic as follows:

22

1

( )O E

Ei i

ii

k

.

The calculated value of this statistic is compared with a critical value found in Table 8. Since any difference between Oi and Ei will cause 2 to increase, we always use a one-tail test, so the critical value for 5% significance will be the value of 2 that cuts off 5% of the probability area in the right tail of the distribution. As in the case of the t-distribution, we have to consider the appropriate number of degrees of freedom for 2 , which is again denoted by .

The basic rule is:No. of degrees of freedom for 2 = (No. of separate classes) (No. of constraints on frequencies).

In the basic2-test, we have k classes and just one constraint, which is that the frequencies for all classes add up to the total number of observations, n , and so = k 1 .

Example (Testing the Fairness of a Die)The following table shows the result of 120 tosses of a single six-sided die:

Face of die 1 2 3 4 5 6Obs.frequency 19 22 17 18 19 25

Carry out a 2-test at the 5% level of significance to check whether the die is fair.

Solution Our classes of outcome here correspond to the various faces of the die. The statement that “the die is fair” implies that each face has an equal probability and hence that this probability should be 1/6. We therefore write our null and alternative hypotheses as follows:H0 : pi = 1/6 for i = 1 to 6 versus H1 : pi 1/6 for at least one value of i .

The test statistic is 2 and the critical value is the 5% point for = 61 = 5, which is 11.07 .We can explain =5 by saying that values of five of the frequencies can be chosen independently but the sixth frequency must be 120 minus the sum of the other five.

To calculate the value of 2 if the null hypothesis is true, we first find the expected frequency for each class, which is Ei = npi = 120 (1/6) = 20 for each class, and so:

calc2

2 2 2 2 2 219 20

20

22 20

20

17 20

20

18 20

20

19 20

20

25 20

20

( ) ( ) ( ) ( ) ( ) ( )

Since this calculated value is less than the critical value of 11.07, we accept the null hypothesis and conclude that the die is fair.

Exactly the same procedure would be followed for any example where there are k classes that are each assumed to have probability 1/k. Another example might be to check whether some kind of event was equally likely to occur on any day of the week; the classes would then correspond to the 7 days of the week and the probabilities would all be 1/7.

WARNINGThis type of 2-test is based on an approximation that is only valid when the expected frequencies are reasonably large. For =1, each expected frequency should be greater than about 10. For > 1, each expected frequency should greater than about 5. In order to obtain expected frequencies that are sufficiently large, it may be necessary to combine classes.

Example (Testing for a Poisson Distribution with Specified Mean)This illustrates a modification of the basic test in which the class probabilities are no longer assumed to be equal but are now given by a particular theoretical distribution. It also illustrates the idea of combining classes to obtain sufficiently large expected frequencies.

The number of phone calls received in one hour by a sales office was monitored over 100 hours at comparable times of day. The results are shown in the following table:

No. of calls, x 0 1 2 3 4 5 6 7 8 >8Obs. frequency 11 30 28 17 12 0 1 0 1 0

Carry out a 2-test at the 5% level to determine whether the hourly number of calls could reasonably be considered to have a Poisson distribution with mean 2.0 .

SolutionWe shall test H0 : distribution is Poisson with mean 2.0 versus H1 : distribution is not Poisson with mean 2.0 .Note that this is a one-tail test because, as explained above, any difference between observed and expected frequencies causes the test statistic 2 to increase.The next step is to define the classes into which the observations will be arranged.Here the classes correspond to the values of x and the class probabilities are found from the Poisson probability function f(x) = x exp(- )/x! . Note however that we must consider the whole distribution, so we include a class that covers all the higher values of x; the probability for this class will be 1 (probability of all the lower values of x). Poisson tables for = 2 show that F(7) = 0.999 and F(8) = 1.000 to 3d.p., so we initially define our last class as “X 8” with probability 0.001.Then, either using tables or the Poisson formula, we find the probabilities for classes corresponding to x = 0, 1, . . . , 7 and multiply by 100 to find the expected frequencies as shown in the table below:

Class 1 2 3 4 5 6 7 8 9x 0 1 2 3 4 5 6 7 8+Prob. .135 .271 .271 .180 .090 .036 .012 .003 .001Ei 13.5 27.1 27.1 18.0 9.0 3.6 1.2 0.3 0.1

Now we see that the Ei values for classes 6 to 9 are all less than 5 but adding them together we would get an expected frequency of 3.6+1.2+0.3+0.1 = 5.2 for a new class 6 which would be “X 5”.The observed frequency for this new class will be 2 = (total no. of observed values 5). Our final table for observed and expected frequencies is therefore:

Class 1 2 3 4 5 6 Total

Oi 11 30 28 17 12 2 100Ei 13.5 27.1 27.1 18.0 9.0 5.2 99.9 (error from using 3 d.p.)

After combining, we have the no. of classes k = 6 , so = k1 = 5and the critical value is the 5% point for 2 with = 5, which is 11.07 (as in the previous example).We now calculate the contribution to 2 = (Oi Ei )2/Ei for each class and sum these to find 2

calc. .

Class Contribution. to 2 1 (1113.5)2/13.5 = 0.463 2 (3027.1)2/27.1 = 0.310 3 (2827.1)2/27.1 = 0.030 4 (1718.0)2/18.0 = 0.056 5 (12 9.0)2 / 9.0 = 1.000 6 (2 5.2) 2 / 5.2 = 0.463

Total = 2calc. .= 3.828

Since 3.828 is less than the critical value, 11.07, we accept H0 and conclude that the distribution could be Poisson with mean 2.0 .

Note: The precise value of 2calc. will depend upon the number of decimal places used for the expected

frequencies. For hand calculation, we recommend calculating these to 1 d.p. and calculating the contributions to 2

to 3 d.p., as shown above.

In this first example of checking whether a Poisson distribution fits the data, we assumed that the value of the parameter was known. The only constraint on the frequencies was that they sum to n, so only one degree of freedom was lost and = k1.If we had to estimate another degree of freedom would be lost, because this imposes the additional

constraint that x O nxi i and so now equals k 2 .

Example (Testing for a Poisson Distribution with Unknown Mean)With the same data as in the previous example, using the observed frequencies before combining, the sample mean is :[0(11) + 1(30) + 2(28) + 3(17) + 4(12) + 5(0) + 6(1) + 7(0) + 8(1)] / 100 = 199 / 100 = 2.0 to 1 d.p.Using this as our estimate of the unknown population mean , we would get the same probabilities as before. Also the calculations for combining classes and evaluating 2

calc would be the same.The only differences would be that our null and alternative hypotheses would refer to “some Poisson distribution” without specifying its mean and our critical value would now be the 5% point for 2 with 6 2 = 4 degrees of freedom, which is 9.488. In this case we would again accept the null hypothesis.

The same basic method can be extended to testing for any kind of theoretical distribution (e.g. the normal) but care must be taken to combine classes corresponding to adjacent values of x so as to get large enough expected frequencies and also to allow for the loss of one degree of freedom for each unknown parameter that is estimated from the data.

Contingency TablesThe basic chi-squared test can also be extended to testing whether two or more ways of classifying data are independent.

Example (Is choice of newspaper associated with gender?)Suppose that a survey of newspaper readership has been carried out and the number of readers tabulated by newspaper title and gender of reader are as follows:

Times Telegraph Guardian Row totalMale 182 215 203 600Female 154 136 110 400Column total 336 351 313 Grand total 1000

We want to find whether there is some statistical association between gender and choice of title.

SolutionWe start with the null hypothesis that the two criteria for classification are independent. The alternative is that they are not independent - in other words they are associated.Our test statistic will be 2 and we test at the usual 5% level, but to find the critical value we first need to consider how to estimate probabilities and expected values.To illustrate the method, consider Male readers. Selecting at random from our grand total of 1000 readers, the probability of selecting a Male reader will be P(Male) = (No.of Males)/1000 = 600/1000.Similarly, the probability of selecting a Times reader will be P(Times) = (No.of Times readers)/1000 = 336/1000If being Male and being a Times reader are independent, the simple multiplication rule tells us that P(Male and Times) = P(Male)P(Times) = (600/1000)(336/1000) = 600336/(1000)2 .The expected number of “Male and Times” is 1000P(Male and Times) = 600336/1000 = 201.6 .

In general, the expected frequency for the class corresponding to row i and column j in the table isEij = (Total for row i)(Total for column j)/(Grand total) .

Now in each row, all but one of the frequencies is independent and the last one must be such as to make the row total come to the figure we use for calculating the expected frequencies. Similarly in each column, all but one of the frequencies is independent but the last one must make the column total come to the figure used to calculate the expected frequencies. The total number of independent frequencies gives us the number of degrees of freedom, which is = (No.of rows 1)(No.of columns 1).

For our newspaper readership example we therefore have = (2 1)(3 1) = 2 and the critical value is the 5% point for 2 with = 2, which is 5.991.

It is convenient to write the expected frequencies in brackets beside or below the corresponding observed frequencies. As usual , we round the expected frequencies to 1 d.p.

Times Telegraph Guardian Row totalMale 182 (201.6) 215 (210.6) 203 (187.8) 600Female 154 (134.4) 136 (140.4) 110 (125.2) 400Column total 336 351 313 Grand total 1000

Contributions to 2 are calculated in the usual way. It is convenient to arrange these by row and column: 2

calc = 1.906 + 0.092 + 1.230 + 2.858 + 0.138 + 1.845 = 8.069.

Since this is greater than the critical value (5.991), we reject H0 and conclude that gender and title of newspaper are associated. It is most important to remember that the null hypothesis for this type of problem is always that there is no association.

Covariance and CorrelationThe categories used in constructing contingency tables may be the values assumed by quantitative variables, in which case the value of 2

calc is a measure of how closely the variables are associated.A more informative measure is the covariance, which is based on the actual values of the variables rather than just the number of values of each of them that corresponds to a particular cell in the contingency table.

Suppose that we are interested in the relationship between two variables, X and Y. We have n pairs of observations, (xi , yi) for i = 1 to n . From these we can calculate the following sums of terms:x , x2 , y , y2 and xy , where denotes summation over i = 1 to n in each case.[Note that some calculators have special facilities for entering the (xi , yi) as pairs and automatically calculate each of the five sums.]The next step for hand calculation is to evaluate each of the following:Sxx = x2 (x)2/n , Syy = y2 (y)2/n , Sxy = xy (x)(y)/n .

Note that Sxx/(n1) is the sample variance for X, denoted by s2x or var(x) ,

similarly Syy/(n1) is the sample variance for Y, denoted by s2y or var(y) ,

and we now define Sxy/(n1) to be the sample covariance for X and Y, denoted by cov(x,y).

Example (Covariance between Height and Weight)The following table shows the height and weight of six people who were selected at random from a particular population.

Height in ft, x 5.1 5.3 5.7 5.8 6.0 6.2Weight in lb, y 70 112 133 154 165 230

From this table we findx = 34.1, x2 = 194.67, y = 864, y2 = 138 974and xy = 5017.9;

then Sxx = x2 (x)2/n = 194.67 (34.1)2/6 = 0.8683,Syy = y2 (y)2/n = 138 974 (864)2/6 = 14 558,Sxy = xy (x)(y)/n = 5017.9 (34.1)(864)/6 = 107.5 .

The sample covariance is thereforecov(x,y) = 107.5 / 5 = 21.5 .

The difficulty about interpreting this as a measure of association between height and weight is that it depends on the units of measurement. If we had calculated the covariance between the heights and weights of the same individuals measured in metres (1m = 3.28ft) and kilograms (1kg = 2.20lb), we would have found that cov(x,y) = 2.98 but the closeness of association would have been just the same.

To overcome this dependence on units, we can calculate the covariance between the standardised values of x and y, which are obtained by subtracting the sample means and dividing by the sample standard deviations. The resulting statistic is called the sample correlation between x and y and is denoted by r(x,y) or simply r. The quick way to calculate r is to use the formula;

rSxy

Sxx Syy

.

For our height and weight data, r 107 5 0 8683 14558 0 956. . . .

The correlation is a measure of linear (i.e. straight-line) relationship between two variables. It can be shown that the value of r must lie between 1 and +1. If r = +1, an increase in one of the variables is exactly matched by a linear increase in the other; if r = 1, an increase in one of the variables is exactly matched by a linear decrease in the other. A value of r near zero indicates that there is no linear relationship between x and y - although there may be some clear nonlinear relationship.

A Test for Zero Correlation between Normal Variables

We have seen that r measures linear relationship between sample values. There is a corresponding quantity denoted by (“rho”) which measures linear relationship between the populations of X and Y. In the same way that the sample mean provides an estimate of the population mean () and the sample variance provides an estimate of the population variance (2), so r provides an estimate of .

The interpretation of this estimate is complicated by the effect of sample size, since small samples quite often give high values of r even when the underlying value of is small.

When X and Y can reasonably be assumed to be normally distributed, we can carry out a simple test using the t-distribution to check whether is really zero.

We test H0 : = 0 versus H1 : 0 (two-tail test).The test statistic is t = r [(n 2) / (1 r2)] with degrees of freedom = n 2 .

Example (Testing for Correlation between Height and Weight)Consider the last example, in which we found that the sample correlation between height and weight for six people was 0.956. Assuming that height and weight are normally distributed (which has often been checked and confirmed), we now want to see whether this result indicates a real (non-zero) correlation between height and weight for the population from which the sample of six was drawn.

SolutionWe test H0 : = 0 versus H1 : 0 (two-tail test).Note that the null hypothesis assumes zero population correlation.The degrees of freedom = n 2 = 6 - 2 = 4, so for a 2-tail test at the 5% levelthe critical values are 2.776 .Then tcalc = r [(n 2) / (1 r2)] = 0.956 [(6 2) / (1 0.9562)] = 6.52 .Since this lies far beyond the upper critical value, we reject H0 and conclude that there is indeed a non-zero correlation between height and weight for the population from which the sample was drawn.

The Distinction between Association and Causal Relationship

When looking at measures of association, such as correlation or 2 in contingency tables, it is most important to understand that association does not necessarily imply cause and effect - thus being male does not cause someone to read a particular newspaper, nor does being tall cause someone to be heavy. Statistical methods can only highlight association; other forms of analysis must then be used to establish causal mechanisms. For example, statistical methods showed that there is an association between smoking tobacco products and developing various forms of cancer, but biomedical research was needed to establish how the cancers are caused.

Decision Analysis

Introduction to Differentiation

Differentiation is the process of finding the rate at which change takes place. For example, the rate of change of distance with respect to time is called speed.

The result of the process of differentiation is called a derivative.Hence, the speed can be found by calculating the derivative of the distance covered with respect to time. In the graph above, it is represented by the slope of the lines.

Business Examples for derivatives are:Marginal Cost – the rate of change of cost with respect to the level of production.Marginal Profit – the rate of change of profit with respect to the level of production.Price Elasticity of Demand - the rate of change in demand with respect to price.

The slope of a straight line

7.5

2.5

Time (minutes)

Distance (km)

5 10

The distance covered in the first 5 minutes is 2.5 km.Constant speed in first 5 minutes:30 km/h = 0.5 km/min

The distance covered in the next 5 minutes is 5 km.Constant speed in next 5 minutes:60km/h = 1km/min

Consider the straight line y = 3x - 5which goes through the points (2,1) and (5,10).

So the change in x is x = 5 - 2 =3 and the corresponding change in y is y = 10 - 1 = 9.

Hence, the slope of the line is given by the ratio

slope =

y

xa

9

33

-5

y

x

5

52

x

y

In general, the slope, a, of a straight line which is given by a function y = ax + b is the rate of change of y with respect to a change in x.

slope =dy

dx

y

xa

where x is the difference between any two values of x and y is difference between the corresponding values of y. The slope of the straight line is equivalent to the derivative of the corresponding linear function.

Finding the slope of a curve

The method that was used for straight lines will not work for curves because the slope of a curve changes. Only a straight line has a constant slope, i.e. the same slope for all values of x.

Instead of looking at the curve itself, we need to look at a linear tangent to the curve:

y

x

P

QCurve y=f(x)

ChordPQ

Tangent at P

Definition of Differentiation

This process of finding the limit of the ratio of the change in y to the change in x, y/x, as the change in x, x, tends towards zero is called differentiation. So, the derivative of y with respect to x is defined as the limit of the gradient of the cord PQ as h, the change in the x value, tends to zero.

dy

dx x

y

x h

f x h f x

h

lim lim

0 0

The result of the process of differentiation is called the derivative of y with respect to x (or simply the derivative if no other variable is involved).

Rules for derivatives

Rather than calculate every derivative from first principles – i.e. calculating the limit value defined above – a number of rules are used. These are summarised in the following table:

f(x) = y a ax axn ln(x) ln(axb) eax beax

A chord is a straight line joining any two points on a curve. A tangent to a curve is a straight line which just touches but does not cross the curve. The slope, or gradient, of the curve at point P is the same as the gradient of the tangent drawn at point P.If we draw shorter chords on the curve then as point Q becomes closer to P the gradient of the chord will increasingly approach the gradient of the tangent at point P. By this method, the slope of the curve can be found analytically.

dy

dx

0 a anxn 1 1

x

b

x

aeax abeax

If the function to be differentiated contains the sum or difference of a number of terms, the derivative is the sum or difference of the derivatives of the individual terms.

Examples

1. Differentiate the function y = 2x3 and hence find the slope of the curve at x = 3.

Recall that if y = axn then dy

dxanxn 1

so, dy

dxx x 2 3 63 1 2

So the slope of the curve at x = 3 is: dy

dx 6 3 542

2. Differentiate the function y = x3 - 2x2 + 15x +76

Recall: if the function to be differentiated contains the sum of a number of terms, the derivative is the sum of the derivatives of the individual terms.Hence:

3. Differentiate the function y = ln(x) + e3x : dy

dx xe x

13 3

Second & higher order derivatives

Recall that the rate of change of distance with respect to time is called speed. It can be found by calculating the derivative of the distance covered with respect to time. Similarly, the acceleration is the rate of change of speed with respect to time. The acceleration is called the second derivative of distance with respect to time.In general, the second derivative gives the rate of change of the first derivative, i.e. the rate by which the rate of change - and hence the slope of the original function - itself changes.

The second derivative is denoted byd y

dx

2

2, the third derivative is denoted by

d y

dx

3

3 and they are

calculated using the same rules as for the first derivative.

ExampleFind the second and third derivatives of the function y = x3 - 2x2 + 15x + 76 with respect to x.

First derivative (see above): Second derivative:

d y

dxx x

2

22 13 2 4 6 4 d y

dx

3

36

Note that for polynomials the degree, i.e. the highest number in the exponent, decreases by one for each derivative that is calculated. I.e., a square function like y = 2x2 has a linear first derivative (4x) and a constant second derivative (4).

Optimisation using Differentiation

In Business and Management, we are often interested in optimising, i.e. in maximising or minimising some objective, as for example:

– Maximising profit– Maximising revenue– Minimising cost (e.g., when producing a fixed amount of goods)– Minimising risk

Hence, it is important to be able to find the output which maximises profit or the sales price which maximises revenue. This can be done by using differentiation.

Maximum and minimum values of a function occur when the slope of the respective curve is zero. That is, the first derivative of the function is zero at the point where the function has a local maximum or minimum:

Moreover, the direction of the change in slope indicates whether the respective point is a maximum or minimum point. That is, the sign of the second derivative at the turning point indicates whether there is a maximum or a minimum. If the slope decreases, i.e. the second derivative is negative, then the function has a maximum at his point, if the slope increases, i.e. the second derivative is positive, then there is a minimum.So the location and type of the turning point can be determined by calculating the first derivative, setting it equal to zero, and then checking for the sign of the second derivative.

Local maxima and minima can be illustrated as follows:

+ slope - slope

slope = 0Maximum

d y

dx

2

20

Decreasing slope Increasing slope

Example

Find the turning points of the function y = x3 - 6x2

+ 9x

+ 6

dy

dxx x 3 12 92

For such a point, we must have: 0 3 12 92 x x

Solve the quadratic equation by using the appropriate formula (see Quantitative Techniques):

x =12 122

4 3 9

2 3

12 144 108

6

12 36

6

12 6

6There are two turning points: x = 1 and x = 3.

Differentiating dy

dxx x 3 12 92 , i.e. calculation of the second derivative gives

d y

dxx

2

26 12

When x = 1, d y

dx

2

26 12 6 , and there is a local maximum at x = 1.

When x = 3, d y

dx

2

218 12 6 , and there is a local minimum at x = 3.

Points of Inflexion

It is possible for a turning or stationary point – i.e. a point at which the slope and hence the first derivative is zero – to be neither a maximum nor minimum. In these cases, the (positive) slope decreases to zero and then increases again or (in the case of a negative slope) it increases to zero and then decreases again. Such points are called points of inflexion. At such a point, the first and the second derivative both take value zero. A point of inflexion may, e.g., look as illustrated below:

Stock Control Example

There is a constant demand for an item of D units per month. It costs £Hc per unit per month to hold the item in stock, and it costs £Rc each time an order is placed for a new delivery of the item in administration charges. If there are no time delays between re-ordering and the delivery of new stock, and shortages are not allowed, find the most economic order quantity.

If the item is ordered in batches of q items, then D/q orders will be placed each month, at a cost of £RcD/q. The average amount of stock will be q/2 and so the holding costs each month will be £Hcq/2. So the total costs per month are Tc(q) = RcD/q + Hcq/2.

Differentiating gives dT q

dqR Dq

Hcc

c 2

2 =! 0

So there are stationary points when R DqH

cc 2

2, that is q

R D

Hc

c

2.

(This can be found by reformulating the above equation in standard format and applying the formula for solving quadratic equations.)

Increasing slope

Slope = 0Decreasing slope

The second derivative is dT q

dqR Dq

cc

2

232 > 0, as all numbers are non-negative.

Clearly negative orders do not have a practical interpretation, and so the minimum of monthly total cost

is reached when qR D

Hc

c

2. This is called the Economic Order Quantity (EOQ).

Partial derivatives

When a function varies with more than one variable, the derivative of the function with respect to one variable can be found holding the other variable(s) constant; this is known as the partial derivative.

Functions in more than one variable occur, e.g., when modelling the profit or the production of multi-product firms. In these cases, we need to introduce one independent variable for the quantity of each of the products.

Example

f(x,y) = 2x + 3xy - y2

When differentiating with respect to x we hold y constant, sof(x,y) = 2x +Ax + B, where A = 3y and B = -y2.

Using the rules of differentiation gives

f x y

xy

( , ) 2 3

Note that a (curly d) is used instead of a d to indicate that a partial derivative has been found.

Similarly, holding x constant and differentiating with respect to y gives

f x y

yx y

( , ) 3 2 .

Partial derivatives can be used to find the stationary points, i.e. the maxima and minima, of functions of more than one variable, by setting the partial derivatives with respect to each of the variables equal to zero. However, the rules for identifying the stationary points for functions of more than one variable are more complicated than for a single variable and are beyond the scope of this course.

The Transportation ProblemWhen a company can supply its customers from a number of warehouses (or factories) it may wish to find a method of doing so which minimises transportation costs, while not exceeding the capacities of the individual warehouses. We will make the following simplifying assumptions:1. Items are shipped “individually”, that is there are no savings (or extra costs) per item when several items are shipped from the same supplier to the same destination.2. The items are identical, that is the customer has no preference as to which warehouse(s) the items are shipped from.3. No shipping from warehouse to warehouse, or to a customer via another customer.4. Multiple “deliveries” are acceptable, i.e. a customer can receive the items he wants from more than one warehouse.5. The scheduling of deliveries will not impose any further constraints.

Some DefinitionsFeasible Solution – This is a solution in which all the constraints are met. That is all the demands are exactly met and all the supplies are used.

Basic Solution – This is a solution in which exactly one less route is used than the number of suppliers, n, plus the number of destinations, m. That is (n+m -1) routes. The optimum solution will never require more routes.Basic Feasible Solution – This is a solution that is both basic and feasible.Degeneracy – This is when we have a solution that involves less routes than the basic n+m -1. It is normally represented by using n+m -1 routes but carrying zero units on some of them.

Solution Method1. Balance supply and demand using a dummy destination.2. Find a basic feasible solution (e.g. Least Cost First).3. Check if the solution can be improved (e.g. Stepping Stone Algorithm).4. If an improvement is possible, make it and repeat from step 3.

ExampleDuff Machine Tools manufacture machine tools at three factories. The Leeds factory can produce 15 machines each month, the Manchester factory 20 machines and the Nuneaton factory 10 machines.In a particular month, there are no machines in stock, and customers in Aston must be supplied with 7 machines, 17 machines each must go to customers in Bradford and Cardiff.The transportation costs (in £) for each machine are given below:

Aston Bradford CardiffLeeds 10 1 20Manchester 12 7 9Nuneaton 2 14 16

1. Balance supply and demand using a dummy destination.The total number demanded by all customers is 41 machines and there is a supply of 45 machines so we introduce a dummy customer with a demand of 4 machines and zero transportation cost to represent the unused supply.

The problem then can be represented in a special tableau form which is shown below. The factories or warehouses are put into the rows, the customers are represented by the columns of the tableau. The amounts available or required are written next to the respective row or column. The costs for transporting an item (here: one machine) from an origin to a destination can be found in the upper right corner of each cell.

2. Find a basic feasible solution (using Least Cost First)

1 Assign as much as possible to the real route with smallest unit cost. (With ties pick one at random.)2 Cross out the row or column which is now satisfied. (If both are satisfied only cross out one of them.)3 Recalculate the supply and demand for the remaining rows and columns.4 Repeat from step 1 until only the dummy column remains.5 Assign the remaining variables with the appropriate amounts.Note that the dummy column is only used in this process after the other columns are already satisfied!

3. Check if the solution can be improved (Stepping Stone Algorithm)

The idea of the Stepping Stone Procedure is to exchange a route which is currently not used for a route which is currently used. Only if taking up the new route gives a saving in cost, this exchange is actually made. Therefore, first it needs to be checked for all unused routes, if (in terms of cost) it is desirable to use one of them. In order to keep the solution feasible, the units which are to be transported through the previously unused route need to be reallocated such that each customer still gets what he wants and each warehouse still delivers the amount it has.

Stepping Stone Procedure: Starting with any basic feasible solution, consider each unused route in turn:

I Assign one item to this route. Indicate this with a +

II Compensate for this item by adjusting the numbers assigned to other used routes to ensure demand is met and supply is not exceeded. Note the increases with + and the decreases with -

III Calculate the net change in cost, by adding the cost of the routes marked with a + and subtracting the cost of the routes marked with a -.

If no unused route gives a negative total then the current solution is optimal. Otherwise: allocate as many units as possible to the route giving the greatest saving, making the compensations indicated by the +’s and –’s. Then repeat the stepping stone method (i.e. steps I to III) until the optimal routes are found.

Note that you need to find a “circle” of routes (tableau cells) which starts at the unused route (empty cell) you are considering, and otherwise only consists of used routes, i.e. of tableau cells which currently have an entry! I.e. you go from the empty cell to a cell with an entry in the same row, then to a cell with an entry in the same column as the cell you are coming from, etc., until you are back at the empty cell you started with. The “circles” for the different unused cells should be marked with different letters, i.e. A for the first unused cell, B for the second, etc.

The cheapest solution, costing £196, is to supply all 7 to Aston from Nuneaton, to supply Bradford with 15 from Leeds and 2 from Manchester, and supply all 17 to Cardiff from Manchester. This will leave 1 machine in Manchester and 3 in Nuneaton.

Stepping Stones: A worked example including improvement step

TaskA car manufacturer has stocks of new cars (one particular model) available as follows: 20 at Liverpool, 25 at Birmingham and 15 at Oxford. He has orders from dealers for 12 cars in Bristol, 25 in London, 13 in Manchester and 10 in Birmingham. How should he arrange the deliveries (each car is driven separately) if the cost of a delivery from each factory to each dealer is as follows:

DealerFactory Bristol London Manchester BirminghamLiverpool 27 31 14 20Birmingham 19 22 19 10Oxford 19 16 25 16

SolutionFirst, a feasible starting solution has to be found by using the Least Cost First Method. This solution is given in the tableau below.Then, choose the first route that is not currently used and put a plus sign in that square.In this case it is the route from Liverpool to Bristol, so we put A+ in that square (the A just means that we do not have to redraw our table when we consider the second route).

A: 10 -1 +7 - 0 + 0 -2 = +14B: 20 - 9 +7 - 1 = +17C: 0 - 0 +7 -1 = +6D: 12 - 2 + 0 -0 = +10E: 14 - 0 + 0 -7 =+7F: 16 - 0 + 0 - 9 = +7Hence these routes are optimal as no costs can be saved.

This plus signifies that we are intending to increase the number of items transported on this route. In order to be able to do this, two things must be true:i) Less must go to the respective customer from somewhere else, i.e., the customers in Bristol

still should get the 12 they want in total, and not more.ii) Less must go from this supplier, i.e. Liverpool, to another destination so that we actually have

at least one car to send to Bristol. (In this case 7 cars were going to London and 13 to Manchester accounting for all 20 cars available at Liverpool.)

Taking the first of these we note that Bristol must receive less from Birmingham as we are currently supplying all 12 from there. So we put a minus sign here to signify for every car we send from Liverpool to Bristol we send one less from Birmingham to Bristol.

Now, because supply and demand exactly match, we know that every car we do not send from Birmingham to Bristol must be going to another customer. Because we only consider introducing one new route at a time, it must go to either the customer in London or the customer in Birmingham as these are the only other customers supplied from Birmingham (at the moment, 3 cars go to London and 10 go to Birmingham respectively). Now if we sent it to Birmingham the customers in Birmingham would receive more than the 10 they want, and because they get them all from Birmingham we can not reduce the number being supplied from elsewhere to compensate. So, the cars which were going from Birmingham to Bristol must now go to London, hence we put a plus sign on the Birmingham to London route.

Now note that both in order for London to get the 25 they want and for us to not send more than 20 from Liverpool we must put a minus on the route Liverpool to London.We now have found that any cars we send from Liverpool to Bristol are ones that previously we would have sent from Liverpool to London.

Now we have come back to where we started and found the only possible set of changes which introduces only the route from Liverpool to Bristol and which ensures that each customer gets exactly the amount he wants without sending more than we have from any supplier. We now calculate how much the transportation cost will change for each car sent on this route.It costs £27 for each car we send from Liverpool to Bristol, but we will save £19 for each not sending them from Birmingham, it will cost £22 each to send a car from Birmingham to London and we will save £31 for each that we do not send from Liverpool to London. This gives a net change of 27-19+22-31=-1 that is we save £1 for each we send on the new route. (Just add the costs of the routes with plus signs and subtract the costs of the routes with minus signs).

As this route saves us money we should send on it as many cars as we can. (Before we actually do that we have to check that currently there is no other route on which we could save more money per car. This can be checked by applying the very same procedure to the other five unused routes. If you do that, you should end up with positive net changes for all other routes, so the only possible improvement is to introduce the route Liverpool-Bristol into the transportation plan!) As for every item we send on this route we send one less on the routes with minus signs, the most that can be sent is determined by the smaller of the numbers using these routes (in this case the smaller of 7 and 12). This is because you cannot send a negative number of items on a route!So our improved set of routes is determined by increasing the numbers going on routes marked with plus signs by 7 and reducing the numbers going on routes with minus signs by 7. At this point it is worth drawing a new table (but not before):

You now apply the same process all over again for each unused route, checking if further improvements can be made. In this case you will find that no further improvements can be found, as all net costs are positive, see below.

Route CostA Liverpool-London 31-27+19-22 = +1 More, so do not use itB Liverpool-Birmingham 20-27+19-10 = +2 More, so do not use itC Birmingham-Manchester 19-14+27-19 = +13 More, so do not use itD Oxford-Bristol 19-19+22-16 = +6 More, so do not use itE Oxford- Manchester 25-14+27-19+22-16 = +25More, so do not use itF Oxford-Birmingham 16-10+22-16 = +12 More, so do not use it

Introduction of any of the unused routes will increase costs, so the last tableau must represent the cheapest way of transportation: Liverpool to Bristol 7 cars, Liverpool to Manchester 13, Birmingham to Bristol 5, Birmingham to London 10, Birmingham to Birmingham 10, Oxford to London 15 cars. The total cost is £1,026.

Linear Programming

Linear Programming (LP) is concerned with looking for the “best” solutions to problems given certain constraints. The decision maker has control over a number of factors which are represented by continuous variables, and the objective is to maximise (or minimise) a function of these decision variables, without exceeding a set of bounds defined by linear constraints.

Formulate a problem as a LP model1. Define decision variables. 2. Define the objective function. 3. Specify the constraints.

Find the solutionIf there are only 2 variables a graphical method can be used, otherwise a specialised computer package such as LINDO can be used (or the solver within Excel).

Example 1: The farmer’s problemA farmer has 100 spare hectares of land, which can be used to plant either wheat or potatoes (or neither). Wheat gives a profit of £90 per hectare and potatoes give a profit of £60 per hectare.

EU regulations limit the amount of potatoes planted to 65 hectares at most.There will be only 480 person-hours available to harvest the crop. A hectare of wheat takes 6 person hours to harvest, while a hectare of potatoes takes only 3 hours. What should the farmer do?

1. Define decision variablesThe farmer has control over the number of hectares of wheat and/or potatoes that are planted. Hence the decision variables are:W the number of hectares of wheat planted,P the number of hectares of potatoes planted.

2. Define the objective functionObjectives in Linear Programming can be either to maximise a linear function of the decision variables or to minimise a linear function of the decision variables.Wheat generates a profit of £90 per hectare and potatoes £60 per hectare. So the profit from planting W hectares of wheat is 90W and the profit from planting P hectares of potatoes is 60P. Hence, the total profit is given by the following linear function of W and P, Profit = 90 W + 60 P.So the objective function is to maximise profit (in £‘s):Maximise Profit = 90 W + 60 P

3. Specify the constraints Constraints in LP are linear equalities or inequalities. This means, a linear function of the variables is either equal to, less than or equal to, or greater than or equal to some value.

The farmer has 100 spare hectares of land. The total amount of land planted must be less than or equal to the available space, so: W + P 100

EU regulations limit the amount of potatoes planted to at most 65 hectares, so: P 65

There will be only 480 person-hours available to harvest the crop. A hectare of wheat takes 6 hours to harvest, while a hectare of potatoes takes only 3 hours. So the time needed to harvest W hectares of wheat is 6W and the time needed to harvest P hectares of potatoes is 3P. Therefore: 6W + 3P 480.

Finally, there are the non-negativity constraints. It is not possible to plant a negative amount, soP 0 and W 0.

The Linear Programming model of the farmer’s problem (completed)

Let W be the number of hectares of wheat planted and P be the number of hectares of potatoes planted.

Maximise Profit = 90 W + 60 P

Subject to the following constraints:Land Constraint W + P 100EU Constraint P 65Harvest Time Constraint 6W + 3P 480Non-negativity P 0 and W 0

Graphical Solution

Plot Constraints on GraphLand Constraint W + P 100EU Constraint P 65Harvest Time Constraint 6W + 3P 480Non-negativity P 0 and W 0Plot trial objective function on the graphe.g. Profit = 5400 = 90 W + 60 P Profit = 7200 = 90 W + 60 P

The maximum profit is realised where a line parallel to the trial objective function just touches the feasible region.

Note the solution to a LP model will always be on a corner (or edge) of the feasible region. Hence, an alternative solution method is to find the profit at each corner of the feasible region and to compare these profits.

So the maximum profit will be where the land constraint crosses the harvest time constraint, that is where W = 60 and P = 40. This gives a profit of £7800.

The land and harvest time constraints are called binding constraints, because there is no slack or surplus; we have used all the land and all the available labour.

Using LINDO

Syntax for a LP model to be solved by LINDO computer package.

1. Start with the objective function. Write MAX (for maximise) or MIN (for minimise) followed by a linear function. Max 90W + 60P2. The end of the objective function and the beginning of the constraints is signified with any of the following: SUBJECT TO, SUCH THAT, S.T. or ST3. This is followed by the constraints. You may, optionally, name constraints in a model. Constraint names make many of LINDO's output reports easier to interpret. Constraint names must follow the same conventions as variable names. To name a constraint you must start the constraint with its name terminated with a right parenthesis. After the right parenthesis, you enter the constraint as a linear function (without any constant terms) followed by either = or >= or <= then a number.

Land) W + P <= 100EC) P <= 65Harvest) 6W +3P <= 480

LINDO will accept < or > in constraints but they will be interpreted as <= and >=. The end of the constraints is signified with the word END.4. LINDO has a limit of eight characters in a variable name. Names must begin with an alphabetic character (A to Z), which may then be followed by up to seven additional characters. These additional characters may include anything with the exception of the following: ! ) + - = < >. So, as an example, the following names would be considered valid: XYZ, My_Var, A12, SHIP$LA,whereas the following would not: THISONESTOOLONG , A-HYPHEN , 1INFRONT5. LINDO will not accept parentheses as indicators of a preferred order of precedence. All operations in LINDO are ordered from left to right. Only , , and inequalities are allowed.6. Comments may be placed anywhere in a model. A comment is denoted by an exclamation mark (!). Anything following the exclamation mark on the current line will be considered a comment.

MAX Profit) 90W + 60P ! Maximise profit (in £)SUBJECT TOLand) W + P <= 100 ! There is only a limited area of land (units are Hectares)EC) P <= 65 ! The Maximum area of Potatoes due to EC regulations (in Hectares)Harvest) 6W +3P <= 480 ! There is a limited amount of labour at harvest time (in person hours)End

Output from LINDO LP OPTIMUM FOUND AT STEP 1

OBJECTIVE FUNCTION VALUE

PROFIT) 7800.000

VARIABLE VALUE REDUCED COST W 60.000000 0.000000 P 40.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES LAND) 0.000000 30.000000 EC) 25.000000 0.000000 HARVEST) 0.000000 10.000000

NO. ITERATIONS= 1

Example 2: Efficient stock portfolio

A fund manager is planning the investment of up to £600,000 so as to maximise its yield. The following stocks are considered:Stock A B C D EEstimated % yield 15 12 7 6 8Stocks A and B are high risk and can compose at most 40% of the total investment. Also neither of the investments in A or B can be greater than half of the total invested in stocks C, D and E. Also, the investment in each of low yield stocks C, D, and E must not exceed the total invested in the other two low yield stocks. Formulate the problem as an LP model.

1. Define decision variablesLet the amount invested in stock A be A thousand pounds.Let the amount invested in stock B be B thousand pounds.Let the amount invested in stock C be C thousand pounds.Let the amount invested in stock D be D thousand pounds.Let the amount invested in stock E be E thousand pounds.

2. Define the objective functionMaximise the overall percentage yield, Y = (15A + 12B +7C +6D +8E)/600.

Max Y = 0.025000A + 0.02000B + 0.011667C +0.01000D + 0.013333E

3. Specify the constraintsTotal investment is less than or equal to available capitalA + B + C + D + E <= 600

Stocks A and B are high risk and can compose at most 40% of the total investmentA + B <= 0.40(A+ B+ C+ D+ E)so 0.60A + 0.60B - 0.40C - 0.40D - 0.40E <= 0

The optimum value of the objective function.In this case the maximum profit is £7800.

The optimum is achieved with these values of the decision variables:W = 60 hectares and P = 40 hectares

All the land is used (Slack is zero).We have planted only 40 hectares of potatoes so another 25 could be planted before the EC directive would be a problem.All the available hours of labour at harvest time have been used (Slack is zero).

The dual price gives the marginal cost / benefit of a change in the right-hand side of the respective constraint.So if the farmer lost a hectare of land because of a road widening this would reduce the profit by £30.If, on the other hand, the farmer had a hectare more on which to plant potatoes or wheat, he would be able to make an extra profit of £30. Therefore, he should be ready to pay a maximum rent of £30 for an additional hectare of land.If the farmer was able to buy extra harvest time, he should be ready to pay a maximum amount of £10 per hour (as he can make an extra profit of £10 per hour).

The reduced costs of an objective function coefficient give the amount by which the coefficient has to change to let the respective variable take a positive value in the optimal solution. I.e., the reduced costs express by how much the profit per hectare needs to increase to make production of a certain sort of crop worthwhile.Variables with positive values have reduced costs of zero, as production of the resp. crop is worthwhile already.

Neither of the investments in A or B can be greater than ½ of the total invested in stocks C, D and E. A <= 0.5(C + D + E) and B <= 0.5(C + D + E) A - 0.5C - 0.5D - 0.5E <= 0 and B - 0.5C - 0.5D - 0.5E <= 0

Also, the investment in each of low yield stocks C, D, and E must not exceed the total invested in the other two low yield stocks.C <= D + E and D <= C + E and E <= C + Dso, C - D - E <=0 and D - C - E <=0 and E - C - D <=0

Non negativity constraintsA >= 0, B >= 0, C >= 0, D >= 0, E >= 0.

Some additional notes about Linear Programming

Convex

Fractional values of decision variables should have meaning (Divisibility)For example, it is possible to plant a fraction of a hectare so the decision variables in the farmer’s problem can take fractional values. With lots of business examples this may not be the case, for example it is not possible to make and sell a fraction of a car (although it may be possible to interpret such figures as averages).

No diminishing (or increasing) returns to scale (Proportionality)In the farmer’s problem, returns from 10 hectares of wheat must be 10 times the return from one hectare. This is because the objective function is a linear function.

AdditivityThe total contribution of all variables in the objective function and their requirements in the constraints are the direct sum of the individual contribution or requirement of each variable. This means, e.g., that the profit contributions of wheat and potatoes are added up in the objective function; the profit resulting from wheat is not influenced by the profit which results from the planting of potatoes.

The feasible region is always convex.That is, it is possible to join any two points in the feasible region with a straight line that is entirely within the feasible region.This is a direct consequence of the constraints being linear inequalities