Revision Notes

Lecture One Principals Payoff = V(d, s) d = decision (Shit we can influence) s = state of nature (Shit we can't influence) Payoff's value is defined not only by it's monetary worth, but by how much relative benefit it shall have to the recipient (aka utility) Product

Contract I (S1)

Contract II (S2)

Product A

250

70

Product B

160

150

Product C

90

170

This can be shown graphically also, as follows: Choose A:d1

s1 s2

Choose B: d2

Start

s1 s2

Choose C: d3

s1

250 70 160 150 90

s2

170 Maximax (optimistic) Find the highest payoff value, in the above case, 250

Maximin (Pessimistic) Find the lowest value of each decision over all the states of nature, then pick the highest value. In layman's terms, the shittest circumstances, but the highest influential performance.

The Hurwicz Criterion H (d

i

) = α V m a x ( d i ) + (1

− α )V

m in

(d i )

This is basically a balance between optimism & pessimism, the value of 'a' is a decimal between 0 & 1, where 0 is ultimate pessimism, and 1 is ultimate optimism.

Minimax Regret R ( d i , si ) = Vmax ( si ) −V ( d i , si ) Basically, the difference in the best & worst outcomes, so for a risk adverse decision maker, you want the lowest value. Basically, difference between that value and the highest vertically (along the state of nature) and then the difference between all the horizontal values. Simple, eh?

Backward Induction P(s1)=0.6 d1 P(s2)=0.4

d2

Start d3

70

P(s1)=0.6

160

P(s2)=0.4

150 90

P(s1)=0.6 P(s2)=0.4

Decimals relate to % chance of corresponding states of nature.

250

170

Perfect Information Could be attained by a shitload of research, but ultimately, what's the point if that cost outweighs the benefit gained from it. To find this we: Take the highest possible gains from all states of nature, Multiply these answers by their estimated probabilities (the decimal corresponding to their state of nature.) Add the values together We already know the best possible outcome, so subtract the worst from that, and there you go, your research to reach perfect information must cost less than that to be worth it.

Lecture Two Easy shit.

Introduction to Differentiation Rule 1

Rule 2

Rule 3

Rule 4

Rule 5

axn

ln(x)

ln(axb)

y

a

ax

dy dx

0

a

anxn-1

1 x

Rule 6

b x

Rule 7

eax

beax

aeax

abeax

Look at that if you need to.

Finding Min/Max points Points occur when Dy/Dx = 0. If second derivative > 0 then = minimum, else, maximum. Nota Bene: If 2nd derivative = 0, then t'is a stationary point of inflection.

Lecture Three Differentiation Example A company produces only one product. The quantity made per week is called Q (measured in tonnes). The production cost depends on the weekly output in the following way: C(Q) = 10 +8Q -2Q2 The achievable selling price also depends on the output, (and hence on the quantity offered) and is:

Pr(Q) =40 -10Q Hence Revenue = Price x quantity sold = (40 -10Q)Q= 40Q -10Q2 And Profit = Revenue – Cost = 40Q -10Q2 -(10 +8Q -2Q2) = 32Q -8Q2 -10

Partial Derivative If you've got more than one variable to work with in a function, you can only find the partial derivative, this is found by holding one of the variables constant.

Lecture Four Transportation Problems (Principals) M = Number of Suppliers N = Number of customers Each index of M / N has it's own amount that can be supplied / demanded This can be shown in a matrix format: Duff Machine Tools manufacture machine tools at three factories. The Leeds factory can produce 15 machines each month, the Manchester factory 25 machines and the Nuneaton factory 5 machines. In a particular month, there are no machines in stock, and customers in Aston must be supplied with 5 machines, 15 machines each must go to customers in Bradford and Cardiff, and a customer in Durham requires 10 machines. The transportation costs per machine are given below:

Aston Bradford Cardiff Durham Leeds 10 1 20 11 Manchester 12 7 9 20 Nuneaton

2

14

16

18

Transportation Problems (Assumptions) 1.Items are shipped “individually”, that is there are no savings (or extra costs) 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 supplier(s) the item is shipped from. 3. No shipping between suppliers or destinations. 4. Multiple “deliveries” are acceptable. 5. The scheduling of deliveries can be ignored.

Supplier

This information can be compounded into a table such as the one shown below:

Transportation cost forDestination/ route Customer Aston Bradford Cardiff DurhamAvailable 10 1 20 11 Leeds 15 15 12 7 9 20 Manchester 25 15 10 Nuneaton Required

5

2

5

14 15

16 15

18

5

10 Total = …...

Amount transported along route How to Solve the Problem To find a minimum cost basic feasible solution, i.e. a set of (m + n - 1) routes between suppliers and customers and the amounts transported there, so that all the demands are met, all the supply is delivered

.

and transportation is done at the lowest possible cost

1. Balance supply and demand using a dummy destination (if needed). 2. Find a basic feasible solution. 3. Check if the solution can be improved. If an improvement is possible, make it and repeat from step 3. If no improvement is possible, the optimal solution has been found.

METHOD: Least Cost First Method 1. Assign as much as possible to the cell with the smallest unit cost. (If two or more tie just pick one at random.) 2. Cross out the row or column which is now satisfied. (If both are satisfied cross out only one of them.)

3. Recalculate the supply and demand for the remaining rows and columns. 4. Repeat from step 1 until only one row and one column remain uncrossed. Assign to remaining cells the appropriate amount. Note: If there is a dummy column, its cells are only used after If a solution is optimal, then no unused route will lower net cost. If a route is an improvement, allocate as many units to it as possible.

Lecture Five Method The Farmer’s Problem A 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 £60 per hectare. EU regulations limit the amount of potatoes planted to at most 65 hectares. 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. Establish Variables Hectares of potatoes = P Hectares of wheat = W Goal Therefore, profit can be defined as = 90W + 60P Constraints are present that: W+P <= 100 P <= 65 6W + 3P <= 480 P >= 0 W >=0

100 80

Feasible Region P

W

Then find a parallel line just touching the feasible region.

100 Maximum Profit

80

Feasible Region

90 W + 60 P = 7800 P 100

160

W

80

W 0 80 7200

P 0 0

Profit 90×0 + 60×0 90×80 + 60×0

60

40

90×60 + 60×40

65 65

90×35 + 60×65 = 7050 90×0 + 60×65 =

60

7800

35

35 0 3900

Feasible Region

=0 =

=

P 65

Lecture 10 Basis of Hypothesis Testing The basic idea behind statistical hypothesis testing is as follows: State a claim or hypothesis Take a random sample and calculate the relevant sample statistic Calculate the probability of getting a result this extreme if the hypothesis is true. Reject the Hypothesis if it is too unlikely By hand method 1 State the null hypothesis, H0 and alternative hypothesis, H1 2 Select the test statistic 3 Specify the level of significance α and find the critical value(s) 4 Calculate the test statistic assuming the null hypothesis is true. Compare the sample statistic with the critical value(s) and make a decision Using Statistics program 1 State the null hypothesis, H0 and alternative hypothesis, H1 2 Select the test statistic 3 Specify the level of significance α 4 Statistics package calculates the probability of a test statistic as extreme

as the one observed (the p-value) assuming the null hypothesis is true. Compare the p-value with the level of significance and make a decision