## Friday, 28 October 2011

### Argand diagram VS cartesian coordinate

For those who lacks knowledge, the problems on complex numbers can be quite hard (Like those who're trying old Pure without some old ref book). That's why we need some simple transformation to make it equivalence to a 2D-co-geom problem.
Basic knowledge:
Complex number can be written in form of $z=x+iy$ where x and y are real numbers. Then $Z(x,y)$ represents the point z on the argand diagram.
The absolute value of the complex number: $|z|=|OZ|=\sqrt{x^2+y^2}$
The principal of complex number: $Arg(z)=\tan ^{-1}(m_{OZ})$ where OZ is a vector such that angle beyond 180 degrees are considered.
Slope between two points $u=a+bi$ and $v=c+di$:
Complex version: $Arg(u-v)$
Cartesian version: $\frac{d-b}{c-a}$
Transform:
Complex version: $zcis\theta$
Cartesian version: $Z\left( \begin{array}{cc}\sin\theta &\cos \theta \\-\cos\theta &\sin \theta\end{array} \right)$

Example 1: Let a, b be solutions of $x^2+x+2=0$ while c and d be solutions of $x^2-2px+2=0$. If the points of a,b,c and d on the argand diagram are concyclic, solve p.

Now consider the values of a,b:$x=-1\pm i$
And the values of c,d: $x=p\pm \sqrt{p^2-1}$
Therefore c and d is $x=cis \theta$ or $x=p\pm \sqrt{p^2-1}\in R$
When p>1, c=d and obviously ABCD isn't a quadrilateral, so p < 1.
Now transform it into geometrical problem:
$A(-1,1), B(-1,-1),C(cos\theta, sin\theta), D(cos\theta,-sin\theta)$
are concyclic.
Note that:
1) ABCD is symmetrical about x-axis.
2) AB is parallel to x-axis.
We can conclude ABCD is a rectangle, therefore , p = 0.

Example 2: HKALE 1983 Q3
a) For complex number w, show that $|w-i|=|w+i|$ if and only if w is real.
Standard ans:
If part: Let $w=x+iy$
$x^2+(y+1)^2=x^2+(y-1)^2$, $y=0$.
Only if part: Let w be real. both sides trivially equal.

Co-geom analyse:
For circle $r=|w+i|$, t's symmetrical along y-axis/Im axis. With $w-i$ being another point on the circle and the line joining w+i and w-i is perpendicular to the symmetrical axis, therefore the midpoint lies on the line joining center and perpendicular to the constructed line. i.e., w is on Re-axis, so w is real.

b) Sketch the locus of $|2u-i|=1$
I think complex number method would be faster here.

Answer:
$|u-\frac{i}{2}|=\frac{1}{2}$, therefore the locus is a circle with center i/2 with radius 1/2.

Co-geom analysis:
$|2x+(2y-1)i|=1$
$x^2+y^2-y=0$
$x^2+(y-\frac{1}{2})^2=\frac{1}{4}$, the same result is given.
c) Show that $|2u-i|=1$ is equivalent to $v=\frac{iu}{i-u}\in R$, hence show that u,v and i are colinear.
The first part of question is easy when you "rationalize" v, you will have:
$v=\frac{-y+ix}{-x-(y-1)i}=a+b(x^2+y^2-y)i$, Im(v) =0 by our deduction before.

The second part is much more difficult, we should analyse the question in a geometrical manner:

Choose a point on $|2u-i|=1$, say U. The center of the circle C1 is X.
rotate the circle along the origin by 90 degress anticlockwisely to make the transform "iu" (Circle C2).

Now UX intersect the circle C1 again at U'. Assume iU intersect Re-axis at V'.
U' represents the point i-u. Note that OU is perpendicular to OU', i.e., the point U''(iu) lies on the line OU'. Therefore U', U'' and O are colinear.
Furthermore, by simple geometry, we can show that U'O is parallel to the line iU and iUOU' forms a rectangle. Then we can show that triangle U'OU and iOV' are in sprial rotation relationship. (similar triangle)

^ Actually we proved the first part of the question here.
Now for the second part of question:
By trigonometry means, and U'O^2 + UO^2 = 1, we can show that U''O:U'O = U'O : UO = 1: OV'. Therefore V lies on iU, Q.E.D.

Example 3: HKALE 1993 Q11
a) Show that $\Delta W_1W_2W_3\sim \Delta Z_1Z_2Z_3$ iff $\frac{z_3-z_1}{z_2-z_1}=\frac{w_3-w_1}{w_2-w_1}$.
b) Show that a triangle is equilateral iff $z_1+cis(\frac{2\pi}{3})z_2+cis(\frac{4\pi}{3})z_3=0$.
c) Disprove the existance of equilateral triangle on argand diagram with all integral point vertices.

We will do the question seperately.

a) In geometrical approach, what can you see from the given information?
Information I: The quotient are equal. It implies that $\angle Z_3Z_1Z_2 = \angle W_3W_1W_2$.
Information II: The quotient are equal. It implies that $\frac{|Z_3Z_1|}{|Z_2Z_1|}=\frac{|W_3W_1|}{|W_2W_1|}$.
Now apply the knowledge learnt in secondary 1. They are similar because angle equal, 2 sides proportional!!!!
We just get 4 mark easily from the AL Pure Maths paper XD

b) Consider another equilateral triangle with vertex $cis(\frac{2k\pi}{3}),k=0,1,2$.
Now $\frac{z_3-z_1}{z_2-z_1}=\frac{cis(\frac{2\pi}{3})-cis 0}{cis(\frac{4\pi}{3})-cis 0}$, since cis 0 = 1,
$(z_3-z_1)(cis(\frac{4\pi}{3})-cis 0)=(z_2-z_1)(cis(\frac{2\pi}{3})-cis 0)$
By simple rearranging, we can obtain the above result.

c) Standard solution:
By equation in b, we have:
Equating real part: $Re(\Delta)=Re(z_1-\frac{z_2}{2}-\frac{z_3}{2})=0$
Equating imaginary part: $Im(\Delta)=Im(z_1+\frac{\sqrt{3}z_2}{2}-\frac{\sqrt{3}z_3}{2})=0$
Due to the closure of rational numbers under multiplication, we can show that the two equation does not have ALL rational solution. Hence there does not exist ALL integral solution.

Disclaimer: The above skills AREN'T answering skills, they are POSSIBLE solutions, but NOT PERFECT PRESENTATIONS. A RIGOROUS PROOF should be respected.

Exercise:
1) Consider example 3a:
a) Proof or disproof the existance of the general form proofing similar triangle on Argand diagram, showing the properties for i) AAA, ii) 3 sides proportional.
b) Write down one more requirement such that two triangles are congruent.
c) Suppose the two triangles are similar, write a general transformation between two triangles in terms of the complex numbers representing the points given.
d) Let triangle A,B,C,D,E be (1+i), (2+i), (3+3i), (4+5i), (-2+4i) respectively. If triangle ABC ~ triangle DEF where F is a point on the Argand diagram, find F and the transformation between two triangles.
2) Consider exmaple 3b:
a) Is the expression symmetrical for all three points of the triangle? Prove your assertion.
b) Equilateral triangle ABC has circumcircle |2u-i|=1. Write the general expression for the triangle.
c) Let A,B represent the point (1+i), (3+0i) respectively.
i) Find two points C and D such that ABC and ABD are equilateral.
ii) Show that CD is perpendicular to AB.
3) Consider example 2, in circle |2u-i|=1,
a) Show that (i-u), (0+i/2) and (u) are colinear.
b) Prove that (i,u), (u), (0) is an right-angled triangle.
c) Show that (iu) lies on line passing through (i-u) and 0.
4) Account the following FACT in terms of complex numbers:
a) Parallel lines do not intersect each other.
b) Angle bisectors of triangle are concurrent.
c) Ptolemy's theorem for both equality and inequality case. i.e. for quadrilateral ABCD, $(AB)(CD)+(AD)(BC)\geq (AC)(BD)$, equality hold iff ABCD is concyclic.
5) HKALE 1984 Q6b: f(z) is defined by $f(z)=\frac{z-a}{bz-1}$ are defined on $D=C/\left\{\frac{1}{b}\right\}$ where $a,b\in C$. Suppose |f(1)|=|f(-1)|=|f(i)|=1.
a) Show that b is the conjugate of a, |f(z)| = 1 for all $z\in D,|z|=1$.
b) Show that f(z) is a constant function if |a|=1.

## Friday, 21 October 2011

### Agressive bidding

(1) Non vul, IMP pair.
C J3
D AKJT983
H K82
S 6
RHS opened 3S, what should you do?
A 3H implies a solid H weak jumping call, you can imagine his/her hand as H AKJxxx with no/useless side power. Then your H K might be a trick and a control. Considering the chance to win on hand, I called a [gambling] 3NT.
LHS called 4S and your partner called a non-invitative 4NT and all pass.
Your partner's hand is
C AK765
D 7
H 7654
S QT9
LHS lead C2, now you'll have to decide the route:
It seems easy but an uneven distribution will make the game risky because you have one and only one H control.The correct way is to keep finessing the D Q. It's safe because LHO probably holds S AK so your S Q is safe. You found the D is in a bad 4-1 distribution, but luckily LHO lacks Heart as well so 4NT is in justmade.
Besides this meaningless play, the controversal point is that whether we should bid this 3NT with two dangerous suit on opponent's hand:
- Gambling 3NT: solid 7 card suit; no side power
- Power 3NT: 25-27points, 8+ tricks with control on every suit
- Unusual NT is for 2NT/4NT, so it's not our interest.
And for me, a defensive gambling 3NT here make sense:
- Against strong opening: very solid 7+ solid suit with 1+ side control, 7.5 tricks
- Against weak opening: solid 7+ solid suit promising control on the weak-opened suit or 2+ side control, 7.5 tricks.
We can make this conclusion becoz the other case fits with a 4X jumping call like:
C K6
D 2
H AKJT9873
S Q2
Such a solid hand deserves a 4H opening.

(2) NS vul, IMP pair
Your partner (N)
C AKJ
D T
H AT842
S AK43
You (S)
C Q9763
D AKJ83
H 975
S ----

A rare 5-5-3-0 distribution; 6C is a good contract, Considering the misfit and poor C AKJ on your partner's hand, it would not be easy to be the final bid.
After N's 1H opening (no doubt for this), how can you describe the 5-5 minor powerful hand? If your partner have some good major power, that will be a 3NT; and with some good minor like:
C AKxxx
D xxxx
H Ax/Kx
S xx
Then 5 or 6C/D can be played.

My approach:
1H - 2D [1]
2S [2] - 3C [3]
4NT [4] - 6C
[1]: 10+P, 5+D, positive
[2]: reversal bid, game force, 4+S
[3]: At least 4-5 minor; do not prefer 3NT; opening power
[4] quantitative invitation, 19~21P
A splinter would not be suitable as your hand still lacks power from a 5X to slam force. Due to the lack of power, other slow approach might not be able to make such calls and stop the bidding at 4H.

P.S. I'm not an expert in bridge but a relaxing player, so I'll need more opinions : D

## Thursday, 20 October 2011

### Chemistry note: Qualitative method to identify chemical species

To access the document version of this chapter of notes click the "Notes Corner" above.

Qualitative analysis on different compounds
Ionic compound: cation (Al3+, Ca2+, Cu2+, Fe2+, Fe3+, Pb2+, Mg2+, K+, Na+, Zn2+, Ag+, NH4+)
1)       Appearance, e.g. Cu2+ is blue-green, Fe3+ is brown.
2)       Solubility: Nitrates and hydrogencarbonate are all soluble
Halides are all soluble except silver halides AgX
Sulphates are all soluble except BaSO4, PbSO4, CaSO4 (slightly soluble)
Carbonates and sulphites are all insoluble except Na+, K+, (NH4)2CO3
3)       Flame test: use a clean Pt wire to dip conc. HCl followed by the sample solid, burn under non-luminous flame (luminous flame has its own colour) to see the characteristic colour. Principle: the compound forms metal chloride under conc. HCl (H2SO4 can’t be used as metal chlorides are more volatile than metal sulphates), and the excited metal cation emit photon at visible spectrum to give the characteristic colour.
Na+: [persistent] golden yellow, K+: lilac, Ca2+: brick red, Ba2+: green, Cu2+: bluish green
4)       Actions on dilute HCl/H2SO4 followed by solubility test, etc.
5)       Actions on NaOH: Metal ion [except Na+, K+] forms hydroxide ppt under NaOH, Al3+, Pb2+, Zn2+ re-dissolve in excess NaOH to give hydroxyl complex.
6)       Actions on NH3: Metal ion [except Na+, K+] forms hydroxide ppt under NH3, Cu2+, Ag+, Zn2+ re-dissolve in excess NH3 to form amine complex. Note:
I)                     [Cu(NH3)4]2+ is in deep blue colour, not the blue colour given by Cu2+.
II)                   Ag form Ag2O rather than AgOH by dehydration 2AgOH Ag2O + H2O, but it forms diamminesilver(I) ion under excess NH3, i.e., [Ag(NH3)2]+
III)                  Al(OH)3 and Pb(OH)2 can be distinguished either by actions in H2SO4 or heat until dryness (Al2O3 VS PbO), PbO change colour when heated.
7)       Test for ammonium: Warm with NaOH to give pungent small/turn wet red litmus paper blue.
Test for gases (NH3, CO2, SO2, HX)
1)       CO2: turning lime water milky and become clear when CO2 is in excess; note that SO2 has similar effect by Ca(OH)2 + SO2 CaSO3 + H2O but CO2 don’t have reducing power.
2)       SO2: It act as an reducing agent so it reduces Cr2O72- (orange) to Cr3+
3)       SO3: equivalent to SO42-
4)       Cl2: turn wet blue litmus paper red then bleached (white) since Cl2 + H2O and HCl + HOCl are in equilibrium. HCl, HOCl are both acidic while OCl- has bleaching effect.
5)       NH3: pungent smell, form dense white fumes (NH4Cl) with HCl(g), there are test each other.
6)       HX: HCl can be tested by NH3, HBr and HI are in equilibrium with H2 + X2 so they gives white fumes and brown liquid (Br2) or black solid (I2) as well. They can also be tested by actions on conc. H2SO4.

Ionic compound: anion (NO3-, CO32-, SO32-, SO42-, OCl-, S2O32-, X-)
1)       Actions of heat: Carbonates decompose upon Bunsen flame: Carbonate oxide + CO2
2)       Sulphates are resistant under lab. flames, but iron(II) sulphates decomposes under heating: 2FeSO4 Fe2O3 + SO2 + SO3.
3)       Actions on HCl: bleaching effect of OCl-: OCl- + H+ Cl2 + H2O
4)       Sulphites and thiosulphates gives SO2 under acid:
SO32- + 2H+ SO2 + H2O, S2O32- + 2H+ SO2 + H2O + S, so thiosulphates gives white ppt besides SO2.
5)       Barium test: sulphates, sulphites and carbonates forms ppt with Ba2+, common reagents includes BaCl2, Ba(NO3)2. BaSO3, BaCO3 dissolve in HCl but BaSO4 is resistant in acid. This serves as an unique test for sulphate ions.
6)       Silver nitrate test: Dissolve the sample in HNO3 followed by AgNO3, X- form ppt AgCl (white) / AgBr (pale yellow) / AgI (yellow). They can be further distinguished by:
(I)                   AgCl dissolve in NH3, AgBr only dissolve in conc. NH3, AgI is insoluble in NH3
(II)                 AgCl turns grey in sunlight, AgBr turns yellowish grey, AgI remains yellow
Tests for organic compounds
1)       Saturated hydrocarbons: blue/clear yellow flame under non-luminous flame
2)       Alkene/alkyne: smoky flame under non-luminous flame since higher carbon-hydrogen ratio leads to more CO produced.
Test for addition: add a few drops of Br2 (in organic solvents)/K2Cr2O7/H+, it decolorize rapidly.
3)       Haloalkanes: Boil with ethanolic KOH (KOH with EtOH), acidify with HNO3 followed by AgNO3, ppt of AgX is expected.
4)       Alcohol:
(I)                   Esterification: Add acyl chloride (RCOOX) to give colourless/white fumes (HCl) (The smell of ester is probably covered by acid)
(II)                 Esterification: Add carboxylic acid and heat with a few drops of conc. H2SO4, a sweet fruity smell is given (ester)
(III)                Lucas test: Add conc. HCl with ZnCl2, cloudiness appears (NOT ppt; the mixture of liquid in different phase) according to the degree of carbon: primary alcohol (1hr), secondary alcohol (5 min) and tertiary alcohol (1 min).
(IV)              Iodoform test for alcohol that the carbon connected to hydroxyl group also connected to a methyl group. Add iodine with NaOH, yellow ppt (CHI3) with characteristic smell is given. The mechanism is as follows:

(V)                Under K2Cr2O7/H+, primary and secondary alcohol oxidizes but tertiary alcohol is resistant to oxidation since CH3- is a poor leaving group. Note that KMnO4 can be used as the oxidizing power is too strong and it may oxidizing the tertiary alcohol.
5)       Carbonyl group (aldehydes/ketones):
(I)                   Test by 2,4-dinitrophenylhydrazine: Aldehydes and ketones react with this reagent to give red~yellow ppt (2,4-dinitrophenylhydrazone). Accurate species can be determined by crystallization of ppt and determine the melting point.

(II)                 Iodoform test (for ketone CH3COR): yellow ppt under I2/NaOH
(III)                Tollens’ reagent (silver mirror test): warm with [Ag(NH3)2]+, the reagent act as oxidizing agent that gives out silver, silver mirror is deposited in the inner surface of the tube
6)       Carboxylic acid
I)                     Esterification with ethanol under conc. H2SO4 as catalyst to give sweet, fruity smell
II)                   RCOOH is the only organic compound to react with NaHCO3. (Most acidic)

## Wednesday, 19 October 2011

### Economics note: International trade VI: Trading situation

To access the document version of this chapter of notes click the "Notes Corner" above.

Trading situation
1)       Globalization
Under economic aspect, globalization is the phenomenon of increasing international trade, goods and money flow among countries, due to the expansion of production line over countries (multinational corporations). Economical globalization is driven by:
-          Free trade policy, which attracts lots of investment.
-          Economic reform, especially changing from planned economy (closed/communist) to market economy (open/capitalism), which creates a large amount of job opportunities. The rapid economic growth attracts lots of foreign investment.
-          Advancement in transportation and communication technology reduces transport and transaction cost, diversifies the market, and promotes global sourcing.
As the international trades are mutually beneficial, international trade increases and different countries become more interdependent. At the same time, increased adoption of free trade policy promotes higher degree of globalization.
However, globalization also lead to several negative effects, including closing of some firms which can't compete with foreign goods (and creating unemployed labour force), and the exploitation of labour (income inequality), and exhausting natural resources much faster.
2)       Trading pattern of Hong Kong
Hong Kong relies on importing since Hong Kong lacks natural resources. External trades are important to Hong Kong, and the total traded value is even larger than the GDP. By external trade, foreign currencies can be earned and for further investment or reserve assets.
Trading patterns: the largest trading partner is mainland China; raw materials and capital goods are imported and re-exported. Hong Kong has been an important port for China. Hong Kong exports several goods especially in the light industry like textile, machinery, etc. Comparing with goods exported, services exported are more important, including merchandising, transportation service, travel services and financial services. Besides China, US, Japan and Taiwan are also major trading partners of Hong Kong.
Challenges faced by Hong Kong including the low negotiation power against trade barriers, trade performance linked to mainland's domestic export, competition from other countries including quality of services and rising production cost, etc. Trade and Industry Department, Oversea Hong Kong Economic and Trade Offices and Hong Kong Trade Development Council (HKTDC) supports the trading in Hong Kong.
3)       Economic growth VS Economic development
Economic growth can be easily measured by per capita real GDP, but economic development includes more consideration like income equality and literacy rate, etc.
per capita GDP measures the productivity per person in the economy, but the cost of living or price level is not taken into account. Then Purchasing power parity (PPP) can be used to compare the living quality between countries.
PPP exchange rate = domestic price of a given basket of goods and services in domestic currency / price of the same basket of goods and services in foreign currency. If the PPP exchange rate < official exchange rate, then it implies that every unit of domestic currency should trade for more foreign currency than the official one, then the domestic currency is undervalued. Oppositely it's called as overvalued. It can measure the living quality/economic growth in a better way as it takes the cost of living into accounts.
The economic development (the living standard/economic welfare) can be measured by Human Development Index (HDI), including the following criteria:
-          per capita GNP(PPP) as it's more suitable for international comparison of national income.
-          Life expectancy at birth reflects a population's health conditions/health care services.
-          Educational attainment (mean/expected years of schooling) since higher education level generally implies higher productivity and hence higher level of economic development.
Economic growth can be promoted by:
-          Increasing investment, including sacrificing current consumption for saving (deposited and banks can used them to invest), foreign investment/aids/loans.
-          Promoting free trade/free competitions that under competition, the quality rises.
-          Improving human capital, protecting property rights and upholding rule of law
-          Develop R&D; specialize in several industries (industrial policy).
4)       Cost of economic development
- Lost in current consumption due to increase in investment
- Pollution and environmental damage; exhaustion of natural resources
- Unemployment and uneven income distribution (capital accumulation)
- Economic instability leads to income instability, worsen the problem of unemployment, and hence the redistributive effect is larger, than the income inequality is worsen.
- Loss in political autonomy due to dependence on trading with foreign countries
Sustainable development is the development that satisfies the current needs without compromising the needs in the next generations, is a way to develop the economy in a better way, including:
- Producing environmentally products
- 4Rs (renew, reuse, refuse, recycle)
- Donating to environmental groups for researches
- Government: law enforcement on related field
5)       Economic growth in different region and economic corporations
-          Developing countries develops slower due to the law of diminishing marginal return, while developing countries has higher growth rate; economy in transition (from command to market economy) has instable growth.
-          Economical integration occurs across countries like ASEAN Free Trade Area in Asia, and EU economic union.
-          There are several international organizations promoting free trade and economic growth, like UN (and WTO), IMF, World Bank and APEC, PECC in Asia.

Yeahhhhhh Economics is finished! >w<

## Saturday, 15 October 2011

### Economics note: International trade V: trading and PPF

To access the document version of this chapter of notes click the "Notes Corner" above.

Trading and PPF
Production possibilities frontier (PPF) shows all combinations of two goods that a country can produce with all given resource and best technology.
Assumption:
1)       Only two goods (X, Y) are produced.
2)       Resource available and is given.
3)       Technology is given.
4)       The country uses all resource and best technology available.
Implication:
1)       Scarcity: the possible combination is limited but the want is unlimited. The good produced can't satisfy all wants.
2)       Choice: Since scarcity occurs, choices have to be made, the more the Y produced, the less the X is produced.
3)       Cost: The cost of producing 1Y = number of X sacrificed, and in general OC = dY/dX. Negative slope implies one must be sacrificed to produce another good. When the PPF is a straight line/linear, then the marginal cost is constant. When the PPF is concave, the marginal cost is increasing. That implies that more cost have to paid for producing further more one kind of the good.

The increasing marginal cost is due to the heterogeneous nature of resources. Some resources are more suitable to produce X while some are better for Y. When more X need to be produced, the one better for Y is less efficient to produce X, hence more Y is sacrificed to produce later unit of X.
Inversely, constant marginal cost implies homogeneous nature of resources.
1)       Efficiency: When the production point is along the frontier, all resources are best used so the economy is efficient. Otherwise the attainable combination is inefficient since the resources are not best used.
2)       Improvement in technology: the curve extends on one side while the maximum output for another good remains unchanged.
3)       Increase in resources: there's a parallel shift in production curve (because the OC remains unchanged)

Advantages and trading
Absolute advantages: when comparing unit of goods per unit of input (resources), the one with higher productivity has the absolute advantages.
Comparative advantages: when the slope is lower than that of another economy; i.e. dYa/dXa > dYb/dxb, the less of X sacrificed per Y produced states the comparative advantage. Note that it's independent of resources available.

For example, A has both the absolute advantage on good X and Y while A has comparative advantage on Y while B has comparative advantage on X.
Under constant marginal cost, total world output is maximized when they specialize in producing in the goods in which the economy has comparative. Then complete specialization is applied. Under a mutual beneficial TOT, the consumption possibilities frontier (CPF) increases when they trade.
Trading for concave PPF

For concave PPF, complete specialization is not preferred.
1)       At Complete specialization of good X
The slope at the production point is greater than the TOT line. i.e., the cost of producing X is higher than the international price (or the cost to produce equivalent amount of Y to trade a specified amount of X) Therefore the economy will decrease production in X and increase production in Y.
2)       At complete specialization of good Y
Similarly, at complete specialization on Y, the slope at the production point is smaller than the TOT line, the cost of producing Y is higher than the international price, so the economy will produce less Y, more X.
3)       The production point that slope = TOT line
At that point, CPF is maximized and can satisfy most wants (indifference curve), so the production point is fixed at that point.
Gain and consumption determination
At the production point B, the economy can consume any point along the TOT=CPF (satisfaction maximization is not our interest). The amount of exported and imported goods is as shown. At point A, if the original production point is along the blue part of PPF, then A boosts both consumption of good X and Y.
﻿
A very tiring topic since many diagrams were drawn. The technical parts ends here so the last one would be the trading situation.﻿