Bi 2: C = 160 + 0,6 I = 150 G = 150 T = 100 a) Tnh GDP cn bng: Gi nh rng xut khu v nhp khu bng khng, ta c: Y=C+I+G Y= Y T) + + G + G c1T Y (1- c1) = c0 + Y= Y=

( c0 + + G c1T) ( 160 + 150 + 150 0,6 . 160 )

Y = 1000 b) Thu nhp kh dng: YD = Y T = 1000 100 = 900 c) Chi tiu dng: C = 160 + 0,6 YD =160 + 0,6 . 900 = 700 Bi sa ca Thy

Bi 2: Cu tinM d = Y $( 0.35 I )

a. Khi li sut 5%M d =Y $( 0.35 I ) = 60000 * (0.35 0.05 ) =18000

Cu tri phiu: 50000+60000-18000=92000 Khi li sut 10%M d =Y $( 0.35 I ) = 60000 * (0.35 0.1) =15000

Cu tri phiu: 50000+60000-15000=95000 b. Khi li sut tng th cu tin gim xungv cu tri phiu tng ln. iu ny ng vi l thuyt c. Gi thit rng li sut l 10% khi thu nhp gim 50% th cu tin ca c ta gim xung 50% d. Gi thit rng li sut l 5% khi thu nhp gim 50% th cu tin ca c ta gim xung 50% e. Khi thu nhp tng th cu tin tng v ngc li trong iu kin li sut khng i Bi sa ca Thy

Bi 4: Md = Y$ (0,25 i) Y$ = 100$ Md = 20$ a. i = ? 20 = 100 (0,25 i) i = 0,05 Vy li sut khi cn bng trn th trng ti chnh l 5%.b.

i = 2% Md = 100 x (0,25 0,02) = 23 $ Cn bng trn th trng ti chnh nn cu tin = cung tin = 23 $ i = 12% Md = 100 x (0,25 0,12) = 13 $ Cn bng trn th trng ti chnh nn cu tin = cung tin = 13 $ Vy cung tin gim 23 13 = 10$ Tng qut: Cn bng trn th trng ti chnh nn cu tin = Md i = 10% cung tin = Md = Y$ x i = 100$ x 10% = 10$ Vy khi li sut tng 10% th cung tin gim 10$. Bi sa ca Thy

Bi 6:

Bi sa ca Thy

Bi 2:a. Chnh ph gim chi tiu G, do G khng xut hin trong mi quan h LM nn G khng tc ng n ng LM. i vi ng IS, G gim tc ng ln sn lng lm gim sn lng do lm ng IS dch chuyn sng bn tri t IS sang IS Gi ng IS mi v ng LM khng i, ti im A, sn lng gim t Y xung Y. Li sut gim t I xung I. Nh vy, khi ng IS dch chuyn nn kinh t dch chuyn dc ng LM, t A n A i vi vic gim G tc ng n I m h bi v Mt mt, sn lng thp hn c ngha l doanh thu thp hn v u t thp hn. Mt khc, li sut thp hn u t li cao hn. Chng ta khng th ni tc ng no tri hn bi nu u t ch ph thuc vo li sut th u t chc chn s tng. Nu u t ch ph thuc vo doanh thu, u t chc chn s gim. Tuy nhin u t ph thuc vo c li sut v doanh thu, nn vic chnh ph gim chi tiu G, khng nht thit dn n u t gim, li hay hi cho I trong ngn hn b. Tnh sn lng cn bng Ta c Y = C (Y-T)+ I(Y, i) + G Y= c0 + c1 (Y-T)+ b0 + bY- b2i +G Y= c0 + (c1+b) Y c1T + b0 b2i + G (1-c1-b)Y = c0 c1T + b0 b2i + G Y= M/P = YL(i) M/P=d1Y d2 i d1Y = M/P +d2i Y= b. iu kin cn bng (1) = (2) (1)

= (c0 c1T +b0 +G)d1 - M/P (1-c1-b) = [d2 (1-c1-b) + b2d1] i i=

(3)

Th cc d liu bi cho vo (3) ta s tim c li sut i

Th i vo I s tnh c u t Th I vo (2) s tnh c Y d. Tnh u t Th Y v i vo I = b0 + b1Y b2i ta s tnh c I e. Khng th kt lun c Vi cc iu kin trn, i vi cc tham s ca m hnh trn, u t tng khi G gim. f. Vi cc iu kin trn, i vi cc tham s ca m hnh trn, u t tng khi G gim ch khi G gim n mc lm Y gim nhiu hn so vi s gim ca li sut. Bi sa ca Thy

Bi 3: A, Tm phng trnh ng IS Y= C + I + G = 200 + 0.25( Y-200) + 150 + 0.25Y -1000i + 250 = 200 + 0.25Y 50 + 150 + 0.25Y 1000i + 250 = 550 1000i + 0.5Y Y= 1100 2000i B, Tm phng trnh ng LM 1600 = 2Y 8000i Y= 800 + 4000i D, Tnh li sut cn bng Li sut cn bng c tnh t phng trnh IS=LM 1100-2000i = 800 + 4000i => i= 5% C, Tnh sn lng cn bng Sn lng cn bng c tnh t phng trnh IS=LM Ycb= 800 + 4000x0.05 = 1000 E, Tnh gi tr cn bng C v I C= 200 + 0.25 (Y-T) = 200 + 0.25(1000-200) = 400 I= 150 + 0.25Y 1000i = 150 + 0.25x1000 -1000x0.05 = 350 Tnh li gi tr Y Y= C + I + G = 400 + 350 + 150 = 1000 F, Cung tin tng ln M/P=1840. Tnh li Y,i,C,I v gii thch bng li ng LM lc ny: 1840 = 2Y 8000i => Y= 920 + 4000i Li sut cn bng i : 1100 2000i = 920 + 4000i => i= 0.03 Sn lng cn bng Y: Y=920 + 4000x0.03 = 1040 Tiu dng h gia nh C : C= 200 + 0.25(Y-T)= 200 + 0.25(1040-200)= 410 u t I : I= 150 + 0.25Y -1000i= 150 + 0.25x1040 -1000x0.03= 380 Gii thch bng li : Khi lng cung tin tng lm cho li sut cn bng i gim xung, do chi ph u vo gim nn u t I tng. Do I tng nn sn lng Y cng tng ln, iu ny lm tng tiu dng C. G, G=400. Tm tt cc chnh sch ngn sch m rng ln Y,i v C

Vi mc M/P nh c, G tng ln 400, th trng cn bng ti : 1250 2000i = 800 + 4000i => i= 0.075 Sn lng cn bng mi : Y= 800 + 4000x0.075= 1100 Tiu dng h gia nh lc ny : C= 200 + 0.25(Y-T)= 200 + 0.25(1100-200)= 425 Tm tt cc chnh sch ngn sch m rng ngn sch : Khi cung tin khng tng, vic tng chi tiu G ca chnh ph trc tip tc ng lm tng tng cu Y, dn n tiu dng C cng tng ln, ko theo li sut i cng tng. Bi sa ca Thy

Bi 2:Cu hi: S dng m hnh AS-AD c xy dng trong chng ny, hy trnh by cho thy cc tc ng ca nhng c sc sau y vi v tr ca cc ng IS, LM, AD v AS trong trung hn. K , cho thy tc ng i vi sn lng v mc gi cng trong trung hn. Gi nh rng trc khi c nhng htay i, nn kinh t mc sn lng t nhin. a) Gia tng lng tin ca ngi tiu dng b) Gia tng thu Bi gii: a) Gia tng lng tin ca ngi tiu dng Khi lng tin ngi tiu dng tng, ngi tiu dng c xu hng tiu dng nhiu hn v tit kim t i do s kh quan v nim tin s lm cho: - C : chi tiu h gia nh tng ng dch chuyn sang phi ca ng IS tng sn lng tng v li sut tng.Li sut -r IS LM

r1 r*

y1

y*

y2

Thu nhp, sn lng -Y

Ban u, khi mc li sut l r* ti im cn bng, mc sn lng s l mc sn lng cn bng y*. Sauk hi nim tin ca ngi tiu dng tng, dn ti li sut s tng ln mc r1>r*. Khi , im thu nhp cn bng cho th trng hng ha l y1, nhng im cn bng trn th trng tin t l y2 lng cung tin cao ti mc no s lm cho li sut gim. S dch chuyn ny s n mt mc no s a th trng tin t v hng ha n li mc cn bng. b) Gia tng thu

S nhn thu trong m hnh Keynes cho thy ng vi mt mc li sut cho trc, tng thu s lm thu nhp gim mt lng bng T x [-MPC/(1 MPC)] do vy s tc ng n kinh t nh sau:

ng IS dch chuyn sang tri mt khong bng gi tr ny. Trng thi cn bng ca nn kinh t di chuyn t im A n im B. Thu tng lm gim li sut t r1 n r2 v lm gim thu nhp quc gia t Y1 n Y2. Tiu dng gim v thu nhp kh dng gim; u t tng v li sut gim.

Li sut -r A B Y IS 1 IS 2 Thu nhp, sn lng -Y Y = YT x [- MPC/(1 MPC)] LM

R1 R2

Bi sa ca Thy

Bi 2: a/ut

-

ut 1

= -0.4 ( g yt - 3%)

nh lut Okun ng Phillips Tng cuun

t t-1 = - ( u t - 5%)g yt

=

g mt

- t

T l tht nghip t nhin ca nn kinh t l Phillips b/ T l tht nghiput

= 5 % theo cng thc ng

= t l tht nghip t nhin

un

=5%

T l lm pht t = 8 % Do t l tht nghip bng t l tht nghip t nhin => t l tng trng thc t bng t l tng trng bnh thngg yt

=3%g yt

Thay s vo cng thc Tng cu ta c => c/g mt

+ t =

g mt

= 3 % + 8 % = 11 %

Gi s nhng iu kin trong cu b vn ng, vy ta c: Nm t - 1:ut 1

=5% = 8%

t-1

Nm t: t = 4%

Theo cng thc ng Phillips: t t-1 = - ( u t - 5%) =>ut

=9%

Theo cng thc nh lut Okun: => 4 % = -0.4 ( g yt - 3% =>g yt

ut

-

ut 1

= -0.4 ( g yt - 3%)

= -7 %g yt

Theo cng thc Tng cu: =>g mt

=

g mt

- t

= -7 + 4 = -3 %

Nm t+1: t+1 = 4%

Theo cng thc ng Phillips: t+1 t = - ( ut + - 5%) 1 =>ut + 1

=5%ut + - u t 11 = -0.4 ( g yt + - 3%)

Theo cng thc nh lut Okun: => -4 % = -0.4 ( g yt - 3% =>g yt + 1

= 13 %g yt

Theo cng thc Tng cu: =>g mt

=

g mt

- t

= 13 + 4 = 16 %

Nm t+2: t+2 = 4%

Theo cng thc ng Phillips: t+2 - t+1 = - ( ut +2 - 5%) =>ut + 2

=5%ut + - ut + 2 1

Theo cng thc nh lut Okun: => 0 = -0.4 ( g yt - 3% =>g yt + 2

= -0.4 ( g yt +2 - 3%)

=3%g yt

Theo cng thc Tng cu: =>g mt

=

g mt

- t

=3+4=7% Bi sa ca Thy

Bi 3: Gi s nn kinh t c th c m t bi 3 ng thc sau: ut ut-1 = - 0.4 ( gyt - 3% ) t t-1 = - (ut 5%) gyt = gmt - t nh lut Okun ng Phillips Tng cu

a. Gim 3 ng thc xung cn 2 bng cch thay gyt trong ng thc tng cu vo nh lut Okun. Gi nh ban u rng ut = ut-1 = 5%, gmt = 13% v t = 10%. By gi gi s rng mc tng tin t nm nay c gim t 13% xung cn 0% mi mi. b. Tnh tc ng vi tht nghip v lm pht nm nay v nm ti. c. Tnh cc gi tr ca tht nghip v lm pht trong trung hn. Bi lm T nh lut Okun v ng Phillips, c th rt ra T l tng trng tim nng gp = 3% T l tht nghip t nhin un = 5%a.

Thay gyt vo nh lut Okun rt gn thnh 2 m hnh (1) (2)

Nn kinh t c m t li vi 2 m hnh sau: ut ut-1 = -0.4 * ( gmt - t - 3%) t t-1 = - (ut 5%) b. T l tht nghip, t l lm pht nm nay v nm ti Phng trnh (1), (2) khi gmt = 0% ut ut-1 = 0.4 * (t + 3%) t t-1 = - (ut 5%) Cc gi nh ban u, nm (t-1) gm,t-1 = 13% gmt = 0% t-1 = 10% ut-1 = 5% b. 1. Nm t thay cc gi tr nm (t-1), v gii h phng trnh (3), (4) ta c ut - 0.4 * t = 6.2% ut + t = 15% (3) (4)

T l lm pht nm nay (nm t) T l tht nghip nm nay (nm t) b. 2. Nm t+1

t = 6.3 % ut = 8.7 %

Thay kt qu t l tht nghip v t l lm pht ca nm t vo h phng trnh (3), (4) v gii h phng trnh ta c: ut+1 - 0.4 * t+1 = 9.9% ut+1 + t+1 = 11.3% T l lm pht nm sau (nm t+1) t = 1 % T l tht nghip nm sau (nm t+1) ut = 10.3 % c. Gi tr ca tht nghip v lm pht trong trung hn. Vi vic ct gim mc tng cung tin t xung cn 0%, lm pht gim dn, tuy nhin, v trung v di hn lm pht khng gim mi mi. Trong nhng nm u, t l tht nghip cn phi tng cao hn mc tht nghip t nhin cho lm pht gim. Sau , lm pht bt u tng tr li. cho tht nghip tr v mc t nhin ca n. Nh vy, v trung hn th t l tht nghip bng tht nghip t nhin u=5% T l tng trng bng t l tng trng tim nng gy = 3% Khi , t l lm pht l t = gm - gy = 0% - 3% = -3% Bi sa ca Thy

Bi 2: Gi s hm sn xut ca nn kinh t l Y = K N 1 (Hm sn xut ny c gi l hm sn xut Cobb Douglas). Chng ta cho =1/2. Gi s by gi =1/3. a. Hm sn xut ny c c trng l sinh li khng i theo quy m phi khng? Gii thch. b. C sinh li gim dn theo vn khng? c. C sinh li gim dn theo lao ng khng? d. Hy chuyn hm sn xut ny thnh mi quan h gia sn lng trn u cng nhn v vn trn u cng nhn. e. Vi t l tit kim (s) v t l khu hao ( ) cho trc, hy cho thy biu thc v vn trn u cng nhn trng thi dng. f. Hy cho thy biu thc v sn lng trn u cng nhn trng thi dng. g. Tm mc sn lng trn u cng nhn trng thi dng khi =0,08 v s = 0,32. h. Gi s rng t l khu hao vn khng i mc =0,08, trong khi t l tit kim gim xung cn mt na s = 0,16; chuyn g s xy ra vi mc sn lng trn u cng nhn trng thi dng. GII a. Hm sn xut ny c c trng l sinh li khng i theo quy m phi khng? Gii thch. Gi s by gi =1/3, nn hm sn xut ca nn kinh t l : Y = K1/3N2/3, t hm sn xut ta thc hin mt php bin i nh sau: (zK)1/3 (zN)2/3 = zK1/3N2/3 = zY (vi z > 0). Nh vy, nu ta gia tng gp i nhp lng (vn v lao ng) th sn lng s tng gp i. Kt lun: y l hm sn xut c c trng sinh li khng i theo quy m. b. C sinh li gim dn theo vn khng? Ta gi s tr lng vn c tng ln 8 ln v gi nguyn lao ng, sn lng s thay i: (8K)1/3 N2/3 = 2K1/3 N2/3 = 2Y. Do , tr lng vn tng gp 8 ln, gi nguyn yu t sn xut cn li, sn lng ch tng ln gp i. iu ny c ngha lm sn xut c sinh li gim dn theo vn. c. C sinh li gim dn theo lao ng khng?

Ta gi s lao ng c tng ln 6 ln v gi nguyn vn, sn lng s thay i: K1/3 (8N)2/3 = (K1/3)4N2/3 = 4Y. Do , lao ng tng gp 8 ln, gi nguyn yu t sn xut cn li, sn lng ch tng ln 4 ln. iu ny c ngha lm sn xut c sinh li gim dn theo lao ng. d. Hy chuyn hm sn xut ny thnh mi quan h gia sn lng trn u cng nhn v vn trn u cng nhn. T hm sn xut Y = K1/3N2/3, ta c th vit li di dng quan h hm sn xut gia sn lng trn u cng nhn ( y hay Y/N) v vn trn u cng nhn (k hay K/N): Y/N = (K1/3N2/3)/N => Y/N = (K/N)1/3 => y = (k)1/3 e. Vi t l tit kim (s) v t l khu hao ( ) cho trc, hy cho thy biu thc v vn trn u cng nhn trng thi dng. trng thi dng ta c: sf(K/N) = K/L, hay l sf(k) = k. Vi hm sn xut cho trn: y = k1/3 ta c k* = (s/ )3/2. f. Hy cho thy biu thc v sn lng trn u cng nhn trng thi dng. Vi hm sn xut cho trn: y = k1/3 ta c k* = (s/ )3/2. Cho nn: y* = (s/ )1/2 g. Tm mc sn lng trn u cng nhn trng thi dng khi =0,08 v s = 0,32. Thay s liu vo cng thc sau: y* = (s/ )1/2 = (0,32/0,08)1/2 = 2. h. Gi s rng t l khu hao vn khng i mc =0,08, trong khi t l tit kim gim xung cn mt na s = 0,16; chuyn g s xy ra vi mc sn lng trn u cng nhn trng thi dng. Thay s liu vo cng thc sau: y* = (s/ )1/2 = (0,16/0,08)1/2 = 21/2 Nh vy, mt t l tit kim thp hn dn n mt mc sn lng bnh qun trn u cng nhn thp hn.

Bi sa ca Thy

Bi 3: Gi s rng hm sn xut ca mt nn kinh t l Y=K1/3 N2/3 v c t l tit kim (s) v t l khu hao ( ) u bng 0,10 a. Mc vn trn u cng nhn trng thi dng l bao nhiu?

Y =K N 3 s = = 0.11 3 2

Chia c 2 v phng trnh cho N3 Y K N K 3= = 1 N N N3 3 3 1 2 1

K = N

Phng trnh m t s tin trin ca vn trn u cng nhn

K t+1 Ns* 3

K s * = K *K t t t 3 N N NK = N3

trng thi dng v tri = 0, mc vn trn u cng nhn l

K * N3

K s3 * = N3

*3

K N0 .1 = 0 .1 1 3

K S = N

=

b. Mc sn lng trn u cng nhn trng thi dng l bao nhiu?

Y = N

3

K N

3

S =

3

S 0.1 = 1 = 0.1

=

Gi s rng nn kinh t t c trng thi dng trong thi gian t v sau trong thi gian t+1 t l khu hao tng gp i bng 0.20 c. Tm mc vn v sn lng trn u cng nhn trng thi dng mi l bao nhiu. S K1 0 .1 1 = = 0 .2 = 8 N 23 3 1/ 2

= 0.354

Y1 3 K1 3 = = 0.354 = 0.707 N N

d. Tnh cc mc vn v sn lng trn u cng nhn trong 3 thi gian u sau khi thay i t l khu hao. Gi GiK2 K3 K4 ; ; N N N Y2 Y3 Y4 ; ; N N N

l mc vn trn u cng nhn thi gian 1,2,3 l mc sn lng trn u cng nhn thi gian 1,2,3

Ta c:K 2 K1 Y K = S 1 1 = 0.1 * 0.707 0.2 * 0.354 = 0.0707 0.0708 = 0.0001 N N N NK2 = 0.354 0.0001 = 0.3539 N

Vy

K3 K Y = (1 ) 2 + S 2 = (1 0.2) * 0.3539 + 0.1 * 3 0.3539 = 0.28312 + 0.0707 = 0.35382 N N N K Y K4 = (1 ) 3 + S 3 = (1 0.2) * 0.35382 + 0.1 * 3 0.35382 = 0.283056 + 0.07073 = 0.353786 N N N

Bi sa ca Thy

Bi 5: a) Tm gi tr trng thi dng ca: * Lng tn vn trn mi cng nhn hiu dng K/AN Ta c: Y= (K)1/2(NA)1/2 Suy ra: Y/AN= (K/AN)1/2(NA/AN)1/2 = (K/AN)1/2 Mt khc: T l khu hao l 10% , gA= 4% v gN = 2% nn u t phi bng 16% lng tn vn I= (10% + 4%+ 2%) K= 16% K Do : I/AN= sY/AN= s(K/AN)1/2 => 16%K/AN= 16%(K/AN)1/2 => K/AN= (16%/16%)2= 1 Vy lng tn vn trn mi cng nhn hiu dng l 2.56 * Sn lng trn mi cng nhn hiu dng Y/AN= (K/AN)1/2= 11/2= 1 * T l tng trng sn lng trn mi cng nhn hiu dng (Y/AN) trng thi dng, trong nn kinh t ny, yu t khng i l sn lng trn mi cng nhn hiu dng. Do , t l tng trng sn lng trn mi cng nhn hiu dng l 0 * T l tng trng sn lng trn mi cng nhn (Y/N) trng thi dng, t l tng trng sn lng trn mi cng nhn bng t l tin b cng ngh gA= 4% * T l tng trng sn lng, trng thi dng, bng t l tin b cng ngh v t l tng trng cng nhn gA + gN = 4%+ 2%= 6% b) gA= 8% v gN = 6%. Tnh li cu a v gii thch * Lng tn vn trn mi cng nhn hiu dng (K/AN) V T l khu hao l 10% , gA= 8% v gN = 6% nn u t phi bng 24% lng tn vn I= (10% + 8%+ 6%) K= 24% K Do : I/AN= sY/AN= s(K/AN)1/2 => 24%K/AN= 16%(K/AN)1/2

=> K/AN= (16%/24%)2= 0.444 * Sn lng trn mi cng nhn hiu dng (Y/AN) Y/AN= (K/AN)1/2= 0.4441/2= 0.666 * T l tng trng sn lng trn mi cng nhn hiu dng (Y/AN) (0.666-1)/1* 100= -33.4% *Nguyn nhn Sn lng trn mi cng nhn hiu dng gim so vi trc v lc ny ng u t s.f(K/AN)= 16%f(K/AN) thp hn ng u t theo yu cu ( + gA + gN)K/AN= 24%K/AN nn Y/AN s st gim t 1 xung cn 0.666. iu ny cng ng ngha u t thc t thp hn mc u t cn thit duy tr vn trn mi cng nhn hiu dng nn K/AN gim t 1 xung 0.444 * T l tng trng sn lng trn mi cng nhn (Y/N) trng thi dng, t l tng trng sn lng trn mi cng nhn bng t l tin b cng ngh gA= 8% * T l tng trng sn lng, trng thi dng, bng t l tin b cng ngh v t l tng trng cng nhn gA + gN = 8%+ 6%= 14% c) gA= 4% v gN = 6%. Tnh li cu a v gii thch * Lng tn vn trn mi cng nhn hiu dng (K/AN) V T l khu hao l 10% , gA= 4% v gN = 6% nn u t phi bng 24% lng tn vn I= (10% + 4%+ 6%) K= 20% K Do : I/AN= sY/AN= s(K/AN)1/2 => 20%K/AN= 16%(K/AN)1/2 => K/AN= (16%/20%)2= 0.64 * Sn lng trn mi cng nhn hiu dng (Y/AN) Y/AN= (K/AN)1/2= 0.641/2= 0.8 * T l tng trng sn lng trn mi cng nhn hiu dng (Y/AN) (0.8-1)/1* 100= - 20% *Nguyn nhn Sn lng trn mi cng nhn hiu dng gim so vi trc v lc ny ng u t s.f(K/AN)= 16%f(K/AN) thp hn ng u t theo yu cu ( + gA + gN)K/AN= 20%K/AN nn Y/AN s st gim t 1 xung cn 0.8. iu ny cng ng ngha u t thc t thp hn mc u t cn thit duy tr vn trn mi cng nhn hiu dng nn K/AN gim t 1 xung 0.64 * T l tng trng sn lng trn mi cng nhn (Y/N)

trng thi dng, t l tng trng sn lng trn mi cng nhn bng t l tin b cng ngh gA= 4% * T l tng trng sn lng, trng thi dng, bng t l tin b cng ngh v t l tng trng cng nhn gA + gN = 4%+ 6%= 10% * Dn chng khm kh hn trong cu a hay cu c v gii thch? Mc d, nn kinh t c, t l tng trng sn lng l 10% cao hn t l tng trng sn lng nn kinh t a l 6% nhng t l tng trng sn lng trn mi cng nhn hai nn kinh t l nh nhau nn dn chng 2 nn kinh t u khm kh nh nhau. iu ny c ngha, sn lng ca nn kinh t c gia tng l do tng trng dn s, nhng chia u u ngi th mc thu nhp ca h vn nh nhau. Hn na, trong nn kinh t c ang xy ra hin tng mt cn i Bi sa ca Thy

Bi 3: a. Tnh hiu qu ca chnh sch tin t ca cc nc theo sau: Di ch T gi hi oi c nh, ngn hng Trung ng khng th gi tng lng tin khng i. Nu lm nh vy th li sut trong nc s tng cao hn li sut nc ngoi, dn ti ni t tng gi. duy tr t gi hi oi, ngn hng Trung ng phi tng cung tin ph hp vi mc tng trong cu tin, do li sut cn bng khng thay i. Coi mc gi P l cho trc, lng tin danh ngha phi iu chnh. S gia tng tin dn n s gia tng M/P, dn n gia tng sn lng. ng tng cu dch chuyn sang phi, dn ti sn lng cao hn v mc gi cng th. Qua thi gian, s iu chnh nhng k vng v gi bt u c tc dng. Khi thy gi cao hn, nhng ngi nh tin cng yu cu tin cng danh ngha cao hn iu ny dn n gi cao hn. Gi tip tc tng ln. Tng ng nh th, nu sn lng vt qu mc sn lng t nhin, ng tng cung dch chuyn ln trn. Nn kinh t di di chuyn ln dc theo ng tng cu mi. Qu trnh iu chnh chm dt khi sn lng tr v mc sn lng t nhin. Trong trung hn, sn lng tr v mc sn lng t nhin v mc gi cao hn. Nu sn lng tr v mc sn lng t nhin, th tng lng tin thc cng phi tr v gi tr ban u ca n. Ni cch khc, t l gia tng trong cc gi phi bng t l gia tng trong tng lng tin danh ngha. Chnh sch tin t m rng c nhng hn ch sau y: By tin xy ra khi Ngn hng Trung ng tng cung tin nhng li sut khng gim, do u t khng tng m vic tng tin ch gy ra lm pht cao. Hin tng ny xy ra khi cu tin qu nhy i vi li sut hoc u t khng nhy i vi li sut. Nu c mt trong hai hn ch ny th chnh sch tin t km hiu qu. b. Nu tt c cc nc c nh t gi hi oi ca h so vi u n, t gi hi oi ca nc u n c th khng c nh, nhng nhng s thay i l rt him. ngha i vi tnh hiu qu ca chnh sch tin t ca nc u n: mc gi trong nc ca cc nc Theo sau tng nhanh hn mc gi ca cc nc u n, dn ti s tng gi u ng ca cc nc Theo sau. Nh vy, cc nc u n s c u th v tng kh nng cnh tranh mt cch nhanh chng. c. Nu cc nc u u gim cung tin chng lm pht, th cc nc Theo sau phi c s iu chnh t t duy tr t gi hi oi ca h bng cch la chn mt t l gim gi ng tin ca h so vi ng tin cc cc nc u n. Tc ng n nn kinh t ca cc nc theo sau: ng ni t gim gi s lm cho cc doanh nghip xut khu gp nhiu thun li hn khi hng ha sn xut c sc cnh tranh hn, trong khi nhp khu s gim i v cn cn thng mi c ci thin. Dng vn gin tip t tin gii ngn: Mt tc dng quan trng khc l iu chnh t gi s thc y thu ht thm cc dng vn u t gin tip, v gip khi nh

u t nc ngoi t tin hn trong vic gii ngn, nh s n nh t gi c th tin on c trong thi gian ti. ng thi, p lc i vi lm pht khng ln. Nu cc nc Theo sau khng lm g c khi cc nc u n gim cung tin chng lm pht th s dn ti mt s tng gi thc u n v lm cho hng ha ca h mt sc cnh tranh mt cch nhanh chng, u t gim v p lc lm pht tng. Bi sa ca Thy