経済研究の基礎(ミクロ経済学)—章末問題解答例

()—
2012
2 1
3 4
4 7
5 9
6 12
1

2
1 :
•
• (
)
(
)
• ETC 1000
()
ETC
•
(peak-load pricing )
2
()
3 (A , B ) (4, 0), (3, 1), (2, 2), (1, 3), (0, 4)
A B 2 C(2, 2) A B 3 1
B
4
3
2
1
C
E
D
O
1 2 3 4
A
1

D(3, 1) A 2 B 1 E(2, 1) E
E ()
E C D
(
() )
4
( M. Friedman
)
5
()
6 A B A B
B A (B )
A
(
A B
)
7
(p.14)
(
)
(
)
2

8
a: (D → D ′ )
(E 0 →E 1 ) (
)
i. () a b c
( a = b + c)b
() P 0
ii. E ′ (E 0 E ′ ) P 0
P 1 c
E ′ E 1 ()
b: c
(a) c E ′ E 1
E 0
c: E ′ E 1
E 1
P
S
P 0
O
b
E ′ c
a
P 1
E 1
D ′ E 0
D
3

3
1 400 3 400×3 = 1200
(1) 600 + 500 = 1100 ()
400
( 2.6 ) 400 () 400
1200 1100 + 400 = 1500
(2) 1200 3
600 + 500 + 350 = 1450 3 400
400
(
——)
2
A B
600
500
A
B
400
350
A
B
A
B
1 2 3 4 5 6
(p.25 3.5 ) 17
(
)
3 (i) A
()
A (q 0 −→ q 1 )
4

P
A
P 0
E 0
d ′ :A
P 1
d:
q 0 q q ′ 1
1
(ii) A A
A (q 0 −→ q ′ 1 )
A (ii) d ′ (i) d
A (ii)
4
()
()
() ()
(
)
5( a d 4
) a. b.
c. (a c)
b. d.
d. (d c)
(a c)
(d c)
()
6 Y SL Y
(
4 )
a4
Y 1 250 SL
5

250 Y Y 4
600 + 500 + 400 + 300 = 1800 250
800 ( 1800 4 1000 )
b4 4 1000 5
5 1100 4 5
5 1100 − 1000 = 100 (Y
4 5
) 5 200
Y 5
5
(5 1100 5 1 1100/5 =
220 5 4 5 1
100 220
4 5 5
5 1 )
c(5
5 5
4 )
Y 5 5
600 + 500 + 400 + 300 + 200 = 2000 5 1100
2000 − 1100 = 900
900 500
900 − 500 = 400 Y
500 SL
5
d()
()
Y
SL 600
(
) Y SL
6

4
1
4 (
)
2
()
3 2000 ()
3000
2000 ()
5000
() 3000
2000
4
( ()
)
5 2
7

()
(
() )
6 ()
Q 0 ΔQ
(= ) Q 0 (
) ()
Q 0
ΔQ
Q 0
ΔQ
()
7
8

5
1
a
()
b
——
()
()
2()
()
(
) (
derived demand )
()
(
)
()
3 1
1
(
1
1 )
() 1 (—
—)
1 1
9

P
S
P 0
1
E
E ′
S ′
1
1
D
P 1
Q 0
O
D ′
Q
1 1
() 1 (
1 )
1
E E ′ (Q 0 ) () (P 0 )
1 (P 1 ) 1
1
4a 800 S ′
F 1
6 G 4
r 2 800
R 1600 ( FGJH )( ()
() S FH 800
)
5 EFHI 900
FKIH (800 )
EKG (100 ) 700
( deadweight loss 200 )
b r 2 2
2 (S ′′ ) ( 2 )
E ′ 4 1200
( E ) 1100 (IEE ′ J )
1100 ( 2 1
S 1
HIEL )
10

()
S ′ F
7
6
H
L
S
2
S ′′
5
I
K
E
4
J
G
E ′
3
D
O
6 8 10 12 14
(100 /)
2400
200
)
5
()
(
)
()
11

6
1. (6.1 )
2.a
Case A
Case B
bCase A = 4/6 = 2/3. = 6/4 = 3/2.
= 2/7. = 7/2.
(2/3 > 2/7)
(3/2 < 7/2)
Case B = 3/4. = 4/3.
= 1/3. = 3/1 = 3.
(3/4 > 1/3)
(4/3 < 3)
c$1 = e
Case A = 400 = 600 = 2e
= 7e
400 ≥ 2e, 600 ≤ 7e. 200 ≥ e ≥ 600/7 = 85.7
Case B = 300 = 400 = e
= 3e
300 ≥ e, 400 ≤ 3e. 300 ≥ e ≥ 400/3 = 133.3
3. 1 1
1 1 2
1 2 1
1/2 1 4/3
1 3/4
4.
12

(
)
5.
1
() 2
(
)
6.
()
7.
5
()
8. 2
2
13

2 case A
2 case B
9. ( p.78
)
(
)
10.
()
()
14