Lecture Cobweb)

1 Instructor Dr Rehana Naz Mathematical Economics I

Lecture 24

Section 17.4 from Fundamental methods of Mathematical Economics, McGraw Hill 2005, 4th Edition. By A. C. Chiang & Kevin Wainwright is covered. The Cobweb model Read details from book. Question: The cobweb model is essentially based on the static market model in which 푄 = 푄 . What economic assumption is the dynamizing agent in the present case? Explain. Answer: In the cobweb model, the supply and demand functions are of the form

푄 =∝ −훽푃 (훼,훽 > 0) (1) 푄 = −훾 + 훿푃 (훾,훿 > 0) (2)

To get first-order difference equation representing the cobweb model, we assume that in each time period the market price is always set at a level which clears the market i.e

푄 = 푄 (3) Now 푄 = 푄 , assumption is same as we do in static analysis. The dynamizing agent is the lag in the supply function. This introduces 푃 term into the model, which together with 푃 , forms a pattern of change. Interpret the solution of cobweb model given as follows:

푃 = (푃 − 푃) −훿훽 + 푃

Note that 푏 = − < 0 implies that the time path is oscillatory. Since

|푏| = −훿훽 =

훿훽

There can be three types of oscillations, explosive, uniform and damped depending on |푏| ⋛ 1.

(i) If |푏| = > 1 ⇒ 훿 > 훽, then time path is divergent and oscillations are

explosive. (ii) If |푏| = = 1 ⇒ 훿 = 훽, then time path is divergent and oscillations are uniform.

(iii) If |푏| = < 1 ⇒ 훿 < 훽, then time path is convergent and oscillations are

damped.

Example 1: Given demand and supply for the cobweb model as follows,


푄 = 18− 3푃 (4)

푄 = −3 + 4푃 (5)

(a) Assuming that in each time period the market price is always set at a level which clears the market find the time path 푃 . At t=0 푃 .

(b) Find the intertemporal equilibrium price. (c) Determine whether the equilibrium is stable. (d) Find the time path of Q and analyze the condition for its convergence.

Solution: (a) Assuming that in each time period the market price is always set at a level which clears

the market yields 푄 = 푄 (6)

Using (4)and (5)in equation (6), we have 18− 3푃 = −3 + 4푃

3푃 + 4푃 = 21 (7)

Letting 푡 → 푡 + 1 in (7), we have

3푃 + 4푃 = 21 Or

푃 +43푃 = 7 (8)

This is first-order linear difference equation with 푎 = , c=7, its solution is

푃 = 퐴(−푎) +푐

1 + 푎

[You can use above formula directly. If you want to do all steps in finding complementary and particular solutions you can solve by general solution method]

푃 = 퐴(−43) + 3 (9)

At t=0 푃 , (9) yields

푃 = 퐴 + 3 ⇒ 퐴 = 푃 − 3 (10)

Using (10) in (9), we have

푃 = (푃 − 3)(−43) + 3 (11)

(b) To find intertemporal equilibrium price set 푃 = 푃 = 푃 in the demand and supply functions and equate them i.e


18− 3푃 = −3 + 4푃 ⇒ 푃 = 3

푃 = 3 is the intertemporal equilibrium price.

(c) We need to check whether the equilibrium is stable. From (11) 푏 = − <0, the time path is oscillatory. But since

|푏| = −43 =

43 > 1

The time path is divergent and oscillations are explosive. (d) Substitution of the time path (11) into the demand equation (4) leads to the time path of

푄 , which we can simply write as 푄 (since 푄 = 푄 by the equilibrium condition):

푄 = 18 − 3 (푃 − 3)(−43) + 3

푄 = 18− 3(푃 − 3)(−43) − 3(3)

표푟

푄 = 9 − 3(푃 − 3)(−43) (12)

Convergence of 푄 depends on the (− ) term, which determines the convergence of 푃 푎푠 푤푒푙푙. Thus 푃 and 푄 must be either both convergent or both divergent.

As for this case the time path 푃 is divergent so 푄 is also divergent and oscillations are explosive.


푄 = 22− 3푃 (13) 푄 = −2 + 푃 (14)

(a) Assuming that in each time period the market price is always set at a level which clears the market find the time path 푃 .


Solution: (a) Assuming that in each time period the market price is always set at a level which

clears the market yields 푄 = 푄 (15)

Using (13)and (14)in equation (15), we have 22− 3푃 = −2 + 푃


3푃 + 푃 = 24 (16) Letting 푡 → 푡 + 1 in (16), we have

3푃 + 푃 = 24 Or

푃 +13푃 = 8 (17)

This is first-order linear difference equation with 푎 = , c=8, its solution is

푃 = 퐴(−푎) +푐

1 + 푎

푃 = 퐴(−13) + 6 (18)


푃 = 퐴 + 6 ⇒ 퐴 = 푃 − 6 (19)


푃 = (푃 − 6)(−13) + 6 (20)


22− 3푃 = −2 + 푃 ⇒ 푃 = 6


(c) We need to check whether the equilibrium is stable. From (20) 푏 = − <0, the time path is oscillatory. But since

|푏| = −13 =

13 < 1

The time path is convergent and oscillations are damped. (d) Substitution of the time path (20) into the demand equation (13) leads to the time path

of 푄 , which we can simply write as 푄 (since 푄 = 푄 by the equilibrium condition):

푄 = 22 − 3 (푃 − 6)(−13) + 6

푄 = 4 − 3(푃 − 6)(−13) (21)

Convergence of 푄 depends on the (− ) term, which determines the convergence of 푃 푎푠 푤푒푙푙. Thus 푃 and 푄 must be either both convergent or both divergent.


As for this case the time path 푃 is convergent so 푄 is also convergent and oscillations are damped.


푄 = 19 − 6푃 (22) 푄 = 6푃 − 5 (23)

(a) Assuming that in each time period the market price is always set at a level which clears the market find the time path 푃 .


Solution:

(e) Assuming that in each time period the market price is always set at a level which clears the market yields

푄 = 푄 (24) Using (22)and (23)in equation (24), we have

19 − 6푃 = −5 + 6푃

푃 + 푃 = 4 (25) Letting 푡 → 푡 + 1 in (25), we have

푃 + 푃 = 4 (26) This is first-order linear difference equation with 푎 = 1, c=4, its solution is

푃 = 퐴(−푎) +푐

1 + 푎

푃 = 퐴(−1) + 2 (27)


푃 = 퐴 + 2 ⇒ 퐴 = 푃 − 2 (28)


푃 = (푃 − 2)(−1) + 2 (29)


19− 6푃 = −5 + 6푃 ⇒ 푃 = 2



(c) We need to check whether the equilibrium is stable. From (29) 푏 = −1<0, the time path is oscillatory. But since

|푏| = |−1| = 1 The time path is divergent and oscillations are uniform.

(d) Substitution of the time path (29) into the demand equation (22) leads to the time path of 푄 , which we can simply write as 푄 (since 푄 = 푄 by the equilibrium condition):

푄 = 19 − 6[(푃 − 2)(−1) + 2 ] 푄 = 7 − 6(푃 − 2)(−1) (30)

As for this case the time path 푃 is divergent so 푄 is also divergent.

Example 4: If

푄 =∝ −훽푃 (훼,훽 > 0) (31)

푄 = −훾 + 훿푃∗ (훾,훿 > 0) (32) where

푃∗ = 푃∗ + 휂(푃 − 푃∗ ), 0 < 휂 ≤ 1 (33) 푄 = 푄 (34)

What happens if 휂 takes its maximum value? Can we consider the cobweb model as a special case of the present model? Solution: Maximum value of 휂 is 1. Taking 휂 = 1 in (33), we have

푃∗ = 푃∗ + 1(푃 − 푃∗ ) ⇒ 푃∗ = 푃 and the model reduces to the cobweb model. Thus the present model includes the cobweb model as a special case.

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