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# Cake Eating in Finite and Infinite Time

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Cake Eating in Finite and Infinite Time

The following was implemented in Maple by Marcus Davidsson (2008) davidsson_marcus@hotmail.com

and is based upon the work by Adamek (2006) The Cake-Eating Problem

1) Cake Eating in Finite Time

Introduction

We assume that we have a cake that in the first period has a size of  CS(1)

 (1)

We now assume that we have five time periods. We assume that we consume a fraction of the cake in each time period denoted by C(1), C(2), C(3), C(4).

Note that we assume that the cake is completly gone in the fifth time period which means that our consumption in period five is zero

The equation of motion for the cake size in each periods are given by

We will now derive the expressions for consumption in the four periods

Period-1

The cake size in the last period is given by

 (2)

where C(1), C(2), C(3) and C(4) are the amount of consumtion in each period

We again assume that all cake is gone in the last period which means that CS(5)=0.

If we plug this into the previous equation then we get the expression for the terminal condition in period one

 (3)

We can now express this equation in C(1) terms only.

Note that we discount consumption over time where B^t is a discount factor. Note that B=1/(1+r) where r is the interest rate.

Note that since B^t < 1 it means that consumption will be decreasing over time

If we plug the all the rewritten expressions into the previous equation we get

 (4)

We now solve for CS(1)

 (5)

We now note that

We know that

 (6)

We now multiply both sides by B so we get

 (7)

Which can be written as

 (8)

Now we know that X and Y are given by

 (9)

 (10)

We now subtract BCS(initial) from both sides of CS(initial) so we get

 (11)

Which can be written as

 (12)

We now divide both sides by (1-B) so we get

 (13)

We can now solve for C(1)

 (14)

Which is the expression for the optimal consumtion in period one

Period-2

Our terminal condition for period two is given by

 (15)

We can now express this equation in C(2) terms only.

If we plug that into the previous equation we get

 (16)

We now solve for CS(2)

 (17)

We now note that

We know that

 (18)

We now multiply both sides by B so we get

 (19)

Which can be written as

 (20)

Now we know that X and Y are given by

 (21)

 (22)

We now subtract BCS(2) from both sides of CS(2) so we get

 (23)

Which can be written as

 (24)

We now divide both sides by (1-B) so we get

 (25)

We can now solve for C(2)

 (26)

Which is the expression for the optimal consumtion in period two

Period-3

Our terminal condition for period three is given by

 (27)

We can now express this equation in C(3) terms only.

If we plug that into the previous equation we get

 (28)

We now solve for CS(3)

 (29)

We now note that

We know that

 (30)

We now multiply both sides by B so we get

 (31)

Which can be written as

 (32)

Now we know that X and Y are given by

 (33)

 (34)

We now subtract BCS(3) from both sides of CS(3) so we get

 (35)

Which can be written as

 (36)

We now divide both sides by (1-B) so we get

 (37)

We can now solve for C(3)

 (38)

Which is the expression for the optimal consumtion in period three

Period-4

Our terminal condition for period three is given by

 (39)

The above equation is already express in only C(4) terms which means that we can solve for consumption directly

 (40)

Which is the expression for the optimal consumtion in period four

Consumtion Dynamics Over Time

We can now visualize the dynamics of consumtion over time

We first note that the equations of motion for the cake size in period two and three are given by

 (41)

 (42)

 (43)

This means that the equation of motions can be written as

 (44)

 (45)

 (46)

 (47)

We now assume that

Which means that the amount of consupmtion in each period are given by

 (48)

 (49)

 (50)

 (51)

We first make sure that all cake is consumed

 (52)

We can now plot the consumption over time

We can now plot the cake size over time

Alternative Formulation

Note that we could have solved the above problem in a much more straight forward way

We assume that our utility from cake consumption at time t is given by

 (53)

We now note that we discount utility from cake consumption over time.

The discounting is done through a discount factor B^t where B=1/(1+r) and r is the interest rate. Note that since B^t < 1 it means that will be decreasing over time.

 (54)

We now note that our objective is to maximize the sum of the discounted utility form consumption over time as seen below

 (55)

If we plug in the previous expression in this equation we get

 (56)

In our case our objective function Ob is therefor given by

 (57)

 (58)

We again assume that

This gives us

 (59)

Which is the same values of , , and we had before

2) Cake Eating in infinite Time

The cake size in the end for a n period example is given by

 (60)

We again assume that all cake is gone in the last period which means that CS(n+1)=0.

If we plug this into the previous equation then we get the expression for the terminal condition in period one

 (61)

We can now express this equation in C(1) terms only.

If we plug the all the rewritten expressions into the previous equation we get

 (62)

which can be written as

 (63)

 (64)

We now note that

##########################################

if we assume that

then the sum

 (65)

can be approximated by

 (66)

##########################################

The sum of   can therefor be approximated by the expression

 (67)

which means that we can write the above expression as

 (68)

which gives us

 (69)

which means that that consumption is given by

 (70)

We now note that

##########################################

when then

We can show this by noting that

 (71)

##########################################

This means that our previous equation is reduced to

 (72)

We can now visualize the consumtion and cake size over time

Consumption vector

 (73)

Cake size vector

 (74)

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