An individual receives an exogenous endowment of cake, denoted So, at time t = 0. You...

  

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An individual receives an exogenous endowment of cake, denoted So, at time t = 0. You know that this individual lives for T periods and has a flow prefer- ence u(c), where c is the size of the slice of cake she consumes. In other words, consuming a piece of cake of size ct generates u(ct) utils at time t. The problem of this individual is to maximize her lifetime utility: T Σβtu(c), max {ct}=0 t=0 (1) where 3 € (0, 1) is the factor by which she discounts future utility streams, i.e. she is impatient. The cake doesn't shrink or grow while it is in the fridge. The only change in the size of the cake is due to consumption. The law of motion for St is therefore: St+1 = St - Ct. (2) Without loss of generality, we can normalize the initial size of the cake to So 1. One interpretation of this normalization is that you start with a whole cake, i.e. you have 100 percent of the cake at the beginning. If the cake has size 0.8 one period later, this means that you consumed 1 - 0.8 = 0.2 = 20% between t = 0 and t = 1. If the size drops to 0.7 after that, you consumed 0.8 -0.7 = 0.1 = 10% of the original size of the cake in the second period. Finally, we will assume that u(c) = ln c and 3 = 0.96. Now we have all the ingredients to set up and solve the dynamic program asso- ciated with this cake-eating problem. 1. Open a spreadsheet and define the state space of the problem for the current and the next period, i.e. write down all possible values of S = {0, 0.01, 0.02, ..., 0.99, 1} in the second row of the first tab and the same values in the first column of the tab. You should be able to simply "recycle" the layout we used in class. To begin with, calculate the value of having a cake of size ST at the start of the final period T: such that VT(ST) = maxcr In CT + BVT+1(ST+1) (3) ST-CT 20 Since this is the final period T, the value of having any cake in the fridge at time T + 1 is zero: VT+1(ST+1) = 0. There is no incentive to keep any left-overs. You eat whatever amount of cake you have at the beginning of period T, i.e. CT ST. Expressed in utils, the value of having a cake of size ST is VT(ST) In ST. = = - Importantly, you still have to compute the value of all possible combina- tions of ST ST+1 = CT> 0. The flow utility from a negative consump- tion stream is, of course, not defined. To deal with this, you need to as- sign a negative value to all cr ≤0. One way to handle this is to use u (ST - ST+1) = ln (max{ST - ST+1, 10-100}). [2 points] Now you know how valuable it is to have a cake of any size in the final period of the model and the final step is to identify the optimal choice of ST+1 for each possible ST. For each column (ST), select the row (ST+1) with the highest value. [2 points] An individual receives an exogenous endowment of cake, denoted So, at time t = 0. You know that this individual lives for T periods and has a flow prefer- ence u(c), where c is the size of the slice of cake she consumes. In other words, consuming a piece of cake of size ct generates u(ct) utils at time t. The problem of this individual is to maximize her lifetime utility: T Σβtu(c), max {ct}=0 t=0 (1) where 3 € (0, 1) is the factor by which she discounts future utility streams, i.e. she is impatient. The cake doesn't shrink or grow while it is in the fridge. The only change in the size of the cake is due to consumption. The law of motion for St is therefore: St+1 = St - Ct. (2) Without loss of generality, we can normalize the initial size of the cake to So 1. One interpretation of this normalization is that you start with a whole cake, i.e. you have 100 percent of the cake at the beginning. If the cake has size 0.8 one period later, this means that you consumed 1 - 0.8 = 0.2 = 20% between t = 0 and t = 1. If the size drops to 0.7 after that, you consumed 0.8 -0.7 = 0.1 = 10% of the original size of the cake in the second period. Finally, we will assume that u(c) = ln c and 3 = 0.96. Now we have all the ingredients to set up and solve the dynamic program asso- ciated with this cake-eating problem. 1. Open a spreadsheet and define the state space of the problem for the current and the next period, i.e. write down all possible values of S = {0, 0.01, 0.02, ..., 0.99, 1} in the second row of the first tab and the same values in the first column of the tab. You should be able to simply "recycle" the layout we used in class. To begin with, calculate the value of having a cake of size ST at the start of the final period T: such that VT(ST) = maxcr In CT + BVT+1(ST+1) (3) ST-CT 20 Since this is the final period T, the value of having any cake in the fridge at time T + 1 is zero: VT+1(ST+1) = 0. There is no incentive to keep any left-overs. You eat whatever amount of cake you have at the beginning of period T, i.e. CT ST. Expressed in utils, the value of having a cake of size ST is VT(ST) In ST. = = - Importantly, you still have to compute the value of all possible combina- tions of ST ST+1 = CT> 0. The flow utility from a negative consump- tion stream is, of course, not defined. To deal with this, you need to as- sign a negative value to all cr ≤0. One way to handle this is to use u (ST - ST+1) = ln (max{ST - ST+1, 10-100}). [2 points] Now you know how valuable it is to have a cake of any size in the final period of the model and the final step is to identify the optimal choice of ST+1 for each possible ST. For each column (ST), select the row (ST+1) with the highest value. [2 points]

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