Intertemporal Utility Optimization For Multiple Goods

Question

I'm building an economic simulation game and I'm trying to solve for the values that a person will spend on each good and the amount they will save in the current period, taking into account all future consumption periods remaining in their life.

Given

• 2 goods (will add more later):
  • food: $U(F) = F^{0.25}$, F = quantity of food consumed
  • luxury goods: $U(L) = L^{0.75}$, L = quantity of luxury goods consumed
• Income I is constant and expected each period
• N is number of periods left in earning/consuming life
• r is interest rate
• β, level of impatience, 0<β<=1
• I am calculating monetary spend here, and will convert to quantities (given prices) later

I have seen and solved a two-good utility function optimization in a 1 period time-frame and separately a same-good, 2-period intertemporal utility function optimization, but have never seen them combined, which is how I am attempting to solve this problem.

Here's what I have done so far:

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Utility -------

I am combining two-good utility optimization and same-good, two-period intertemoral utility optimization. I am keeping two periods by consolidating all future consumption periods into a present value "future consumption" utility function for each good, but I'm not confident that I'm doing this correctly, or if it makes sense:

$U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) = U(F_{current}) + U(F_{future}) + U(L_{current}) + U(L_{future})$

$U(F_{current}) = F_{current}^{0.25}$

$U(L_{current}) = L_{current}^{0.75}$

$U(F_{future}) = \sum_{t=1}^N (β^t)*(F_{future}^{0.25})$ Using sum of finite geometric series formula, $S_n = a_1\frac{1−r^n}{(1−r)}$, r≠1 I think I can simplify to:

• $a_1 = (β^1)(F_{future}^{0.25})$

• $r = β$

$U(F_{future}) = \frac{(β^1)(F_{future}^{0.25})(1-β^N)}{(1-β)}$

$U(L_{future}) = \frac{(β^1)(L_{future}^{0.75})(1-β^N)}{(1-β)}$

Then,

$U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) = F_{current}^{0.25} + \frac{β^1(F_{future}^{0.25})(1-β^N)}{(1-β)} + L_{current}^{0.75} + \frac{β^1(L_{future}^{0.75})(1-β^N)}{(1-β)}$

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Budget Constraint -----------------

In a two-period 1 good intertemporal consumption utility optimization, there are two budget contraints, one for each period, that become consolidated into 1. So here I will do the same:

Current period budget constraint

$F_{current} + L_{current} + S_{current} = I, S_{current}$ = savings of current period

Future period budget constraint

$F_{future} + L_{future} = I_{future} + (1+r)*S_{current}$

Value of future income

$I_{future} = \sum_{t=1}^N \frac{I}{(1+r)^t}$ finite geo series --> $\frac{I}{1+r}*\frac{1-(\frac{1}{(1+r)})^N}{1-\frac{1}{1+r}}$

**Wolphram Alpha simplifies to**

$I_{future} = \frac{I-I(1+r)^{-N}}{r}$

eliminate current savings variable

$S_{current} = I - F_{current} - L_{current}$

plug into future budget constraint equation and isolate constants I, N, and r

$F_{future} + L_{future} = \frac{I-I(1+r)^{-N}}{r} + (1+r)*(I - F_{current} - L_{current})$

$\frac{F_{future}}{1+r} + \frac{L_{future}}{1+r} + F_{current} + L_{current} = \frac{I-I(1+r)^{-N}}{r(1+r)} + I$

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Problem Summary ---------------

Max $U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) = F_{current}^{0.25} + \frac{β^1F_{future}^{0.25}*(1-β^N)}{1-β} + L_{current}^{0.75} + \frac{β^1L_{future}^{0.75}(1-β^N)}{1-β}$

Subject to budget constraint $\frac{F_{future}}{1+r} + \frac{L_{future}}{1+r} + F_{current} + L_{current} = \frac{I-I(1+r)^{-N}}{r(1+r)} + I$

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Solution Start --- Lagrangian $L = U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) + λ(\frac{F_{future}}{1+r} + \frac{L_{future}}{1+r} + F_{current} + L_{current} - \frac{I-I(1+r)^{-N}}{r(1+r)} - I)$

Will do:

• find partial derivatives of L with respect to each consumption variable and set equal to λ.
• set λ(budget constraint) = 0
• solve for variables

Are my assumptions and solution on the right track? Should I be doing anything differently? Thank you!

Answer 1

I think you are implicitly assuming that the level of consumption from period 2 onwards will remain constant. In general, this will not be the case (e.g. if $\beta$ is lower than $1/(1+r)$ you will decrease consumption over time as you are more impatient relative to the the interest rate).

Putting $F_t$ and $L_t$ to be the consumption of $F$ and $L$ at period $t$, you need to solve the following problem: $$ \max_{F_0, \ldots, F_T, L_0, \ldots, L_T} \sum_{t = 0}^T \beta^t ((F_t)^{0.25} + (L_t)^{0.75}),\\ \text{ subject to } \sum_{t = 0}^T \frac{L_t + F_t}{(1+r)^t} = \sum_{t = 0}^T \frac{I_t}{(1+r)^t},\\ F_t, L_t \ge 0 $$ The optimal levels of $F_t$ and $L_t$ can be found using the usual optimality conditions.

If you have a lot of time periods, it might be better to approach the problem using dynamic programming. In your case, the Bellman equation takes the form: $$ V_t(S_t) = \max_{L_t, F_t} \left[(F_t)^{0.25} + (L_t)^{0.75} + \beta V_{t+1}(S_{t+1})\right] \text{ s.t. } S_{t+1} = (1+r)(S_t + I_t - L_t - F_t). $$ Together with the terminal value $V_{T+1}(S_{T+1}) = 0$ (and initial value $S_0 = 0$).

The functions $V_T(.), \ldots V_1(.)$ can be solved recursively from period $T$ to period $1$ (numerically).