<p>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 <span class="math-container">$\beta$</span> is lower than <span class="math-container">$1/(1+r)$</span> you will decrease consumption over time as you are more impatient relative to the the interest rate).</p>
<p>Putting <span class="math-container">$F_t$</span> and <span class="math-container">$L_t$</span> to be the consumption of <span class="math-container">$F$</span> and <span class="math-container">$L$</span> at period <span class="math-container">$t$</span>, you need to solve the following problem:
<span class="math-container">$$
\max_{F_0, \ldots, F_T, L_0, \ldots, L_T} \sum_{t = 0}^T \beta^t ((F_t)^{0.25} + (L_t)^{0.75}),\\
\text{ subject to } \sum_{t = 0}^T \frac{L_t + F_t}{(1+r)^t} = \sum_{t = 0}^T \frac{I_t}{(1+r)^t},\\
F_t, L_t \ge 0
$$</span>
The optimal levels of <span class="math-container">$F_t$</span> and <span class="math-container">$L_t$</span> can be found using the usual optimality conditions.</p>
<p>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:
<span class="math-container">$$
V_t(S_t) = \max_{L_t, F_t} \left[(F_t)^{0.25} + (L_t)^{0.75} + \beta V_{t+1}(S_{t+1})\right] \text{ s.t. } S_{t+1} = (1+r)(S_t + I_t - L_t - F_t).
$$</span>
Together with the terminal value <span class="math-container">$V_{T+1}(S_{T+1}) = 0$</span> (and initial value <span class="math-container">$S_0 = 0$</span>).</p>
<p>The functions <span class="math-container">$V_T(.), \ldots V_1(.)$</span> can be solved recursively from period <span class="math-container">$T$</span> to period <span class="math-container">$1$</span> (numerically).</p>

CC BY-SA 4.0
2023-08-01T08:19:58.697Z
2023-08-01T08:19:58.697Z
tdm
<p>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.</p>
<h2>Given</h2>
<ul>
<li>2 goods (will add more later):
<ul>
<li>food: <span class="math-container">$U(F) = F^{0.25}$</span>, F = quantity of food consumed</li>
<li>luxury goods: <span class="math-container">$U(L) = L^{0.75}$</span>, L = quantity of luxury goods consumed</li>
</ul>
</li>
<li>Income I is constant and expected each period</li>
<li>N is number of periods left in earning/consuming life</li>
<li>r is interest rate</li>
<li>β, level of impatience, 0&lt;β&lt;=1</li>
<li>I am calculating monetary spend here, and will convert to quantities (given prices) later</li>
</ul>
<p>I have seen and solved a two-good utility function optimization in a 1 period time-frame <strong>and separately</strong> 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.</p>
<p>Here's what I have done so far:</p>
<hr><br>
Utility
-------
<p>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 &quot;future consumption&quot; utility function for each good, but I'm not confident that I'm doing this correctly, or if it makes sense:</p>
<p><span class="math-container">$U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) = U(F_{current}) + U(F_{future}) + U(L_{current}) + U(L_{future})$</span></p>
<p><span class="math-container">$U(F_{current}) = F_{current}^{0.25}$</span></p>
<p><span class="math-container">$U(L_{current}) = L_{current}^{0.75}$</span></p>
<p><span class="math-container">$U(F_{future}) = \sum_{t=1}^N (β^t)*(F_{future}^{0.25})$</span>
Using sum of finite geometric series formula, <span class="math-container">$S_n = a_1\frac{1−r^n}{(1−r)}$</span>, r≠1
I think I can simplify to:</p>
<ul>
<li><p><span class="math-container">$a_1 = (β^1)(F_{future}^{0.25})$</span></p>
</li>
<li><p><span class="math-container">$r = β$</span></p>
</li>
</ul>
<p><span class="math-container">$U(F_{future}) = \frac{(β^1)(F_{future}^{0.25})(1-β^N)}{(1-β)}$</span></p>
<p><span class="math-container">$U(L_{future}) = \frac{(β^1)(L_{future}^{0.75})(1-β^N)}{(1-β)}$</span></p>
<p>Then,</p>
<p><span class="math-container">$U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) = F_{current}^{0.25} + \frac{β^1(F_{future}^{0.25})(1-β^N)}{(1-β)} + L_{current}^{0.75}  + \frac{β^1(L_{future}^{0.75})(1-β^N)}{(1-β)}$</span></p>
<hr><br>
Budget Constraint
-----------------
<p>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:
<br><br>
<strong>Current period budget constraint</strong></p>
<p><span class="math-container">$F_{current} + L_{current} + S_{current} = I,  S_{current}$</span> = savings of current period
<br><br></p>
<p><strong>Future period budget constraint</strong></p>
<p><span class="math-container">$F_{future} + L_{future} = I_{future} + (1+r)*S_{current}$</span>
<br><br></p>
<p><strong>Value of future income</strong></p>
<p><span class="math-container">$I_{future} = \sum_{t=1}^N \frac{I}{(1+r)^t}$</span> finite geo series --&gt; <span class="math-container">$\frac{I}{1+r}*\frac{1-(\frac{1}{(1+r)})^N}{1-\frac{1}{1+r}}$</span></p>
<br>
**Wolphram Alpha simplifies to**
<p><span class="math-container">$I_{future} = \frac{I-I(1+r)^{-N}}{r}$</span></p>
<p><br><br></p>
<p><strong>eliminate current savings variable</strong></p>
<p><span class="math-container">$S_{current} = I - F_{current} - L_{current}$</span>
<br><br></p>
<p><strong>plug into future budget constraint equation and isolate constants I, N, and r</strong></p>
<p><span class="math-container">$F_{future} + L_{future} = \frac{I-I(1+r)^{-N}}{r} + (1+r)*(I - F_{current} - L_{current})$</span></p>
<p><span class="math-container">$\frac{F_{future}}{1+r} + \frac{L_{future}}{1+r} + F_{current} + L_{current} = \frac{I-I(1+r)^{-N}}{r(1+r)} + I$</span></p>
<hr><br />
Problem Summary
---------------
<p>Max <span class="math-container">$U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) = F_{current}^{0.25} + \frac{β^1F_{future}^{0.25}*(1-β^N)}{1-β} + L_{current}^{0.75}  + \frac{β^1L_{future}^{0.75}(1-β^N)}{1-β}$</span></p>
<p>Subject to budget constraint <span class="math-container">$\frac{F_{future}}{1+r} + \frac{L_{future}}{1+r} + F_{current} + L_{current} = \frac{I-I(1+r)^{-N}}{r(1+r)} + I$</span></p>
<hr><br>
Solution Start
---
Lagrangian
<span class="math-container">$L = U_{globalIntertemporal}(F_{current}, F_{future}, L_{current}, L_{future}) + λ(\frac{F_{future}}{1+r} + \frac{L_{future}}{1+r} + F_{current} + L_{current} - \frac{I-I(1+r)^{-N}}{r(1+r)} - I)$</span>
<p>Will do:</p>
<ul>
<li>find partial derivatives of L with respect to each consumption variable and set equal to λ.</li>
<li>set λ(budget constraint) = 0</li>
<li>solve for variables</li>
</ul>
<p>Are my assumptions and solution on the right track? Should I be doing anything differently?
Thank you!</p>

<utility><optimization><consumption><intertemporal-choice>
Intertemporal Utility Optimization For Multiple Goods
Aidan Loten
# Intertemporal Utility Optimization For Multiple Goods

(Asked by: username_0 on 2023-07-29)

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:

---

  
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,

$U\_{globalIntertemporal}(F\_{current}, F\_{future}, L\_{current}, L\_{future}) = F\_{current}^{0.25} + \frac{β^1(F\_{future}^{0.25})(1-β^N)}{(1-β)} + L\_{current}^{0.75} + \frac{β^1(L\_{future}^{0.75})(1-β^N)}{(1-β)}$

---

  
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**

$F\_{future} + L\_{future} = \frac{I-I(1+r)^{-N}}{r} + (1+r)\*(I - F\_{current} - L\_{current})$

$\frac{F\_{future}}{1+r} + \frac{L\_{future}}{1+r} + F\_{current} + L\_{current} = \frac{I-I(1+r)^{-N}}{r(1+r)} + I$

---

  
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 $\frac{F\_{future}}{1+r} + \frac{L\_{future}}{1+r} + F\_{current} + L\_{current} = \frac{I-I(1+r)^{-N}}{r(1+r)} + I$

---

  
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!

**Tags:** utility, optimization, consumption, intertemporal-choice


---

## 1 answers

### Answer by username_1 on 2023-08-01. Score: 3 | Accepted answer

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).


---

