Solving utility-maximization selection problems with Multinomial Logit demand: Is the First-Choice model a good approximation?

Abstract

We investigate First-Choice (FC) assignment models, a simple type of choice model where customers are allocated to their highest utility option, as a heuristic or starting point for the Multinomial Logit (MNL) model in the context of selection problems with a utility maximization objective. This type of problem occurs in a variety of applications, from location problems to assortment planning or transportation planning. FC assignment models are less refined but computationally more tractable than the more commonly used MNL. MNL suffers from tractability issues due to its nonlinear structure when used within a large size optimization problem with binary decision variables. We design the first comparison of the two modeling frameworks in a context of customer utility maximization for selection problems with binary variables. We provide a probabilistic analysis of the expected customer choice probabilities, document the computational challenges faced by the MNL model in our setting and show in numerical experiments that the FC model exhibits excellent performance as an approximation of the MNL model with an average gap for instance of at most 2.2% for uniformly distributed utilities and of at most 1.4% for normally distributed utilities (and below 1% in a majority of test cases). The key contribution of this paper is to build the case for the FC model as a tractable, high-quality approximation of the MNL model for binary selection problems with utility maximization.

Appendices

A Proof of Lemma 1

Proof

We are looking for $\max _{(v_1,v_2) \in [a,b]^2, v_1\le v_2} f(v_1,v_2) = \dfrac{e^{v_1} +e^{v_2} }{e^{v_1}+e^{v_2} \frac{v_2}{v_1}}$.

Let us note $x=v_1$ and $y=v_2$ (although notation y is already associated with the y-vector solution, we use $y=v_2$ for this lemma only, in order to ease notation). Therefore, we consider the function:

$$\begin{aligned} f(x,y)= \frac{e^x +e^y }{e^x+e^y \frac{y}{x}} \end{aligned}$$

To check whether there exists a candidate point strictly inside (a, b) for being a local maximum of f, we apply the first-order conditions:

$$\begin{aligned} \frac{\delta f(x,y)}{\delta x}= & {} \frac{e^{x}(e^{x}+e^{y} \frac{y}{x})- (e^{x}+e^{y})(e^{x} -\frac{y}{x^2}e^{y})}{( e^{x} + e^{y} \frac{y}{x} )^2} = 0 \end{aligned}$$

(14)

$$\begin{aligned} \frac{\delta f(x,y)}{\delta y}= & {} \frac{e^{y}(e^{x}+e^{y} \frac{y}{x})- (e^{x}+e^{y})(y e^{y} +e^{y})\frac{1}{x}}{( e^{x} + e^{y} \frac{y}{x} )^2} =0 \end{aligned}$$

(15)

Summing (14) and (15) and factorizing by $(e^{x}+e^{y})$, we get $e^y(-\frac{1}{x} + \frac{y}{x^2})=0$, therefore $y=x$. Plugging that into Eq. (15) we obtain $1/x=0$ which is impossible. Therefore the point $(x^*,y^*)$ maximizing f(x, y) with $x\ge y$ cannot be strictly inside interval (a, b), i.e., it is necessarily a point (b, y) or (x, a).

We now prove that the maximum of f is obtained for a point (x, a). For this, we prove that for any point (b, y), $y \in [a,b]$, there exists a point (x, a) such that $f(x,a) \ge f(b,y)$. We take $x=a+b-y$ which is indeed in [a, b] if $y \in [a,b]$. We show that

$$\begin{aligned}&f(a+b-y,a)-f(b,y)=\frac{e^{a+b-y} +e^{a} }{e^{a+b-y}+e^{a} \frac{a}{a+b-y}} - \frac{e^{b} +e^{y} }{e^{b}+e^{y} \frac{y}{b}}\\&\quad = \frac{e^{b-y}+1}{e^{b-y}+ \frac{a}{a+b-y}} - \frac{e^{b-y} + 1 }{e^{b-y}+ \frac{y}{b}} \ge 0 \end{aligned}$$

Indeed, the last expression above is positive if and only if

$$\begin{aligned} \frac{a}{a+b-y} \le \frac{y}{b}\Longleftrightarrow & {} ab \le ay + by - y^2 \\\Longleftrightarrow & {} a(b-y) \le y(b-y)\\\Longleftrightarrow & {} a \le y \end{aligned}$$

which is true. Therefore, we have for any $y \in [a,b]$, $f(a+b-y,a) \ge f(b,y)$ and we conclude that there always exists a point (x, a) which is optimal for f. $\square$

B Proof of Proposition 1

Proof

Let f(x, y) be the joint density function of two uniform random variables:

$$\begin{aligned} E(\max (X_{1j},X_{2j}))= & {} \int _a^b \left( \int _{a}^b f(x,y) \frac{\max (e^x,e^y)}{e^x+e^y} dy \right) dx \\= & {} \frac{1}{(b-a)^2} \int _{a}^b \left( \int _a^x \frac{e^x}{e^x+e^y} dy+ \int _{x}^b \frac{e^y}{e^x+e^y} dy \right) dx \\= & {} \frac{1}{(b-a)^2} \int _{a}^b \left( \int _a^x \left( 1- \frac{e^y}{e^x+e^y}\right) dy+ \int _{x}^b \frac{e^y}{e^x+e^y} dy \right) dx \\= & {} \frac{1}{(b-a)^2} \int _{a}^b \left( x-a - \int _a^x \frac{e^y}{e^x+e^y} dy+ \int _{x}^b \frac{e^y}{e^x+e^y} dy \right) dx \\= & {} \frac{1}{2} + \frac{1}{(b-a)^2} \int _a^b \left( -[\ln (e^x+e^y)]^x_a + [\ln (e^x+e^y)]^b_x \right) dx \\= & {} \frac{1}{2} + \frac{1}{(b-a)^2} \int _a^b \left( \ln (e^x+e^a)+ \ln (e^x+e^b) - 2 \ln (2e^x) \right) dx\\= & {} \frac{1}{2} + \frac{1}{(b-a)^2} \int _a^b \left( \ln \left( \frac{1}{2}+\frac{1}{2} e^{a-x}\right) + \ln \left( \frac{1}{2}+\frac{1}{2} e^{b-x}\right) \right) dx \end{aligned}$$

$\square$

C Proof of Proposition 2

Proof

The goal is to find an approximation of $\int _a^b \left( \ln (\frac{1}{2}+\frac{1}{2} e^{a-x})+ \ln (\frac{1}{2}+\frac{1}{2} e^{b-x}) \right) dx$.

For a twice-differentiable function f, the Taylor formula gives:

$$\begin{aligned} f(x) = f(x_0) + (x-x_0) f'(x_0) + \frac{(x-x_0)^2}{2} f''(x_0) + \frac{(x-x_0)^3}{6} f'''(c) \end{aligned}$$

(16)

where c is between x and $x_0$. Let us apply (16) to $f_{x_0}(x) = \ln (\frac{1}{2}+\frac{1}{2} e^{x_0-x})$ with first (i) $x_0=a$, and then (ii) $x_0=b$. We have that

• $f_{x_0}(x_0)=\ln (1)=0$,

• $f'_{x_0}(x)= \dfrac{-\frac{1}{2}e^{x_0-x} }{\frac{1}{2}+\frac{1}{2} e^{x_0-x}}=-\dfrac{e^{x_0}}{e^x+e^{x_0}}$ so $f'_{x_0}(x_0)=-\frac{1}{2}$,

• $f''_{x_0}(x) = \dfrac{e^{x_0-x}}{(1+e^{x_0-x})^2}$ so $f''_{x_0}(x_0)=\frac{1}{4}$,

• $f'''_{x_0}(x)= \dfrac{e^{x_0+x}(e^{x_0}-e^x)}{(e^{x_0}+e^x)^3}$ so $f'''_{x_0}(x_0)=0$,

• $f''''_{x_0}(x) = \dfrac{e^{x_0+x}(-4 e^{x_0+x}+e^{2x_0}+e^{2x})}{(e^{x_0}+e^x)^4}$

(i) So taking $x_0=a$ in (16), there exists $c\in (a,x)$ such that for $x\in (a,b)$:

$$\begin{aligned} \ln \left( \frac{1}{2}+\frac{1}{2}e^{a-x}\right)= & {} - \frac{(x-a)}{2}+\frac{(x-a)^2}{8}+\frac{(x-a)^3}{6}f'''_a(c) \end{aligned}$$

We deduce as $f'''_a(x^*) \le f'''_a(c) \le 0$, with $x^*={{\,\mathrm{arg\,min}\,}}_{a \le x \le b} f'''_a(x)$:

$$\begin{aligned} - \frac{(x-a)}{2}+\frac{(x-a)^2}{8} +\frac{(x-a)^3}{6} f'''_a(x^*)\le & {} \ln \left( \frac{1}{2}+\frac{1}{2}e^{a-x}\right) \le - \frac{(x-a)}{2}+\frac{(x-a)^2}{8} \end{aligned}$$

(17)

Let us now find $x^*={{\,\mathrm{arg\,min}\,}}_{a \le x \le b} f'''_a(x)$ and $f'''_a(x^*)$ to get the lower bound. The first-order condition for minimizing $f'''_a(x)$ is to have $f''''_a(x)=0$, i.e.,

$$\begin{aligned} -4 e^{a+x} + e^{2a}+e^{2x}= (e^x)^2 -4 e^a e^x +e^{2a}=0 \end{aligned}$$

(18)

which is a quadratic equation in $e^x$. Its discriminant is $\Delta = (-4e^a)^2 - 4e^{2a} = 12 e^{2a}>0$, so the quadratic equation has too roots $e^a(2+\sqrt{3})$ and $e^a(2-\sqrt{3})$, which gives for x: $x= a+\ln (2+\sqrt{3})$ and $x=a + \ln (2-\sqrt{3})$. The latter is smaller than a so the minimum of $f'''_a$ is obtained for

$$\begin{aligned} x^* = \min (a+\ln (2+\sqrt{3}),b). \end{aligned}$$

We have

$$\begin{aligned} f'''_a(a+\ln (2+\sqrt{3})) = \frac{(e^a-e^{a+\ln (2+\sqrt{3})})e^{2a+\ln (2+\sqrt{3})}}{(e^a+e^{a+\ln (2+\sqrt{3})})^3} = -\frac{(1+\sqrt{3})(2+\sqrt{3})}{(3+\sqrt{3})^3}=-\alpha \end{aligned}$$

with $\alpha \approx 0.09623$, which gives

$$\begin{aligned} f'''_a(x^*)= & {} -\alpha \text { if } a+\ln (2+\sqrt{3}) \le b \\= & {} \dfrac{(e^a-e^b)(e^{a+b})}{(e^a+e^b)^3} > -\alpha \text { if } b< a+\ln (2+\sqrt{3}) \end{aligned}$$

With (17) and $f'''_a(x^*) \ge -\alpha$ we obtain:

$$\begin{aligned}&- \frac{(b-a)^2}{4}+\frac{(b-a)^3}{24} -\alpha \frac{(b-a)^4}{24} \le \int _a^b \ln \left( \frac{1}{2}+\frac{1}{2} e^{a-x}\right) dx \nonumber \\&\quad \le - \frac{(b-a)^2}{4}+\frac{(b-a)^3}{24} \end{aligned}$$

(19)

(ii) Now, let us apply the Taylor formula for $f_b(x)$, we get

$$\begin{aligned} \ln (\frac{1}{2}+\frac{1}{2}e^{b-x})= & {} - \frac{(x-b)}{2}+\frac{(x-b)^2}{8}+\frac{(x-b)^3}{6}f'''_b(c') \end{aligned}$$

with $c'\in (x,b)$. As $(x-b)^3\le 0$ and $f'''_b(x)\ge 0$ for $x\le b$ we have

$$\begin{aligned}&- \frac{(x-b)}{2}+\frac{(x-b)^2}{8} +\frac{(x-b)^3}{6} f'''_b(x^{**})\le & {} \ln \left( \frac{1}{2}+\frac{1}{2}e^{b-x}\right) \nonumber \\&\quad \le - \frac{(x-b)}{2}+\frac{(x-b)^2}{8} \end{aligned}$$

(20)

where $x^{**}={{\,\mathrm{arg\,max}\,}}_{a\le x \le b} f'''_b(x)$. Just like for $f_a(x)$, we have to find $f''''_b(x)=0$ i.e.,

$$\begin{aligned} -4 e^{b+x} + e^{2b}+e^{2x}= (e^x)^2 -4 e^b e^x +e^{2b}=0 \end{aligned}$$

(21)

which is again a quadratic equation in $e^x$ with discriminant $12 e^{2b}>0$ and roots: $e^x= e^b(2+\sqrt{3})$ or $e^x=e^b(2-\sqrt{3})$. This gives $x= b+\ln (2+\sqrt{3})$, which is discarded as larger than b, or $x=b + \ln (2-\sqrt{3}) <b$. We have:

$$\begin{aligned} f'''_b(b+\ln (2-\sqrt{3})) = \frac{(e^b-e^{b+\ln (2-\sqrt{3})})e^{2b+\ln (2-\sqrt{3})}}{(e^b+e^{b+\ln (2-\sqrt{3})})^3} = \frac{(\sqrt{3}-1)(2-\sqrt{3})}{(3-\sqrt{3})^3}=\alpha \end{aligned}$$

which gives

$$\begin{aligned} f'''_b(x^{**})= & {} \alpha \text { if } a\le b+\ln (2-\sqrt{3}) \\= & {} \dfrac{(e^b-e^a)(e^{a+b})}{(e^a+e^b)^3} < \alpha \text { if } a >b+\ln (2-\sqrt{3}) \end{aligned}$$

With (20) and $f'''_b(x^{**}) \le \alpha$ we obtain:

$$\begin{aligned} \frac{(b-a)^2}{4}+\frac{(b-a)^3}{24} - \alpha \frac{(b-a)^4}{24} \le \int _a^b \ln \left( \frac{1}{2}+\frac{1}{2} e^{a-x}\right) dx \le \frac{(b-a)^2}{4}+\frac{(b-a)^3}{24} \nonumber \\ \end{aligned}$$

(22)

Summing (19) and (22) we get

$$\begin{aligned} \frac{(b-a)^3}{12} - \alpha \frac{(b-a)^4}{12} \le \int _a^b \ln (\frac{1}{2}+\frac{1}{2} e^{a-x}) + \ln (\frac{1}{2}+\frac{1}{2} e^{b-x}) dx \le \frac{(b-a)^3}{12} \end{aligned}$$

and finally,

$$\begin{aligned}&\frac{1}{2}+\frac{(b-a)}{12} - \alpha \frac{(b-a)^2}{12} \le \frac{1}{2} + \frac{1}{(b-a)^2} \int _a^b \ln \left( \frac{1}{2}+\frac{1}{2} e^{a-x}\right) \\&\quad + \ln (\frac{1}{2}+\frac{1}{2} e^{b-x}) dx \le \frac{1}{2}+ \frac{(b-a)}{12} \end{aligned}$$

From (13) we obtain the claimed result of the proposition. $\square$

D Proof of Proposition 3

Proof

For the discrete uniform distribution we have

$$\begin{aligned}&E(\max (X_{1j},X_{2j})= = \sum _{t=1}^T \sum _{t'=1}^t P(U_{1j}=t) P(U_{2j}=t') \frac{\max (e^{(b-a)t/T},e^{(b-a)t'/T} ) }{e^{(b-a)t/T}+e^{(b-a)t'/T}} \\&\quad = \frac{1}{T^2} \sum _{t=1}^T \left( \sum _{t'=1}^t \frac{e^{(b-a)t/T}}{e^{(b-a)t/T}+e^{(b-a)t'/T}} + \sum _{t'=t+1}^T \frac{e^{(b-a)t'/T}}{e^{(b-a)t/T}+e^{(b-a)t'/T}} \right) \\&\quad = \frac{1}{T^2} \sum _{t=1}^T \left( \sum _{t'=1}^{t-1} \frac{1}{1+e^{(b-a)(t'-t)/T}} +\frac{1}{2}+ \sum _{t'=t+1}^T \frac{1}{1+e^{(b-a)(t-t')/T}} \right) \\&\quad = \frac{1}{T^2} \left( T \times \frac{1}{2} + 2 \sum _{t=1}^{T-1} (T-t) \frac{1}{1+e^{-(b-a)t/T}} \right) \end{aligned}$$

Letting T tend to infinity we get Proposition 1 for a continuous uniform distribution. $\square$

E Proof of Proposition 4

Proof

We rewrite $E \max (X_{1j},X_{2j})$ as

$$\begin{aligned} \frac{1}{2}+ \frac{1}{(b-a)^2} \int _a^b \left[ \ln (1+e^{a-x}) + \ln (1+e^{b-x}) \right] dx - 2\, \frac{\ln 2}{b-a} \end{aligned}$$

The third term tends toward 0 as b goes to $\infty$. For the second term, we will derive lower and upper bounds that both converge toward 1/2.

From $e^{a-x} \ge 0$, we have $\ln (1+e^{a-x}) \ge \ln (1)=0$ so

$$\begin{aligned} \frac{1}{(b-a)^2} \int _a^b \ln (1+e^{a-x}) dx \ge 0 \end{aligned}$$

(23)

Further, $x \mapsto f(x) = \ln (1+e^{a-x})$ is convex in x, as can be seen from computing the second derivative, so an upper bound to the function is provided by the line that connects (a, f(a)) with (b, f(b)). Here:

$$\begin{aligned} \ln (1+e^{a-x}) \le \ln 2 + (x-a) \cdot \frac{\ln (1+e^{a-b}) - \ln 2}{b-a}. \end{aligned}$$

Integrating between a and b and dividing by $(b-a)^2$, we obtain:

$$\begin{aligned} \frac{1}{(b-a)^2} \int _a^b \ln (1+e^{a-x}) dx \le \frac{\ln 2}{b-a} + \frac{1}{2} \cdot \frac{\ln (1+e^{a-b}) - \ln 2}{b-a}. \end{aligned}$$

The right hand side converges to 0 as b goes to $\infty$. Combining with Eq. (23), we obtain that the first term in the integral converges to 0.

For the second term, we use that $1+e^{b-x} \ge e^{b-x}$ so $\ln (1+e^{b-x}) \ge b-x$, we have

$$\begin{aligned} \frac{1}{(b-a)^2} \int _a^b \ln (1+e^{b-x}) dx \ge \frac{1}{2} \end{aligned}$$

(24)

On the other hand, again invoking convexity of the function and using the line connecting the extremities as an upper bound:

$$\begin{aligned} \ln (1+e^{b-x}) \le \ln (1+e^{b-a}) + (x-a) \cdot \frac{\ln 2 - \ln (1+e^{b-a})}{b-a}. \end{aligned}$$

Integrating between a and b and dividing by $(b-a)^2$, we observe that the first term in the sum in the right hand side converges to 1 as b goes to $\infty$ and the second one goes to $-1/2$ (because $\ln (1+e^{b-a}) \approx b-a$ when b large). So the lower and upper bound of this second term in the integral converge toward 1 / 2.

Bringing everything together, we obtain convergence to 1 when b goes to $\infty$. $\square$