Valuing information technology infrastructures: a growth options approach

Abstract

Decisions to invest in information technology (IT) infrastructure are often made based on an assessment of its immediate value to the organization. However, an important source of value comes from the fact that such technologies have the potential to be leveraged in the development of future applications. From a real options perspective, IT infrastructure investments create growth options that can be exercised if and when an organization decides to develop systems to provide new or enhanced IT capabilities. We present an analytical model based on real options that shows the process by which this potential is converted into business value, and discuss middleware as an example technology in this context. We derive managerial implications for the evaluation of IT infrastructure investments, and the main findings are: (1) the flexibility provided by IT infrastructure investment is more valuable when uncertainty is higher; (2) the cost advantage that IT infrastructure investment brings about is amplified by demand volatility for IT-supported products and services; (3) in duopoly competition, the value of IT infrastructure flexibility increases with the level of product or service substitutability; and (4) when demand volatility is high, inter-firm competition has a lower impact on the value of IT infrastructure.

Appendices

Appendix

Mathematical representation of the investment value function

Several derivations illustrate how we arrived at the main results. Substituting Eq. (7) for ${\pi^{\hbox {INVESTMENT}}}$ in Eq. (8), we represent the value function as:

$$ \begin{array}{l} V^{\hbox {INVESTMENT}}(\theta _0,\sigma)\ =\ E(\pi ^{\hbox {INVESTMENT}}\vert \theta\ \ge\ \theta_1)\cdot \hbox {prob}(\theta\ \ge\ \theta _1)-K,\\ =\int_{\theta _1}^\infty {\frac{(p-c)^2\theta ^2-4fp(p-c)}{4f}\hbox {d}[F(\theta )]}\ -\ K\\ =\int_{\theta _1}^\infty {\frac{(p-c)^2\theta ^2\ -\ 4fp(p-c)}{4f}\cdot f(\theta)\cdot \hbox {d}\theta}\ -\ K\\ \end{array} $$

(A1)

where θ₁ is the same as in Eq. (6). F(θ) is the cumulative density function and f(θ) is the probability density function of θ. When θ follows the log-normal distribution ${\ln(\theta)\sim N\left({\ln(\theta_0)-\frac{1}{2}\sigma ^2,\sigma ^2}\right)}$, we substitute ${\theta =\hbox {e}^x}$ in Eq. (A1) to arrive at:

$$ \begin{array}{l} V^{\hbox {INVESTMENT}}(\theta _0,\sigma)\ =\ \int_{\ln \theta _1}^\infty {\frac{(p-c)^2(\hbox {e}^x)^2-4fp(p-c)}{4f}\cdot \hbox {d}[N(x)]}\ -\ K \\ =\int_{\ln \theta _1}^\infty {\frac{(p-c)^2(\hbox {e}^x)^2-4fp(p-c)}{4f}\cdot \varphi (x)\cdot \hbox {d}x}\ -\ K \\ \end{array} $$

(A2)

where x is normally distributed as ${x\sim N\left({\ln (\theta_0)-\frac{1}{2}\sigma ^2,\sigma^2}\right)}$, N(x) is the cumulative density function and ϕ(x) is the probability density function of x, $\varphi(x)=\frac{1}{\sigma \sqrt{2\pi}} \cdot \hbox {e}^{-\frac{1}{2\sigma^2}\left[x-(\ln\theta_0-\sigma^2 /2)\right]^2}$

Derivation of the value function when infrastructure investment is made at Stage 1

Equation (A2) can be extended as follows:

$$ \begin{array}{l} V^{\hbox {INVESTMENT}}(\theta _0,\sigma)=\int_{\ln \theta _1}^\infty {\frac{(p-c)^2(\hbox {e}^x)^2-4fp(p-c)}{4f}\cdot \hbox {d}[N(x)]} -K \\ =\int_{\ln \theta _1}^\infty {\frac{(p-c)^2}{4f}\cdot \hbox {e}^{2x}\cdot \hbox {d}[N(x)]-\int_{\ln \theta _1}^\infty {p(p-c)\cdot \hbox {d}[N(x)]} -K} \\ =\frac{(p-c)^2}{4f}\int_{\ln \theta _1}^\infty {\hbox {e}^{2x}\cdot \hbox {d}[N(x)]} -p(p-c)\int_{\ln \theta _1}^\infty {\hbox {d}[N(x)]} -K \\ \end{array}\quad. $$

(A3)

Substituting ${y=\frac{x-(\ln \theta _0 -\frac{\sigma ^2}{2})}{\sigma}}$ in the integral in the first term in Eq. (A3), we get:

$$ \begin{array}{l} \int_{\ln \theta _1}^\infty {\hbox {e}^{2x}\hbox {d}[N(x)]=\int_{y_1}^\infty {\hbox {e}^{2\sigma y+2(\ln \theta _0 -\sigma^2/2)}\hbox {d}[\Phi (y)]} =\int_{y_1}^\infty {\hbox {e}^{2\sigma y+2(\ln \theta _0 -\sigma^2/2)}\cdot \phi (y)\cdot \hbox {d}y}} \\ =\hbox {e}^{2(\ln \theta _0 -\sigma ^2/2)}\int_{y_1 }^\infty {\hbox {e}^{2\sigma y}\cdot \phi (y)\cdot \hbox {d}y} \\ \end{array}, $$

(A4)

In Eq. (A4), $y_1 =\frac{\ln \theta _1 -(\ln \theta_0-{\sigma^2}/2)}{\sigma}$, Φ(y) is the cumulative density function and φ(y) is the probability density function of the standard normal distribution. Using the formula for the normal integral [42], we compute the integral in Eq. (A4) as:

$$ \begin{array}{l} \int_{y_1}^\infty {\hbox {e}^{2\sigma y}\cdot \phi (y)\cdot \hbox {d}y} =\hbox {e}^{4\sigma ^2/2}\cdot \Phi (y-2\sigma)\vert _{y_1}^\infty =\hbox {e}^{2\sigma ^2}\cdot \Phi (2\sigma -y_1)\\ =\hbox {e}^{2\sigma ^2}\cdot \Phi \left({\frac{2\ln (\theta _0 /\theta_1)+3\sigma ^2}{2\sigma}}\right) \\ \end{array}. $$

(A5)

Plugging Eq. (A5) into Eq. (A4) gives:

$$ \begin{array}{l} \int_{\ln \theta _1}^\infty {\hbox {e}^{2x}\hbox {d}[N(x)]}\ =\ \hbox {e}^{2(\ln \theta _0 -\sigma ^2/2)}\int_{y_1 }^\infty {\hbox {e}^{2\sigma y}\cdot \phi (y)\cdot \hbox{d}y} \\ =\ \hbox {e}^{2(\ln \theta _0 -\sigma^2/2)}\cdot \hbox {e}^{2\sigma ^2}\cdot \Phi \left({\frac{2\ln ({\theta _0}\mathord{\left/ {\vphantom {{\theta _0} {\theta _1}}} \right. \kern-\nulldelimiterspace} {\theta _1})+3\sigma ^2}{2\sigma}} \right)\ =\ \hbox {e}^{\ln \theta _0^2}\cdot \hbox {e}^{\sigma ^2}\cdot \Phi \left({\frac{\ln ({\theta _0} \mathord{\left/ {\vphantom {{\theta _0} {\theta _1}}} \right. \kern-\nulldelimiterspace} {\theta _1})^2+3\sigma ^2}{2\sigma}} \right) \\ =\theta _0^2 \cdot \hbox {e}^{\sigma ^2}\cdot \Phi \left({\frac{\ln ({\theta _0} \mathord{\left/ {\vphantom {{\theta _0} {\theta _1}}} \right. \kern-\nulldelimiterspace} {\theta _1})^2+3\sigma ^2}{2\sigma}} \right) \\ \end{array}. $$

(A6)

Substituting ${y=\frac{x-(\ln \theta _0 -\frac{\sigma ^2}{2})}{\sigma }}$ in the integral’s second term in Eq. (A3) gives:

$$ \begin{array}{l} \int_{\ln \theta _1}^\infty {\hbox {d}[N(x)]=\int_{y_1}^\infty {\hbox {d}[\Phi (y)]}} =\Phi (-y_1)=\Phi \left({-\frac{\ln \theta _1 -(\ln \theta _0 -\frac{\sigma ^2}{2})}{\sigma}} \right) \\ =\Phi \left({\frac{\ln ({\theta _0} \mathord{\left/ {\vphantom {{\theta _0} {\theta _1}}} \right. \kern-\nulldelimiterspace} {\theta _1 })^2-\sigma ^2}{2\sigma}} \right) \\ \end{array}, $$

(A7)

where $y_1 =\frac{\ln \theta _1 -(\ln \theta _0 -\sigma ^2/2)}{\sigma}$, and Φ(y) is the standard normal cumulative density function.

Plugging Eqs. (A6) and (A7) into Eq. (A3) yields Eq. (11), as we saw earlier in this article:

$$\begin{array}{l} V^{\hbox {INVESTMENT}}(\theta _0,\sigma)=\frac{(p-c)^2}{4f}\int_{\ln \theta _1 }^\infty {\hbox {e}^{2x}\hbox {d}[N(x)]} -p(p-c)\int_{\ln \theta _1}^\infty {\hbox {d}[N(x)]} -K \\ =\frac{(p-c)^2\cdot \theta _0^2 \cdot \hbox {e}^{\sigma ^2}}{4f}\cdot \Phi \left({\frac{\ln ({\theta _0} \mathord{\left/ {\vphantom {{\theta _0} {\theta _1}}} \right. \kern-\nulldelimiterspace} {\theta _1 })^2+3\sigma ^2}{2\sigma}} \right)-p\cdot (p-c)\cdot \Phi \left( {\frac{\ln ({\theta _0} \mathord{\left/ {\vphantom {{\theta _0} {\theta _1}}} \right. \kern-\nulldelimiterspace} {\theta _1})^2-\sigma ^2}{2\sigma}} \right)-K\\ \end{array} $$

Derivation of the value function when no infrastructure investment made at Stage 1

Following the same procedures, we can derive the value function for the case when no infrastructure investment is made at Stage 1, based on Eqs. (1)–(6). We can obtain the optimal attribute level, ${s^{\hbox {NO INVESTMENT*}}}$, and optimal Stage 2 profits, ${\pi ^{\hbox {NO INVESTMENT*}}}$, both modified with λ:

$$ s^{\hbox {NO INVESTMENT*}}=\frac{p-c}{2\lambda f}\theta $$

(A8)

$$ \pi ^{\hbox {NO INVESTMENT*}}=\frac{(p-c)^2\theta ^2-4\lambda fp(p-c)}{4\lambda f} $$

(A9)

The Stage 2 profit when no IT infrastructure investment was made at Stage 1 is conditioned on whether customer demand for the product or service, θ, goes beyond a base level. The base level demand is different from the base demand when the firm makes an investment at Stage 1. We represent this base level as ${\theta _1^\prime=\sqrt {\frac{4\lambda fp}{p-c}}}$ . The firm’s optimal profit at Stage 2 becomes:

$$ \pi ^{\hbox {NO INVESTMENT}}=\left\{ {\begin{array}{ll} 0;&\theta < \theta _1^\prime\\ \frac{(p-c)^2\theta ^2-4\lambda fp(p-c)}{4\lambda f};&\theta \ge \theta _1^\prime\\ \end{array}}\right. $$

(A10)

In this case, the expected value of IT investment becomes Eq. (12):

$$ \begin{array}{l} V^{\hbox {NO INVESTMENT}}(\theta _0,\sigma)\ =\ \int_{\theta _1^\prime}^\infty {\frac{(p-c)^2\theta ^2-4\lambda fp(p-c)}{4\lambda f}} \hbox {d}[F(\theta)]\\ =\frac{(p-c)^2}{4\lambda f}\int_{\ln \theta _1^\prime}^\infty {\hbox {e}^{2x}\cdot \hbox {d}[N(x)]} -p(p-c)\int_{\ln \theta _1^\prime}^\infty {\hbox {d}[N(x)]}\\ =\frac{(p-c)^2\cdot \theta _0^2 \cdot \hbox {e}^{\sigma ^2}}{4\lambda f}\cdot \Phi \left({\frac{\ln ({\theta _0} \mathord{\left/ {\vphantom {{\theta _0} {\theta _1^\prime}}} \right. \kern-\nulldelimiterspace} {\theta _1^\prime})^2+3\sigma ^2}{2\sigma}} \right)-p\cdot (p-c)\cdot \Phi \left({\frac{\ln ({\theta _0} \mathord{\left/ {\vphantom {{\theta _0} {\theta _1^\prime}}} \right. \kern-\nulldelimiterspace} {\theta _1^\prime})^2-\sigma ^2}{2\sigma}} \right) \\ \end{array} $$

Derivation of the value function in the case of imperfect competition

Similarly, we derive the value functions in the case of imperfect competition. Firm 1’s value function for an infrastructure investment at Stage 1 is shown in an expansion of Eq. (16):

$$ \begin{array}{l} V_1^{\hbox {INVESTMENT}} (\theta _0,\sigma)=\int_{\theta _1}^{\theta _2} {\frac{(p-c)^2\theta ^2-4fp(p-c)}{4f}\hbox {d}[F(\theta)]+\int_{\theta _2}^\infty {\frac{(\lambda -2z)(p-c)^2\theta ^2-4\lambda fp(p-c)}{4\lambda f}}} \hbox {d}[F(\theta)]-K\\ =\frac{(p-c)^2}{4f}\int_{\ln\theta _1}^{\ln\theta _2}{\hbox {e}^{2x}} \hbox {d}[N(x)]+\frac{(\lambda -2z)(p-c)^2}{4\lambda f}\int_{\ln\theta _2}^\infty {\hbox {e}^{2x}\hbox {d}[N(x)]} -p(p-c)\int_{\ln\theta _1}^\infty{\hbox {d}[N(x)]} -K\\ =\frac{(p-c)^2\theta _0^2\cdot \hbox {e}^{\sigma ^2}}{4f}\Phi\left({\frac{\ln ({\theta _0}\mathord{\left/{\vphantom{{\theta _0}{\theta _1}}} \right.\kern-\nulldelimiterspace}{\theta _1})^2+3\sigma ^2}{2\sigma}} \right)-p(p-c)\Phi\left({\frac{\ln ({\theta _0}\mathord{\left/ {\vphantom{{\theta _0}{\theta _1}}}\right.\kern-\nulldelimiterspace} {\theta _1})^2-\sigma ^2}{2\sigma}}\right)\\ -\frac{2z(p-c)^2\theta _0^2\cdot \hbox {e}^{\sigma ^2}}{4\lambda f}\Phi\left({\frac{\ln ({\theta _0}\mathord{\left/{\vphantom{{\theta _0}{\theta _2}}}\right.\kern-\nulldelimiterspace}{\theta _2})^2+3\sigma ^2}{2\sigma}}\right)-K\\ \end{array} $$

Firm 1’s value function with no IT investment at Stage 1 is an expansion of Eq. (17):

$$ \begin{array}{l} V_1^{\hbox {NO INVESTMENT}} (\theta _0,\sigma)=\int_{\theta _3}^\infty {\frac{(1-2z)(p-c)^2\theta ^2-4\lambda fp(p-c)}{4\lambda f}\hbox {d}[F(\theta)]}\\ =\frac{(1-2z)(p-c)^2}{4\lambda f}\int_{\ln\theta _3}^\infty{\hbox {e}^{2x}} \hbox {d}[N(x)]-p(p-c)\int_{\ln\theta _3}^\infty{\hbox {d}[N(x)]}\\ =\frac{(1-2z)(p-c)^2\theta _0^2\cdot \hbox {e}^{\sigma ^2}}{4\lambda f}\Phi\left({\frac{\ln ({\theta _0}\mathord{\left/{\vphantom{{\theta _0}{\theta _3}}}\right.\kern-\nulldelimiterspace}{\theta _3})^2+3\sigma ^2}{2\sigma}}\right)-p(p-c)\Phi\left({\frac{\ln ({\theta _0}\mathord{\left/ {\vphantom{{\theta _0}{\theta _3}}}\right.\kern-\nulldelimiterspace} {\theta _3})^2-\sigma ^2}{2\sigma}}\right)\\ \end{array} $$