Analysis of the stochastic cash balance problem using a level crossing technique

Ben A. Chaouch

doi:10.1007/s10479-018-2822-2

Annals of Operations Research

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Analysis of the stochastic cash balance problem using a level crossing technique

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Ben A. Chaouch

Original Research

First Online: 04 April 2018

Abstract

The simple cash management problem includes the following considerations: the opportunity cost of holding too much cash versus the penalty cost of not having enough cash to meet current needs; the cost incurred (or profit generated) when making changes to cash levels by increasing or decreasing them when necessary; the uncertainty in timing and magnitude of cash receipts and cash disbursements; and the type of control policy that should be used to minimize the required level of cash balances and related costs. In this paper, we study a version of this problem in which cash receipts and cash disbursements occur according to two independent compound Poisson processes. The cash balance is monitored continuously and an order-point, order-up-to-level, and keep-level $ \left( {s, S, M} \right) $ policy is used to monitor the content, where $ s \le S \le M $. That is, (a) if, at any time, the cash level is below s, an order is immediately placed to raise the level to S; (b) if the cash level is between s and M, no action is taken; (c) if the cash level is greater than M, the amount in excess of M is placed into an earning asset. We seek to minimize the expected total costs per unit time of running the cash balance. We use a level-crossing approach to develop a solution procedure for finding the optimal policy parameters and costs. Several numerical examples are given to illustrate the tradeoffs.

Keywords

Cash balance problem Stationary analysis Level-crossing method

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Notes

Acknowledgements

The author thanks the referee for constructive comments and suggestions that have led to many improvements in the paper’s content.

Appendix

Let us assume that the distributions of disbursement and receipt quantities are given by $ G(x) = 1 - e^{ - \mu x} ,x \ge 0 $ and $ H(x) = 1 - e^{ - \nu x} ,x \ge 0 $, respectively. Then, in order to find the stationary distribution of the cash balance, the system of rate balance equations is obtained by equating the exit rate to the entrance rate for each state-space interval and can be stated as follows.

$$ \begin{aligned} & \lambda \left[ {e^{ - \mu (M - S)} - e^{ - \mu (M - x)} } \right]\Pi_{M} + \lambda \int\limits_{S}^{M} {\left[ {e^{ - \mu (y - S)} - e^{ - \mu (y - x)} } \right]f_{2} (y)dy} + \lambda \left[ {1 - e^{ - \mu (S - x)} } \right]\Pi_{S} \\ & \quad + \,\theta \int\limits_{s}^{x} {\left[ {e^{ - \nu (x - y)} - e^{ - \nu (S - y)} } \right]f_{1} (y)dy = \lambda \int\limits_{x}^{S} {e^{ - \mu (y - x)} f_{1} (y)dy + \theta \int\limits_{x}^{S} {e^{ - \nu (S - y)} f_{1} (y)dy} } ,} \quad x \in (s,S). \\ \end{aligned} $$

(A.1)

$$ \begin{aligned} & \lambda \left[ {1 - e^{ - \mu (M - x)} } \right]\Pi_{M} + \theta \int\limits_{S}^{x} {\left[ {e^{ - \nu (x - y)} - e^{ - \nu (M - y)} } \right]f_{2} (y)dy + \theta \left[ {e^{ - \nu (x - S)} - e^{ - \nu (M - S)} } \right]\Pi_{S} } \\ & \quad + \,\theta \int\limits_{s}^{S} {\left[ {e^{ - \nu (x - y)} - e^{ - \nu (M - y)} } \right]f_{1} (y)dy} = \lambda \int\limits_{x}^{M} {e^{ - \mu (y - x)} f_{2} (y)dy + \theta \int\limits_{x}^{M} {e^{ - \nu (M - y)} f_{2} (y)dy} } ,\quad x \in (S,M). \\ \end{aligned} $$

(A.2)

$$ \lambda \int\limits_{s}^{S} {e^{ - \mu (x - s)} f_{1} (x)dx + \lambda \int\limits_{S}^{M} {e^{ - \mu (x - s)} f_{2} (x)dx + } } \lambda e^{ - \mu (S - s)} \Pi_{S} + \lambda e^{ - \mu (M - s)} \Pi_{M} = (\lambda + \theta )\Pi_{S} , \, x = S. $$

(A.3)

$$ \theta \int\limits_{S}^{M} {e^{ - \nu (M - x)} f_{2} (x)dx} + \theta \int\limits_{s}^{S} {e^{ - \nu (M - x)} f_{1} (x)dx} + \theta e^{ - \nu (M - S)} \Pi_{S} = \lambda \Pi_{M} , \, x = M. $$

(A.4)

The normalizing condition is:

$$ \int\limits_{s}^{S} {f_{1} (x)dx} + \int\limits_{S}^{M} {f_{2} (x)dx} + \Pi_{S} + \Pi_{M} = 1. $$

(A.5)

First, let us consider equation (A.1). After some manipulations, (A.1) can be changed to the following integral equation.

$$ \lambda \int\limits_{x}^{S} {e^{ - \mu (y - x)} f_{1} (y)dy - \theta \int\limits_{s}^{x} {e^{ - \nu (x - y)} f_{1} (y)dy} } - K_{1} e^{\mu x} - K_{2} = 0, \, x \in (s,S) , $$

(A.6)

where K₁ and K₂ are constants that must satisfy:$ K_{1} + \lambda \left( {\Pi_{S} e^{ - \mu S} + \Pi_{M} e^{ - \mu M} } \right) + \lambda \int\limits_{S}^{M} {e^{ - \mu y} f_{2} (y)dx = 0,} $ and $ e^{\mu S} K_{1} + K_{2} + \theta \int\limits_{s}^{S} {e^{ - \nu (S - y)} f_{1} (y)dy = 0.} $

Next, we turn integral equation (A.6) with unknown function $ f_{1} \left( x \right) $ into an ordinary differential equation (ODE) by differentiating with respect to x. This yields the following ODE of the first-order: $ (\lambda + \theta )f_{1}^{\prime } (x) + (\lambda \nu - \theta \mu )f_{1} (x) = \mu \nu K_{2} . $ A general solution to this equation can be given as follows: $ f_{1} (x) = C_{1} e^{ - \alpha x} + \frac{{\mu \nu K_{2} }}{(\lambda \nu - \theta \mu )}, $ where $ \alpha = \frac{\lambda \nu - \theta \mu }{\lambda + \theta } $ and C₁ is a constant to be determined. Substituting this expression for $ f_{1} (x) $ back into equation (A.6) and working out the operations, it can be shown that $ C_{1} = - \frac{{\mu (\nu - \alpha )e^{\alpha s} K_{2} }}{(\lambda \nu - \theta \mu )}. $

Second, let us now consider equation (A.2). Again, after some manipulations, (A.2) can be transformed into the following integral equation.

$$ \lambda \int\limits_{x}^{M} {e^{ - \mu (y - x)} f_{2} (y)dy - \theta \int\limits_{S}^{x} {e^{ - \nu (x - y)} f_{2} (y)dy} } + \lambda e^{ - \mu (M - x)} \Pi_{M} - K_{3} e^{ - \nu x} - K_{4} = 0, \, x {\in} (S,M), $$

(A.7)

where K₃ and K₄ are constants that must be determined and that must satisfy: $ - K_{3} + \theta \left[ {\Pi_{S} e^{\nu S} + \int\limits_{s}^{S} {e^{\nu y} f_{1} (y)dy} } \right] = 0, $ and $ K_{3} + e^{\nu M} K_{4} - \lambda \Pi_{M} e^{\nu M} + \theta \int\limits_{S}^{M} {e^{\nu y} f_{2} (y)dy = 0.} $

In similar fashion, we turn integral equation for (A.7) with unknown function $ f_{2} \left( x \right) $ into an ODE by differentiating with respect to x. It can be checked that $ f_{2} \left( x \right) $ is the solution to the following ODE: $ (\lambda + \theta )f_{2}^{\prime } (x) + (\lambda \nu - \theta \mu )f_{2} (x) = \mu \nu K_{4} . $ A general solution to this equation can be given as follows: $ f_{2} (x) = C_{2} e^{ - \alpha x} + \frac{{\mu \nu K_{4} }}{(\lambda \nu - \theta \mu )}, $ where C₂ is a constant to be determined. Substituting this expression for $ f_{2} (x) $ back into equation (A.7), it can be shown that $ C_{2} = \frac{{(\nu - \alpha )K_{3} e^{(\alpha - \nu )S} }}{\theta } - \frac{{(\nu - \alpha )\mu K_{4} e^{\alpha S} }}{(\lambda \nu - \theta \mu )}. $

Next, using these expressions for $ f_{1} (x) $ and $ f_{2} (x) $, plugging the results into equations (A.3) and (A.4) and after some lengthy algebra, it can be verified that:

$$ \begin{aligned} & K_{2} = (\lambda + \theta )\Pi_{S} , \\ & C_{1} = - \frac{{\mu (\nu - \alpha )e^{\alpha s} \Pi_{S} }}{\alpha }, \\ & K_{3} = \frac{{\theta \left( {\lambda (\mu + \nu ) - \mu (\lambda + \theta )e^{ - \alpha (S - s)} } \right)e^{\nu S} \Pi_{S} }}{{\left( {\lambda \nu - \theta \mu } \right)}}, \\ & K_{4} = 0, \\ & C_{2} = \frac{{(\nu - \alpha )\left( {\alpha + \mu \left( {1 - e^{ - \alpha (S - s)} } \right)} \right)e^{\alpha S} \Pi_{S} }}{\alpha },{\text{ and}} \\ & \Pi_{M} = \frac{{\theta \left( {\alpha + \mu \left( {1 - e^{ - \alpha (S - s)} } \right)} \right)e^{ - \alpha (M - S)} \Pi_{S} }}{\lambda \alpha }. \\ \end{aligned} $$

With the above expressions on hand, it can then be checked that $ f_{1} \left( x \right) $ and $ f_{2} \left( x \right) $ can be given by

$$ \begin{aligned} & f_{1} (x) = \frac{{\Pi_{S} \mu \nu \left( {1 + \left( {{\alpha \mathord{\left/ {\vphantom {\alpha {\nu - 1}}} \right. \kern-0pt} {\nu - 1}}} \right)e^{ - \alpha (x - s)} } \right)}}{\alpha },\quad x \in (s,S) \\ & f_{2} (x) = \frac{{\Pi_{S} \mu \nu \left( {1 - {\alpha \mathord{\left/ {\vphantom {\alpha \nu }} \right. \kern-0pt} \nu }} \right)\left( {1 + {\alpha \mathord{\left/ {\vphantom {\alpha {\mu - e^{ - \alpha (S - s)} }}} \right. \kern-0pt} {\mu - e^{ - \alpha (S - s)} }}} \right)e^{ - \alpha (x - S)} }}{\alpha },\quad x \in (S,M). \\ \end{aligned} $$

Finally, using the normalizing condition (A.5) together with some algebra, it can be shown that $ f_{1} \left( x \right), f_{2} \left( x \right), \Pi_{S} $ and $ \Pi_{M} $ can be presented as shown in expression (4.1).

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Ben A. Chaouch
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1.Odette School of BusinessUniversity of WindsorWindsorCanada

Analysis of the stochastic cash balance problem using a level crossing technique

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