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A system model of work flow in the patient room of hospital emergency department

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Abstract

Modeling and analysis of patient flow in hospital emergency department (ED) is of significant importance. In a hospital ED, the patients spend most of their time in the patient room and most of the care delivery services are carried out during this time period. In this paper, we propose a system model to study patient (or work) flow in the patient room of an ED when the resources are partially available. A closed and re-entrant process model is developed to characterize the care service activities in the patient room with limited resources of doctors, nurses, and diagnosis tests. Analytical calculation of patient’s length of stay in the patient room is derived, and monotonic properties with respect to care service parameters are investigated.

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Author information

Authors and Affiliations

  1. Laureate Education Inc., Baltimore, MD, 21202, USA

    Junwen Wang

  2. Department of Industrial and Systems Engineering, University of Wisconsin, Madison, WI, 53706, USA

    Jingshan Li

  3. Trauma and Emergency Services, University of Kentucky Chandler Medical Center, Lexington, KY, 40506, USA

    Patricia K. Howard

Authors
  1. Junwen Wang
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  2. Jingshan Li
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  3. Patricia K. Howard
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Corresponding author

Correspondence to Jingshan Li.

Additional information

This paper is supported in part by NSF Grants No. CMMI 1063671 and 1233807.

Appendix: Proofs

Appendix: Proofs

Proof of Proposition 1

By adding the balance equations and eliminating the identical terms, the process throughput TP follows immediately. The patients may leave the system when they are ready for the second service and the provider is available, i.e.,

$$TP=c_{2}[P(1;0,1)+P(1;1,1)].$$

From the Little’s law, the length of stay is obtained through the inverse of TP since only one patient is in the system. □

Proof of Corollary 1

Define the steady state probabilities as follows:

$$ \begin{array}{rll}\pi_{0}=P(0;0,0), & \pi_{1}=P(0;0,1), & \pi_{2}=P(0;1,0), \\\pi_{3}=P(0;1,1), & \pi_{4}=P(1;0,0), & \pi_{5}=P(1;0,1), \\\pi_{6}=P(1;1,0), &\quad \pi_{7}=P(1;1,1).\end{array}$$

By adding Eqs. 1, 2, 3, and 4 with Eqs. 5, 6, 7, and 8, respectively, we obtain

$$ \begin{array}{rll} (\pi_{0}+\pi_{4})(\mu_{1}+\mu_{2})&=&(\pi_{2}+\pi_{6})\lambda_{1}+(\pi_{1}+\pi_{5})\lambda_{2}, \\(\pi_{1}+\pi_{5})(\mu_{1}+\lambda_{2})&=&(\pi_{3}+\pi_{7})\lambda_{1}+(\pi_{0}+\pi_{4})\mu_{2}, \\(\pi_{2}+\pi_{6})(\lambda_{1}+\mu_{2})&=&(\pi_{0}+\pi_{4})\mu_{1}+(\pi_{3}+\pi_{7})\lambda_{2}, \\(\pi_{3}+\pi_{7})(\lambda_{1}+\lambda_{2})&=&(\pi_{1}+\pi_{5})\mu_{1}+(\pi_{2}+\pi_{6})\mu_{2}.\end{array} $$

Then, it follows that

$$ \begin{array} {rll} L_{0}&:=&\pi_{0}+\pi_{4}=\frac{\lambda_{1} \lambda_{2}}{(\lambda_{1}+\mu_{1})(\lambda_{2}+\mu_{2})}, \\[5pt] L_{1}&:=&\pi_{1}+\pi_{5}=\frac{\lambda_{1} \mu_{2}}{(\lambda_{1}+\mu_{1})(\lambda_{2}+\mu_{2})}, \\[5pt] L_{2}&:=&\pi_{2}+\pi_{6}=\frac{\mu_{1} \lambda_{2}}{(\lambda_{1}+\mu_{1})(\lambda_{2}+\mu_{2})}, \\[5pt] L_{3}&:=&\pi_{3}+\pi_{7}=\frac{\mu_{1} \mu_{2}}{(\lambda_{1}+\mu_{1})(\lambda_{2}+\mu_{2})}.\end{array} $$

Replacing with \(\pi _{i},i=0,1,2,3\) to rewrite Eqs. 1–4, we have

$$ \begin{array}{rll}\pi_{0}(\mu_{1}+\mu_{2})&=&\pi_{1} \lambda_{2}+\pi_{2} \lambda_{1}, \\\pi_{1}(\mu_{1}+\lambda_{2}+c_{2})&=&\pi_{0} \mu_{2}+\pi_{3} \lambda_{1}+L_{1} c_{2}, \\\pi_{2}(\lambda_{1}+\mu_{2}+c_{1})&=&\pi_{0} \mu_1+\pi_{3} \lambda_{2}, \\\pi_{3}(\lambda_{1}+\lambda_{2}+c_{1}+c_{2})&=&\pi_{1} \mu_{1}+\pi_{2} \mu_{2}+L_{3}c_{2}.\end{array} $$

Then, it implies that

$$ X_{2}=A_{2}^{-1}B_{2}.$$

where

$$ A_{2} = \left(\begin{array}{llll}\mu_{1}+\mu_{2} & -\lambda_{2} & -\lambda_{1} & 0 \\-\mu_{2} & \mu_{1}+\lambda_{2}+c_{2} & 0 & -\lambda_{1} \\-\mu_{1} & 0 & \lambda_{1}+\mu_{2}+c_{1} & -\lambda_{2} \\0 & -\mu_{1} & -\mu_{2} & \lambda_{1}+\lambda_{2}+c_{1}+c_{2} \\\end{array}\right).$$
$$ \begin{array}{rll} X_{2}&=&[\pi_{0},\pi_{1},\pi_{2},\pi_{3}]^{T}, \\[5pt]B_{2}&=&[0,L_{1}c_{2},0,L_{3}c_{2}]^{T}\\[5pt]&=&\bigg[0,\frac{\lambda_{1}\mu_{2}c_{2}}{(\lambda_{1}+\mu_{1})(\lambda_{2}+\mu_{2})},0, \frac{\mu_{1}\mu_{2}c_{2}}{(\lambda_{1}+\mu_{1})(\lambda_{2}+\mu_{2})}\bigg]^{T}. \end{array} $$

Since now the matrix dimension is small, it is possible to express its inverse in an explicit format. Thus, after some manipulation, we obtain

$$ \begin{array}{rll} \pi_{0}&=&\lambda_{2}c_{2}L_{1}\frac{(c_{1}+\lambda_{1}+\mu_{1})E+F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\\\pi_{1}&=&c_{2}L_{1}\frac{(c_{1}+\lambda_{1}+\mu_{1})\mu_{2}E+c_{1}\mu_{1}E+\mu_{2}F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\\\pi_{2}&=&\lambda_{2}c_{2}L_{3}\frac{(\lambda_{1}+\mu_{1})E+F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\\\pi_{3}&=&c_{2}L_{3}\frac{(\lambda_{1}+\mu_{1})\mu_{2}E+c_{1}\mu_{1}E+(c_{1}+\mu_{2})F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\\\pi_{4}&=&\lambda_{1}c_{1}L_{2}\frac{(c_{2}+\lambda_{2}+\mu_{2})E-F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\\\pi_{5}&=&\lambda_{1}c_{1}L_{3}\frac{(\lambda_{2}+\mu_{2})E-F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\\\pi_{6}&=&c_{1}L_{2}\frac{(c_{2}+\lambda_{2}+\mu_{2})\mu_{1}E+\mu_{2}c_{2}E-\mu_{1}F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\\\pi_{7}&=&c_{1}L_{3}\frac{(\lambda_{2}+\mu_{2})\mu_{1}E+\mu_{2}c_{2}E-(c_{2}+\mu_{1})F}{c_{1}\mu_{1}E^{2}+(c_{1}+\lambda_{1}+\mu_{1})EF+F^{2}},\end{array} $$

where

$$ \begin{array}{rll} E&=&c_{1}+c_{2}+\lambda_{1}+\lambda_{2}+\mu_{1}+\mu_{2}, \\ F&=&c_{2}\mu_{2}-c_{1}\mu_{1}.\end{array} $$

Then, \(P(1;0,1)=\pi _{5}\) and \(P(1;1,1)=\pi _{7}\) are obtained and \(T_{s}=\frac {1}{c_{2}[P(1;0,1)+P(1;1,1)]}\) can be calculated. □

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Wang, J., Li, J. & Howard, P.K. A system model of work flow in the patient room of hospital emergency department. Health Care Manag Sci 16, 341–351 (2013). https://doi.org/10.1007/s10729-013-9235-1

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