1. calculation of the real personnel requirement after time optimisation
The reduction of personnel requirements is carried out in two steps:
a) Reduction in inpatient staffing requirements = time saved for ward work. Surgeon goes to the operating theatre later and returns earlier (no 2nd full-time employee required) = 5.7 knee TEP/T = 2.5 hours. The principle is applied to 28 operating theatres and calculated per month. The average time saving for all operations is 44 min / operation. From this follows:
44 x 28 = 1,232 min ≈ 20.5 hrs/D. ► 20.5 x 5 x 4.2 = 431 hrs/month ≈1 VK / month
i.e.: 1 Surgeon can be saved per month for ward work
b) Reduction of personnel requirements in the entire operating theatre.
For operating theatre scheduling, time probability is more suitable than time distribution.
In contrast to the usual, fixed ‘OP scheduling’, OP scheduling now results from the highest time probabilities of the confidence interval. This eliminates all idle times and makes it possible to ‘shrink’ the non-genuine personnel requirements by 33.4%. By calculating the activity density, the 8 operating theatres ‘shrink’ to just 5 operating theatres/day with the same performance.
Conclusion: 6,75 x 5 ≈ 34 Employees / operating theatre/T:
saving 20 Employees /operating theatre + 1 Surgeon
Total reduction in personnel requirements: 21 Employees / day (--38.1%)
2. Calculation of personnel cost savings
/ yr (balance sheet data; 1.3 employer flat rate)
For 8 operating theatres + 1 stat; ≈ € 3,400,000 / yr
For 5 operating theatres: 26 x 1.3 x 12 ≈ 2,100,000 € / Y ►3,4 – 2,1 ≈ € 1,3 million
Reduction in personnel costs: € 1.3 million / year (-- 35.2%)
3. Profitability of the investment
Licence agreement for 5 years secures the new personnel requirements digitally. Total value: One-off 75% of new personnel costs/year: 1,300,000 x 0.75 = € 975,000
→ € 14,300 / month licence fee
Investment ratios:
Net present value: Co (8%) = 3.027; → Internal rate of return: ro: 42.7%
Net present value rate (5 years) = 571%; average annual excess return k* = 46.3%