In the September 2000 issue of APCO's monthly Public Safety Communications, Editor Jennifer Hagstrom gave a 16-variable formula for determining the number of dispatchers required to cover a set number of positions 24x7x365. It does not determine how many positions you need to perform a certain number of tasks (call-taking, radio dispatching, warrants, tows, etc.). Rather, it's a staffing calculation, taking into account vacation, sick leave, days-off and other vacancies, to come up with the number of total positions to provide 24-hour staffing. math corrected 1-22-2001

The calculations basically figure two factors, one for breaks and the other for leave. The factors are then multiplied times the number of required positions (seats) on a shift to give the number of persons on-staff that are required to keep those seats filled 24 hours a day, 365 days a year.

We've simplified the formula below, and modified it to allow for different staffing on each shift (the original calculation assumes staffing is the same for each of three shifts), and more than three different types of shifts.

• break factor (BF) = (365 x (8~10~12))÷ (365 x ([8~10~12] - breaks per shift))
• BF x number of positions (seats) required per shift = positions to cover breaks (PcB)
• leave factor (LF) = 2080 ÷ (2080 - average hours worked)
• total staff needed for one shift= LF x PcB

Note: average hours worked = 2080 ­ (vacation, sick leave, jury duty, death leave, disability, etc.)

For example, calculating one hypothetical shift's requirement, assume these conditions:

• two breaks of 15 minutes each, and one 30-minute meal break (1 hour total, making a 7-hour workday)
• 8-hour shifts
• average of 160 hours of vacation, sick leave, death leave, jury duty, etc. per year, per employee (based on past records)
• 3 call-taking, 1 radio dispatch, and 1 service radio position required to be staffed on the shift

Here are the calculations based on the above hypothetical comm center shift:

• 2920 ÷ 2555 = 1.142 (BF)
• 1.142 x 6 = 6.852 positions (PcB)
• leave factor = 1.0833 [2080 ÷ (2080 - 160)]
• total staff needed to cover 6 positions on one shift= 1.0833 times 6.852 or 7.442 for shift of six positions
• repeat for each different shift configuration, total the results for staffing requirement for the entire comm center

Here's a worksheet to get you started. Print this out and fill in the blanks.

A. __________ length of shift (8, 10 or 12 hours)

B. __________ total break time per shift in minutes per shift

C. __________ total meal time per shift in minutes per shift

D. __________ total break & meal time in hours per shift (B+C), and multiplied by 365

E. __________ average amount of leave used (vacation, sick, death, jury, disability) per year, per employee in hours

F. __________ number of work positions to be staffed on the shift (call-taker, radio, etc.)


Step 1. Multiply 365 times the length of shift (A) ______________

Step 2. Subtract total break & meal times (C) from result in Step 1 ______________

Step 3. Divide result in Step 1 by result in Step 2 ______________

Step 4. Multiply result in Step 3 by number of personnel required on the shift (F) ______________

Step 5. Subtract average leave (E) from 2080 ______________

Step 6. Divide 2080 by result in Step 5 ______________

Step 7. Multiple the result in Step 6 by result in Step 4 ______________. You're done! Repeat for each shift.

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