(Nuclear plant staffing problem) South Central Utilities has just announced the

Question

(Nuclear plant staffing problem) South Central Utilities has just announced the August 1 opening its second nuclear generator at its Baton Rou, Louisiana, nuclear power plant. Its personnel department has been directed to determine how many nuclear technicians need to be hired and trained the remainder of the year. The plant currently employs 350 fully trained technicians and projects the following personnel needs: MONTH PERSONNEL HOURS NEEDED August 40,000 September 45,000 October 35,000 November 50,000 December 45,000 By Louisiana law, a reactor employee can actually work no more than 130 hours per month. (Slightly over one hour per day is used for check-in and checkout, recordkeeping, and for daily radiation health scans.) Policy at South Central Utilities also dictates that layoffs are not acceptable in those months when the nuclear plant is overstaffed. So, if more trained employees are available than are needed in any month, each worker is still fully paid, even though It or she is not required to work the 130 hours. Training new employees is an importantly costly procedure. It takes one month of one-to-one classroom instruction before a new technician is permitted to work alone in the reactor facility. Therefore South Central must hire trainees one month before they are actually needed. Each trainee teams up with a skilled nuclear technician and requires 90 hours of that employee's time, meaning that 90 hours less of the technician's time are available that month for actual reactor work. Personnel department records indicate a turnover rate of trained technicians at 5% per month. In other words, about 5% of the skilled employees at the start of any month resign by the end of that month. A trained technician earns an average monthly salary of $2,000 (regardless of the number of hours worked, as noted earlier). Trainees are paid $900 during their one month of instruction. - Formulate this staffing problem using LP. Could you solve using excel solver; no graphs needed.

Explanation / Answer

Let the decision variables be:

e1 = Number of employees for the month of August

e2 = Number of employees for the month of September

e3= Number of employees for the month of October

e4 = Number of employees for the month of November

e5 = Number of employees for the month of December

t1 = Number of trainees for the month of August

t2 = Number of trainees for the month of September

t3 = Number of trainees for the month of October

t4 = Number of trainees for the month of November

t5 = Number of trainees for the month of December

The objective function can be stated as MINIMIZING total cost = Employee cost + Trainee cost =

$2000e1 + $2000e2 + $2000e3 + $2000e4 + $2000e5 + $2000t1 + $900t2 + $900t3 + $900t4 + $900t5

Subject to the following constraints

Balance equations or constraints

(Number of employees) + (number of trainees) – (employee turnover) = (number of employees and trainees at end of current period)

e1 = 350                               (Number of employees in August)

e1 + t1 – (0.05*e1) = e2 (Number of employees in September)

e2 + t2 – (0.05*e2) = e3 (Number of employees in October)

e3 + t3 – (0.05*e3) = e4 (Number of employees in November)

e4 + t4 – (0.05*e4) = e5 (Number of employees in December)

Demand constraints

(e1*130) – (t1*90) >= 40,000 (Demand for employees in August)

(e2*130) – (t2*90) >= 45,000 (Demand for employees in September)

(e3*130) – (t3*90) >= 35,000 (Demand for employees in October)

(e4*130) – (t4*90) >= 50,000 (Demand for employees in November)

(e5*130) – (t5*90) >= 45,000 (Demand for employees in December)

e1, e2, e3, e4, e5, t1, t2, t3, t4, t5 >= 0 (non-negativity)

LP Model Formulation

MIN Z =

$2000e1 + $2000e2 + $2000e3 + $2000e4 + $2000e5 + $900t1 + $900t2 + $900t3 + $900t4 + $900t5

s.t.

e1 = 350                              

e1 + t1 – (0.05*e1) = e2

e2 + t2 – (0.05*e2) = e3

e3 + t3 – (0.05*e3) = e4

e4 + t4 – (0.05*e4) = e5

Demand constraints

(e1*130) – (t1*90) >= 40,000

(e2*130) – (t2*90) >= 45,000

(e3*130) – (t3*90) >= 35,000

(e4*130) – (t4*90) >= 50,000

(e5*130) – (t5*90) >= 45,000

e1, e2, e3, e4, e5, t1, t2, t3, t4, t5 >= 0 (non-negativity)

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