Application to Optimization Problems
(Nuclear plant staffing problem) South Central Utilities has just announced the August 1 opening of its second nuclear generator at its Baton Rouge, Louisiana, nuclear power plant. Its personnel department has been directed to determine how many nuclear technicians need to be hired and trained over the remainder of the year.
The plant currently employs 350 fully trained technicians and projects the following personnel needs:
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 check-out, 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 through he or she is not required to work the 130 hours.
Training new employees is an important and costly procedure. It takes one month of one-on-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 trams up with a skilled nuclear technician and requires 90 ours 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 % 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.
a) Formulate this staffing problem using LP.
b) Solve the problem. How many trainees must begin each month?
Solution: We have the following decision variables
_{}
For this problem we have the following restrictions:
_{}
These restrictions are rewritten as:
_{}
The cost function is given by:
_{}
This function can be rewritten as:
_{}
Finally, the problem is written as:
_{}
We use Excel’s solver to get the following:
Target Cell (Min) |
||||||
Cell |
Name |
Original Value |
Final Value |
|||
$C$12 |
Z c1 |
3629143.5 |
3629143.5 |
|||
Adjustable Cells |
||||||
Cell |
Name |
Original Value |
Final Value |
|||
$C$4 |
x1 |
14 |
14 |
|||
$D$4 |
x2 |
0 |
0 |
|||
$E$4 |
x3 |
72 |
72 |
|||
$F$4 |
x4 |
0 |
0 |
|||
$G$4 |
x5 |
0 |
0 |
|||
Constraints |
||||||
Cell |
Name |
Cell Value |
Formula |
Status |
Slack | |
$C$16 |
c1 |
3740 |
$C$16>=0 |
Not Binding |
3740 | |
$C$17 |
c1 |
45 |
$C$17>=0 |
Not Binding |
45 | |
$C$18 |
c1 |
1312.75 |
$C$18>=0 |
Not Binding |
1312.75 | |
$C$19 |
c1 |
13.11 |
$C$19>=0 |
Not Binding |
13.11 | |
$C$20 |
c1 |
2511.634 |
$C$20>=0 |
Not Binding |
2511.634 | |
$C$4 |
x1 |
14 |
$C$4=integer |
Binding |
0 | |
$D$4 |
x2 |
0 |
$D$4=integer |
Binding |
0 | |
$E$4 |
x3 |
72 |
$E$4=integer |
Binding |
0 | |
$F$4 |
x4 |
0 |
$F$4=integer |
Binding |
0 | |
$G$4 |
x5 |
0 |
$G$4=integer |
Binding |
0 |
This means that the optimal solution corresponds to training 14 people in August and 72 people in October. The cost of this solution is $3,629,144.