# The workforce scheduling problem

This model determines a set of workforce levels that will most economically meet demands and inventory requirements over time. The formulation is motivated by the experiences of a large producer in the United States. The data are for three products and 13 periods.

Problem taken from the Appendix C of the expanded version of Fourer, Gay, and Kernighan, A Modeling Language for Mathematical Programming

Originally contributed by Louis Luangkesorn, February 26, 2015.

using JuMP
import HiGHS
import Test

function example_prod(; verbose = true)
# PRODUCTION SETS AND PARAMETERS
prd = ["18REG" "24REG" "24PRO"]
# Members of the product group
numprd = length(prd)
pt = [1.194, 1.509, 1.509]
# Crew-hours to produce 1000 units
pc = [2304, 2920, 2910]
# Nominal production cost per 1000, used
# to compute inventory and shortage costs
#
# TIME PERIOD SETS AND PARAMETERS
firstperiod = 1
# Index of first production period to be modeled
lastperiod = 13
# Index of last production period to be modeled
numperiods = firstperiod:lastperiod
# 'planning horizon' := first..last;
# EMPLOYMENT PARAMETERS
# Workers per crew
cs = 18
# Regular-time hours per shift
sl = 8
# Wage per hour for regular-time labor
rtr = 16.00
# Wage per hour for overtime labor
otr = 43.85
# Crews employed at start of first period
iw = 8
# Regular working days in a production period
dpp = [19.5, 19, 20, 19, 19.5, 19, 19, 20, 19, 20, 20, 18, 18]
# Maximum crew-hours of overtime in a period
ol = [96, 96, 96, 96, 96, 96, 96, 96, 96, 96, 96, 96, 96]
# Lower limit on average employment in a period
cmin = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
# Upper limit on average employment in a period
cmax = [8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8]
# Penalty cost of hiring a crew
hc = [
7500,
7500,
7500,
7500,
15000,
15000,
15000,
15000,
15000,
15000,
7500,
7500,
7500,
]
# Penalty cost of laying off a crew
lc = [
7500,
7500,
7500,
7500,
15000,
15000,
15000,
15000,
15000,
15000,
7500,
7500,
7500,
]
# DEMAND PARAMETERS
d18REG = [
63.8,
76,
88.4,
913.8,
115,
133.8,
79.6,
111,
121.6,
470,
78.4,
99.4,
140.4,
63.8,
]
d24REG = [
1212,
306.2,
319,
208.4,
298,
328.2,
959.6,
257.6,
335.6,
118,
284.8,
970,
343.8,
1212,
]
d24PRO = [0, 0, 0, 0, 0, 0, 0, 0, 0, 1102, 0, 0, 0, 0]
# Requirements (in 1000s) to be met from current production and inventory
dem = Array[d18REG, d24REG, d24PRO]
# true if product will be the subject of a special promotion in the period
pro = Array[
[0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
]
# INVENTORY AND SHORTAGE PARAMETERS
# Proportion of non-promoted demand that must be in inventory the previous
# period
rir = 0.75
# Proportion of promoted demand that must be in inventory the previous
# period
pir = 0.80
# Upper limit on number of periods that any product may sit in inventory
life = 2
# Inventory cost per 1000 units is cri times nominal production cost
cri = [0.015, 0.015, 0.015]
# Shortage cost per 1000 units is crs times nominal production cost
crs = [1.1, 1.1, 1.1]
# Inventory at start of first period; age unknown
iinv = [82, 792.2, 0]
# Initial inventory still available for allocation at end of period t
iil = [
[
max(0, iinv[p] - sum(dem[p][v] for v in firstperiod:t)) for
t in numperiods
] for p in 1:numprd
]
# Lower limit on inventory at end of period t
function checkpro(
product,
timeperiod,
production,
promotionalrate,
regularrate,
)
if production[product][timeperiod+1] == 1
return promotionalrate
else
return regularrate
end
end
minv = [
[dem[p][t+1] * checkpro(p, t, pro, pir, rir) for t in numperiods]
for p in 1:numprd
]
# DEFINE MODEL
prod = Model(HiGHS.Optimizer)
# VARIABLES
# Average number of crews employed in each period
@variable(prod, Crews[0:lastperiod] >= 0)
# Crews hired from previous to current period
@variable(prod, Hire[numperiods] >= 0)
# Crews laid off from previous to current period
@variable(prod, Layoff[numperiods] >= 0)
# Production using regular-time labor, in 1000s
@variable(prod, Rprd[1:numprd, numperiods] >= 0)
# Production using overtime labor, in 1000s
@variable(prod, Oprd[1:numprd, numperiods] >= 0)
# a numperiods old -- produced in period (t+1)-a --
# and still in storage at the end of period t
@variable(prod, Inv[1:numprd, numperiods, 1:life] >= 0)
# Accumulated unsatisfied demand at the end of period t
@variable(prod, Short[1:numprd, numperiods] >= 0)
# CONSTRAINTS
# Hours needed to accomplish all regular-time production in a period must
# not exceed hours available on all shifts
@constraint(
prod,
[t = numperiods],
sum(pt[p] * Rprd[p, t] for p in 1:numprd) <= sl * dpp[t] * Crews[t]
)
# Hours needed to accomplish all overtime production in a period must not
# exceed the specified overtime limit
@constraint(
prod,
[t = numperiods],
sum(pt[p] * Oprd[p, t] for p in 1:numprd) <= ol[t]
)
# Use given initial workforce
@constraint(prod, Crews[firstperiod-1] == iw)
# Workforce changes by hiring or layoffs
@constraint(
prod,
[t in numperiods],
Crews[t] == Crews[t-1] + Hire[t] - Layoff[t]
)
# Workforce must remain within specified bounds
@constraint(prod, [t in numperiods], cmin[t] <= Crews[t])
@constraint(prod, [t in numperiods], Crews[t] <= cmax[t])
# 'first demand requirement
@constraint(
prod,
[p in 1:numprd],
Rprd[p, firstperiod] + Oprd[p, firstperiod] + Short[p, firstperiod] -
Inv[p, firstperiod, 1] == max(0, dem[p][firstperiod] - iinv[p])
)
# Production plus increase in shortage plus decrease in inventory must
# equal demand
for t in (firstperiod+1):lastperiod
@constraint(
prod,
[p in 1:numprd],
Rprd[p, t] + Oprd[p, t] + Short[p, t] - Short[p, t-1] +
sum(Inv[p, t-1, a] - Inv[p, t, a] for a in 1:life) ==
max(0, dem[p][t] - iil[p][t-1])
)
end
# Inventory in storage at end of period t must meet specified minimum
@constraint(
prod,
[p in 1:numprd, t in numperiods],
sum(Inv[p, t, a] + iil[p][t] for a in 1:life) >= minv[p][t]
)
# In the vth period (starting from first) no inventory may be more than v
# numperiods old (initial inventories are handled separately)
@constraint(
prod,
[p in 1:numprd, v in 1:(life-1), a in (v+1):life],
Inv[p, firstperiod+v-1, a] == 0
)
# New inventory cannot exceed production in the most recent period
@constraint(
prod,
[p in 1:numprd, t in numperiods],
Inv[p, t, 1] <= Rprd[p, t] + Oprd[p, t]
)
# Inventory left from period (t+1)-p can only decrease as time goes on
secondperiod = firstperiod + 1
@constraint(
prod,
[p in 1:numprd, t in 2:lastperiod, a in 2:life],
Inv[p, t, a] <= Inv[p, t-1, a-1]
)
# OBJECTIVE
# Full regular wages for all crews employed, plus penalties for hiring and
# layoffs, plus wages for any overtime worked, plus inventory and shortage
# costs. (All other production costs are assumed to depend on initial
# inventory and on demands, and so are not included explicitly.)
@objective(
prod,
Min,
sum(
rtr * sl * dpp[t] * cs * Crews[t] +
hc[t] * Hire[t] +
lc[t] * Layoff[t] +
sum(
otr * cs * pt[p] * Oprd[p, t] +
sum(cri[p] * pc[p] * Inv[p, t, a] for a in 1:life) +
crs[p] * pc[p] * Short[p, t] for p in 1:numprd
) for t in numperiods
)
)
# Obtain solution
optimize!(prod)
Test.@test termination_status(prod) == OPTIMAL
Test.@test primal_status(prod) == FEASIBLE_POINT
Test.@test objective_value(prod) ≈ 4_426_822.89 atol = 1e-2
if verbose
println("RESULTS:")
println("Crews")
for t in 0:length(Crews.data)-1
print(" $(value(Crews[t])) ") end println() println("Hire") for t in 1:length(Hire.data) print("$(value(Hire[t])) ")
end
println()
println("Layoff")
for t in 1:length(Layoff.data)
print(" \$(value(Layoff[t])) ")
end
println()
end
return
end

example_prod()
Presolving model
177 rows, 230 cols, 675 nonzeros
166 rows, 218 cols, 695 nonzeros
Presolve : Reductions: rows 166(-56); columns 218(-17); elements 695(-45)
Solving the presolved LP
Using EKK dual simplex solver - serial
Iteration        Objective     Infeasibilities num(sum)
0    -2.6189817551e+02 Ph1: 9(8.2635); Du: 3(261.898) 0s
111     4.4268228908e+06 Pr: 0(0); Du: 0(1.42109e-13) 0s
Solving the original LP from the solution after postsolve
Model   status      : Optimal
Simplex   iterations: 111
Objective value     :  4.4268228908e+06
HiGHS run time      :          0.00
RESULTS:
Crews
8.0  6.439849038461538  5.947339134615384  5.947339134615384  5.947339134615384  5.947339134615384  6.162946095647774  6.340057203947365  7.805664375000001  7.805664375000001  7.805664375000001  7.805664375000001  8.0  8.0
Hire
0.0  0.0  0.0  0.0  0.0  0.21560696103238985  0.17711110829959065  1.4656071710526364  0.0  0.0  0.0  0.194335624999999  0.0
Layoff
1.5601509615384623  0.49250990384615356  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0  0.0

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