EPQcoo(n=3,a=600,d=c(500,300,400),h=c(9.6,11,10),r=rep(600,3),b=c(100,150,200))
#EPQ model
#Cooperative case
#$`Optimal order
# 1 2 3 Costs
# 0.0000 0.0000 0.0000 0.0000
# 641.0928 0.0000 0.0000 935.9019
# 0.0000 265.0557 0.0000 1358.2049
# 0.0000 0.0000 388.8444 1234.4268
# 363.7611 218.2567 0.0000 1649.4341
# 387.3208 0.0000 309.8566 1549.1036
# 0.0000 196.1473 261.5297 1835.3556
# 291.2332 174.7399 232.9866 2060.2045
#
#$`Optimal shortages`
# 1 2 3
# 0.000000 0.000000 0.000000
# 9.359019 0.000000 0.000000
# 0.000000 9.054699 0.000000
# 0.000000 0.000000 6.172134
# 5.310381 7.455973 0.000000
# 5.654318 0.000000 4.918359
# 0.000000 6.700683 4.151265
# 4.251580 5.969377 3.698200
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