Python 3.x 获取所用资源的价值

给定以下代码（示例来源：）

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由于使用的资源介于1和10之间，有没有办法知道给定解决方案使用的资源的确切数量？

您可以使用一些间接决策变量：

from docplex.cp.model import CpoModel, INTERVAL_MAX
import docplex.cp.utils_visu as visu

mdl = CpoModel()

a1=mdl.integer_var(name="a1")
a=mdl.interval_var(size=(1,10),end=(0,14),name="a")
b=mdl.interval_var(size=(1,10),end=(0,14),name="b")
c=mdl.interval_var(size=(1,10),end=(0,14),name="c")
d=mdl.interval_var(size=(1,10),end=(0,14),name="d")
#Horizon of the schedule is 14
# Cumul function resourceUse represents the level of a discrete resource.
# Activities a1,a2,a3,a4 require between 1 and 10 units of resource
resourceUse = mdl.pulse(a, 1,10) + mdl.pulse(b,1,10) + mdl.pulse(c, 1,10) + mdl.pulse(d, 1,10)

mdl.add(mdl.size_of(a)*mdl.height_at_start(a,resourceUse) >= 22)
mdl.add(mdl.size_of(b)*mdl.height_at_start(b,resourceUse) >= 22)
mdl.add(mdl.size_of(c)*mdl.height_at_start(c,resourceUse) >= 22)
mdl.add(mdl.size_of(d)*mdl.height_at_start(d,resourceUse) >= 22)
mdl.add(resourceUse<=7)

height_a = mdl.integer_var(0,10)
mdl.add(height_a == mdl.height_at_start(a,resourceUse))

height_b = mdl.integer_var(0,10)
mdl.add(height_b == mdl.height_at_start(b,resourceUse))

height_c = mdl.integer_var(0,10)
mdl.add(height_c == mdl.height_at_start(c,resourceUse))

height_d = mdl.integer_var(0,10)
mdl.add(height_d == mdl.height_at_start(d,resourceUse))


# Solve model

print("Solving model....")
msol = mdl.solve(FailLimit=100000, TimeLimit=10)
print("Solution: ")
msol.print_solution()



ha=msol[height_a]
hb=msol[height_b]
hc=msol[height_c]
hd=msol[height_d]

print(ha," ",hb," ",hc," ",hd)

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谢谢Alex，我这里有个新问题：，你能帮我看看吗？

a: (start=6, end=14, size=8, length=8)
b: (start=0, end=8, size=8, length=8)
c: (start=0, end=6, size=6, length=6)
d: (start=8, end=14, size=6, length=6)

from docplex.cp.model import CpoModel, INTERVAL_MAX
import docplex.cp.utils_visu as visu

mdl = CpoModel()

a1=mdl.integer_var(name="a1")
a=mdl.interval_var(size=(1,10),end=(0,14),name="a")
b=mdl.interval_var(size=(1,10),end=(0,14),name="b")
c=mdl.interval_var(size=(1,10),end=(0,14),name="c")
d=mdl.interval_var(size=(1,10),end=(0,14),name="d")
#Horizon of the schedule is 14
# Cumul function resourceUse represents the level of a discrete resource.
# Activities a1,a2,a3,a4 require between 1 and 10 units of resource
resourceUse = mdl.pulse(a, 1,10) + mdl.pulse(b,1,10) + mdl.pulse(c, 1,10) + mdl.pulse(d, 1,10)

mdl.add(mdl.size_of(a)*mdl.height_at_start(a,resourceUse) >= 22)
mdl.add(mdl.size_of(b)*mdl.height_at_start(b,resourceUse) >= 22)
mdl.add(mdl.size_of(c)*mdl.height_at_start(c,resourceUse) >= 22)
mdl.add(mdl.size_of(d)*mdl.height_at_start(d,resourceUse) >= 22)
mdl.add(resourceUse<=7)

height_a = mdl.integer_var(0,10)
mdl.add(height_a == mdl.height_at_start(a,resourceUse))

height_b = mdl.integer_var(0,10)
mdl.add(height_b == mdl.height_at_start(b,resourceUse))

height_c = mdl.integer_var(0,10)
mdl.add(height_c == mdl.height_at_start(c,resourceUse))

height_d = mdl.integer_var(0,10)
mdl.add(height_d == mdl.height_at_start(d,resourceUse))


# Solve model

print("Solving model....")
msol = mdl.solve(FailLimit=100000, TimeLimit=10)
print("Solution: ")
msol.print_solution()



ha=msol[height_a]
hb=msol[height_b]
hc=msol[height_c]
hd=msol[height_d]

print(ha," ",hb," ",hc," ",hd)

a: (start=0, end=6, size=6, length=6)
b: (start=8, end=14, size=6, length=6)
c: (start=0, end=8, size=8, length=8)
d: (start=6, end=14, size=8, length=8)
4   4   3   3