How to use the pyscipopt.multidict function in PySCIPOpt

def make_data():
    I,d = multidict({0:80, 1:270, 2:250, 3:160, 4:180})            # demand
    J,M,f = multidict({0:[500,1000], 1:[500,1000], 2:[500,1000]}) # capacity, fixed costs
    c = {(0,0):4,  (0,1):6,  (0,2):9,    # transportation costs
         (1,0):5,  (1,1):4,  (1,2):7,
         (2,0):6,  (2,1):3,  (2,2):4,
         (3,0):8,  (3,1):5,  (3,2):3,
         (4,0):10, (4,1):8,  (4,2):4,
         }
    return I,J,d,M,f,c

def make_data():
    I,d = multidict({1:80, 2:270, 3:250, 4:160, 5:180})            # demand
    J,M,f = multidict({1:[500,1000], 2:[500,1000], 3:[500,1000]}) # capacity, fixed costs
    c = {(1,1):4,  (1,2):6,  (1,3):9,    # transportation costs
         (2,1):5,  (2,2):4,  (2,3):7,
         (3,1):6,  (3,2):3,  (3,3):4,
         (4,1):8,  (4,2):5,  (4,3):3,
         (5,1):10, (5,2):8,  (5,3):4,
         }
    return I,J,d,M,f,c

def make_data():
    I,d = multidict({1:80, 2:270, 3:250, 4:160, 5:180})            # demand
    J,M,f = multidict({1:[500,1000], 2:[500,1000], 3:[500,1000]}) # capacity, fixed costs
    c = {(1,1):4,  (1,2):6,  (1,3):9,    # transportation costs
         (2,1):5,  (2,2):4,  (2,3):7,
         (3,1):6,  (3,2):3,  (3,3):4,
         (4,1):8,  (4,2):5,  (4,3):3,
         (5,1):10, (5,2):8,  (5,3):4,
         }
    return I,J,d,M,f,c

def example():
    J,v = multidict({1:16, 2:19, 3:23, 4:28})
    a = {(1,1):2,    (1,2):3,    (1,3):4,    (1,4):5,
         (2,1):3000, (2,2):3500, (2,3):5100, (2,4):7200,
         }
    I,b = multidict({1:7, 2:10000})
    return I,J,v,a,b

K: set of resources
    P: set of items
    f[t,p]: set-up costs
    g[t,p]: set-up times
    c[t,p]: variable costs
    d[t,p]: demand values
    h[t,p]: holding costs
    a[t,k,p]: amount of resource k for producing product p in period. t
    M[t,k]: resource upper bounds
    UB[t,p]: upper bound of production time of product p in period t
    phi[(i,j)] : units of i required to produce a unit of j (j parent of i)
    """
    T = 5
    K = [1]
    P = [1,2,3,4,5,6,7,8,9,10]
    _, f, g, c, d, h, UB = multidict({
        (1,1): [10, 1, 2,  0, 0.5, 24],
        (1,2): [10, 1, 2,  0, 0.5, 24],
        (1,3): [10, 1, 2,  0, 0.5, 24],
        (1,4): [10, 1, 2,  0, 0.5, 24],
        (1,5): [10, 1, 2,  0, 0.5, 24],
        (1,6): [10, 1, 2,  0, 0.5, 24],
        (1,7): [10, 1, 2,  0, 0.5, 24],
        (1,8): [10, 1, 2,  0, 0.5, 24],
        (1,9): [10, 1, 2,  0, 0.5, 24],
        (1,10):[10, 1, 2,  0, 0.5, 24],
        (2,1): [10, 1, 2,  0, 0.5, 24],
        (2,2): [10, 1, 2,  0, 0.5, 24],
        (2,3): [10, 1, 2,  0, 0.5, 24],
        (2,4): [10, 1, 2,  0, 0.5, 24],
        (2,5): [10, 1, 2,  0, 0.5, 24],
        (2,6): [10, 1, 2,  0, 0.5, 24],

def make_data():
    I,d = multidict({1:80, 2:270, 3:250, 4:160, 5:180})            # demand
    J,M,f = multidict({1:[500,1000], 2:[500,1000], 3:[500,1000]}) # capacity, fixed costs
    c = {(1,1):4,  (1,2):6,  (1,3):9,    # transportation costs
         (2,1):5,  (2,2):4,  (2,3):7,
         (3,1):6,  (3,2):3,  (3,3):4,
         (4,1):8,  (4,2):5,  (4,3):3,
         (5,1):10, (5,2):8,  (5,3):4,
         }
    return I,J,d,M,f,c

def make_1r():
    J, p = multidict({       # jobs, processing times
        1 : 1,
        2 : 3,
        3 : 2,
        4 : 2,
        })
    P = [(1,2), (1,3), (2,4)]
    R = [1]
    T = 6
    c = {}
    for j in J:
        for t in range(1,T-p[j]+2):
            c[j,t] = 1*(t-1+p[j])
    a = {
        (1,1,0):2,
        (2,1,0):2, (2,1,1):1, (2,1,2):1,
        (3,1,0):1, (3,1,1):1,

def example(n):
    """
    Data generator for the one machine scheduling problem.
    """
    J,p,r,d,w = multidict({
        1:[1,4,0,3],
        2:[4,0,0,1],
        3:[2,2,0,2],
        4:[3,4,0,3],
        5:[1,1,0,1],
        6:[4,5,0,2],
        })
    return J,p,r,d,w