“…Now let S(H) be the family of subdivisions of H; it follows that sat(n, M(C r )) = sat(n, S(C r )). It is known that there exists a constant c so that 5 4 n ≤ sat(n, S(C r )) ≤ ( 5 4 + c r 2 )n + O(1) [1]. Finally, sat(n, M(K 3 )) = n − 1, sat(n, M(K 4 )) = 2n − 3, and sat(n, M(K 5 )) = 11 6 n + O(1); these values are easily obtained via well known characterizations of graphs which don't contain K 3 , K 4 , or K 5 as minors.…”