The graph. The signal flows from the input node R through three cascaded branches of gains 5, 3 and 2 to the output node C, with a feedback branch of gain −3 placed around the middle (gain-3) branch.

Tool — Mason's gain formula.

\(T=\dfrac{\sum_k P_k \Delta_k}{\Delta}, \qquad \Delta = 1-\sum L_i+\sum L_iL_j-\ldots\)

where P_k is a forward-path gain, L_i are the individual loop gains, and Δ_k is the determinant with all loops touching path k removed.

Step 1 — forward path. There is exactly one path from R to C that visits no node twice:

\(P_1 = 5\times 3\times 2 = 30\)

Step 2 — loops. One loop exists, formed by the gain-3 forward branch and the −3 feedback branch:

\(L_1 = 3\times(-3) = -9\)

Step 3 — determinant. With a single loop and no non-touching loops:

\(\Delta = 1-L_1 = 1-(-9) = 10\)

Step 4 — path cofactor. The forward path passes through the loop, so no loop remains untouched:

\(\Delta_1 = 1\)

Step 5 — combine.

\(T=\dfrac{P_1\Delta_1}{\Delta}=\dfrac{30\times1}{10}=3\)

Cross-check by block-diagram algebra. The inner loop is a negative-feedback block with forward gain 3 and feedback 3, giving \(3/(1+3\times3)=3/10=0.3\); cascading with the outer gains 5 and 2 gives \(5\times0.3\times2=3\) ✓ — the same answer, which is a good habit for confirming a Mason's-formula result.

Note the distractor. "Thirty" is the raw forward-path gain, i.e. the answer you get if the feedback loop is ignored; the feedback divides it by Δ = 10.

Hence, the transfer function of the signal flow graph is 3 (Three).