Operations Research, Closed Manufacturing Loops & Computer Performance
Gordon-Newell Closed Queueing & Buzen Convolution Formula
Calculate Exact 2-Station Gordon-Newell Normalization Constant G(N) via Buzen’s Convolution Sum, Exact Closed-Loop System Throughput G(N-1)/G(N), Bottleneck Asymptote 1/D_max, and Bottleneck Server Utilization.
Compute exact closed-pallet-loop throughput via Gordon-Newell & Buzen’s G(N-1)/G(N) theorem!
Flexible Manufacturing Systems with a fixed number of N circulating pallet fixtures—and database servers with a fixed N-connection thread pool—form Closed Gordon-Newell Queueing Networks where a new job enters ONLY when a completed job finishes! Unlike open Jackson networks, exact steady-state probabilities require Buzen’s Normalization Constant G(N) = SUM_{n1 + n2 = N} (X_1^n1 * X_2^n2), which for two single-server exponential stations with visit-weighted service demands X_1 and X_2 evaluates in closed form to (X_1^(N+1) - X_2^(N+1)) / (X_1 - X_2) when X_1 <> X_2, and (N + 1) * X_1^N when X_1 = X_2! Buzen’s celebrated ratio theorem then gives Exact System Throughput TH(N) = G(N - 1) / G(N) and Station-1 Utilization U_1(N) = X_1 * TH(N)!
Verify on a 2-Station Automated Machining Cell—Machining_Demand X_1 = 2.0 sec/part (B2), Optical_Inspection_Demand X_2 = 1.0 sec/part (B3), and Circulating_Fixtures N = 3 pallets (B4): Buzen Normalization Constant G(3) is (2^4 - 1^4) / (2 - 1) = 15.0000 (E2), while G(2) = 2^3 - 1 = 7; Exact Closed-Loop Throughput is G(2) / G(3) = 7 / 15 = 0.4667 parts/sec (E3) vs the Infinite-WIP Bottleneck Ceiling 1 / MAX(2, 1) = 0.5000 parts/sec (E4); and Bottleneck Machining Utilization reaches 2.0 * (7 / 15) = 0.9333 (93.33% in E5)!
| Gordon-Newell Closed Loop (X1=2.0s, X2=1.0s, N=3 Circulating Pallets) | Formula | Verified Output |
|---|---|---|
| Buzen Constant G(N) | IF(X1=X2, (N+1)*X1^N, (X1^(N+1)-X2^(N+1))/(X1-X2)) | 15.0000 (32.0000 @ X1=X2=2) |
| Exact System Throughput TH(N) | G(N - 1) / G(N) | 0.4667 /s (0.3750 @ X1=X2=2) |
| Infinite-WIP Bottleneck Cap | 1 / MAX(X_1, X_2) | 0.5000 /s |
| Station-1 Utilization U_1(N) | X_1 * TH(N) | 0.9333 (0.7500 @ X1=X2=2) |
Exact Buzen G(N-1) / G(N) Closed-Network Throughput Formula
Validate X_1 > 0, X_2 > 0, and integer-valued circulating population N >= 1, handling both unbalanced (X_1 <> X_2) and balanced (X_1 = X_2) demands smoothly!
=LET(x_1, B2, x_2, B3, n_p, B4, IF(OR(x_1 <= 0, x_2 <= 0, n_p < 1, INT(n_p) <> n_p), "Invalid Closed Network Input", LET(g_n, IF(ABS(x_1 - x_2) < 0.000001, (n_p + 1) * (x_1 ^ n_p), ((x_1 ^ (n_p + 1)) - (x_2 ^ (n_p + 1))) / (x_1 - x_2)), g_m, IF(ABS(x_1 - x_2) < 0.000001, n_p * (x_1 ^ (n_p - 1)), ((x_1 ^ n_p) - (x_2 ^ n_p)) / (x_1 - x_2)), ROUND(g_m / g_n, 4))))Balanced-Line Singularity Check (Set X_1 = 2.0 s and X_2 = 2.0 s)
Set Inspection_Demand B3 = 2.0 s (perfectly balanced with X_1 = 2.0 s at N = 3): the X_1 = X_2 branch avoids 0/0 division, giving G(3) = 4 * 8 = 32 in E2, Exact Throughput E3 = 12 / 32 = 0.3750 parts/s, and Utilization E5 = N / (N + 1) = 3 / 4 = 0.7500!
=IF(ISTEXT(E3), E3, LET(x_1, B2, x_2, B3, n_p, B4, g_m, IF(ABS(x_1 - x_2) < 0.000001, n_p * (x_1 ^ (n_p - 1)), ((x_1 ^ n_p) - (x_2 ^ n_p)) / (x_1 - x_2)), ROUND(x_1 * g_m / E2, 4)))Try the example
Use the original values or change them to test your understanding.