Bracket Cell at 108% of Capacity: Choosing Between SMED, Batch Size, and Capital
A job shop machines a family of 6061 aluminum brackets through three steps: saw, CNC mill on two identical vertical machining centers, and manual deburr. Monthly demand just rose 25% to 9,400 good pieces and is expected to hold at that level. Each VMC is scheduled 2 shifts × 7.5 h × 21 days = 315 h per month, less 5% planned maintenance, giving 299 available hours per machine.
Milling runs at 3.2 minutes per piece with 4% scrap. The average batch is 350 pieces with a 4.5 hour changeover. Deburr has spare capacity and the saw is not constrained. Overtime is already at 6%. The customer lead-time commitment is 5 working days and the shop currently quotes 7. A third VMC is quoted at $185,000 with a 14-week lead time. The plant manager wants to double the batch size to 700 pieces because 'setups are killing us,' and the sales team wants a third machine ordered this week.
1. Confirm the constraint with arithmetic and state the size of the capacity gap in hours and percent.
Read the worked answer
Required starts. 9,400 good pieces at 4% scrap = 9,400 / 0.96 = 9,792 starts per month.
Run hours. 9,792 × 3.2 min = 31,334 min = 522.2 h of run time per month across both machines, or 261.1 h per machine.
Setup hours. 9,792 / 350 = 28 batches per month, split evenly = 14 batches per machine × 4.5 h = 63.0 h per machine.
Load per machine. 261.1 + 63.0 = 324.1 h against 299 available hours. That is a shortfall of 25.1 h per machine, or 108.4% utilization.
Interpretation. The milling pair is the constraint and it is already infeasible on paper - which means the 6% overtime, the 7-day actual lead time against a 5-day promise, and any expediting are all symptoms of that gap, not separate problems. Deburr and saw capacity are irrelevant until the mill is relieved; adding capacity anywhere else buys nothing.
Why this matters for the decision. The gap is 8.4%, which is small enough that a process improvement can close it and large enough that no improvement should be assumed without arithmetic. It also means a third machine would be justified at roughly 108% load only if demand growth is certain: a $185,000 machine at 14-week lead time arrives after the current quarter and would sit at roughly 72% utilization once installed. The decision should be sequenced: close the 8.4% with low-cost levers now, and place the capital order only against a demand forecast that keeps the cell above about 85% after the improvement.
2. Quantify three levers - SMED to 1.5 h changeover, batch size doubled to 700, and a 10% run-time reduction - and recommend an order of implementation.
Read the worked answer
Lever 1: SMED, changeover 4.5 h → 1.5 h. Saves 3.0 h × 14 setups = 42.0 h per machine. New load = 324.1 − 42.0 = 282.1 h, or 94.3% utilization. Closes the gap with 17 h per machine of headroom. Cost is fixture carts, pre-set tooling, standardized work, and externalizing the die/fixture change - typically a fraction of capital and available in weeks, not 14 weeks.
Lever 2: batch size 350 → 700. Batches fall to 14 total, 7 per machine, so setup = 7 × 4.5 = 31.5 h. New load = 261.1 + 31.5 = 292.6 h, or 97.8% utilization. It technically fits, but it leaves essentially no recovery margin for a tool failure or a quality hold, and it does nothing to build resilience. Note it buys 31.5 h versus SMED's 42.0 h - less benefit for a much larger inventory and lead-time penalty.
Lever 3: 10% run-time reduction (toolpath, feeds and speeds validated in aluminum, or high-feed cutters). Run hours fall to 235.0 h; load = 235.0 + 63.0 = 298.0 h, or 99.7%. Marginal on its own and it carries a real risk of pushing tool life or finish out of spec - treat it as a supplement, not a solution.
Recommendation. SMED first (largest gain, lowest capital, fastest). Verify with a 30-day load chart. Then a modest run-time program only if validated by a trial. Hold the $185,000 machine as a contingency against confirmed demand growth, not against this month's overtime.
3. Explain why doubling the batch size is the weakest option here, using Little's Law. What should the manager be told instead?
Read the worked answer
Little's Law. WIP = throughput × flow time, so flow time = WIP / throughput. At the mill, average WIP waiting for a machine is roughly half a batch plus the pieces in cut. Per machine, throughput is about 4,896 good pieces over 261 run hours, or about 18.8 pieces per hour.
Doubling the batch from 350 to 700 adds roughly 175 pieces of average queue per machine. ΔFlow time ≈ 175 / 18.8 ≈ 9.3 hours, about 1.2 working days, on the milling step alone. Because the bracket passes through three steps, queue time added at the constraint propagates: the shop's quoted 7-day lead time gets worse, not better, at exactly the moment the customer is being promised 5 days. Inventory carrying cost, floor space, and the cost of a quality escape caught late in a 700-piece batch all rise with it. And the gain is smaller than SMED's.
What the manager should hear. Setup time is indeed consuming 63 of 324 loaded hours - 19% of the mill's capacity - so the instinct is correct but the remedy is wrong. The goal is to reduce setup time, not to amortize it over more parts. Reducing changeover to 1.5 h recovers 42 h per machine per month with no inventory penalty and no lead-time penalty; doubling the batch recovers 31.5 h while making lead time worse. If batch size must rise, do it after SMED, when the economic batch quantity has genuinely moved - and size the new batch from the changeover cost and holding cost, not from a desire to feel busier.
Also worth stating plainly: overtime at 6% and a 2-day lead-time miss are the same constraint showing two symptoms. Fixing either symptom alone will not hold.