4.3 Capacity Analysis: Machine Requirements, Scrap Allowances & Personnel Requirements

4.3 Capacity Analysis: Machine and Personnel Requirements

Facilities planning starts with a capacity question: how many machines and people does the production plan require? The answer sets the floor space in the space-requirement models, the number of material handling moves, and the budget. The NCEES specification lists "calculation of personnel requirements" and "calculation of machine requirements" under capacity analysis.

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1. Step One: Work Backward Through Scrap

Every operation loses some units to scrap. To deliver $O_k$ good units from operation $k$ with scrap fraction $d_k$, the operation must start with:

$I_k = \frac{O_k}{1 - d_k}$

The input to operation $k$ is the required output of operation $k - 1$, so the calculation runs backward from the final operation.

Example: A plant must ship 1,000 good shafts per day. The routing has three operations:

The plant must start about 1,119 blanks per day. The overall yield is $0.96 \times 0.98 \times 0.95 = 0.894$, and $1{,}000 / 0.894 = 1{,}119$ gives the same answer. Note that you divide by $(1 - d)$; multiplying 1,000 by 1.05 understates the need, because 5% of the larger input is also lost.

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2. Step Two: Machine Requirements

A common facilities-planning form of the machine requirement is:

$F = \frac{S \cdot Q}{E \cdot H \cdot R}$

Example (Operation 20, milling): $S = 0.12$ hr per unit, $Q = 1{,}074.1$ units per day, $E = 0.90$, two 8-hour shifts so $H = 16$ hr, and $R = 0.95$:

$F = \frac{0.12 \times 1{,}074.1}{0.90 \times 16 \times 0.95} = \frac{128.9}{13.68} = 9.42 \text{ machines}$

Rounding. Ten machines cover the load with about 6% spare capacity. Nine machines would need roughly 6% overtime or a third shift on some machines. On an exam, round up unless the problem allows overtime or subcontracting. In design work, compare the cost of a tenth machine against overtime premiums.

Including Setup Time

When an operation runs several products or lots, setups consume machine time:

$\text{Machine hours required} = \sum_j \left( n_j \cdot t_{\text{setup},j} + Q_j \cdot t_{\text{run},j} \right)$

where $n_j$ is the number of setups for product $j$ in the period. Divide by $E \cdot H \cdot R$ to get machines. Cutting lot sizes in half doubles the number of setups. Without setup reduction, smaller lots increase the machines required, which is the capacity side of the JIT tradeoff.

Example: A press runs two parts per week. Part X needs 4 setups of 1.5 hr and 6,000 pieces at 0.01 hr each. Part Y needs 6 setups of 2 hr and 3,000 pieces at 0.02 hr each. Machine hours $= 4(1.5) + 60 + 6(2) + 60 = 138$ hours. With $H = 80$ hr per week, $E = 0.95$, and $R = 0.92$, $F = 138 / 69.9 = 1.97$, so two presses are needed.

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3. Step Three: Personnel Requirements

Direct Labor for Manual Operations

$\text{Operators} = \frac{\sum (\text{units} \times \text{standard hours per unit})}{\text{productive hours per worker} \times \text{efficiency}} \times \frac{1}{1 - \text{absenteeism}}$

Example: A manual assembly area must build 1,200 units per day at a standard of 0.40 hr per unit (allowances already included). Each worker has 7.5 productive hours per shift and works at 95% efficiency, and absenteeism averages 5%.

$\frac{1{,}200 \times 0.40}{7.5 \times 0.95} = \frac{480}{7.125} = 67.4 \text{ workers present}, \qquad \frac{67.4}{0.95} = 70.9 \Rightarrow 71 \text{ on the roster}$

Machine Operators

For automated equipment, staffing depends on how many machines one person can tend. That comes from worker–machine (man–machine) analysis, covered in the methods engineering chapter. If each operator can tend three machines, ten mills need four operators per shift (10 ÷ 3 rounded up), not ten.

Indirect and Support Staff

Material handlers, setup technicians, inspectors, and maintenance staff are sized from their own workloads: moves per shift times minutes per move, setups per week times setup hours, and so on. Support staffing is often checked against ratios from comparable plants.

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4. Capacity Concepts That Tie It Together

• Design capacity is the maximum output under ideal conditions. Effective capacity allows for setups, maintenance, and product mix. Actual output is what is achieved.
• A capacity cushion (planned spare capacity) protects against demand variability and breakdowns. Queueing theory shows that waiting time rises sharply as utilization nears 100%, so do not plan machines at 100% load.
• The machine count for each operation reveals the bottleneck: the operation with the highest load relative to its capacity sets plant throughput.