Computer Aided Manufacturing, Automation and Production Control
Axes, Loops and the Smallest Move an Axis Can Make
The axis convention is fixed by the spindle, not by the operator. Z always lies along the spindle axis, positive in the direction that increases the distance between tool and work; X is the longest travel perpendicular to Z and lies horizontal where a horizontal surface exists; Y completes a right-handed set. On a turning centre there is no Y, so the slides are X and Z, and X is programmed as a diameter rather than a radius on almost every control — a commanded X40.0 puts the tip 20 mm from the centre line. A vertical machining centre uses X, Y and Z. Rotary axes about those three are named A, B and C. Adding a rotary table gives a fourth axis and lets one setup reach several faces; adding two rotary axes gives five-axis machining, which is built either as table-table, where both rotations are carried by the work, or as head-head, where both are carried by the spindle, with the mixed arrangement in between. The motion-control classes are separate from the axis count: point-to-point control cares only where a move ends, which suits drilling and spot welding; straight-cut control feeds along one axis at a time; contouring control coordinates every axis continuously so the resultant tip path is a specified curve, and that is what milling a profile requires.
An open-loop axis commands a position and hopes; a closed-loop axis measures one. In the open-loop drive a stepping motor is given a train of pulses and advances a fixed angle per pulse, and nothing on the machine ever reports back where the slide actually is. It is cheap, it holds position without power dissipation in a servo amplifier, and it fails in one specific way: if the cutting load exceeds the motor torque the motor loses steps, the control does not know, and every coordinate for the rest of the programme is offset by the lost amount. In the closed-loop drive a rotary encoder on the screw or a linear scale on the slide reports actual position, the control forms the difference against the commanded position and drives the error to zero. The feedback may be indirect, taken from the leadscrew, in which case screw wear and backlash are outside the loop and are not corrected, or direct, taken from a scale on the slide, in which case they are inside the loop and are.
| Feature | Open loop | Closed loop |
|---|---|---|
| Position feedback | None at all | Encoder, resolver or linear scale |
| Usual motor | Stepping motor | DC or AC servo motor |
| Behaviour on overload | Loses steps silently and stays offset | Lags, then catches up or raises an alarm |
| Backlash and screw wear | Never corrected | Corrected only if the scale is on the slide |
| Typical use | Light duty, low cost, plotters and small routers | Production machine tools |
Control resolution, usually called the basic length unit, is the smallest increment of table motion the control can command. It is set by the transmission, not by the electronics alone. Take an axis whose leadscrew has a pitch of 6 mm and is driven through a 2 : 1 reduction, the motor carrying an encoder that issues 500 pulses per motor revolution. Two motor revolutions turn the screw once, so the screw receives 500 × 2 = 1000 pulses per revolution, and one screw revolution advances the table by one pitch. The basic length unit is therefore 6/1000 = 0.006 mm. A commanded move of 30 mm is 30/0.006 = 5000 pulses, and a feed of 360 mm/min is 6 mm/s, which is 6/0.006 = 1000 pulses per second. Every one of those three numbers changes if the gear ratio is misread, which is why the ratio is worth writing down before the arithmetic starts.
Interpolation is how a control makes a curve out of axes that can only be told to move. The interpolator takes the endpoint of a block and issues pulses to each axis at ratios that keep the resultant on the wanted path: a fixed ratio gives a straight line, a continuously varying ratio gives a circle, and a circle combined with a steady third axis gives a helix, which is how a thread or a helical ramp into a pocket is cut. The path a control actually produces is a chain of very short straight chords, and the chord error is what the basic length unit and the block processing rate together decide. One control feature belongs here because it exists only on CNC machines: constant surface speed. Under G96 the programmed S value is a cutting speed in m/min and the control varies the spindle speed to hold it as the diameter changes, using N = 1000V/(πD). Facing a disc at 150 m/min, the spindle runs at 1000 × 150/(π × 80) = 596.83 rev/min at 80 mm diameter and must reach 1000 × 150/(π × 20) = 2387.32 rev/min by the time the tool is at 20 mm — four times the speed, because the diameter has fallen to a quarter. That is also why every G96 programme carries a spindle speed clamp: without one the control would ask for an unbounded speed as the tool reaches the centre.
A Part Programme, Block by Block
A part programme is written in word-address format: each block is a line, each word is an address letter followed by a number, and the order of words within a block does not matter to the control. N numbers the block, G gives a preparatory function that sets a mode or a motion type, X Y Z give coordinates, I J K give arc centre offsets, R gives an arc radius, F the feed, S the spindle speed, T the tool and M a miscellaneous machine function. Most G words are modal: once G01 is active every following block with coordinates is a feed move until some other motion code replaces it, so a repeated code is omitted rather than repeated. A few are non-modal and act only in their own block, G04 dwell being the usual example. The same distinction applies to M words, and it matters because the commonest beginner defect is a block that inherits a mode nobody intended.
| Block | What it commands |
|---|---|
| N10 G21 G17 G90 G94 | Millimetre input, the XY plane for arcs, absolute coordinates, feed in mm per minute |
| N20 G54 G00 X0 Y0 Z25.0 | Select the first work offset and rapid to a safe height above the part origin |
| N30 T01 M06 | Change to tool one, here a four-flute end mill of 16 mm diameter |
| N40 S1200 M03 | Spindle on clockwise at 1200 rev/min |
| N50 M08 | Coolant on |
| N60 G00 X10.0 Y10.0 | Rapid over the start of the cut, still clear of the work |
| N70 G43 H01 Z5.0 | Apply the length offset stored in register one and come to 5 mm above the face |
| N80 G01 Z-3.0 F100 | Plunge 8 mm at 100 mm/min to a depth of 3 mm below the face |
| N90 G01 X70.0 Y10.0 F240 | Straight cut of 60 mm along the X direction at the working feed |
| N100 G02 X90.0 Y30.0 R20.0 | Clockwise quarter circle of 20 mm radius, its centre at X90 Y10 |
| N110 G01 Y70.0 | Straight cut of 40 mm along Y, the motion code and feed both inherited |
| N120 G00 Z25.0 M09 | Rapid clear and coolant off |
| N130 M05 | Spindle stop |
| N140 M30 | End of programme and reset to the first block |
Now read the cutting conditions out of the programme rather than off a chart. The spindle word and the tool together give the cutting speed: V = πDN/1000 = π × 16 × 1200/1000 = 60.32 m/min. The feed word is in mm per minute, so the load on each tooth has to be recovered: with four flutes at 1200 rev/min, the feed per tooth is 240/(1200 × 4) = 0.05 mm. Those two numbers, not S1200 F240, are what decide whether the cut is sensible, and they are the two a control never checks. The geometry gives the time. Block N90 cuts 60 mm; block N100 is a quarter circle of radius 20, so its arc length is 2π × 20/4 = 31.42 mm; block N110 cuts 40 mm. The profile is 60 + 31.42 + 40 = 131.42 mm at 240 mm/min, which is 0.5476 min, or 32.85 s. The plunge of N80 runs 8 mm at 100 mm/min, another 0.08 min or 4.80 s. The cutting time for the part is therefore 37.65 s, and the rapid moves add machine time but no cutting time at all.
| Move | Length, mm | Feed, mm/min | Time, s |
|---|---|---|---|
| Plunge from Z5.0 to Z-3.0 | 8.00 | 100 | 4.80 |
| Linear cut along X | 60.00 | 240 | 15.00 |
| Clockwise quarter arc | 31.42 | 240 | 7.85 |
| Linear cut along Y | 40.00 | 240 | 10.00 |
| Total cutting time | 139.42 | mixed | 37.65 |
I and J as the vector from the arc start point to the centre, always incremental even in G90. In the block above the tool starts at X70 Y10 and the centre is X90 Y10, so the equivalent block is G02 X90.0 Y30.0 I20.0 J0. The radius form gives R instead, and it is shorter but ambiguous: two arcs of the same radius join the same pair of points, one shorter than a semicircle and one longer. The convention resolves it with a sign — a positive R selects the arc of 180 degrees or less, a negative R selects the major arc — so R20.0 here is the quarter circle and R-20.0 would be the three-quarter circle sweeping round the other way. Two practical consequences follow. A full circle cannot be programmed with R at all, because the start and end points coincide and every radius satisfies them, so a full circle must use I and J. And when the chord is nearly 2R the radius form becomes numerically fragile, since a small rounding in the endpoints moves the computed centre a long way; I and J state the centre outright and do not suffer from it.G00rapid traverse,G01linear feed,G02clockwise arc,G03anticlockwise arc,G04dwellG17 G18 G19select the XY, ZX and YZ planes, which is what tells an arc block which two axes it lives inG20andG21select inch and millimetre input;G90andG91select absolute and incremental coordinatesG40 G41 G42cancel, left and right cutter radius compensation, which lets the programme carry the finished profile rather than a path offset by handG43applies a tool length offset from a register, so a tool of a different length needs a new offset and not a new programmeM03andM04start the spindle clockwise and anticlockwise,M05stops it,M06changes the toolM08andM09switch coolant on and off,M30ends the programme and rewinds it
Cylinders and Valves: Force from Pressure, Speed from Flow
A fluid power actuator converts pressure into force and flow into speed, and those are two independent statements. Pressure is set by the load, not by the pump or the compressor: a cylinder pushing against nothing sits at almost no pressure however hard the supply is trying. Flow is set by the valve and the supply, and it decides how fast the piston travels. A single-acting cylinder is pressurised on one side only and is returned by a spring or by the weight of the load, so it does useful work in one direction and needs only one supply line; a double-acting cylinder is pressurised on both sides in turn and can do work both ways. Two important variants exist. A double-rod cylinder has a rod out of both ends so the two areas are equal and the extend and retract behaviour is symmetric. A telescopic cylinder nests several stages to get a long stroke out of a short body, at the price of a force that falls stage by stage as the effective area shrinks.
Directional control valves are named by two numbers: ports first, then positions. A 3/2 valve has three ports and two positions and is what a single-acting cylinder needs — supply, cylinder, exhaust, switching the cylinder between supply and exhaust. A 5/2 valve has five ports and two positions and is the standard valve for a double-acting cylinder, because each cylinder port needs its own exhaust so that the two sides can be throttled independently. A 4/3 valve has four ports and three positions and appears mostly in hydraulics, where the centre position matters: a closed centre locks the actuator, a tandem centre locks it while unloading the pump to tank, and a float or open centre lets the actuator be moved by hand. Valves are actuated by a push button, a roller lever, a pilot pressure or a solenoid, and it is the solenoid that makes the system electro-pneumatic, because an electrical signal from a relay, a limit switch or a controller output can now command fluid. A single-solenoid valve is spring returned and so has a defined rest state when power fails; a double-solenoid valve is bistable and holds its last position, which is safer for some machines and much less safe for others.
Put numbers on one cylinder and every comparison in this section falls out of it. Take a double-acting cylinder of 80 mm bore with a 25 mm rod, supplied at 6 bar gauge. The full piston area is A = π × 80²/4 = 5026.55 mm², and 6 bar is 0.6 N/mm², so the extend force is 0.6 × 5026.55 = 3015.93 N. On the retract stroke the rod occupies part of the face, so the effective annulus is π × (80² − 25²)/4 = 4535.67 mm² and the retract force is 0.6 × 4535.67 = 2721.40 N, which is 4535.67/5026.55 = 90.23 per cent of the extend force. Note what that percentage is a ratio of: areas, not diameters. The rod is 31.25 per cent of the bore in diameter but only 9.77 per cent of it in area, and a candidate who subtracts diameters gets 68.75 per cent and is wrong by a wide margin. Speed comes from the same two areas and the flow. Feeding 20 L/min, which is 333 333 mm³/s, into the full bore gives an extend speed of 333 333/5026.55 = 66.31 mm/s, while the same flow into the annulus gives a retract speed of 333 333/4535.67 = 73.49 mm/s. A rodded cylinder always retracts faster than it extends and always pushes harder than it pulls, by exactly the same area ratio of 1.108.
| Quantity | Extend | Retract |
|---|---|---|
| Effective area, mm^2 | 5026.55 | 4535.67 |
| Force at 6 bar, N | 3015.93 | 2721.40 |
| Speed at 20 L/min, mm/s | 66.31 | 73.49 |
| Force at 150 bar, kN | 75.40 | 68.04 |
The choice between air and oil is decided by that last row and by compressibility, and almost never by anything else. The same cylinder at a hydraulic pressure of 150 bar develops 15 × 5026.55 = 75.40 kN, twenty-five times the pneumatic force, because pressure is twenty-five times higher; getting the same force from air would need a bore of 400 mm. Against that, air is compressible, which is why a pneumatic cylinder cannot hold an intermediate position accurately under a varying load and why it lurches when a stiff load suddenly yields — the air behind the piston is a spring that has been storing energy. Air is also cheap to exhaust to atmosphere, needs no return line, tolerates leaks as a nuisance rather than a hazard, and is safe in a wet or explosive area. Oil is stiff, so position and speed are controllable, and it carries away heat, but it needs a return line, a reservoir, filtration and a leak-free installation. Air is not free, though, and the cost is in the compressor. Sizing that compressor means counting free air, not swept volume, because the cylinder is filled with compressed air and the compressor is rated in air drawn from the atmosphere.
Work it through for the same cylinder on a 300 mm stroke running 15 complete cycles a minute. The extend stroke swallows 5026.55 × 300 = 1.508 litres of compressed air and the retract stroke 4535.67 × 300 = 1.361 litres, so one full cycle takes 2.869 litres at working pressure. The absolute pressure is 6 + 1 = 7 bar, so each litre at working pressure came from seven litres of free air: 2.869 × 7 = 20.08 litres of free air per cycle, and at 15 cycles a minute the demand is 301.2 litres of free air per minute. Three slips are worth naming because each gives a plausible wrong number. Forgetting the compression ratio entirely gives 43.0 L/min. Using the gauge pressure 6 instead of the absolute 7 gives 258.2 L/min, an error of one seventh that hides easily. Counting only the extend stroke gives 158.3 L/min, and it is the one that most often reaches a purchase order.
Electro-pneumatics is the layer where the fluid circuit meets the electrical one, and it is worth seeing what that layer actually contains before a controller is introduced. The power side stays as it was: compressor, receiver, a filter-regulator-lubricator unit, a 5/2 valve, a cylinder. The signal side becomes electrical: reed switches clamped to the cylinder barrel sense the magnetic piston at each end of stroke, push buttons and limit switches supply operator and position signals, and relays with normally-open and normally-closed contacts combine those signals and energise the valve solenoids. A two-cylinder sequence such as A+ B+ A- B- is then built by making each step trigger the next: the reed switch at the extended end of A energises the relay that fires B, and so on around the cycle, with a latching relay holding the start condition and a normally-closed emergency-stop contact breaking the whole control supply. That circuit works, and it is exactly the circuit whose weaknesses a programmable controller was invented to remove, since a change of sequence means rewiring a panel rather than editing a file.
The Programmable Controller and Its Scan
A programmable logic controller is a computer with its input and output hardened for a factory and its operating system reduced to one repeated loop. Five parts make it up. The processor holds the user program and executes it. Memory is divided in a way worth being precise about: program memory holds the rungs, while data memory holds the input image table, the output image table, and the timer, counter and internal bit files that the rungs read and write. The power supply feeds the rack and often the input field devices. The input and output modules are the whole reason the machine survives a factory: every channel is optically isolated, so a 240 V surge on a field wire does not reach the processor, and each module converts between field levels and logic levels. Finally a communications port and programming terminal load the program, force bits during commissioning and read the live image tables, which is how a PLC is debugged without stopping the plant. Physically these come as a modular rack, where a backplane accepts whatever mix of modules a machine needs, or as a compact brick with a fixed I/O count for small machines.
Everything surprising about a PLC follows from the scan, so learn the order before the instructions. One scan runs four stages and then starts again. First housekeeping: the processor checks memory, the watchdog timer and the I/O modules. Second the input scan: every input channel is read at that instant and copied into the input image table, and it is that frozen copy — not the field terminal — which the rungs will examine for the rest of the scan. Third the program scan: the rungs are solved top to bottom and, within a rung, left to right, and output instructions write to the output image table rather than to the terminals. Fourth the output scan: the whole output image is transferred to the output modules at once. Then communications are serviced and the loop repeats. The frozen images are the design decision that makes a PLC deterministic: a rung solved late in the scan sees exactly the same inputs as one solved early, so the logic cannot depend on how long the processor took.
| Stage | Contribution | Time, ms |
|---|---|---|
| Input and output scan | Fixed transfer of both image tables | 2.00 |
| Program scan | 5000 instructions at 0.08 microsecond each | 0.40 |
| Housekeeping and communications | Watchdog, diagnostics, port service | 0.50 |
| One complete scan | Sum of the three above | 2.90 |
| Worst case input to output | Two scans, when the input changes just after it was read | 5.80 |
Read the last row of that table carefully, because it is the number a machine designer needs and the number a beginner never computes. An input that changes state one microsecond after the input scan has passed it will not be seen at all this scan; it is read at the start of the next scan, acted on during that scan, and written to the terminal at the end of it. The delay from the physical event to the physical output is therefore up to two full scans, 5.8 ms here, plus the hardware filter delay of the input module, which is typically several milliseconds more and is there deliberately to reject contact bounce. The same arithmetic sets the shortest event a PLC can catch on an ordinary input: a pulse narrower than one scan may fall entirely between two input scans and be missed. Consider a proximity sensor watching a shaft at 1500 rev/min with one 30-degree target on it. The shaft turns once every 40 ms, and the target covers 30/360 of that, so the sensor output is high for 40 × 30/360 = 3.33 ms. Against a 2.90 ms scan that pulse is seen, but with no margin worth having, and at 2000 rev/min it would be 2.5 ms and would be missed intermittently — which is the worst possible failure, because it works on the bench. The fix is never a faster program; it is a high-speed counter module or a latching input, which catches the edge in hardware and holds it until the processor asks.
- Discrete DC input modules are wired as sinking or sourcing, which is a statement about the direction of conventional current at the module terminal and must match the sensor that drives it
- Relay outputs switch AC or DC and tolerate a wide load range but wear out mechanically; transistor outputs switch DC only, are silent and fast, and survive millions of operations
- Analogue input modules digitise 4 to 20 mA or 0 to 10 V signals; the current loop is preferred on a long run because a broken wire reads as zero current and is detectable, while a broken voltage wire reads as zero volts and looks like a real measurement
- A watchdog timer resets the processor if one scan overruns its limit, which is what stops a program fault from leaving the outputs frozen in a dangerous state
- An emergency stop is wired to drop power to the output devices directly and is never left to the program alone, because a processor fault must not be able to defeat it
Ladder Logic: Latches, Interlocks, Timers and Counters
Ladder logic is drawn as a relay schematic because it was meant to be read by people who already maintained relay panels. Two vertical rails carry notional power from left to right through horizontal rungs. A rung is a series and parallel arrangement of contacts ending in one or more coils: if a continuous path of true contacts exists from the left rail to the coil, the coil is energised. The one idea that must be right before anything else works is that a contact is a question about a bit in memory, not a picture of the field device. The symbol --] [--, called examine-if-closed, is true when its bit is a one; --]/[--, examine-if-open, is true when its bit is a zero. Whether the switch out in the plant is normally open or normally closed is a separate, electrical fact, and the two must be reasoned about together: a stop button wired as a normally-closed field contact holds its input bit at one while all is well, so the rung examines it with --] [--, and breaking the wire drops the bit to zero and stops the machine. That is a fail-safe design, and writing --]/[-- there instead would give a machine that runs correctly until the day its stop wire is cut.
| Rung | Logic | What it does |
|---|---|---|
| 1 | ( I0.0 OR Q0.0 ) AND I0.1 AND NOT Q0.1 -> Q0.0 | Forward contactor, sealed in by its own contact and blocked while reverse is on |
| 2 | ( I0.2 OR Q0.1 ) AND I0.1 AND NOT Q0.0 -> Q0.1 | Reverse contactor, the mirror image of rung one |
| 3 | Q0.0 OR Q0.1 -> TON T4:0, 10 ms x 750 | Runs while the motor runs and times out 7.5 s after it starts |
| 4 | I0.3 -> CTU C5:0, PRE 200, RES I0.4 | Counts parts past the sensor and sets its done bit at 200 |
Rung one is the latch, and the whole of it is in one parallel branch. I0.0 is a momentary start button, so it is true only while a finger is on it; Q0.0 is the output itself, examined as a contact in parallel with the button. Press the button and the rung is true, so Q0.0 goes to one; from the next scan onwards the parallel branch through Q0.0 is true on its own and the output holds even though the button has been released. That is the seal-in or holding contact, and it is how a momentary signal becomes a maintained state without any special instruction. I0.1 is the stop input in series with the whole branch, so it can break the path from either route, which is why the stop must be in series and not in parallel with the start. Some controllers offer set and reset coils instead, which latch without the seal-in branch; the behavioural difference is worth knowing, because a set coil is retentive across a power failure on most processors while a seal-in latch is not, and a machine that restarts itself when the supply returns is a different machine from one that waits to be told.
The interlock is the NOT Q0.1 in rung one and the NOT Q0.0 in rung two, and the sense of those contacts is the whole point. Each output is examined for the off state in the other output's rung, so energising either one immediately makes the other rung false and it cannot be energised until the first drops out. Putting an examine-if-closed contact there instead would produce the opposite machine, one in which neither contactor can start until the other is already running, and it would simply never start. Two cautions belong with this. A software interlock is not a safety function. A short across a contactor coil, a welded main contact or a processor fault defeats it, so a forward-reverse starter also carries a mechanical interlock on the contactors and, where the risk warrants it, a hardwired safety relay in the coil circuit. And the output image table is written only at the end of the scan, so if two different rungs write the same output coil, the one solved last silently decides the result and the earlier rung appears to have no effect at all — the commonest logic bug in a large program, and the reason a coil address is written in exactly one place.
| Instruction | Behaviour | Typical duty |
|---|---|---|
| TON | Accumulates while the rung is true and sets its done bit at the preset; resets to zero the moment the rung goes false | Delay a conveyor start until the dust extractor is up to speed |
| TOF | Done bit is set while the rung is true and stays set for the preset after it goes false | Keep a cooling fan running after the motor stops |
| RTO | Accumulates while true and retains the accumulated value when false, needing an explicit reset | Total running hours before a service is due |
| CTU | Increments once per false-to-true transition of its rung, not once per scan | Count parts into a carton of 200 |
| CTD | Decrements on each transition, and is paired with CTU to track a level that both fills and empties | Track how many pallets remain in a buffer |
A timer is set by two numbers and the product of them is the delay: the time base is the tick, and the preset is how many ticks. Rung three above uses a 10 ms base with a preset of 750, so the done bit sets 750 × 0.010 = 7.5 s after the rung became true. Choosing a coarse base and a small preset saves nothing and loses resolution, while an unnecessarily fine base can overflow a sixteen-bit accumulator on a long delay, so the base is chosen to suit the delay rather than by habit. Counters are simpler but carry one trap of their own: a counter increments on the transition of its rung, not on the condition being true. A rung that is true for two hundred consecutive scans increments the counter once, which is what allows a slow sensor to be counted correctly and what makes a counter useless for measuring how long something lasted.
Work Study: Method First, Then the Stopwatch
Productivity is a ratio of output to input, and production is only the numerator. A plant that doubles its output by doubling everything it consumes has raised production and left productivity where it was. Because there are several inputs, there are several ratios: partial productivity divides output by one input, so labour productivity is pieces per labour hour and machine productivity is pieces per machine hour; total factor productivity divides output by labour and capital together; total productivity divides it by every input valued in money. Each answers a different question, and they can move in opposite directions — installing a robot raises labour productivity and may lower total productivity if the machine is idle half the shift. Work study is the pair of techniques that improves the ratio without buying anything: method study attacks the numerator and the input together by changing how the job is done, and work measurement establishes how long the job should take once it is done that way.
Method study runs in a fixed order, and the order is the method. Select the job worth studying — usually a bottleneck, a high-cost operation or an unsafe one. Record what is done now, using a chart chosen to match the question: an operation process chart for the sequence of operations and inspections on a whole product, a flow process chart for one item through operations, transports, delays, inspections and storages, a string diagram or flow diagram for distance travelled on the floor, a two-handed process chart for the movements of one operator's hands, and a man-machine chart for the idle time of each when they work in turn. Examine the record by challenging every element in the order purpose, place, sequence, person, means. Develop the improved method, define it in writing, install it, and maintain it, since a method left unaudited reverts. Underneath sit the principles of motion economy — symmetrical two-handed motions, materials pre-positioned within a normal working area, gravity used to deliver and remove, hands relieved of holding by a jig — and beneath those the therbligs, the elementary motions such as reach, grasp, move and release into which any manual task can be decomposed.
Work measurement turns a stopwatch reading into a standard, and it takes two corrections to get there. The analyst times an element over many cycles and takes the mean, which is the observed time. That figure describes the operator who was watched, not a qualified average operator, so it is corrected by a rating factor: a rating above 100 means the observer judged the worker faster than standard pace, below 100 slower. Observed time times rating gives the normal time, sometimes called basic time. The normal time still assumes a person who never rests, so allowances are added — relaxation allowance covering personal needs and fatigue, contingency allowance for small irregular delays, and where applicable a policy or interference allowance. Normal time plus allowances is the standard time, and it is the number that feeds capacity planning, costing, line balancing and incentive schemes.
| Step | Working | Result |
|---|---|---|
| Observed time, mean of the timed cycles | given | 0.75 min |
| Normal time, after rating at 120 per cent | 0.75 x 1.20 | 0.90 min |
| Standard time, allowances 20 per cent of normal time | 0.90 x 1.20 | 1.08 min |
| Standard output in a 480 minute shift | 480 / 1.08 | 444 pieces |
| Labour efficiency when 400 pieces are actually made | 400 x 1.08 / 480 | 90.0 per cent |
NT × (1 + a), here 0.90 × 1.20 = 1.08 min, giving 444 pieces in a 480 minute shift. If instead it is expressed as a percentage of the standard time, meaning the fraction of the paid day that is allowed to be non-productive, then the normal time must occupy the remaining 80 per cent and standard time is NT/(1 − a) = 0.90/0.80 = 1.125 min, giving 427 pieces. The same stopwatch data and the same twenty per cent produce a four per cent difference in the standard, which is enough to matter in an incentive scheme and enough to make an answer wrong. The defence is to read the sentence in the question rather than the number: the phrase to look for is what the percentage is of. Where a question says only allowances of 20 per cent and nothing more, take the normal-time convention, and state the assumption. The same care is needed with the rating factor, which multiplies the observed time and never divides it — a rating of 120 means the operator was fast, so a qualified worker needs more time than was observed, and the normal time must come out larger than the observed time, which is the sanity check that catches an inverted rating instantly.Time study is not the only way to measure work, and the alternatives exist because it is expensive and intrusive. Work sampling takes a large number of instantaneous random observations and estimates the proportion of time spent in each activity rather than the duration of a cycle; it suits long, irregular and non-repetitive work such as maintenance, it needs no stopwatch, and its precision improves only as the square root of the number of observations, so halving the error costs four times the observations. Predetermined motion time systems such as MTM dispense with observation altogether by summing published times for elementary motions, which allows a standard to be set for a job that does not exist yet — the decisive advantage when a line is being designed. Standard data reuses element times already established for similar work. The choice is a trade between cost and directness, and the honest ordering is that a synthetic standard is cheaper and quicker while a timed standard is more defensible in front of the people it will be applied to.
Six Sigma: DMAIC, Defects per Million and Capability
Six sigma is a programme for reducing variation, and its name is a target expressed in standard deviations. The improvement cycle is DMAIC and each phase has a deliverable rather than an attitude. Define fixes the problem, the customer requirement it violates and the boundaries of the process, and leaves a charter. Measure establishes how the characteristic is measured, proves the measurement system itself is capable, and records the current performance as a baseline. Analyse finds the few causes that actually drive the variation, using the data rather than opinion. Improve changes the process and proves with a trial that the change worked. Control puts the new setting under a control chart and a revised procedure so the gain survives the departure of the team. A separate cycle, DMADV, applies when the process does not exist yet and has to be designed. The roles are borrowed from martial arts — green belts run projects part time, black belts full time, master black belts train and coach, champions remove obstacles — and the point of naming them is that the method is a resourcing commitment, not a technique.
The sigma level of a process is how many standard deviations fit between the process mean and the nearer specification limit, and the published table that converts it to defects assumes the mean has drifted 1.5 standard deviations off centre. That drift allowance is a convention, not a law of nature; it exists because a process observed over a long period wanders more than a short capability study suggests, and it is why six sigma corresponds to 3.4 defects per million rather than to the two parts per billion a perfectly centred normal distribution would give. Defects are counted as defects per million opportunities, which normalises for complexity: a unit with many chances to go wrong is not being compared unfairly with a simple one. If 5 defects are found in 400 units inspected and each unit offers 12 opportunities, then DPMO = 5/(400 × 12) × 10⁶ = 1041.7, which sits between the four and five sigma rows of the table, at about 4.6 sigma.
| Sigma level | Defects per million opportunities | Yield per cent |
|---|---|---|
| 2 | 308 537 | 69.15 |
| 3 | 66 807 | 93.32 |
| 4 | 6 210 | 99.38 |
| 5 | 233 | 99.977 |
| 6 | 3.4 | 99.99966 |
Process capability compares the voice of the process with the voice of the customer, and it takes two indices because a process can fail in two different ways. C_p = (USL − LSL)/6σ compares the specification width with the spread and says nothing whatever about where the process is sitting. C_pk = min[(USL − μ)/3σ, (μ − LSL)/3σ] takes the nearer limit and so folds the centring in. Work a case. A diameter is specified as 50.00 ± 0.15 mm, so USL = 50.15, LSL = 49.85 and the tolerance is 0.30 mm. The process has σ = 0.04 mm and a mean of 50.04 mm. Then C_p = 0.30/(6 × 0.04) = 1.25, the upper index is (50.15 − 50.04)/(3 × 0.04) = 0.9167 and the lower is (50.04 − 49.85)/0.12 = 1.5833, so C_pk = 0.9167. Read the pair together: the spread is comfortably inside the tolerance, and the process is still producing scrap, because it is sitting off centre and the upper limit is close. Recentring the mean on 50.00 with no reduction in variation at all would lift C_pk to 1.25, the value of C_p, which is the ceiling — C_pk can never exceed C_p, and the two are equal exactly when the process is centred.
| Index | Expression | Mean at 50.04 | Mean at 50.00 |
|---|---|---|---|
| C_p | (USL - LSL)/6s | 1.2500 | 1.2500 |
| Upper one sided index | (USL - mean)/3s | 0.9167 | 1.2500 |
| Lower one sided index | (mean - LSL)/3s | 1.5833 | 1.2500 |
| C_pk | the smaller of the two above | 0.9167 | 1.2500 |
C_pk from 0.9167 to 1.2500 in the right-hand column required no new machine, no better tooling and no reduction in variation at all — only an adjustment of the tool offset by 0.04 mm, which is an afternoon. Getting the same C_pk by attacking the spread instead, with the mean left at 50.04, would mean solving 0.11/3σ = 1.25, so σ = 0.02933 mm: the standard deviation would have to fall by 27 per cent, and reducing variation by a quarter is a machine rebuild, a tooling change or a material specification, all of them capital. Centring is an adjustment; spread is a capability, and the useful habit is to compute both indices and compare them before proposing anything, because a large gap between C_p and C_pk is a sign saying the cheap fix has not been done yet. The same reading runs the other way. If C_p itself is below the target then no amount of adjustment will help, since even a perfectly centred process cannot beat C_p, and any plan that consists of nudging the setting is going to fail. One caution on the numbers: indices are computed from a sample, so a C_pk of 0.92 from thirty parts carries a confidence interval wide enough to contain 1.0, and acting on the point estimate as though it were the truth is its own error.Lean and six sigma are often bolted together and they are not the same idea. Six sigma attacks variation and its instruments are statistical — capability studies, control charts, designed experiments. Lean attacks waste, meaning any activity the customer would not pay for, and its instruments are physical — pull scheduling with kanban cards, single-minute exchange of dies to make small batches economic, standard work, levelled production, and a stop-the-line rule so a defect is fixed rather than passed on. The waste categories usually listed are overproduction, waiting, transport, over-processing, inventory, motion and defects, with unused human ingenuity added later. They complement each other because they fail differently: a lean line with an unstable process starves the moment a defect appears, since there is no buffer to hide behind, and a statistically capable process inside a batch-and-queue factory still has a long lead time and a warehouse full of work in progress. Lean makes the problems visible by removing the inventory that was concealing them; six sigma is what fixes the problems lean exposes.
Reliability, Failure Rate and Availability Across a Product Life
Reliability is a probability with a mission attached: the probability that an item performs its function, under stated conditions, for a stated period. Drop the period and the number means nothing. The failure rate λ is the proportion of the surviving population failing per unit time, and when it is constant the reliability is R(t) = e^(−λt). Constant failure rate is the one case in which the arithmetic is easy, because the mean time between failures is then simply 1/λ. Take a pump with λ = 2 × 10⁻⁴ failures per hour. Its MTBF is 1/2 × 10⁻⁴ = 5000 hours, and its reliability over a 1000 hour campaign is e^(−0.2) = 0.8187. Two readings of that pair are worth fixing. The MTBF is not a life: at t = MTBF the reliability is e^(−1) = 0.368, so only 37 per cent of a population survives to its own mean time between failures. And a constant failure rate means the item is memoryless — a unit that has run 4000 hours is exactly as likely to fail in the next hour as a new one, which is why replacing a good component early buys nothing at all in this regime.
The constant failure rate is only the middle of the story, and the bathtub curve is the whole of it. Plot failure rate against age and three regimes appear. Infant mortality comes first, with a failure rate that falls with age, caused by manufacturing defects, bad assembly and weak components; it is removed by burn-in, that is by running the population under stress and shipping the survivors, and it is the reason a burn-in improves a delivered batch without improving any individual unit. The useful life follows, with a roughly constant failure rate dominated by random overloads rather than by age, and this is the only region where the exponential model and the MTBF arithmetic apply. Wear-out comes last, with a failure rate that rises with age through fatigue, abrasion, corrosion and creep; here age does predict failure, so scheduled preventive replacement pays, and it is exactly the region a memoryless model would advise against replacement in. The Weibull distribution parameterises all three with one shape factor: β < 1 is infant mortality, β = 1 reduces to the exponential, and β > 1 is wear-out.
Systems are built out of components, and two configurations cover most of the arithmetic. In series every unit must work, so reliabilities multiply: R_s = R₁R₂R₃…. That has an unpleasant consequence worth feeling — a series system is always less reliable than its worst component, and the more components there are the worse it gets, which is why adding a monitoring device can reduce the reliability of the thing it monitors. In parallel, or active redundancy, the system works if any one unit works, so it is the failures that multiply: R_p = 1 − (1 − R₁)(1 − R₂)…. Take three units in series at 0.95, 0.90 and 0.98 over the mission. The system is 0.95 × 0.90 × 0.98 = 0.8379. Duplicate the weakest unit so that the pair is in parallel: the pair is now 1 − 0.10² = 0.99, and the system becomes 0.95 × 0.99 × 0.98 = 0.9217. One duplicated component moved the system from 0.838 to 0.922, and it did so because redundancy pays where the failures are, not where the components are cheapest. When the failure rates are constant there is a second shortcut for series systems: the rates add, so units at 2 × 10⁻⁴, 3 × 10⁻⁴ and 5 × 10⁻⁴ per hour give λ_s = 10⁻³ and a system MTBF of 1000 hours — half the 2000 hours of the worst unit alone.
| Arrangement | Working | System reliability |
|---|---|---|
| All three in series | 0.95 x 0.90 x 0.98 | 0.8379 |
| The 0.90 unit duplicated in parallel | 0.95 x 0.99 x 0.98 | 0.9217 |
| The 0.95 unit duplicated instead | 0.9975 x 0.90 x 0.98 | 0.8798 |
| The whole chain duplicated | 1 - (1 - 0.8379)^2 | 0.9737 |
1/Σλ, so three units of MTBF 5000, 3333 and 2000 hours give a system MTBF of 1000 hours, not 3444 and not 2000 — adding MTBFs or averaging them are both wrong, and the correct route is always through the rates. And a parallel formula must be applied to the failures, not to the reliabilities. Two units at 0.90 in parallel give 1 − 0.10 × 0.10 = 0.99, not 0.90 × 0.90 = 0.81; multiplying the reliabilities there is using the series rule on a parallel arrangement and produces a redundancy that has made the system worse, which should be the signal that something is upside down. One further caution on redundancy, because it is the assumption that fails first in practice: the parallel formula assumes the units fail independently. Two pumps on one power supply, two sensors in one flooded pit or two channels compiled from one piece of software share a common cause, and against that cause the redundancy contributes nothing at all.Reliability alone does not tell an owner what fraction of the year a machine will actually be producing, and availability does. Availability is uptime divided by total time, and in its steady-state form it is A = MTBF/(MTBF + MTTR), where the mean time to repair measures maintainability — spares held, access designed in, diagnostics fitted, crew trained. The pump above with an MTBF of 5000 hours and an MTTR of 50 hours has A = 5000/5050 = 0.9901, so across a 8760 hour year it is down for about 86.7 hours. Now notice which lever is cheaper. Doubling the MTBF to 10 000 hours takes the availability to 0.9950; halving the MTTR to 25 hours takes it to 0.9950 as well — identical benefit, and the second is usually a stores decision rather than a redesign. That is the practical content of the whole topic, and it is what links it to lifecycle analysis: over a product life the owner pays for acquisition, energy, spares, repair labour and disposal, and the reliability decisions made at design time fix most of the cost that will be incurred after the sale. The design-side tools follow from that — FMEA, which walks each failure mode and scores severity, occurrence and detection into a risk priority number so that attention goes to the worst rather than to the most recent, and fault tree analysis, which works backwards from one undesired event to the combinations of causes that could produce it.
Key takeaways
- A neutral file decides what survives the trip.
STEPcarries solids with topology and tolerances,IGEScarries surfaces without topology,DXFis two-dimensional, andSTLkeeps only a triangular facet mesh — right for additive work, wrong for milling a curved surface, because the analytic surface is gone. - The basic length unit comes from the whole transmission, not from the encoder. A 6 mm pitch screw driven through a 2 : 1 reduction from a 500 pulse encoder gives
6/(500 × 2) = 0.006 mm; a 30 mm move is 5000 pulses and 360 mm/min is 1000 pulses per second. Resolution is a ceiling on accuracy, never a measure of it. - Read the cutting conditions out of the programme. With a 16 mm four-flute cutter at
S1200 F240, the cutting speed is 60.32 m/min and the feed per tooth is240/(1200 × 4) = 0.05 mm; a profile of60 + 31.42 + 40 = 131.42 mmat that feed takes 32.85 s, and the plunge adds 4.80 s for 37.65 s in all. - A rodded cylinder is asymmetric by area, not by diameter. An 80 mm bore with a 25 mm rod at 6 bar pushes with 3015.93 N and pulls with 2721.40 N, which is 90.23 per cent — and at 300 mm stroke and 15 cycles a minute it draws
2.869 × 7 × 15 = 301.2litres of free air a minute, the seven being the absolute pressure ratio that is so easy to drop. - A controller reads frozen images, so the scan sets the response. With 2.0 ms of I/O, 0.4 ms of logic and 0.5 ms of overhead the scan is 2.9 ms and the worst-case input-to-output delay is two scans, 5.8 ms; a pulse narrower than one scan can be missed entirely, and the cure is a high-speed input rather than a shorter program.
- Rating multiplies, allowances are added, and the wording decides which. Observed 0.75 min at a rating of 120 gives a normal time of 0.90 min; allowances of 20 per cent of the normal time give a standard time of 1.08 min and 444 pieces in a 480 minute shift, while the same 20 per cent taken of the standard time gives 1.125 min and 427 pieces.
C_pkcan never beatC_p, and reliabilities never average. For 50.00 ± 0.15 mm atσ = 0.04and a mean of 50.04,C_p = 1.25whileC_pk = 0.9167, and recentring alone lifts it to 1.25. In the same spirit, units at 0.95, 0.90 and 0.98 in series give 0.8379, not their mean; duplicating the 0.90 unit gives 0.9217, and the failure rates, not the MTBFs, are what add.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A design office must send the curved body of an impeller to an outside toolmaker who will mill it on a machining centre. Sending it as a triangular facet mesh rather than as a solid exchange file is a poor choice mainly because such a mesh
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Answer: C — replaces every curved face by flat triangles, so the exact surface that a tool path should be generated against no longer exists
A facet mesh is an approximation of the skin and nothing more: it carries triangles, and it discards the analytic surfaces, the topology, the units and the tolerances. A milling tool path has to be generated against the true surface, because the cutter contact point and the scallop height both depend on the local curvature, and a mesh has replaced that curvature by flat patches whose size was fixed when the mesh was exported. Refining the mesh shrinks the error without ever removing it, and the facet edges can print through onto a finished surface. The tempting wrong answer is the one about file size, and it is tempting because a fine mesh of a large model genuinely is a big file; but size is a nuisance that compression and a faster link solve, while lost curvature cannot be recovered from the mesh at any price. The claim about a single body per file is simply untrue, since a mesh file can carry several disjoint shells, and the claim about inches is the opposite of the real defect — the format records no units at all, which is why a mesh is sometimes imported at a thousandth of its intended size. A solid exchange file such as STEP keeps the surfaces and the topology and is what should have been sent; the mesh is the right choice only when the receiving process slices a mesh anyway, as additive manufacturing does.A machine slide is driven by a leadscrew of 6 mm pitch through a reduction gear of ratio 2 : 1 between the motor and the screw. The motor carries a rotary encoder issuing 500 pulses in one motor revolution. While the slide is being fed at 360 mm/min, the rate at which the control receives encoder pulses, in pulses per second, is
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Answer: B — 1000
Work down the transmission before touching the feed. The reduction is 2 : 1 from motor to screw, so the motor turns twice for one turn of the screw and the screw therefore receives 500 x 2 = 1000 pulses per revolution. One revolution of the screw advances the slide by one pitch, 6 mm, so the basic length unit is 6/1000 = 0.006 mm of travel per pulse. Now convert the feed to the same time unit: 360 mm/min is 6 mm/s, and 6/0.006 = 1000 pulses per second. The value 500 is the answer obtained by ignoring the gearbox and dividing the 6 mm pitch by the 500 pulses of the motor, which gives a basic length unit of 0.012 mm and exactly half the true pulse rate; it is the most tempting of the wrong answers because every number in the question is used and only the gear ratio is left out. The value 250 comes from applying the ratio the wrong way round, as though one motor revolution turned the screw twice, giving 0.024 mm per pulse. The value 60 000 is the arithmetic done correctly but left in minutes, dividing 360 mm/min by 0.006 mm without converting to seconds, and it is worth noticing that it is exactly sixty times the right answer. A useful check on the final figure: 1000 pulses per second at 0.006 mm each is 6 mm/s, which is the stated feed.In a milling programme running in absolute coordinates with the XY plane selected, the cutter centre is standing at X70.0 Y10.0 when the block
G02 X90.0 Y30.0 R20.0is executed. The path the cutter centre then follows isShow answer
Answer: B — a clockwise quarter circle whose centre lies at X90.0 Y10.0
Two points and a radius admit two centres, and the code words settle which one. The chord from X70.0 Y10.0 to X90.0 Y30.0 has length 28.28 mm, and the two points that lie 20 mm from both ends are X90.0 Y10.0 and X70.0 Y30.0. About the first, the start point sits at 180 degrees and the end point at 90 degrees, so travelling start to end is a decrease in angle, which is clockwise; about the second the same journey runs anticlockwise. SinceG02is clockwise, the centre is X90.0 Y10.0 and the arc sweeps a quarter circle of 31.42 mm. The sign of the radius does the rest of the work: a positive R selects the arc of 180 degrees or less, soR20.0is the quarter circle, whileR-20.0would have selected the clockwise 270 degree arc about the other centre. The anticlockwise option is the most tempting wrong answer, because it uses the correct alternative centre and the correct quarter sweep and fails only on the direction thatG02had already fixed; writingG03in that block is exactly what would produce it. The straight-line option is wrong because the radius form is a standard and complete way of writing an arc, with the centre offsetsIandJbeing an alternative rather than a requirement — hereI20.0 J0would express the same arc, measured incrementally from the start point to the centre even in absolute mode.A double-acting cylinder has a bore of 80 mm and a piston rod of 25 mm diameter, and the same supply pressure acts on it during both strokes. Its pulling force while retracting, expressed as a percentage of its pushing force while extending, is closest to
Show answer
Answer: D — 90.2
Force is pressure times area, so at one pressure the ratio of the two forces is simply the ratio of the two areas. Extending, the whole piston face is acted on: A = pi x 80 squared / 4 = 5026.55 square millimetres. Retracting, the rod occupies part of that face, so the effective area is the annulus pi x (80 squared minus 25 squared)/4 = 4535.67 square millimetres. The ratio is 4535.67/5026.55 = 0.9023, that is 90.2 per cent, and the pressure cancels so the answer is the same at 6 bar as at 150 bar. The tempting wrong answer is 68.8 per cent, which comes from taking the ratio of diameters rather than of areas: the rod is 25/80 = 31.25 per cent of the bore in diameter, and subtracting that from 100 gives 68.75. Area goes as the square of diameter, so the rod removes only (25/80) squared = 9.77 per cent of the face, which is the figure that appears as the 9.8 option, this time as the loss rather than as the remaining force. The answer 100 would require a cylinder with a rod out of both ends, whose two areas are genuinely equal. Two consequences of the same asymmetry are worth carrying: at one flow rate the cylinder retracts faster than it extends, by the reciprocal ratio 1.108, and the rod is therefore sized for buckling on the extend stroke where the force is largest.A motor is started and stopped from momentary push buttons through a programmable controller, and a second contactor reverses it. Which of the following statements about that program and its wiring are correct?
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Answer: A — A contact of the output itself, examined for its on state and placed in parallel with the momentary start contact, keeps the output energised after the button is released; C — Examining each contactor output for its off state in the other contactor's rung prevents the two from being energised together; D — Because the output image is transferred to the terminals only at the end of a scan, when two rungs write the same output coil it is the rung solved later that decides the state at the terminal
Three of the four describe the controller as it actually behaves. The parallel contact of the output is the seal-in or holding contact, and it is what turns a momentary press into a maintained state: the rung is true while the button is held, the output goes to one, and from the next scan onwards the parallel branch through that output holds the rung true on its own. Examining each output for its off state in the other rung is the correct interlock, because energising either output immediately makes the other rung false. And the output image is written to the terminals in one transfer at the end of the scan, so two rungs writing the same coil leave the later one in control and the earlier one with no visible effect at all, which is why a coil address should appear in exactly one place in a program. The false statement is the one claiming that a pulse between two input scans is still detected, and it is the most tempting because it describes how a microcontroller polling its own port would behave. A programmable controller deliberately does not work that way: every input is sampled once, at the input scan, and copied into a frozen image table that the rungs then examine for the whole of the program scan. That freeze is what makes the logic deterministic, and its price is exactly this blindness, so a pulse narrower than one scan can fall between two samples and never be seen. The remedy is a latching input or a high-speed counter module, which catches the edge in hardware.Forty timed cycles of a bench operation average 0.75 minutes each. The analyst rates the operator at 120 per cent, and allowances amounting to 20 per cent of the normal time are granted. In a shift offering 480 working minutes, the standard output, in pieces, is closest to
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Answer: B — 444
Three steps, taken in order. The stopwatch average is the observed time, 0.75 minutes, and it describes the operator who happened to be watched. Rating corrects it to a qualified worker at standard pace, and rating multiplies: a rating of 120 means the operator was faster than standard, so a standard worker needs more time, and the normal time is 0.75 x 1.20 = 0.90 minutes. Allowances are then added for relaxation and contingency, here 20 per cent of the normal time, giving a standard time of 0.90 x 1.20 = 1.08 minutes. The shift then yields 480/1.08 = 444.4, that is 444 pieces. The closest wrong answer is 427, and it comes from a genuine ambiguity rather than a blunder: if the 20 per cent were expressed as a fraction of the standard time instead of the normal time, the normal time would have to fit into the remaining 80 per cent and the standard time would be 0.90/0.80 = 1.125 minutes, giving 426.7 pieces. The wording of the question settles which convention applies, and the habit worth forming is to read what the percentage is taken of before dividing anything. The figure 533 is 480/0.90, the output of a worker granted no allowances at all, and 640 is 480/0.75, which uses the raw stopwatch reading and so ignores both corrections. A quick sanity check on the chain: the rating was above 100, so the normal time must exceed the observed time, and the standard time must exceed the normal time, which 0.75, 0.90 and 1.08 satisfy in order.A turned diameter is specified as 50.00 ± 0.15 mm. The process is stable, its standard deviation is 0.04 mm and its mean has settled at 50.04 mm. The process capability index
C_pk, rounded to two decimals, isShow answer
Answer: C — 0.92
The limits are 50.15 and 49.85, so the tolerance band is 0.30 mm wide. Compute both one-sided indices, becauseC_pkis defined as the smaller of them. Towards the upper limit, (50.15 − 50.04)/(3 × 0.04) = 0.11/0.12 = 0.9167; towards the lower limit, (50.04 − 49.85)/0.12 = 0.19/0.12 = 1.5833. The smaller governs, soC_pk= 0.92. The value 1.25 is the most tempting wrong answer because it is a real index of this process and a correctly computed one — it isC_p= 0.30/(6 × 0.04), which compares the tolerance width with the spread and deliberately says nothing about where the process is sitting. Quoting it here amounts to claiming the process is centred when it is 0.04 mm high, and the whole reasonC_pkexists is to refuse that claim. The value 1.58 is the other one-sided index, taken towards the far limit instead of the near one, which is the error of choosing the larger rather than the smaller. The value 0.63 comes from treating the half tolerance, 0.15 mm, as the full band in theC_pexpression. Two general facts follow from the numbers and are worth carrying:C_pkcan never exceedC_p, and the two are equal exactly when the process is centred, so recentring this mean on 50.00 with no reduction in spread whatever would liftC_pkfrom 0.92 to 1.25.Three units whose reliabilities over one mission are 0.95, 0.90 and 0.98 are connected so that all three must work. A fourth unit identical to the 0.90 one is then added in active redundancy beside it, the two failing independently. The reliability of the modified arrangement over the same mission is
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Answer: A — 0.9217
Deal with the redundant pair first, then with the chain. Two independent units in active redundancy fail only if both fail, so the pair survives with 1 − (1 − 0.90)(1 − 0.90) = 1 − 0.01 = 0.99. That pair now sits in a chain with the other two units, all of which must work, so the reliabilities multiply: 0.95 × 0.99 × 0.98 = 0.9217. Notice how much one duplicated unit bought — the chain was 0.95 × 0.90 × 0.98 = 0.8379 before, so the gain is more than eight points, and it came from duplicating the weakest member rather than the cheapest one. The value 0.8379 is the unmodified chain and is the answer of someone who computed the series correctly and then forgot to apply the redundancy at all. The most instructive wrong answer is 0.7541, which comes from multiplying the two redundant reliabilities together, 0.90 × 0.90 = 0.81, and then chaining: that applies the series rule to a parallel arrangement, and the giveaway is that it makes the system worse than before the spare unit was added, which no redundancy can do. The value 0.9737 is what you get by duplicating the entire chain instead of the single weak unit, 1 − (1 − 0.8379) squared, a different and more expensive design. One assumption is doing real work here and is the first to fail in practice: the parallel formula requires the two units to fail independently, so a common cause such as a shared power supply or a shared calibration would remove most of the benefit.