Man-machine chart: work, waiting and machine allocation
In one sentence
A man-machine chart shows readiness and waiting times together for machines served by the same worker; output may not rise in proportion to machine count.
By the end of this lesson
- Close the service, automatic processing and waiting intervals of two machines on a shared axis.
- Calculate operator waiting and additional machine waiting separately.
- Compare the two whole-number options adjacent to the theoretical machine count by output and resource use.
- Identify an assignment for which multiplying capacity linearly by machine count is invalid.
Why it matters
While one machine runs automatically, K01 can serve another. This possibility does not mean that every additional machine will provide the same output. K01 must unload the part, load a new one, complete a safe start and walk between machines. All of these use human time. A man-machine chart gives the person and each machine a separate lane to show what happens on a shared timeline. If the machine is ready while the worker is away, the machine waits. If the worker has returned but automatic processing has not finished, the worker waits. These are not two names for the same event; they are unavailable time for different resources and must be calculated separately.
Service time here is the total of unloading, loading and starting tasks performed while the machine is stopped. Automatic time starts after service. Walking covers travel from one machine to the next, including the return from the last machine to the first. The theoretical machine count formula is unreliable unless these definitions are preserved. Including service in automatic time and then adding it again in the formula counts the same work twice. Ignoring walking moves the worker to a location earlier than they can actually arrive. The chart is where these definitions are checked. A short formula does not mean it includes all working conditions. The machine's ability to run safely unattended and the need for timely worker intervention must be assessed separately.
Two limits are compared for a route through identical machines with constant times. The time needed for a machine to complete service and automatic processing is the lower limit for machine readiness. The time needed for the worker to complete service and travel for all machines is the lower limit for the route. Steady repetition cannot be shorter than the larger of these limits. Test on the chart whether a suitable sequence and staggered starts can actually achieve this value. Two machines requiring service simultaneously or different processing times may invalidate the simple model. The theoretical result is therefore a starting point for generating options, rather than a definitive assignment order. Events must be placed individually, especially with different machines, a shared fixture or additional quality work.
If machine count is fractional, examine the whole-number options below and above it. More machines can increase output while also increasing waiting per machine. No operator waiting alone does not indicate a good arrangement; filling all the worker's time with service and walking can make the arrangement sensitive to variability. Conversely, operator waiting does not require adding work to a safe supervision interval. Assess demand, quality, maintenance access and feasible assignment limits together. The calculations in this lesson compare ideal steady repetition. Without separately including breakdowns, setup, periodic work and nonconforming products, calculated output cannot be presented as actual shift production. The aim is to distinguish options numerically and show which assumptions support the result.
Formula
Theoretical machine count = (Service time + Automatic time) ÷ (Service time + Travel time)
Repetition time C for identical machines (s) = max(Service + Automatic time, Machine count × (Service + Travel))
Operator waiting (s/repetition) = C − Machine count × (Service + Travel)
Additional waiting per machine (s/repetition) = C − (Service + Automatic time)
Ideal output rate (units/h) = Machine count × 3,600 s/h ÷ C
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- Step-by-step instructions
- Worked example with real numbers
- Where the method misleads
- Common mistakes
Lab instructions: Operator and automatic process
- Measure 8 cycles at every station.
- Enter the elements with the names below and in this order. Each element starts at its start point and ends at its end point.
- Enter the operator's elements and waits on the Operator track and the machine's running on the Machine track; a machine row runs from the moment the machine starts until it stops.
- Enter a wait or a machine run as a row of its own every time you see it. Assign every row to its station.
1.Waiting: machine runningWait
2.Open the door
Starts: right hand reaches for the door handle · Ends: door is fully open
3.Take the machined part out of the chuck
Starts: right hand reaches for the chuck · Ends: part comes out of the chuck
4.Clamp the raw part in the chuck
Starts: left hand reaches for the chuck with the raw part · Ends: part is clamped in the chuck, hand lets go
5.Close the door
Starts: left hand reaches for the door handle · Ends: door is closed, hand lets go
6.Start the cycle
Starts: operator turns to the panel · Ends: start button is pressed
7.Machine runsMachine track
8.Deburr
Starts: operator walks to the side table · Ends: file is put down on the side table
9.Measure the part
Starts: left hand reaches for the caliper · Ends: caliper is put down on the side table
10.Put the part in the crate
Starts: right hand grips the part · Ends: part is put down in the crate
11.Take a raw part
Starts: operator turns to the cage · Ends: raw part comes out of the cage
12.Turn back to the lathe
Starts: operator turns to the lathe · Ends: operator stands at the lathe