9.2 Worker–Machine (Man–Machine) Charts, Multiple Activity Analysis & Synchronous Servicing

Key Takeaways

  • A worker–machine (man–machine) chart plots the worker and each machine on a common time scale, showing concurrent, independent, and idle time in one cycle.

  • For synchronous servicing, the ideal number of machines per operator is n' = (l + m) ÷ (l + w), where l is load/unload time, m is automatic machine time, and w is the worker's walk and other independent time per machine.

  • If n ≤ n', the cycle time is l + m and the worker has idle time; if n > n', the cycle time is n(l + w) and the machines wait for the worker.

  • Cost per unit is (K1 + n × K2) × cycle time ÷ n, where K1 is the operator cost rate and K2 is the machine cost rate; compare the integers on each side of n'.

  • When machines stop at random rather than on a fixed cycle, machine interference must be estimated with probability models or simulation rather than the synchronous formula.

Last updated: October 2026

9.2 Worker–Machine Charts & Synchronous Servicing

The specification lists man–machine charts with therbligs and motion study as methods for analysis and improvement. Process charts follow one subject, such as a part or a worker. A multiple activity chart puts several subjects side by side on one time scale, so you can see who waits for whom.


1. Building a Worker–Machine Chart

Draw one vertical column per subject (the worker and each machine), with time running down the page. Each block is shaded by type:

  • Combined (concurrent) work: the worker and machine are both engaged, such as loading or unloading.
  • Independent work: the worker inspects, packs, or walks while the machine runs automatically, or the machine runs while the worker is elsewhere.
  • Idle: waiting for the other party.

At the bottom, summarize the cycle time, the working and idle time of each subject, and the utilization (working time ÷ cycle time).

Example: one operator, one CNC lathe. Loading takes 1.0 min (worker and machine together), the automatic cut takes 4.0 min, and the operator spends 0.5 min deburring the previous part.

Time (min)   OPERATOR                 LATHE
0.0 - 1.0    Load / unload (with M)   Being loaded
1.0 - 1.5    Deburr previous part     Cutting (automatic)
1.5 - 5.0    IDLE                     Cutting (automatic)
Cycle = 5.0 min | Operator busy 1.5 min (30%) | Lathe busy 5.0 min (100%)

The operator is idle 70% of the cycle. That idle time is the opportunity to have one person tend several machines.


2. Synchronous Servicing

Synchronous servicing applies when machines run identical, regular cycles and the worker visits them in a fixed order. Define:

  • ll = loading and unloading time per machine (worker and machine both occupied)
  • mm = automatic machine run time
  • ww = worker's independent time per machine (walking to the next machine, inspecting, packing)

The ideal number of machines per operator, where neither worker nor machines wait, is:

n′=l+ml+wn' = \frac{l + m}{l + w}

Usually n′n' is not an integer, so compare the integers on either side:

CaseCycle time TcT_cWho waits
n≤n′n \le n' (fewer machines)Tc=l+mT_c = l + mThe worker is idle Tc−n(l+w)T_c - n(l + w) per cycle
n>n′n > n' (more machines)Tc=n(l+w)T_c = n(l + w)Each machine is idle Tc−(l+m)T_c - (l + m) per cycle

Each cycle produces nn units. With operator cost rate K1K_1 and machine cost rate K2K_2 (per minute), the cost per unit is:

TECn=(K1+nK2) TcnTEC_n = \frac{(K_1 + n K_2)\, T_c}{n}

Worked Example

Using the lathe data, l=1.0l = 1.0, m=4.0m = 4.0, and w=0.5w = 0.5 min:

n′=1.0+4.01.0+0.5=3.33n' = \frac{1.0 + 4.0}{1.0 + 0.5} = 3.33

Labor costs $30/hr ($0.50/min) and each machine costs $60/hr ($1.00/min).

nnCycle timeUnits per cycleCost per unit
3l+m=5.0l + m = 5.0 min3(0.50+3.00)(5.0)/3=$5.83(0.50 + 3.00)(5.0)/3 = \text{\textdollar}5.83
44(1.5)=6.04(1.5) = 6.0 min4(0.50+4.00)(6.0)/4=$6.75(0.50 + 4.00)(6.0)/4 = \text{\textdollar}6.75

Assign 3 machines. The worker is idle 5.0−3(1.5)=0.55.0 - 3(1.5) = 0.5 min per cycle, but that costs less than leaving four expensive machines idle 1 min each cycle. When machine time is cheap relative to labor, the answer often shifts to the higher integer, so always compute both.

Tip

Throughput also matters. With 3 machines the cell makes 3 parts per 5.0 min (36 per hour); with 4 it makes 4 per 6.0 min (40 per hour). If demand requires 40 per hour, the cost comparison must include the cost of the extra capacity found another way.


3. Random Servicing and Machine Interference

When machines stop at random (jams, tool breaks, random material runouts), several may need the operator at the same moment. A machine that waits while the operator serves another is suffering machine interference.

  • A simple binomial model treats each machine as independently down with probability pp at a random instant. The chance that kk of nn machines are down is (nk)pk(1−p)n−k\binom{n}{k} p^k (1-p)^{n-k}, and the expected number waiting beyond the one being served estimates interference.
  • Finite-source queueing models (the calling population equals the number of machines) and simulation give more exact answers.

Example: Four machines are each down 10% of the time at random. P(2 or more down)=1−0.94−4(0.1)(0.9)3=1−0.6561−0.2916=0.052P(\text{2 or more down}) = 1 - 0.9^4 - 4(0.1)(0.9)^3 = 1 - 0.6561 - 0.2916 = 0.052. About 5% of the time a stopped machine waits for the operator, which is usually acceptable.


4. Related Multiple Activity Charts

  • Left-hand/right-hand (operator) charts compare the two hands of one worker; see therblig analysis in the methods section.
  • Gang (crew) charts show several workers and machines, such as a crew loading a press or a pit crew, to balance their work.
  • Service versions apply the same logic to people and equipment in services: one technician running several analyzers in a laboratory, or one agent handling several chat sessions.
Improvement ideaEffect on the chart
Quick-load fixturesShorter ll raises n′n' and shortens the cycle
Place machines in a U-cellShorter walk time ww raises n′n'
Automatic unload and part chutesTurns combined work into machine-only work
Inspect while the machine runsMoves inspection into the worker's waiting time
Test Your Knowledge

For identical automatic machines, loading and unloading takes 1.2 min, the automatic run takes 6.0 min, and the operator needs 0.6 min per machine to walk and inspect. What is the ideal number of machines per operator for synchronous servicing?

A

3.0

B

5.0

C

4.0

D

10.0

Test Your Knowledge

With l = 0.8 min, m = 5.0 min, and w = 0.4 min, an operator costs $0.60 per minute and each machine costs $0.50 per minute. Which assignment minimizes cost per unit?

A

4 machines, at about $3.77 per unit

B

5 machines, at about $3.72 per unit

C

4 machines, at about $3.10 per unit

D

5 machines, at about $4.60 per unit

Test Your Knowledge

On a worker–machine chart for one operator and one machine, the cycle is 5 minutes and the operator is busy 1.5 minutes of it. Which change would most directly let the operator tend more machines?

A

Shorten the machine's automatic cycle so the machine is never idle

B

Increase the operator's pace rating in the time study

C

Add an inspection step during the automatic cycle

D

Place the machines in a compact U-cell to cut walking time between them

Sections you finish are checked off in the contents.