OPIM 201 Operations Management
Session 3: Process Analysis II
Bhavani Shanker UPPARI
OPIM 201, SMU
Bhavani Shanker Uppari
1
Last Week
• Defined process performance measures.
• Flow rate (R):
• rate at which the process delivers output.
• Flow time (T):
• time a flow unit spends in the process.
• Inventory (I):
• number of flow units in the process.
• Capacity:
• maximum rate at which process can produce output.
• Cycle time (CT):
• avg. time between successive output units; equal to 1/(flow rate).
• Work-in-progress (WIP):
• same as inventory.
• Manufacturing lead time (MLT):
• same as flow time.
• Little’s law (in terms of R, I and T):
• I=RxT
• Little’s law (in terms of CT, MLT and WIP):
• MLT = WIP x CT
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Last Week
• Two process representation methods:
• Process flow diagram and Gantt chart.
• Bottleneck:
• resource with lowest capacity or longest cycle time.
• Flow rate as a function of demand and supply constraints:
• flow rate = min{available input, demand, process capacity}.
• Sequential system:
• Important formulas:
• Examples: ?
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Last Week
• Parallel system with similar tasks:
• Important formulas:
• Examples: ?
• Parallel system with different tasks:
• Important formulas:
• Examples: ?
• Why do we do analysis under steady state?
• We are mostly interested in average performance measures.
• Averages makes sense (i.e., useful/informative) under steady state.
• What is steady state?
• State where avg. inflow = avg. outflow.
• Visually: same pattern repeats again and again in Gantt chart.
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How to Track MLT and CT on Gantt Chart?
• In the plot below:
• Start times of jobs: Dashed green lines (
• End times of jobs: Dashed red lines (
).
).
• For MLT:
• Track from the start time of a job to the end time of the same job.
• In the plot: blue arrows (
).
• For CT:
• Track from end time of one job to the end time of next job.
• In the plot: yellow arrows (
).
18 mins.
Task 1
Task 2
1
18 mins.
18 mins.
2
3
4
1
2
3
10 mins.
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Bhavani Shanker Uppari
10 mins.
4
10 mins.
5
Last Week: Practice Question 7
• There are two machines (M-1, M-2) in parallel at stage 1.
Stage 1
Compute
Stage 2
M-1
10 min
M-3
8 min
M-2
10 min
• One input schedule: Feed a new unit every 8 minutes (why?)
M1
M2
M3
1
3
2
1
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5
4
2
3
7
6
4
5
9
8
6
7
8
MLT
18 mins.
WIP M1
0.625 units
WIP M2
0.625 units
WIP M3
1 unit
Total WIP
2.25 units
CT Stage 1
5 mins./unit
CT Stage 2
8 mins./unit
Overall CT
8 mins./unit
10
Verify Little’s Law
9
WIP x CT
= 2.25 units x 8 mins./unit
= 18 mins. = MLT
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6
How to Compute MLT and CT from Process Flow Diagram?
• Replace the complex systems with representative single boxes.
• MLT of the representative box = MLT of the complex system.
• CT of the representative box = CT of the complex system.
• To compute MLT and CT of the complex systems:
• Identify if that system is sequential, or parallel with similar or different tasks.
• Use the formulas for MLT and CT for the corresponding identified system.
(Similar Parallel Tasks)
CT =
Stage 1
Stage 2
1
1 1
+
10 10
= 5 mins./unit
Min. MLT = 10 mins.
M-1
10 min
M-3
8 min
M-3
8 min
M-2
10 min
CT = max {5,8} = 8 mins./unit
Min. MLT = 10 + 8 = 18 mins.
(Sequential Tasks)
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How to Compute WIP from Gantt Chart?
• Identify repetitive patterns for each resource in the Gantt chart.
• In each such pattern, identify
• the total duration of that pattern (Total time).
• time for which the resource is working on the job unit (Working time).
• Then WIP = Working time/Total time.
WIP(M1) = 10/16
= 0.625 units.
WIP(M2) = 0.625 units.
WIP(M3) = 1 unit.
Total WIP = 0.625 + 0.625 + 1 = 2.25 units.
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Session Outline
More Practice Problems
Effect of Randomness
Work Method Design
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9
Practice Question 8 (Complex System)
• Consider the process diagram for a product shown below.
• Component “AA” is processed either on M1 or M2 and then sent to the buffer.
Component “BB” is first processed on X1 then on Y1, and then sent to the buffer. At
P1, one unit of “AA” and one unit of “BB” are assembled. The assembly is then
processed on one of the three parallel machines - R1 or R2 or R3.
M1: 9 min
R1: 16 min
M2: 6 min
X1: 3 min
P1: 6 min
R2: 18 min
R3: 18 min
Y1: 5 min
• What cycle time can we expect for the entire process? Ans: 6 mins./unit
• What capacity (units per hour) can we expect? Ans: 10 units/hour
• What is the minimum MLT? Ans: 30 mins.
• Which step is the bottleneck? Ans: P1
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Solution to Problem 8
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Practice Question 9 (Impact of Schedule)
• Consider the following process. Processing times at stage 1 (on M1 and
M2) are 12 minutes each and processing times at stage 2 (on P1) are 4
minutes.
M1
P1
M2
Buffer
• Compute
• minimum MLT for the process: 12 + 4 = 16 mins.
• cycle time of Stage 1 (M1, M2 in parallel): 6 mins./unit
• cycle time of Stage 2: 4 mins./unit
• cycle time of the entire process: 6 mins/.unit
• Which stage is the bottleneck? Ans: Stage 1
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Input Rate vs. Input Schedule
M1
P1
Cycle time of the entire
process = 6 mins./unit
M2
Buffer
• For the system to be in steady state, input rate should be same as output rate.
• Input rate = 1 unit in every 6 mins.
• Several ways to achieve this input rate:
• Feed 1 unit in every 6 mins.
• Feed 2 units in every 12 mins.
• Alternate between 4 mins. per unit and 8 mins. per unit.
• On average, the time between units is again 6 mins.
• The exact manner in which we feed the system is called input schedule.
• Multiple input schedules can achieve the same input rate.
• Pros and cons of different schedules?
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Practice Question 9a (Schedule 1)
• Schedule 1: Feed M1 and M2 at the same time.
Average cycle time
Average MLT
WIP
Machine M1
Machine M2
Buffer
Machine P1
Total WIP
Verify Little’s law
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(4+8)/2 = 6 mins./unit
(16+20)/2 = 18 mins.
1 unit
1 unit
4/12 = 1/3 units
8/12 = 2/3 units
3 units
6 mins./unit x 3 units = 18 mins.
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Practice Question 9a (Schedule 1)
20 mins.
Avg. MLT = (20+16)/2 = 18 mins.
16 mins.
4 mins.
8 mins.
Avg. CT = (4+8)/2 = 6 mins./unit
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Practice Question 9a (Schedule 1)
WIP(M1) = WIP(M2) = 1 unit
4 mins.
WIP(Buffer) = 4/12 = 1/3 units
8 mins.
WIP(P1) = 8/12 = 2/3 units
12 mins.
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16
Practice Question 9b (Schedule 2)
• Schedule 2: Delay the starting of job 2 on machine 2 by 4 minutes.
Average cycle time
Average MLT
WIP
Machine M1
Machine M2
Buffer
Machine P1
Total WIP
Verify Little’s law
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(4+8)/2 = 6 mins./unit
16 mins.
1 unit
1 unit
0 unit
8/12 = 2/3 units
8/3 units
6 mins./unit x (8/3) units = 16 mins.
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Practice Question 9c (Schedule 3)
• Schedule 3: Delay the starting of job 2 on machine 2 by 6 minutes.
Average cycle time
Average MLT
WIP
Machine M1
Machine M2
Buffer
Machine P1
Total WIP
Verify Little’s law
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6 mins./unit
16 mins.
1 unit
1 unit
0 unit
4/6 = 2/3 units
8/3 units
6 mins./unit x (8/3) units = 16 mins.
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18
Input Rate vs. Input Schedule
• What did you learn from the performance metrics of previous three schedules?
• Does CT change across the three schedules?
• No. It’s same as input rate under steady state.
• Does MLT change across the three schedules?
• Yes. Depends on the time spent in the buffer.
• When there is a gap between input units:
• MLT is lower, and
• buffer is empty.
• Will we have the same effect if we give a gap of 2 mins. between input units?
• No, minimum gap is 4 mins. Why?
• Difference between schedule 2 and schedule 3?
• Schedule 3 has a smoother outflow, whereas schedule 2 (and schedule 1)
has outflow in bursts.
• More is the gap between input units, smoother is the outflow.
• Any real-world applications?
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19
Session Outline
More Practice Problems
Effect of Randomness
Work Method Design
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Practice Question 10 (Effect of Randomness)
• Question 10(a): Base case: No randomness. What is the CT of the system?
M1
1 min
M2
1 min
CT = 1 mins./unit
• Question 10(b): Now let’s consider processing times to be random. Suppose
that task 1 requires 1 minute, and task 2 requires 1 minute on average but the
actual times can fluctuate uniformly between 0.5 to 1.5 minutes. We denote this
as follows: Task 2 ~ U(0.5,1.5). “U” indicates uniform distribution.
M1
1 min
M2
U(0.5,1.5)
Why could a process
step have random
processing time?
• Compute:
Max. value
Min. value
Avg. value
MLT
1+1.5 = 2.5 mins.
1+0.5 = 1.5 mins.
1+1 = 2 mins.
CT
max{1,1.5} = 1.5 mins./unit
max{1,0.5} = 1 mins./unit
??
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Computing Average Cycle Time Under Uniform Distribution
• Let 𝑥 be the cycle time of M2.
CT of M2 = 𝑥 mins./unit
M1
1 min
M2
U(0.5,1.5)
• Then cycle time of entire system as a function of 𝑥 is
CT 𝑥 = max{1, 𝑥}
=ቊ
1,
𝑥,
𝑥 ∈ [0.5,1]
𝑥 ∈ [1,1.5]
• Under uniform distribution:
Area under CT curve
• Average CT = Length of uniform distribution
• Why?
1
• Here: Average CT =
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1×1+2×0.5×0.5
1.5−0.5
= 1.125 mins./unit.
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Practice Question 10 (Contd.)
• Question 10(c): If we change the task time of task 2 to be U(0.2,1.8):
M1
1 min
M2
U(0.2,1.8)
• Compute:
Max. value
Min. value
Avg. value
MLT
1+1.8 = 2.8 mins.
1+0.2 = 1.2 mins.
1+1 = 2 mins.
CT
max{1,1.8} = 1.8 mins./unit
max{1,0.2} = 1 mins./unit
??
• Average CT using the method described on previous slide:
1
Average CT =
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1.6×1+2×0.8×0.8
1.8−0.2
= 1. 2 mins./unit.
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Practice Question 10 (Contd.)
• Question 10(d): What would happen if we increase the number of sequential
operations from 2 to 3 to 4,…, etc., each requiring processing time U(0.5,1.5)?
M0
1 min
M1
U(0.5,1.5)
M2
U(0.5,1.5)
M100
U(0.5,1.5)
1
1.5
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Practice Question 10 (Contd.)
• Question 10(e): Now buffers are allowed between the two workstations with
random processing times.
• When there is no buffer space between two stages, the stages are tightly
coupled.
• When there is a very large buffer space between two stages, the stages
become decoupled.
M2
U(0.5,1.5)
M1
U(0.5,1.5)
• How do you think buffers affect the production rate of the system?
• Buffers bring some independence between the process steps.
• The bigger the buffer, more is the independence, and higher is
production rate (or lesser is the inefficiency due to randomness).
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Practice Question 10 (Summary)
• What did you observe from:
• Question 10(b):
• Randomness increases avg. cycle time and decreases avg. production rate.
• Question 10(c):
• More randomness further increases avg. cycle time and further decreases
avg. production rate.
• Question 10(d):
• As the number of sequential steps with randomness increase, avg. cycle
time increases and avg. production rate decreases.
• Question 10(e):
• Buffers can help decrease avg. cycle time and increase avg. production rate.
• What are different ways to deal with randomness?
• Introduce buffers.
• Reduce randomness
• Training of employees, standardization of procedures, maintenance of
machines.
• Reduce the number of sequential tasks with randomness.
• Examine process flow diagram to see if some tasks can be done parallelly.
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Session Outline
More Practice Problems
Effect of Randomness
Work Method Design
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27
Practice Question 11 (Work Method Design)
• A product assembly process consists of four tasks requiring different skills. The
company operates for one 8-hour shift per day. Details of tasks are as follows.
• Productive time of workers is 7.5 hours per day (with half hour break).
• Material cost is $7.15 per unit.
• The average salary of a worker is $12 per hour (paid for 8 hours per day) and there is
a 75% overhead cost.
• The overhead cost is assumed to be a percentage increase on the labor costs.
• Total production cost is the sum of material cost and labor cost.
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Practice Question 11 (Work Method Design)
• Questions 11(a): If the four tasks are done sequentially (one task at each station)
by 4 operators in an assembly line, what will be (i) the rate of production per
hour, and (ii) production cost per day?
• Solution:
• Cycle time of the line (CT) = max {20,32,36,24} = 36 sec./unit.
• Rate of production = 1/CT = 1 unit./(36 sec.) = 100 units/hour.
• Labor cost = $12/hour x 8 hours/day x 4 x 1.75 = $672/day.
• Material cost = $7.15/unit x 100 units/hour x 7.5 hours/day = $5362.5/day.
• Production cost = 672 + 5362.5 = $6034.5/day.
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Design Your Factory
• Now the product became famous in the market. It is estimated that, in the next few
months, the demand is going to be 6-7 times of the current value.
• The company is seeking your advice on how to meet this demand.
• Can the current assembly line meet such surge in demand or is there a need to grow?
• What will be your advice? What is your advice based on?
• Discuss in groups of 2-3.
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Practice Question 11 (Contd.)
• Questions 11(b): To produce 5000 items per day, how many parallel lines will
you need assuming that the productive time is 7.5 hours per day. Also calculate
per day production cost of this system.
• Solution:
• Say there are 𝑛 lines.
• Each line’s CT is: 36 sec./unit.
• The 𝑛 lines are parallelly connected, so CT of the entire system is:
• Actual CT of the system =
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36
𝑛
36
𝑛
sec./unit.
sec./unit.
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31
Practice Question 11 (Contd.)
• Questions 11(b): To produce 5000 items per day, how many parallel lines will
you need assuming that the productive time is 7.5 hours per day. Also calculate
per day production cost of this system.
• Solution:
• Requirement from management: 5000 units in 7.5 hours.
• Required CT of the system = 7.5 × 60 × 60/5000 = 5.4 sec./unit.
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Practice Question 11 (Contd.)
• Questions 11(b): To produce 5000 items per day, how many parallel lines will
you need assuming that the productive time is 7.5 hours per day. Also calculate
per day production cost of this system.
• Solution:
• We need Actual CT ≤ Required CT.
Why?
• Otherwise, the system will operate slower than what management expects,
and hence cannot meet the production requirement.
•
36
𝑛
36
≤ 5.4 ⟹ 𝑛 ≥ 5.4 = 6.67 ⟹ 𝑛 = 7 lines.
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Practice Question 11 (Contd.)
• Questions 11(b): To produce 5000 items per day, how many parallel lines will
you need assuming that the productive time is 7.5 hours per day. Also calculate
per day production cost of this system.
• Actual CT of the system =
36
7
sec./unit.
• Production rate = 7/36 units/sec. = 700 units/hour.
• Labor cost = ($12/hour x 8 hours/day x 4 x 1.75) x 7 = $4704/day.
• Material cost = $7.15/unit x 700 units/hour x 7.5 hours/day = $37537.5/day.
• Production cost = 4704 + 37537.5 = $42241.5/day.
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Practice Question 11 (Contd.)
• Question 11(c): Suppose you form 4 departments, each for one task. Assuming
7.5 hours of work per day per person, how many total people would you need to
produce 5000 items per day? What is the production cost per day?
• Solution:
• Say there are 𝑛1 workers in Dept.1, 𝑛2 in Dept.2, 𝑛3 in Dept.3, and 𝑛4 in Dept.4.
• Actual CT of the system = max
20 32 36 24
, , ,
𝑛1 𝑛2 𝑛3 𝑛4
sec./unit.
• Required CT of the system = 5.4 sec./unit.
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Practice Question 11 (Contd.)
• Question 11(c): Suppose you form 4 departments, each for one task. Assuming
7.5 hours of work per day per person, how many total people would you need to
produce 5000 items per day? What is the production cost per day?
• We need Actual CT ≤ Required CT:
max
20 32 36 24
20
32
36
24
, , ,
≤ 5.4 ⟹
≤ 5.4 and
≤ 5.4 and
≤ 5.4 and
≤ 5.4
𝑛1 𝑛2 𝑛3 𝑛4
𝑛1
𝑛2
𝑛3
𝑛4
⟹ 𝑛1 ≥ 3.70, 𝑛2 ≥ 5.93, 𝑛3 ≥ 6.67, 𝑛4 ≥ 4.44
⟹ 𝑛1 = 4, 𝑛2 = 6, 𝑛3 = 7, 𝑛4 = 5.
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Practice Question 11 (Contd.)
• Question 11(c): Suppose you form 4 departments, each for one task. Assuming
7.5 hours of work per day per person, how many total people would you need to
produce 5000 items per day? What is the production cost per day?
• Actual CT of the system = max
20 32 36 24
, , ,
4 6 7 5
=
32
6
sec./unit.
• Production rate = 6/32 units/sec. = 675 units/hour.
• Labor cost = $12/hour x 8 hours/day x 22 x 1.75 = $3696/day.
• Material cost = $7.15/unit x 675 units/hour x 7.5 hours/day = $36196.875/day.
• Production cost = 3696 + 36196.875 = $39892.875/day.
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Practice Question 11 (Contd.)
• Question 11(d): Suppose all operators are multi-skilled. Each one can perform
all four operations and assembles the complete product. Assuming 7.5 hours of
work per day per person, how many total people would you need to produce
5000 items per day? What is the production cost per day?
• Solution:
• Say there are 𝑛 multi-skilled labor.
• Actual CT of the system
=
20+32+36+24
𝑛
=
112
𝑛
sec./unit.
• Required CT of the system = 5.4 sec./unit.
• We need Actual CT ≤ Required CT:
𝑛≥
112
5.4
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= 20.74 ⟹ 𝑛 = 21 workers.
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Practice Question 11 (Contd.)
• Question 11(d): Suppose all operators are multi-skilled. Each one can perform
all four operations and assembles the complete product. Assuming 7.5 hours of
work per day per person, how many total people would you need to produce
5000 items per day? What is the production cost per day?
• Actual CT of the system =
112
21
sec./unit.
• Production rate = 21/112 units/sec.
= 675 units/hour.
• Labor cost
= $12/hour x 8 hours/day x 21 x 1.75 = $3528/day.
• Material cost
= $7.15/unit x 675 units/hour x 7.5 hours/day
= $36196.875/day.
• Production cost = 3528 + 36196.875
= $39724.875/day.
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Practice Question 11 (Summary)
• Given the target of producing 5000 items per day, discuss the pros and cons of
building the product using processes described in parts (b), (c) and (d).
(b)
(c)
(d)
28
22
21
Production rate
700 units/hour
675 units/hour
675 units/hour
Cost per day
$42241.5/day
$39892.875/day
$39724.875/day
Cost per unit
$8.046/unit
$7.88/unit
$7.85/unit
Product-based
Process-based
Cellular
Low
Low
High
100 x 600/700 = 86%
100 x 540/675 = 80%
100 x 642.8/675 = 95%
Low
Low
High
No. of people
Layout
Skills
Robustness
Job satisfaction
• Cost per unit = Cost per day / Actual capacity.
• Think of what happens if one employee is absent.
Robustness calculations
on the next slide
• Any thoughts?
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What Happens When One Employee is Absent?
• In part (b)
• If an employee is absent (doesn’t matter which employee), the corresponding line
collapses entirely, resulting in only 6 working lines.
• Capacity with 6 lines = 600 units/hour.
• % of original capacity satisfied = 100 x 600/700 = 85.7%.
• In part (c)
• If an employee from Dept. 1 is absent:
• CT = max{20/3, 32/6, 36/7, 24/5} = 20/3 sec./unit ⟹ Capacity = 540 units/hour.
• % of original capacity satisfied = 100 x 540/675 = 80%.
Worst case
• If an employee from Dept. 2 is absent:
• CT = max{20/4, 32/5, 36/7, 24/5} = 32/5 sec./unit ⟹ Capacity = 562.5 units/hour.
• % of original capacity satisfied = 100 x 562.5/675 = 83.3%.
• If an employee from Dept. 3 is absent:
• CT = max{20/4, 32/6, 36/6, 24/5} = 36/6 sec./unit ⟹ Capacity = 600 units/hour.
• % of original capacity satisfied = 100 x 600/675 = 88.9%.
Best case
• If an employee from Dept. 4 is absent:
• CT = max{20/4, 32/6, 36/7, 24/4} = 24/4 sec./unit ⟹ Capacity = 600 units/hour.
• % of original capacity satisfied = 100 x 600/675 = 88.9%.
• In part (d)
• If an employee is absent, then remaining 20 multi-skilled employees are still working.
• Capacity with 20 workers = 642.8 units/hour.
• % of original capacity satisfied = 100 x 642.8/675 = 95.2%.
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Summary
• More practice with Little’s law and process analysis.
• Analyzed the impact of schedules and randomness on process performance.
• Analyzed the efficacy of different process designs.
• Now you should be able to solve homework sets 1 and 2.
• Next time: Assembly line balancing.
• How we can increase capacity and achieve higher efficiency without
incurring any additional costs?
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