# homeowork submission ITOM ```ITOM HOWEWORK
Question 1:
a) Draw a process flow diagram and identify the bottleneck in the repair process.
Diagnosis,
60 Min
Ticketing,
15 Min
Spare parts
identify. 30 Min
Repair, 120 Min
Assembly, 90 Min
Bottleneck is Repair process which takes 120 minutes to complete
b) Suppose there are 800 machines all over campus. Each machine, on average, fails 9 times every
year (300 working days). What should be the minimum capacity in Joe’s shop? What is the
corresponding cycle time?
Scenario 1: When all demand of repair is fulfilled
Average Total number of failures = 800 * 9=7200 failures per year
Repair per day needed = 7200/300= 24 machines
Repair Capacity per hours = 24/8=3 machines per hour
Cycle time = 60/3 = 20 minutes
Scenario 2: When only 5 persons are working
Process cycle time = 120 minutes in steady state condition
Capacity = 60/120= 0.5 machine per hour
Hence total machines repairs per day = 0.5*8 = 4 machines per day (8 hours shift)
Total repair capacity per year in existing setup =300*4 = 1200 machines
Backlog of machine repair = 7200 – 1200= 6000 machines
c) How many people (in round numbers) does Joe need for each of the above 5 steps?
1. Just to complete demand, 7200/300=24 machines must be repaired each day. There are 8 hours =
480 minutes, hence 480/24= every 20 minutes should be cycle time of process. Which is possible
only if we hire additional manpower at each stage. Hence, 1-person ticketing + 3 persons at
diagnosis + 1.5(round to 2) person at spare parts identification, 6 persons at repair, and 4.5(round
to 5) persons at assembly which totaled to 17 individuals.
•
•
•
•
•
Persons at ticket = 1
Persons at Diagnosis = 60/20=3
Persons at spare parts identification = 30/20= 1.5 round to 2
Persons at Repair = 120/20= 6
Persons at Assembly = 90/20 =4.5 rounded to 5
2. If we don’t combine activities, set up multiple lanes for repair and hire a complete team of 5
individuals for each lane then 30 individuals will be needed.
3. If we don’t combine activities and just hire additional manpower in the existing setup so that no
person is sitting idle, then we will be needing (60/15) + (30/15) + (120/15) + (9/15) = 4+2+8+6
=20 individuals. In this way, our cycle time will be reduced to 15 minutes and our capacity per
year will be 9600 machines, hence we are going way beyond than demand.
4. A simple Gantt. Chart is drawn below which shows major time is taken by person who will be
repairing machines, while rest of individuals will be sitting idle, therefore they can be trained to
do multiple tasks, hence leading to proper utilization of manpower. Like Assembly (90 min.) can
be combined with spare parts identification activity (30 min.) which totaled to 120 mins equal to
repair time. Similarly, Ticketing and Diagnostic can be combined. This activity combining will
make the whole system efficient and only 3 members team will be needed per lane instead of 5.
While in order to repair 7200 machines per year, 6 such teams will be needed. 6 teams mean 18
individuals.
Ti
.
Diagnosis
Spare
Repair
Question 2:
Assembly
a) What is the capacity of the Step 1 if the batch size is 40 parts?
Total time for step 1 = Set-up time + 0.25* batch size = 30+0.25*40=40 minutes
Capacity in steady state= 60/10=6 units per hour
Capacity by considering set-up time = 60/40 = 1.5 unit per hour
b) What is the capacity of the process if batch size is 40 parts?
Total time for step 1 = Set-up time + 0.25* batch size = 30+0.25*40=40 minutes
Total time for step 2 = Set-up time + 0.25* batch size = 20+0.20*40=28 minutes
Total time for step 3 = Set-up time + 0.25* batch size = 45+0.15*40=51 minutes
Bottle neck will be step 3, hence cycle time of process = 51 minutes
Capacity with set-up time included = 60/51= 1.17 units per hour
c) For what batch sizes will Step 1 be the bottleneck?
30+0.25*x = 45+0.15*x, While x is batch size
15/0.1=x → x &gt; 150 units
Step 1 will be bottle neck when batch size is equal to greater than 150. At batch size of exact
150, both step 1 and step 3 will be bottleneck but to clearly say that step 1 is the only bottleneck
batch size should be greater than 150.
Question 3:
a) What is the product with the lowest rush order flow time?
Product Family
A
B
C
Drilling (min)
10
15
12
Cutting (min)
10
8
10
Grinding (min)
10
9
13
Total Time (min)
30
32
35
Increase of machines will not impact overall rush order flow time. Lowest ROFT is 30 minutes and it’s
for product family A.
b) If you were to select one of the three products to produce, which of the three would you select?
Why?
As per below table, I will select Product B as it is giving highest profit amongst all. Both capacity and
profit margins matter for a firm and Product B is giving best combination of capacity and margins.
Product Family Drilling (min)
# of mach.
A
B
C
2
5
7.5
6
Cutting (min)
1
10
8
10
P&amp;L of company ABC
Cycle Time
Grinding (min)
Capacity pe hour Margin per unit
(min)
1
10
10
6.00
5.00
9
9
6.67
6.00
13
13
4.62
5.50
# of hours per
day
8.00
8.00
8.00
Total Profit
\$
\$
\$
240.00
320.00
203.08
Question 4:
a) What is the average time a car coming for car wash spends in the system?
Average time for car wash service + Average time for cash payment = 36.17+20= 56.17 minutes
b) What is the average time a car that comes to get an oil change spends in the queue?
Probability of path (enter → oil change → cash payment) * (average waiting time in queue for oil change
+ average waiting time in queue for cash payment)
+
Probability of path (enter → oil change → Repair → cash payment) * (average waiting time in queue for
oil change + average waiting time in queue for repair + average waiting time in queue for cash payment)
= 0.8 * (40+16) + 0.2 * (40+7.5+16) = 57.5 minutes
Hand written calculations for following table are attached
Station
Car wash
Oil change
Repair
Cash paymnent
Entrance rate
(per hour)
10
2
0.4
12
Average Waiting Average queue Service time
time (min)
time (min)
(min)
36.17
21.17
15
60
40
20
37.5
7.5
30
20
16
4
# of people in
queue
2.1261
1.33
0.05
3.2
# of people in
service
2.5
0.67
0.2
0.8
Question 5:
a) In February, there is a very high demand for Centipedes, and no demand for Millipedes. How will
you set up the four cells in February? In other words, how many cells will you dedicate to each of
these steps? In your configuration, what is the maximum number of Centipedes that Arthropod can
produce in an 8-hour shift?
Because step B is bottleneck in centipede production, therefore we will install 2 machines at step B.
Second machine installation for step B will reduce our overall cycle time (from 45 to 40 minutes), hence
increase capacity. Capacity per 8 hours shift is 12 units of centipedes.
Centipede Production Plan
Cycle
Capacity CapacityProduct A (min) B (min) C (min) Time
per hour 8 hour
(min)
# of cells
1
2
1
Centipede
20
22.5
40
40
1.5
12
b) In September, the scenario would have switched completely. There is no demand for Centipedes,
and there is high demand for Millipedes. Will you change the configuration of the factory? In your
configuration, what is the maximum number of Millipedes that Arthropod can produce in an 8hour shift?
Because step A is bottleneck in centipede production, therefore we will install 2 machines at step A.
Second machine installation for step A will reduce our overall cycle time (from 40 to 30minutes), hence
increase capacity. Capacity per 8 hours shift is 16 units of centipedes.
Millipede Production Plan
Cycle
Capacity CapacityProduct A (min) B (min) C (min) Time
per hour 8 hour
(min)
1
1
# of cells
2
20
30
20
30
2
Millipede
16
c) During May and June, Arthropod would like to maintain a 50-50 mix of Centipedes and Millipedes
to meet the even demands for the two. What is the most number of Centipedes and Millipedes that
can be produced in 8 hours, and how should the factory be configured to achieve that?
By having 50-50 demand means each from each cell 1 unit of centipede should be followed by 1 unit of
millipede. You cannot produce multiple units of centipedes and then multiple units of millipedes.
Therefore, if we combine time of both products then we will get to know that Step B will be bottleneck,
hence we will install 2nd machine at step B and this installation will reduce our cycle time from 75 to 60
minutes, hence capacity will be 1 unit of each product per hour. Overall capacity for 50-50 will be 8
units of each product, total 16 units.
50-50 Production Plan
Product
# of cells
Centipede only
Millipede only
Combined time before
Combined time after
Cycle
Capacity CapacityA (min) B (min) C (min) Time
per hour 8 hour
(min)
1
2
1
20
45
40
45
1.33
10.67
40
30
20
40
1.50
12.00
60
75
60
75
0.80
6.40
60
37.5
60
60
1.00
8.00
```