REPORT OF ANALYSIS ASSIGNMENT 3
LAB WORK OF COMPUTER SIMULATION
Arranged by:
Wildan Adiwena (12522107)
Imam Mukhri (12522276)
Group IP-02
ASSISTANT CODE : D – 121
INDUSTRIAL ENGINEERING DEPARTMENT
FACULTY OF INDUSTRIAL TECHNOLOGY
UNIVERSITAS ISLAM INDONESIA
YOGYAKARTA
2015
1. CASE STUDY 1
a. Printscreen Layout
Figure 1.1 Layout Case Study 1
b. Description of Case Study
In this Post Office, there are queuing systems, the researcher make a queuing
model using Flexsim software. The location of this post office is contain from
door, exit 1, queue, teller, exit 2. The probability of arrival customer is one
customer come every 60 seconds, so the researcher type exponential (0.60.1)
in source. The new customer will directly go if see in queue there are more
than 20 customers waiting, then that new customer will indicated “unsatisfied
customer”. Then the time of customer when come in and process in teller is
lognormal2 (31,3.1,0.5).
c. Analysis of Report based on question
Tabel 1.1 Summary Report Study Case 1
Object Classstats_
content
stats_
content
min
stats_
content
max
stats_
content
avg
stats_
input
stats_
output
stats_
stay
time
min
stats_
stay
time
max
stats_
stay
time
avg
state_
current
state_
since
Exit 2 Sink 0 0 0 0 0 0 0 0 0 7 0
Queue Queue 0 0 3 0.22 104 104 0 102.84 14.90 6 7118.81
TellerProcesso
r0 0 1 0.50 104 104 32.10 43.44 34.55 1 7151.97
Exit 2 Sink 1 1 1 0 104 0 0 0 0 7 0
Source5 Source 0 0 0 1 0 104 0 0 0 5 7318.81
Table 1.2 State Report Study Case 1
Object Class idle processing Busy Blocked generating empty collecting releasing
Waiting
for
operator
Waiting
for
transporter
breakdown
Exit 2 Sink
Queue Queue 0.00% 0.00% 0.00% 0.00% 0.00% 83.74% 0.00% 16.26% 0.00% 0.00% 0.00%
Teller Processor 49.75% 50.25% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Exit 2 Sink
Source5 Source 0.00% 0.00% 0.00% 0.00% 100.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Object Classscheduled
downconveying
travel
empty
travel
loaded
offset
travel
empty
offset
travel
loaded
loading unloading down setup utilize
Exit 2 Sink
Queue Queue0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
0.00
% 0.00% 0.00% 0.00%
Teller Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00 0.00% 0.00% 0.00%
%
Exit 2 Sink
Source5 Source0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
0.00
% 0.00% 0.00% 0.00%
1. How much customers who aren’t satisfied and is served until finish?
Answer: Based on summary report, there is no unsatisfied customer because the
number in queue never reach more than 20 customers. This conclusion get from
object queue, class queue in stat_contentmax only get 3 and this argument
supported also by object Exit 2 in class sink which stats_input 0 in summary
report. It means that there is no customer aren’t satisfied before processing by
teller then there are 104 satisfy customer until finish.
2. How much customer was served by the post office? (teller)
Answer: Based on summary report in object Teller, class processor in stat_output
are 104. It means that there are 104 customer was served by the post office.
3. How is the performance of server (teller)?
Answer: We can see the performance of server based on state report from idle and
processing the server, which are idle 49.75%, and processing 50.25%. It can be
concluded that the performance of the server (teller) into the less well because the
server has a large idle.
4. What is the queue condition in the system?
Answer: From the state report, the percentage of allowing customer from queue to
teller is 16.26%, then percentage queue empty is 83.74% because customer wait in
queue not reached 20 customers and not busy.
d. Summary
In this model, there is no unsatisfied customer; 104 customers are the total
customers that can be served until finish. In the teller there are idle time is
49.75%, and processing time 50.25%. The process of queue is smoothly because
customer wait in queue not reached 20 customers and not busy.
2. CASE STUDY 2
a. Printscreen Layout
Figure 2.1 Layout Case Study 2
b. Description of Case Study
In this Post Office, there are queuing systems, the researcher make a queuing
model using Flexsim software. The location of this post office is contain
from door, exit 1, queue, teller, exit 2. The probability of arrival customer is
one customer come every 60 seconds, so the researcher type exponential
(0.60.1) in source. The new customer will directly go if see in queue there are
more than 20 customers waiting, then that new customer will indicated
“unsatisfied customer”. Then the time of customer when come in and process
in teller is lognormal2 (31,3.1,0.5).and there are additional teller in this
model, so after queue, the model set the 40% customer will go to Teller 1 and
60% customer will go to Teller 2.
c. Analysis of Report based on question
Table 2.1 Summary Report Case 2
Object Classstats_
content
stats_
content
min
stats_
content
max
stats_
content
avg
stats_
input
stats_
output
stats_
stay
time
min
stats_
stay
time
max
stats_
stay
time
avg
state_
current
state_
since
Exit 2 Sink 0 0 0 0 0 0 0 0 0 7 0
Queue Queue 0 0 2 0.031517 112 112 0 33.038128 1.985143 6 7054.554144
Teller 1 Processor 0 0 1 0.216634 44 44 31.971191 39.37447 34.902947 1 7089.036798
Exit 1 Sink 1 1 1 0 112 0 0 0 0 7 0
Source5 Source 0 0 0 1 0 112 0 0 0 5 7054.554144
Teller 2 Processor 0 0 1 0.101788 68 68 10 10 10 1 6680.541074
Table 2.2 State Report Case 2
Object Class idleprocessin
gbusy blocked
Generatin
gEmpty
collectin
greleasing
waiting
for
operato
r
Waiting
for
transporter
Break
down
scheduled
down
Exit 2 Sink
Queue Queue 0.00% 0.00% 0.00% 0.00% 0.00% 96.85% 0.00% 3.15% 0.00% 0.00% 0.00% 0.00%
Teller 1 Processor 78.34% 21.66% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Exit 1 Sink
Source
5 Source 0.00% 0.00% 0.00% 0.00% 100.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Teller 2 Processor 89.82% 10.18% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Object Class conveying travel emptytravel
loaded
offset travel
empty
offset
travel
loaded
loading Unloading down setup utilize
Exit 2 Sink
Queue Queue 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Teller 1 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Exit 1 Sink
Source5 Source 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Teller 2 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
1. What is the performance difference after added on server? (compare with
case study 1)
Answer: The different is, in case study 2, there are 2 teller that can make
services process more optimal, with total customer served is 112 customer,
which teller 1 are services 44 customers, and teller 2 services 68 customers.
2. How much costumer type 2 was served?
Answer: To determine the number of customer that can be served on type
2, can be seen from the comparison object, and then select a total of two
outs and teller to see its results. From the chart picture, it is known to result
in type 2 28 people that can be served by tellers 1, and 37 people can be
served by tellers 2.
Figure 3.2 Comparison Objective Step
Figure 3.3 Comparison Objective customer type 2 diagram in teller 1
and 2
3. How much is the minimal and maximal wait time in the server?
Answer: Based on summary report, the stats_staytimemin is 0 and
stats_staytimemax is 33.038128.
4. What will happen when the queue capacity is reduced to 1?
Answer: If the queue capacity reduced to 1, it will make disappoint the
customer because based on summary report in object queue, class queue in
stat_contentmax are 2. It means that since the simulation occurs with 2
teller there are probability that there are 2 customer will wait in queue.
d. Summary
from all of the answer above in case study 2, additional Teller can make services
and queuing process more optimal, with total customer served is 112 customer,
which teller 1 are services 44 customers, and teller 2 services 68 customers. And
the stats_staytimemax in case study 2 is more decrease than stats_staytimemax
in case study 1.
3. CASE STUDY 4
a. Printscreen Layout
Figure 4.1 Layout Case Study 4
b. Description of Case Study
The component of electonic will be testing firstly after manufacturing
process. There are 2 type component arrived in Queue randomly where
40% type 1 and 60% type others. The time between arrival from
component is exponential(0,30,0). There are 2 inspection machine for
component 1, 3 inspection machine for component 2. The component
will go to machine that firstavailable. The all inspection time between
120 until 150 seconds and uniform distribution.
c. Analysis of Report based on question
Object Classstats_
content
stats_
content
min
stats_
content
max
stats_
content
avg
stats_
input
stats_
output
stats_
stay
time
min
stats_
stay
time
max
stats_
stay
time
avg
state_
current
state_
since
Source1 Source 0 0 0 1 0 253 0 0 0 5 7128.256977
Queue2 Queue 8 0 19 9.202494 253 245 0 997.155334 258.960789 8 7128.256977
Processor3 Processor 1 0 1 0.977671 53 52 120.193741 149.608994 134.693596 2 7164.034762
Processor4 Processor 1 0 1 0.975462 51 50 122.331902 149.85228 138.192801 2 7083.451468
Processor5 Processor 1 0 1 0.894973 49 48 120.083626 149.46434 133.1863 2 7143.169775
Processor6 Processor 1 0 1 0.886759 47 46 120.243378 149.484329 136.508907 2 7081.300107
Processor7 Processor 1 0 1 0.838482 45 44 121.320023 149.959534 135.490249 2 7109.952991
Sink8 Sink 1 1 1 0 240 0 0 0 0 7 0
Table 4.1 Summary Report Case 4
Table 4.2 State Report Case 4
Object Class idleprocessin
gbusy
blocked
generating
emptycollectin
greleasin
g
Waitingfor
operator
waitingfor
transporter
Breakdown
scheduled
down
Source1 Source 0.00% 0.00% 0.00% 0.00% 100.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Queue2 Queue 0.00% 0.00% 0.00% 0.00% 0.00% 1.56% 0.00% 98.44% 0.00% 0.00% 0.00% 0.00%
Processor3
Processor 2.23% 97.77% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Processor4
Processor 2.45% 97.55% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Processor5
Processor 10.50% 89.50% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Processor6
Processor 11.32% 88.68% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Processor7
Processor 16.15% 83.85% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Sink8 Sink
Object Class conveying
travel empty
travel loaded
offset travel empty offset travel loaded loading Unloading down setup utilize
Source1 Source 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Queue2 Queue 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Processor3 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor4 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor5 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Processor6 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor7 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
Sink8 Sink
1. Are the three inspection machines for component 2 working in
balance?
Answer: Yes, because in state report processing percentage for the
second component inspection machine doesn’t show significant
difference (89.5%; 88.68%; 83.85%). This is also supported by the
summary report which is also show there is no significant difference
for input (49; 47; 45) and output (48; 46; 44) of the second
component inspection machines.
2. How is the last condition of inspection machine when the simulation
ended?
Answer:
Object state_current Condition
Processor3 2 Processing
Processor4 2 Processing
Processor5 2 Processing
Processor6 2 Processing
Processor7 2 Processing
3. How much item type 1 and 2 has been inspected? Is there a
significant difference?
Answer: Item type 1 has been processed is 102 item while item type
2 is 138. It gets based on summary report in class processor which
are for type 1 the objective is processor 3 and 4 with stats_output 52
and 50 and type 2 the objective is processor 5, 6, and with
stats_output 48, 46, and 44 . There is a quite significant difference
because the comparison of type 1:type 2 is 40:60.
4. How is the system performance based on total items in and out of
the system?
Answer: Based on the total input and output of system there is a
slightly difference where total input are 253 and total output are 240.
It gets from summary report in the objective Source1, class Source
in stats_output that are 253 and the objective Sink8, class Sink in
stats_input are 240. It means that the performance of the system
based on total items in and out is not quite optimal yet.
d. Summary
It can be inferred that the process is running quite smoothly and the
process is still running after 2 hours with quite proportional input and
output. The allocation of burden in 3 inspection machine in component
type 2 is quite balance. The last 2 hours still remain process in every
inspection machine.
5. CASE STUDY 5
a. Printscreen Layout
Figure 5.1 Layout Case Study 5
b. Description of Case Study
3 kinds of product (uniform distribution) enter to 5 work station cell.
Product will be porcess at station that have determined as:
a. Product 1 will be processing at station 1,3, and 2
b. Product 2 will be processing at station 1,2, and 4
c. Product 3 will be processing at station 2,4,3, and 5
c. Analysis of Report based on question
Object Classstats_content
stats_contentmin
stats_contentmax
stats_contentavg
stats_input
stats_output
stats_staytimemin
stats_staytimemax
stats_staytimeavg
state_current
state_since
Source1 Source 0 0 0 1 0 718 0 0 0 5 7197.953489
Queue2 Queue 6 0 12 4.601184 3088 3082 085.72810
4 10.687929 8 7197.953489Processor3 Processor 1 0 1 0.663253 478 477 10 10 10 2 7191.826516Processor4 Processor 1 0 1 0.985172 710 709 10 10 10 2 7196.712723Processor5 Processor 1 0 1 0.666034 480 479 10 10 10 2 7191.826516Processor6 Processor 0 0 1 0.650297 468 468 10 10 10 1 7196.712723Processor7 Processor 0 0 1 0.329317 237 237 10 10 10 1 7196.712723Sink8 Sink 1 1 1 0 709 0 0 0 0 7 0
Table 5.1 Summary Report Case 5
Table 5.2 State Report Case 5
Object Class conveyingtravel empty
travel loaded
offset travel empty
offset travel loaded loading unloading down setup utilize
Source1 Source 0.00% 0.00% 0.00% 0.00% 100.00% 0.00% 0.00% 0.00% 0.00% 0.00%Queue2 Queue 0.00% 0.00% 0.00% 0.00% 0.00% 2.65% 0.00% 97.35% 0.00% 0.00%Processor3 Processor 33.67% 66.33% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor4 Processor 1.48% 98.52% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor5 Processor 33.40% 66.60% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor6 Processor 34.97% 65.03% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor7 Processor 67.07% 32.93% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Sink8 Sink
Object Class conveyingtravel empty
travel loaded
offset travel empty
offset travel loaded loading unloading down setup utilize
Source1 Source 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Queue2 Queue 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor3 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor4 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor5 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor6 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Processor7 Processor 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%Sink8 Sink
1. Why stat_input dan stat_output on queue is not proportional to
total flowitem entering the system?
Answer:
Because there is a global time in this model where the sequence of
every type routing should enter queue before going to another
sequence. Hence, There are double input enter to the queue.
2. How is the final condition of each location?
Answer:
Object
state_curr
ent
Conditio
n
Source1 5
Generati
ng
Queue2 6 Empty
Processo
r3 1
Idle
Processo
r4 2
Processi
ng
Processo
r5 2
Processi
ng
Processo
r6 2
Processi
ng
Processo
r7 2
Processi
ng
Sink8 7
Collecti
ng
3. Why station 2 has higher percentage than other stations?
Answer: Because station 2 must process all types of items while the
other station only process 2 types of items. It means that from 3
different routing of this production process, station 2 include in all
type routing while other station not include in all type routing.
4. Why total flowitem coming into the system is not proportional with
the flowitem going out? How to overcome this problem?
Answer: Because capacity in all station is limited (only 1 item for
each process). It can get from summary report in class Processor
which is the stats_contentmax only 1. Solution for this problem is
increasing capacity of workstation 2 because in workstation 2 has
high processing time and low idle time that is 98.52% and 1.48%
respectively. Hence, it can give chance to another routing to join in
that station to accelerate the process production with reducing
stats_contentmax, stat_timemax in objective Queue2 class Queue.
d. Summary
There is a bias information in summary report in objective queue in
stats_input because of the routing that made in global table. And other
consequences from this global table is the station 2 has higher
percentage than other stations because of station 2 include in all type
routing. For the flowitem that going out and in is not proportional
because the processor only have one capacity therefore there is a state
where every type routing should wait other routing finish their process.
Hence, it needs to add the capacity worksation especially in worksation
2 because of the idle that 1.48% and processing 98.52%/
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