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7/25/2019 System Dynamics, Home task Mr. Wang
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Yevhenii Usenko, Faculty of Economics-4, group 1
System Dynamics
Home Assignment 2
Task 1
Increases in Backlog, as well as in Service time (which are undoubtedly connected),
seem to pose a big problem for Mr. Wang because such delays in customer service may
cause dissatisfaction of Mr. Wangs customers, who, in turn, will prefer another bicycle
repair shop. As a consequence, Mr. Wang may experience a decline of his business or, at
least, deterioration of its financial position.
Task 2
a) It seems that Mr. Wang reacts to changes in Order rate with some lag: he needs time
to perceive new demand. Moreover, Mr. Wang has no opportunity to tune his production
line immediately; he needs some more time to adjust his capacity to new order rate.
Therefore, after each rise in repair demand both backlog and service time increase at first,
then fluctuate and stabilize only in the long run.
b) No, the backlog stabilizes at slightly higher level after each increase in order rate.
We even can infer from the graphs given that after the first event equilibrium backlog rose
from 1000 to 1100, and after the last one to 1331. These values are exactly the same as
the respective order rates, so we can conclude that a directly proportionate relationship
exists between the equilibrium backlog and the order rate.
c) There is a cumulative relationship between Order rate and Backlog because
immediately after each increase in the former (when production rate is not adjusted)
Backlog rises linearly, while Order rate is a constant. However, Service time and Backlog
are connected by a directly proportionate relationship: the bigger is Backlog, the higher is
Service time.
Moreover, Service time equilibrates each time on the same level 1.0, because Service
time is defined as Backlog divided by Repair rate; and for Backlog to stabilize Repair rate
should be exactly the same as the Order rate, and, therefore, the same as the Backlog itself.
So under equilibrium conditions, Service time could be expressed as:
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Task 3
a) The following model was built using iThink software in accordance with equations
given.
The documentation is as such:
a stock; initial value is set to 1000; unit of measurement (UoM): bicycles;
an inflow into the stock ; UoM: bicycles per day;
an outflow of the stock ; UoM: bicycles per day;
a stock; initial value is set to be equal the ; UoM:
bicycles per day;
a biflow of ; UoM: (bicycles
per day) per day;
a variable (an auxiliary); UoM: days;
a constant; UoM: days.
b) No, in my view, determining Expected Order Rate as a stock is correct because
adjustments in it (Change in Expected Order Rate) are accumulated over time. This
accumulative relationship is enabled by using a stock and flow structure.
Moreover, it seems infeasible to create such a structure with auxiliaries, because a
circular connection would arise.
Backlog
Expected Order Rate
Change in Expected OR
Order RateProdution Rate
Time Corr Expected Order Rate
Service Time
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c)
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After simulation of the model, it can be seen that it takes approximately 85 days for
Mr. Wang to fully acknowledge his new demand for repairs. Therefore, the backlog rises in
logarithmic growth mode up to 2600 bicycles, as well as service time do up to 2.36 days.
Such behavior arises due to the underlying model structure. It involves negative
balancing loop which includes expected order rate, its change and adjustment time. Such a
structure leads to stabilization of the stock Backlog, but only in the long run.
d) No, the model does not adequately resemble reality because Mr. Wangs company
does not try to increase production in order to reduce the backlog.
To illustrate, let us change the order rate step in time 10 to 200 bicycles per day. As a
consequence, the backlog equilibrium value rises to 4200. So, Mr. Wang has no
mechanisms of control over his backlog in such a model.
Task 4
a) New model structure is presented below. Documentation added:
a constant; UoM: days;
a variable; UoM: bicycles;
a constant; UoM: days;
a variable; UoM: bicycles per day;
, described previously, changed its initial value to .
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b) I have identified two negative balancing loops in the model: one for expected order
rate adjustment and the other for backlog adjustment. The causal loop diagram is given
below:
In both cases, the bigger the main stock (Expected Order Rate and Backlog
respectively), the higher would be the negative change to them. First of this balancing
loops reproduce the behavior of Mr. Wang with respect to perception of the real order rate
that he needs time to adjust his expectations to new demand. The second one resembles
adjustment of production to changes in demand the need to increase repairs to
compensate for imperfection of Mr. Wangs expectations.
Backlog
Expected Order Rat e
Change i n Expected OR
Order RateProdution Rate
Time Corr Expected Order Rate
Service Time
Target Backlog
Target Service Time
Desired Backlog Adjus tment
Time to Adjust Backlog
BacklogOrder RateProduction Rate
Expected Order Rate
Change in Expected Order Rate
Target Service Time
Target Backlog
Target Backlog Adjustment
+
-
-
+
+
+
+
-
+
+
+
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From my point of view, the goal of the first loop (for Expected Order Rate) is explicit
the actual Order Rate, but the goal of the second loop is actually stated in Target Service
Time, which does not directly participate in this loop, so it might be considered as implicit
goal.
The mechanisms here operate without delays.
c) As we see from the following graphs, Backlog and Service Time initially increase to
the level of 1428 bicycles and 1.3 days respectively, and fall to the equilibrium values of
1100 and 1 since.
The reason for such behavior lies in the structure of the model. Expected Order Rate
balancing loop provides us with some means of stabilization, but at much higher levels, as
we have seen in the previous model. Here, the Backlog balancing loop rules. For
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approximately 12 days after increase of demand, Desired Backlog Adjustment, given our
time for adjustment, is insufficient to totally offset the effects of this rise. But then, when
Backlog becomes very high, this adjustment also becomes high and drives the system back
to the equilibrium of 1100 bicycles.
d) For Time to Adjust Backlog = 16:
For Time to Adjust Backlog = 4:
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The result is that with higher adjustment time equilibrium values are attained much
later than with lower one: approximately 166 against 108 days after the initial growth of
demand for repairs. Moreover, the highest backlog and service time are bigger for AT=16:1617 bicycles and 1.47 days respectively against 1276 and 1.16. The reason is that higher
adjustment time means lower desired backlog adjustments, as seen from the formula.
Therefore, backlog needs to get much higher values for backlog adjustment to become
sufficiently huge to offset Order Rate inflow, which also requires more time.
e) No, the model does not fully reproduce the real behavior of Mr. Wangs production
due to the absence of oscillations in Backlog and Service Time. To see the reason, lets
assume that the model built can oscillate. Oscillations in Backlog can be caused only by
oscillations in Production Rate, because Order Rate is stable after the increase on 10thday.
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However, the structure does not allow for Production Rate to oscillate. We have only
negative balancing loops, which work in a very direct manner they decrease the stock
until it reaches equilibrium level, and then nothing changes.
f) The model definitely behaves more favorably for Mr. Wang than reality does,
because there are no unwanted oscillations.
g) iBelieves efforts are unsuccessful, because they have not reproduced key part of the
reality, which constitutes a major problem for Mr. Wang these oscillations, which he
wants to be reduced.
Task 5
a) So we have introduced all the structures asked for. Specifically, workforce and its
recruitment/attrition were introduced as a stock, named Workforce, with initial value of
50 workers, and a biflow Net Hiring (units: workers/day). The latter is dependent upon
time to train or to fire employees, which is set to be 25 days, Workforce itself, and Desired
Workforce (unit of measurement (UoM): workers):
Backlog
Expected Order Rate
Change in Expected OR
Order RateProdution Rate
Time Corr Expected Order Rate
Service Time
Target Backlog
Target Service Time
Desired Backlog Adjustment
Time to Adjust Backlog
Workforce
Net Hiring
Time to Train or Dismi ssDesired Workforce
Productivity
Desired Production
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b) To define Desired Workforce, we initially define Desired Production, which is to be
equal to:
Then, with Productivity given as 20 (bicycles per worker per day), we are able to write
that:
c) Finally, Production Rate is equal to the quantity of bicycles each worker repairs a
day times the number of workers:
d) The following graphs were obtained after the simulation of the last model:
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e) The model fully replicates all movements in backlog and service time as given by the
data.
f) There are two negative feedback loops, affecting Expected Order Rate (discussed
previously), and Workforce. The second one replicates the way in which Mr. Wang adjusts
his labor supply: he needs time to hire employees when there is a need for them, and
dismiss them when labor is redundant. Its goal is implicit it is defined a long sequence of
actions, but the actual goal is Target Service Time. There are no delays in this mechanism.
But a very new type of loop introduced a major feedback loop (marked by green
lines). It allows for oscillations in the model because material delays are present here.
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1: Workf orce 2: Desired Workforce
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2
2 2
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g) A major feedback loop with material delays is introduced in our model. Due to this
structure, the behavior is as follows: after the demand shock in time 10, Mr. Wang firstly
perceives new order rate, then adjusts his desired overproduction to offset already
accumulated backlog, and finally decides how many students to hire. All of these actions
need time, as expressed by time to correct variables. So backlog rises up to approximately
24th day after the shock, and then begins to decline. When the backlog reaches the
potential point of equilibrium, Mr. Wangs reaction is to fire students. But once again,
there is a delay between such necessity and Mr. Wangs ability to do so. As a consequence,
backlog falls under the equilibrium level up to time 81. Then the process begins from the
beginning and repeats for a few times, causing oscillations in Backlog and Service Time.
h) Time to Correct Expected Order Rate is doubled (=32):
Backlog
Order Rate
Production Rate
Expected Order Rate
Chang e in Expected Order Rate
Target Service Time
Target Backlog
Target Backlog Adjustment
Desired Production
Desired Workforce
Workforce
Net Hiring
Module 1
+
-
-
+
+
++
-
+
+
+
+
++
+
-
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Time to Correct Expected Order Rate is halved (=8):
Time to Adjust Backlog is doubled (=16):
Time to Adjust Backlog is halved (=4):
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Time to Train is doubled (=50):
Time to Train is halved (=12.5):
Time to Correct Expected Order Rate and Time to Train are doubled, and Time to
Adjust Backlog is halved:
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Time to Correct Expected Order Rate and Time to Adjust Backlog are doubled, and
Time to Train is halved:
Conclusions from sensitivity analysis:
Changes in Time to Correct Expected Order Rate affect Backlog only on the first
stage, rising or decreasing its peak value, because if Mr. Wang perceives new demand,
e.g., faster, desired production and workforce will adjust more rapidly, so backlog will takeon lower values after the shock. Once Mr. Wang fully acknowledges his new demand,
nothing is changed.
Changes in Time to Adjust Backlog cause Backlog fluctuations to change frequency
and amplitude. If Mr. Wang adjusts Backlog by a small amount every day (Time to Adjust
Backlog is high), Backlog must become sufficiently large in order to Production Rate to
offset its increase, and vice versa. Naturally, it takes more time, which means lower
frequency of oscillations.
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Changes in Time to Train and Dismiss affect oscillations. If Time to Train is high,
Mr. Wang is unable to vary his workforce timely, so there will be over- and under-
production multiple times.
From combined simulations we can conclude that Time to Train is the most
important time variable, because halving it and doubling Time to Correct Expected Order
Rate and Time to Adjust Backlog leads to comparatively fast equilibrium.
Task 6
a) The maximum schedule pressure is found to be 1.14 in time 26.
b) The table given shows how much more (less) will be produced is workers worked
more (less) by some margin.
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c) The previously introduced loops are unchanged, but a new one is introduced (partly
drawn in turquoise color). This loop says that increased backlog increases desired
production, which, in turn, increases desired productivity level, which leads to higher
schedule pressure, higher effects on it, higher production rates, and, finally, lower backlog.
Its main difference from the major negative feedback loops lies in the absence of a stock in
it, and, therefore, the absence of material delays.
The major feedback loop and newly introduced one interplay in the following way:
1. The second one reduces the initial need to hire a lot of employees after a shock
(immediate increase in production when backlog increases).
2. The first one allows hiring some additional workers to return pressure schedule to
the normal level of 1.
d) The model:
Backlog
Order Rate
Production Rate
Expected Order Rate
Change in Expected Order Rate
Target Service Time
Target Backlog
Target Backlog Adjustment
Desired Production
Desired Workforce
Workforce
Net Hiring
Module 1
Desired Productivity Schedule Pressure
Effect on Schedule Pressu re
+
-
+
++
+
+
+
-
+
+
+
+
-
+
-
+
- +
+
+
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The results of simulation show that the problem of oscillations is solved by the
willingness of employees to work overtime. Maximum values for backlog and service time
are 1428 bicycles and 1.30 days respectively.
Backlog
Expected Order Rate
Change in Expected OR
Order RateProdution Rate
Time Corr Expected Order Rate
Service Time
Target Backlog
Target Service Time
Desired Backlog Adjus tment
Time to Adjust Backlog
Workforce
Net Hiring
Tim e to Train or Dism iss
Desired Workforce
Productivity
Desired Production
Schedule Pres sure
Effect of Schedule Pressure
~
Desired Productivity
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e) The workforce increases to 55 people in the long run (approximately on 160 thday).
Overtime is a short-term solution and hiring a long-term one because, first of all,
hiring policy, as we have seen, is unable to tackle the problem effectively in the short-term
(resulting oscillations), so overtime provide some kind of support for hiring to offset
material delay problem. On the other hand, overtime cannot be a permanent solution in
THIS model, because after a demand shock not only overtime need is adjusted, but also
desired workforce (and real workforce, consequently) grows, offsetting the need of high
schedule pressure and leading to a stable long-term equilibrium.
f) Maximum schedule pressure is 1.07. It is lower because we introduced the effect of
schedule pressure, thus making backlog adjustment faster.
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Task 7
We modify the causal loop diagram to include the structure Mr. Wang is concerned
about (marked in blue). This is a positive (self-reinforcing) loop. That is, the higher is the
backlog, the higher would be target backlog adjustment, desired production and
productivity, and schedule pressure, which, in turn, increases the inflow of workers
fatigue. Higher levels of exhaustion lead to lower production rates, and, consequently, to
higher values of backlog.
Note that accordingly with Mr. Wangs concerns, possibility to hire additional workers
is eliminated from the model. So, workers will inevitably experience high and persistent
levels of schedule pressure (1.1 in the long term), which will cause great fatigue, and
backlog will eventually rise.
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Backlog
Order Rate
Production Rate
Expected Order Rate
Change in Expected Order Rate
Target Service Time
Target Backlog
Target Backlog Adjustment
Desired Production
Fatigue Change in Fatigue
Module 1
Desired Productivity Schedule Pres sure
Effect on Schedule Pressure
+
+
-
+
-
+
+
+
+
+
-
+
+
-
+
+
+
+