Instruction  4,  October  14  

Response of Material Delays to Steps, Ramps, and Cycles (pp. 425 & 426) Subquestion 1 & 2: “In all cases assume that prior to time zero the delay is in equilibrium with the outflow and inflow both equal to 100 units/day.” Because of this statement you can determine that the initial value of the stock equals 500 units (100 units/day * 5 days average delay time).

Material inTransitinflow outflow

step input

ramp input

exponential growth

oscillation

average delay time

<Time>

Step-input: STEP(100,0) Ramp-input: IF THEN ELSE (Time >=0, 5*Time, 0) Exponential growth: IF THEN ELSE (Time>=0, -100+100*(1.05)^Time, 0) Oscillation: IF THEN ELSE (Time>=0, 100*sin(Time/(10/(2*3.14159))), 0) Note that: 3.14159 = π To simulate the Step-input, use the following equation for the inflow: Inflow = 100+step input+0*ramp input+0*exponential growth+0*oscillation Outflow = Material in Transit/average delay time

inflow400

300

200

100

0-5 -3 -1 1 3 5 7 9 11 13 15 17 19 21 23 25

Time (Day)

units

/Day

inflow : normalinflow : stepinflow : ramp

inflow : exponentialinflow : oscillation

outflow400

300

200

100

0-5 -3 -1 1 3 5 7 9 11 13 15 17 19 21 23 25

Time (Day)

units

/Day

outflow : normaloutflow : stepoutflow : ramp

outflow : exponentialoutflow : oscillation

Subquestion 3: Third-order delay:

Material inTransit 1inflow outflow 1

step input

ramp input

exponential growth

oscillation

average delay time

<Time>

Material inTransit 2

Material inTransit 3outflow 2 outflow 3

Outflows of third-order delays (assuming that average delay time is 5 days for each of the three outflows):

outflow 3200

170

140

110

80-5 -3 -1 1 3 5 7 9 11 13 15 17 19 21 23 25

Time (Day)outflow 3 : normal 3rdoutflow 3 : step 3rdoutflow 3 : ramp 3rd

outflow 3 : exponential 3rdoutflow 3 : oscillation 3rd

Pipeline delay:

Material inTransitinflow outflow

step input

ramp input

exponential growth

oscillation

average delay time

<Time>

Outflow = DELAY FIXED(inflow, average delay time, 100) Outflows of pipeline delay:

outflow400

300

200

100

0-5 -3 -1 1 3 5 7 9 11 13 15 17 19 21 23 25

Time (Day)

units

/Day

outflow : normal pipelineoutflow : step pipelineoutflow : ramp pipelineoutflow : exponential pipelineoutflow : oscillation pipeline

Response of Delays to Changing Delay Times (pp. 435) Subquestion 1: The Postoffice.

average delay time

Letters inTransitmailing delivery

mailing = 100 Letters in Transit(0) = 0 delivery = DELAY3(mailing, average delay time) average delay time = 5 Change “average delay time” into:

a) A longer delay: 5 + STEP(5, 5) b) A shorter delay: 5 – STEP (2.5, 5)

delivery200

150

100

50

0-5 -3 -1 1 3 5 7 9 11 13 15 17 19 21 23 25

Time (Day)delivery : postoffice normaldelivery : postoffice shorter delaydelivery : postoffice longer delay

Letters in Transit600

300

0

-300

-600-5 -3 -1 1 3 5 7 9 11 13 15 17 19 21 23 25

Time (Day)Letters in Transit : postoffice normalLetters in Transit : postoffice shorter delayLetters in Transit : postoffice longer delay

Subquestion 2: If the actual order rate (O) is equal to the expected order rate (Ô), then the error (see Figure 11-10) is zero. Zero divided by the adjustment time is still zero. So, when the adjustment time suddenly decreases, nothing happens.

Exercise 1 of handout Assume that initially 30 employees have a contract and 10 future employees are in training. The average contract duration C is 5 years and the training period takes 1 year. Suppose the desired number changes from 30 to 40.

a) Determine the number of employees P(t) in the cases A=1 year and A=10 years. The situation is described by the following model:

O PF2F1

G

F3

A B

C

Figure 1-1. Vensim diagram for trainees and employees. where: O = the number of future employees in training, 10)( =tO at 0=t ; P = the number of actual employees working on a contract basis, 30)( =tP at 0=t ; G = the desired number of employees, G = 40 employees; A = the system’s adjustment time; B = the training period, B=1 year; C = the contract duration, C= 5 years. The given task can be solved by a system of differential equations describing the model:

BtO

AtPG

dtdO )()(

−−

= (1)

CtP

BtO

dtdP )()(

−= (2)

After a certain number of transformations of (2) we can substitute it into the equation (1) and obtain the following second order differential equation for )(tP :

AG

BtP

ACBdtdP

BCdtPd 1)()11(1)11(2

2

=++++ (3)

Equation (3) can be solved by finding a general solution to the homogeneous differential equation and partial solution for inhomogeneous one derived from (3).

1. The homogeneous differential equation is:

0)()11(1)11(2

2

=++++ tPACBdt

dPBCdt

Pd (4)

Substituting atetP =)( and multiplying the right-hand side of equation (4) by ABC , we will receive:

0))()(( 2 =++++ ateCAaCBAABCa (5) The solutions for (5) are:

ABCABCCBACBA

a2

4)()( 222

2,1−−±+−

= (6)

For the positive discriminant in (6), the solution of equation (4) has the following form: tata eDeDtP 21

21)( += (7) where: 1D and 2D are arbitrary constants. If the discriminant in equations (6) has a negative value, then the system experiences an oscillation. For this case, the following function provides the solution to (4):

)sin()cos()( 21 wteCwteCtP ktkt += (8) where: 1C and 2C are arbitrary constants, and

ABCABCCBA

wBCCBk

24)(

2)( 222 +−−

=+−

= (9)

2. The inhomogeneous differential equation: The partial solution for the inhomogeneous equation is a constant 0C which satisfies equation (3). By differentiating (3), we will find the value of :0C

CAGC+

=1

0 (10)

Thus, the general solution for equation (3) is the sum of expressions (7) and (10) for the positive discriminant, or (8) and (10) for the negative discriminant.

In order to determine the constants 1D and 2D or 1C and 2C we need to explode the initial conditions for 0=t , which leads to: For the positive discriminant:

22110

021)0(

aDaDdtdP

CDDP

t

+=

++=

=

(11)

For the negative discriminant:

CAGCCCP+

+=+=1

)0( 101

kCwCdtdP

t12

0

+==

(12)

Through substitution in equations (11) and (12) we will find the following expressions for 1D and 2D or 1C and 2C : For the positive discriminant:

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=

=022

0211 )0(1 CaPa

dtdP

aaD

t

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=

=011

0122 )0(1 CaPa

dtdP

aaD

t

(13)

For the negative discriminant:

CAGPCPC+

−=−=1

)0()0( 01

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=

kCdtdP

wC

t1

02

1 (14)

Now, we are ready to solve this exercise numerically for A=1 and 10 years. For A=1 year:

From (6): 10

8462,1

−±−=a ,

i.e. an oscillation is undergone by the system. Then, to find )(tP we need to refer to equation (8), where:

92.010846.0 ==−= wk

To determine 1C and 2C , we calculate 0C using equation (10):

33.3351140

0 =+

=C

Given the values of )0(P , 0C , w and k , the parameters 1C and 2C from (14) will be determined as follows: 33.333.33301 −=−=C

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=

292.01

02

tdtdPC

From equation (2) we can find the value of the derivative of )(tP at the initial moment 0=t :

4530

110

0

=−==tdt

dP

Hence, we will calculate: 17.22 =C . Substituting the obtained parameters into equation (8) and summing up the solutions of (8) and (10), we will determine the solution for )(tP :

)92.0sin(17.2)92.0cos(33.333.33)( 6.06.0 tetetP tt −− +−=

Graph for P40

30

200 4 8 12 16 20

Time (Year)

P : Current empl/Year

Figure 1-2. Number of employees as a function of time (for A = 1 year). For A=10 years:

From (6): 10

662,1

±−=a ,

i.e. the system does not go through oscillations.

From (10): 33.135

10140

0 =+

=C

Substituting 1a and 2a in expressions (13), we will receive:

⎟⎟⎠

⎞⎜⎜⎝

⎛×−+

+−=

=

33.1384.0)0(84.084.036.0

1

01 P

dtdPD

t

⎟⎟⎠

⎞⎜⎜⎝

⎛×−+

+−=

=

33.1336.0)0(36.036.084.0

1

02 P

dtdPD

t

The initial values of )(tP and its derivative are the same as for A=1 year, i.e. 30)0( =P and

40

==tdt

dP .

Hence, we will calculate 1D and 2D as: 51.371 =D 84.202 −=D

Thus, joining the solutions for equations (7) and (10), the solution for )(tP will be the following:

tt eetP 84.036.0 84.2051.3733.13)( −− −+= .

Graph for P40

20

00 4 8 12 16 20

Time (Year)

P : Current empl/Year

Figure 1-3. Number of workers in time (for A = 10 years).

b) Which values of A result in an oscillating system?

If the discriminant in equations (6) is negative, then the system experiences an oscillation. For this case, A should satisfy the following inequality:

04)( 222 <−− ABCCBA Since 0<A is unrealistic, the only possible solution is:

25.6)(

42

2

<

−<

ACB

BCA

c) If we want to limit overshoot to a maximum of 1 employee, what is the best choice for A? The point of a maximal overshoot is defined by the maximal value of )(tP . At this point the first derivative of )(tP is equal to 0, and the second one is negative. To determine the derivatives, we need to consider the function )(tP corresponding to oscillating behavior:

)sin()cos()( 210 wteCwteCCtP ktkt ++= (see exercise 1.a).

0))sin()()cos()(()( 1221 =−++=ʹ′ wtwCkCwtwCkCetP kt 0))sin())(2()cos())(2(()( 22

2122

12 <−+−+−+=ʹ′ʹ′ wtwkCkwCwtwkCkwCetP kt The solution of the above mentioned formulas will be:

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎥⎦

⎤⎢⎣

⎡

−

+= π

kCwCwCkCarctg

wt

21

21* 1

Hence, )sin()( **

21

22

21

0*

max wtewCkC

CCwCtPP kt⎟⎟⎠

⎞⎜⎜⎝

⎛

+

++==

Since the overshoot shows the difference between maxP and )(∞P , the maximal overshoot is equal to:

)sin( **

21

22

21 wte

wCkCCCwOvershoot kt

⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+=

Setting the overshoot to 1 and using equations (9), (10) and (14) yield a non-linear equation for A. This equation can be solved numerically by Mathematica for Windows 5.1, resulting in A=0.840691 years.

d) Determine the function describing the number of employees in training in case of A=1 year? The number of employees in training is described by the following equation, derived from equation (2) (exercise 1.a):

⎟⎠

⎞⎜⎝

⎛ +=CtP

dtdPBtO )()(

Next, we need to substitute the values of )(tP and its derivative, obtained for A=1 year in exercise 1.a and 1.c. The final expression )(tO is:

)92.0sin(18.2)92.0cos(33.367.6)( 6.06.0 tetetO tt −− ++=

Graph for O10

8

60 4 8 12 16 20

Time (Year)

O : Current

Figure 1-4. Number of trainees in time.