HW7 Soln S2K - user.engineering.uiowa.edu

57:022 Principles of Design II HW#7 page 1 of 4

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57:022 Principles of Design IIHomework #7

Due Wednesday, March 8, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

1. A system contains 4 types of devices, with the system reliability represented schematically by

It has been estimated that the lifetime probability distributions of the devices are as follows:

A: Weibull, with mean 2000 days and standard deviation 1200 days (u=2243.03, k=1.71708)B: Normal, with mean 1200 days and standard deviation 200 daysC: Exponential, with mean 2000 daysD: Exponential, with mean 4000 days

a. Compute the reliabity of a unit of each individual device for a designed system lifetime of 1000 days:

Device Reliability

A 0.778958 RA(1000) = 1 – FA(1000) = 1 - ( 1 - ( ) 71708.103.2243/1000e − ) B 0.840882 RB(1000) = From CDF table of Normal Distribution C 0.606531 RC(1000) = 1 – FC(1000) = 2000/1000e−

D 0.778801 RD(1000) = 1 – FD(1000) = 4000/1000e−

b. For each situation, indicate whether the system has failed:

Component failures System failure?B1 & C2 fail: Yes or NoA & B2 fail: Yes or NoC1 & C2 fail: Yes or NoB1 & B2 fail: Yes or No

D fails: Yes or No

c. Using the reliabilities in (a), compute the system reliability:

Subsystem ReliabilityB1B2 RB×RB = 0.707082C1C2 1 − (1 −RC)( 1 −RC)= 1− 0.15482 = 0.845182 B1B2+ C1C2 1 − (1 −RBB)( 1 −RCC) = 1 − (0.292917)(0.15482) = 1 − 0.04535 = 0.954650Total system: RA×RBBCC×RD = (0.778958)(0.954650)(0.778801) = 0.579142


2. A system has 6 components which are subject to failure, each having lifetimes with exponential distributions.The average lifetimes are:

Component Average LifetimeA 2000 daysB 3000 daysC 800 daysD 800 daysE 500 daysF 500 daysG 500 days

The system will fail if any one of the following occur:

• Either A or B fails• Both C and D fail• Components E, F, and G all fail.

a. Draw a diagram showing the hybrid series/parallel configuration of the components, as in the Hypercard Stack"System Reliability".

b. Suppose that the system is required to survive for 1000 days. What is the reliability of each component, i.e., theprobability that it survives 1000 days?Solution: R(t) = e−λt where λ is the failure rate, i.e., 1/(2000days), 1/(3000days), 1/(800days), etc.

RA = 2000/1000e− = 0.6065306597RB = 3000/1000e − = 0.7165313106RC = RD = 800/1000e − = 0.2865047969RE = RF = RG = 500/1000e − = 0.1353352832

c. What is the reliability of the system, i.e., the probability that the system survives 1000 days?RCD = 1 − (1 − RC )( 1 − RD ) = 0.490925REFG = 1 − (1 − RE )( 1 − RF ) )( 1 − RG ) = 0.353538Reliability of system = RA × RB × RCD × REFG = 0.0754291

d. Construct an ARENA model which can simulate the lifetime of this system.

e. Run the ARENA simulation model, using 500 runs, collecting statistics on the time of system failure. Requestthat a histogram be printed. Select about 15 cells, with HLOW and HWID parameters which will give you a"nice" histogram with the mean approximately in the center and with small tails.


Output Summary for 500 Replications

Project: Reliability Run execution date : 3/13/2000Analyst: DLB Model revision date: 3/13/2000

OUTPUTS

Identifier Average Half-width Minimum Maximum # Replications_______________________________________________________________________________

TNOW 507.51 29.594 1.3470 2041.4 500

The histogram can be obtained by performing “Output Analyzer” in ARENA as follows :

Output AnalyzerData File : Rel.dataReplications : LumpedHistogram Cells :

Number(Interior) : 20Lower Limit : 0Width : 25


Histogram Summary System Lifetime

Cell Limits Abs. Freq. (Time) Rel. Freq. Cell From To Cell Cumul. Cell Cumul. 1 -Infinity 0 0 0 0 0 2 0 25 4 4 0.008016 0.008016 3 25 50 3 7 0.006012 0.01403 4 50 75 11 18 0.02204 0.03607 5 75 100 12 30 0.02405 0.06012 6 100 125 12 42 0.02405 0.08417 7 125 150 12 54 0.02405 0.1082 8 150 175 14 68 0.02806 0.1363 9 175 200 16 84 0.03206 0.1683 10 200 225 14 98 0.02806 0.1964 11 225 250 21 119 0.04208 0.2385 12 250 275 23 142 0.04609 0.2846 13 275 300 22 164 0.04409 0.3287 14 300 325 12 176 0.02405 0.3527 15 325 350 13 189 0.02605 0.3788 16 350 375 14 203 0.02806 0.4068 17 375 400 15 218 0.03006 0.4369 18 400 425 15 233 0.03006 0.4669 19 425 450 15 248 0.03006 0.497 20 450 475 13 261 0.02605 0.523 21 475 500 9 270 0.01804 0.5411 22 500 +Infinity 229 499 0.4589 1

f. Suppose you will offer a warranty on this system, such that 95% of the systems will survive past the length of thewarranty. According to the ARENA model, what should be the length of the warranty?

Solution: After 50 days, about 3.6% of the systems have failed, while after 75 days, about 6% have failed.If we perform linear interpolation, we get about 25+14.5 = 39.5 days. That is, a system has about 95%probability of surviving a 39.5-day warranty period.

Note: You may wish to consult the web pagehttp://www.alf.ie.engineering.uiowa.edu/bricker/Reliability_ARENA.html

57:022 Principles of Design II -- HW1 Solns page 1 of 2

57:022 Principles of Design II

♦♦♦♦♦♦♦ 57:022 Principles of Design II ♦♦♦♦♦♦♦

☺☺☺ Homework Solutions ☺☺☺

1. In preparation for a game of "Craps", Nathan Detroit has asked you the followingquestions:a. What is the probability of throwing a 7 or 11 at least twice in six tosses of a pairof dice?

Solution: Out of the 62 = 36 possible outcomes of a toss of a pair of dice, there are six waysto throw a seven: [1,6], [2,5], [3,4], [4,3], [5,2], and [6,1]. There are two ways to throw an11, namely [5,6] and [6,5]. Thus, there are 6+2=8 ways to obtain the outcome "7 or 11" in atoss of the dice, and the probability of this outcome is 8/36 = 2/9.Define the random variable

N6 = the number of 7’s & 11’s obtained in 6 throws of a standard pair of dice.N6 has the binomial distribution with parameters n=6 and p=8/36

{ } { }161060

66 368

1368

16

368

1368

06

12NP12NP−−

−

−

−

−=<−=≥ = 0.3991

b. What is the expected number of 7's & 11’s obtained in 6 throws of a pair of dice? Solution: The expected number of outcomes "7 or 11" in six tosses of the dice isE(N6) = np = 6(8/36) = 4/3

c. What is the expected number of throws of the dice to obtain a 7 or 11? Solution: Define the random variable T1 = the number of throws of a standard pair of dice in

which the 1st 7 or 11 is obtained.T1 has the geometric distribution with parameter p=8/36, and E(T1) = 1/P = 36/8 = 4.5

2. The foreman of a casting section in a certain factory finds that on the average, 1 inevery 5 castings made is defective.

a. If the section makes 8 castings a day, what is the probability that 2 of these will bedefective?

Solution: One in five castings are defective, so (assuming the defects are independent andidentically dependent) the probability of a casting being defective is p=1/5. Define therandom variable N8 = the number of defects in 8 castings.

N8 has the binomial distribution with parameters n=8 and p=1/5

{ }282

8 51

151

28

2NP−

−

== = 0.2936

b. What is the probability that 5 or more defective castings are made in one day?

57:022 Principles of Design II -- HW1 Solns page 2 of 2

Solution:

{ }888787686585

8 51

151

88

51

151

78

51

151

68

51

151

58

5NP−−−−

−

+

−

+

−

+

−

=≥

= 0.00917504 + 0.00114688 + 0.00008192 + 0.00000256 = 0.0104064 ≈ 0.0104

3. A recent survey suggested that if the presidential election was held today, Al Gorewould receive 40% of the popular vote. If the Principles of Design II class conducts asurvey of 9 students, what is the probability that our results will suggest Mr. Gorewould be elected president if the election were held today, i.e., what is the probabilitythat at least 5 of those surveyed would vote for Gore.

Define the random variable N9 = the number of votes for Gore in a survey of 9 students.N9 has the binomial dist. with parameters n=9 and p=0.4

{ } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 8987976965959 4.014.0

89

4.014.079

4.014.069

4.014.059

5NP −−−− −

+−

+−

+−

=≥

( ) ( ) ( ) ( ) 999898 4.014.099

4.014.089 −− −

+−

+

= 0.2667

57:022 HW#2 Spring 2000 page 1 of 1

❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍57:022 Principles of Design II

Homework #2Due Wednesday, February 2, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

1. A light bulb in an apartment entrance fails randomly, with an expected lifetime of 15 days, and is replacedimmediately by the custodian. Assume that this bulb's lifetime has an exponential distribution.

a. What is the probability that a bulb lasts longer than its expected lifetime?b. If the current bulb was inserted 10 days ago, what is the probability that its lifetime (since it was inserted)

will exceed the expected lifetime of 15 days?c. If you were to test 10 of these bulbs, what is the probability that more than half will exceed the expected

lifetime?d. If the custodian has 2 spare bulbs, what is the probability that these (including the one currently in use) will

be sufficient for the next 30 days?

2. The probability that each car stops to pick up a hitchhiker is p=3%; different drivers, of course, make theirdecisions to stop or not independently of each other.

(a) Given that a hitchhiker has counted 30 cars passing him without stopping, what is the probability that hewill be picked up by the 40th car or before?

Suppose that the cars arrive according to a Poisson process, at the average rate of 20 per minute. Then"success" for the hitchhiker occurs at time t provided that both an arrival occurs at t and that car stops to pickhim up. Let T be the time (in seconds) that he finally gets a ride, when he begins his wait at time zero.

(b) What is the distribution of T? (Give both name & parameters of distribution.) What are E(T) and Var(T)?(c) Given that after 3 minutes (during which 53 cars have passed by) he is still there waiting for a ride, compute

the expected value of T (his total waiting time, including the 3 minutes he has already waited).

3. (a.) Using the last four digits of your ID# as the "seed" for the "Midsquare" technique, generate a sequence of 5pseudo-random numbers uniformly distributed in the interval [0,1].(b.) Using the "Inverse Transformation" technique and the first 5 numbers generated in (a.), generate theinterarrival times for 5 vehicles which form a Poisson process with arrival rate λ=2/minute.(c.) What is the expected number of arrivals during the first minute? What is the actual # of arrivals in yoursimulation? What is the probability that you would observe exactly this number of arrivals in this Poissonprocess?(d.) Why cannot the Rejection Technique be used in (b)?

57:022 HW#2 Solutions Spring 2000 page 1 of 3



1. A light bulb in an apartment entrance fails randomly, with an expected lifetime of 15 days, and is replacedimmediately by the custodian. Assume that this bulb's lifetime has an exponential distribution.

a. What is the probability that a bulb lasts longer than its expected lifetime?

Solution: Let Nt be the random variable defined as the cumulative number of bulb failures at time t. Then {Nt,t≥0} is a Poisson process with rate λ = 1/15 (failures/day). If Ti is the random variable defined as the timebetween failures (i-1) and i, then Ti has exponential distribution, with mean value 1/λ = 15 days.

{ } 3679.0e15TP 15)15/1( ==> −

b. If the current bulb was inserted 10 days ago, what is the probability that its lifetime (since it was inserted) willexceed the expected lifetime of 15 days?

Solution: Because of the memoryless property of the exponential distribution, we wish the probability that arandom variable with exponential distribution (with parameter λ=1/15) exceeds 5 days, which is

{ } 7165.0ee5TP 3/15)15/1( ===> −−

c. If you were to test 10 of these bulbs, what is the probability that more than half will exceed the expectedlifetime?

Solution: The testing of the bulbs is a discrete-event Bernouilli process, with "success" defined as a bulb'slifetime exceeding 15 days. The number of successes in 10 trials (n=10) with probability of success p= 0.3679(from (a) above) has Binomial distribution ( n=10, p=0.3679):

∴ { } 1176.05NP 10 => , the required probability.

d. If the custodian has 2 spare bulbs, what is the probability that these (including the one currently in use) will besufficient for the next 30 days?

Solution: We wish to compute P{N30 ≤3} (accounting for the bulb which is currently installed), where N30 hasthe Poisson distribution:

P Nt = x =λt x

x! e– λt

In this case, λ = 1/15 per day and t = 30 days, so λt = 2 is the expected number of bulb failures. Hence,

P N30 ≤ 3 =2 x

x! e– 2Σn = 0

3

Computations:

x P[x]0 0.135335281 0.270670572 0.270670573 0.18044704Sum: 0.857123

2. The probability that each car stops to pick up a hitchhiker is p=3%; different drivers, of course, make theirdecisions to stop or not independently of each other.

(a) Given that a hitchhiker has counted 30 cars passing him without stopping, what is the probability that he willbe picked up by the 40th car or before?

Solution:


Consider this to be a discrete-time Bernouilli process, where each car is a "trial" and "success" (with p=0.03) isdefined as a car stopping. Then T1, defined as the number of trials until the first success, has geometricdistribution:

{ } { } )97.0...97.097.0(03.0P)P1(nTP10TP 9109

1n

1n9

1n11 +++=−===≤ ∑∑

=

−

=

= 0.2626

Suppose that the cars arrive according to a Poisson process, at the average rate of 20 per minute. Then"success" for the hitchhiker occurs at time t provided that both an arrival occurs at t and that car stops to pickhim up. Let T be the time (in seconds) that he finally gets a ride, when he begins his wait at time zero.

(b) What is the distribution of T? (Give both name & parameters of distribution.) What are E(T) and Var(T)?

Solution: T has exponential distribution with parameter (rate) λ = (20/minute)×(0.03) = 0.6/minute.E(T) = 1/λ = 1/0.6 = 1.6667 (minutes)V(T) = 1/λ2 = 1/(0.6)2 = 2.7778

(c) Given that after 3 minutes (during which 53 cars have passed by) he is still there waiting for a ride, compute theexpected value of T (his total waiting time, including the 3 minutes he has already waited).

Solution: Because of the memoryless property of the exponential distribution, the time between now (when 3minutes have passed) until the next car which stops has the same distribution as the original T. Therefore, theexpected value is the expected value of T plus the 3 minutes which have already passed, i.e.,

3+ E(T) = 3 + (1.6667) = 4.6667

3. (a.) Using the last four digits of your ID# as the "seed" for the "Midsquare" technique, generate a sequence of 5pseudo-random numbers uniformly distributed in the interval [0,1].

Solution: Suppose that the last four digits of Hansuk's ID# is 1250. Then X0=1250 serves as the "seed" for therandom number generator:

i Xi (Xi)2 Ri1 1250 01562500 0.56252 5625 31640625 0.64063 6406 41036836 0.03684 0368 00135424 0.13545 1354 01833316 0.8333

(b.) Using the "Inverse Transformation" technique and the first 5 numbers generated in (a.), generate theinterarrival times for 5 vehicles which form a Poisson process with arrival rate λ=2/minute.

Solution: Tk is the time of the k th arrival (which has k-Erlang distribution) and the times between arrivals haveExponential distribution with λ=2/minute. Since the CDF of the Exponential distribution is F(t) = 1 - e −λt , theinverse transformation method requires using a random number R uniformly distributed in [0,1] and finding τ suchthat 1 - e −λτ = R, namely

τ = –ln 1 – R

λUsing the sequence generated above (0.5625, 0.6406, ….), we obtain

i Ri τi Ti1 0.5625 0.413339 0.4133392 0.6406 0.511660 0.9249993 0.0368 0.018747 0.943746


4 0.1354 0.072744 1.0164905 0.8333 0.895780 1.912270

where the arrival times T1, T2, … are obtained by summing the appropriate interarrival times, e.g.,T1 = τ1, T2 = τ1 + τ2, T3 = τ1 + τ2+ τ3 , etc.

(c.) What is the expected number of arrivals during the first minute? What is the actual # of arrivals in yoursimulation? What is the probability that you would observe exactly this number of arrivals in this Poissonprocess?

Solution: Nt, the number of arrivals which have occurred at time t, has the Poisson distribution. Its expectedvalue is E(Nt) = λt and so E(N1) = (2/min)(1min) = 2.

The actual # of arrivals in my simulation is 3, i.e., 3 arrivals in the first minute, since the 4th arrival occurs at 1.01649minutes which is later than 1 minute.

The probability that exactly x cars arrive during the first t minutes is

P Nt = x =λt x

x! e– λt

and so

P N1 = 3 =2 3

3! e– 2 = 0.1805

(d.) Why cannot the Rejection Technique be used in (b)?

Solution: The interarrival times have the Exponential distribution which does not have its probability restricted toa finite interval [a,b] (since f(0)=+∞ and f(t)>0 for all t>0). That is, the graph of the density function cannot becontained within a rectangle. Thus, the Rejection Method cannot be used in (b).




1. Generating random numbers by rejection method. Consider the triangular distribution with density functionshown below.

a. Write the expressions for the density function: f(t) = _______ if t≤3 = _______ if 3<t<4

= 0 for t>4 & t<0b. Write the expression for the CDF (cumulative distribution function) corresponding to this density function.

F(t) = P{T≤ t} = 0 if t<0= _________ for 0≤t≤3= _________ for 3<t≤4= 1 for t>4

c. What is the value of C?

d. Write the inverse of the CDF: i.e., write F-1(R) by solving F(t) = R for t, where R is in the interval [0,1].F-1(R) = _____________ if 0 ≤ R ≤ _?__

= _____________ if _?__ ≤ R ≤ 1

e. Generating random numbers by rejection method. Use the table of random numbers distributed in class, startingwith the first row and continuing with as many additional rows as might be necessary, use the rejection method togenerate ten random numbers having the triangular distribution shown.

f. Generating random numbers by the inverse transformation method. Using the random number table again,starting with the first column, use the inverse transformation method to generate ten random numbers having thissame distribution.

g. The CDF of the normal distribution N(µ , σ) is

F(t) = 1σ 2π

1 21 2exp –

t – µ2

2σ2

– ∞

t

Which, if any, of the two methods above would you suggest using to generate a sequence of numbers having theNormal distribution? (Explain why you would or would not choose each of the 2 methods above.)


2. Manual simulation of Centerville bank drive-up window:a. Use column A of the right half of the page (starting with 234903), generate the arrival times of twenty customers

having exponential distribution with mean 5 minutes (i.e., λ = 0.2/minute). Use the inverse transformation methodas we did in class: T = − (ln R)/λ.

b. Using the 20 inter-arrival times which you generated in (a), and the 20 service times which you generated in (1e)and (1f) above, simulate the operation of the system with a single teller window. Consider the entry in the"Length of Queue" column to be the number of cars waiting (not including that being served) after the arriving carhas joined the queue.

Car # Random#

Inter-arrivaltime

Random#

Servicetime

Arrivaltime

TimeServiceBegins

Departuretime

Waitingtime

Lengthofqueue

1234567891011121314151617181920

c. What is the maximum length of the waiting line (not including the car currently being served at the window)?

d. What is the maximum waiting time?

e. What is the average waiting time?



Homework #3 SolutionSpring 2000

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1. Generating random numbers by rejection method. Consider the triangular distribution with density functionshown below.

a. Write the expressions for the density function:

Solution: f(t) = f1(t) = (1/6)t , if 0 ≤ t ≤ 3 = f2(t) = (-1/2)t + 2 , if 3 < t ≤ 4

= 0 for t>4 & t<0Note: This requires knowledge of the height C from part c below.

b. Write the expression for the CDF (cumulative distribution function) corresponding to this density function.

Solution: The Cumulative Distribution Function (CDF) is the integral of the density function:F(t) = P{T≤ t} = 0 if t<0

= 1 for t>4

F(t) = P T ≤ t = f(v)dv0

t

if 0 ≤ t ≤ 4

= f1(v)dv0

t

= v6dv

0

t

= t 2

12 if 0 ≤ t ≤ 3

For 3 ≤ t ≤ 4,

F(t) = f1(v)dv0

3

+ f 2(v)dv3

t

if 3 ≤ t ≤ 4

I.e., if 3 ≤ t ≤ 4,

F(t) = 9

12+ 2 – v2 dv

3

t

= 34 + 2v – v2

43

t

= 34 + 2t – t 2

4 – 2× 3 – 32

4 = 2t – t 2

4 – 3

c. What is the value of C?Solution: The area under the density function, i.e., the triangle, must have the value 1.0. Therefore, 0.5×4×C=1

which implies that C=0.5.


d. Write the inverse of the CDF: i.e., write F-1(R) by solving F(t) = R for t, where R is in the interval [0,1].F-1(R) = _____________ if 0 ≤ R ≤ _?__

= _____________ if _?__ ≤ R ≤ 1Solution: F(0)=0, F(3)= 0.75, and F(4) = 1. Therefore, if 0≤R≤0.75, we must find t such thatt 2

12 = R ⇒ t 2 =12R ⇒ t = 12R = 2 3R

while if 0.75 ≤ R ≤ 1.0, R3t2t41 2 =−+− ⇒ 0)R3(t)2(t

41 2 =++−+

⇒ R124

412

)3R(4142)2(

t

2

−±=

−

−±−−

=

Clearly, the root which we seek is that corresponding to "−", i.e., R124t −−= . Summarizing, then, we generate a random number R distributed uniformly in the interval [0,1], and

if 0≤R≤0.75, obtain the random number T = 2 3R , while if 0.75≤R≤1, we obtain the number

R124T −−=

e. Generating random numbers by rejection method. Use the table of random numbers distributed in class, startingwith the first row and continuing with as many additional rows as might be necessary, use the rejection method togenerate ten random numbers having the triangular distribution shown.

Solution: Step1 :Generate 2 random numbers R1 and R2 uniformly distributed in [0, 1], in this case, selected from the table.

Step2 :M = ½, a = 0, b = 4We must scale R1 so as to be distributed uniformly in [0,4] and R2 so as to be distributed uniformly in [0,C] = [0, 0.5].Let t* = a + (b-a)R1 = 0 + (4-0) R1 = 4 R1 and y* = M R2 = (1/2) R2

to get a point (t,y) uniformly distributed in the rectangle.

If 0 ≤ t* ≤ 3⇒ f(t*) = f1(t*) = (1/6)t* = (1/6)( 4 R1) = (2/3) R1

If 3 ≤ t* ≤ 4⇒ f(t*) = f2(t*) = (-1/2)t* + 2 = (-1/2)( 4 R1) + 2 = - 2 r1 + 2 = 2(1 - R1)

Step3 :Accept t* if y* (= M r2 ) ≤ f(t*) is satisfied…i.e., the point lies in the shaded region

under the graph of y = f(t).otherwise, reject t and return to step1.

i.e., for 0 ≤ t* ≤ 3⇒ Accept t* if y* (= M r2 = (1/2) r2) ≤ f1(t*) = (2/3) r1

for 3 ≤ t* ≤ 4⇒ Accept t* if y* (= M R2 = (1/2) R2) ≤ f2(t*) = 2(1 - R1)

The results, using random numbers R selected from the table distributed in class, are:r1 = 0.142582 r1 = 0.948535 r1 = 0.621651 r1 = 0.014694


r2 = 0.838531 r2 = 0.204547 r2 = 0.329695 r2 = 0.097086t* = 0.570328 t* = 3.794140 t* = 2.486604 t* = 0.058776y* = 0.419266 y* = 0.102274 y* = 0.164848 y* = 0.048543

f1 (t*) = 0.095055 f2(t*) = 0.102930 f1 (t*) = 0.414434 f1(t*) = 0.009796y* > f1(t*) y* < f2 (t*) y* < f1 (t*) y* > f1 (t*)

Thus,reject t* Thus,Accept t* Thus,Accept t* Thus,reject t*

r1 = 0.190024 r1 = 0.170674 r1 = 0.124008 r1 = 0.358324r2 = 0.666521 r2 = 0.144070 r2 = 0.702818 r2 = 0.034739t* = 0.760096 t* = 0.682696 t* = 0.496032 t* = 1.433296y* = 0.333261 y* = 0.072035 y* = 0.351409 y* = 0.017370

f1(t*) = 0.126683 f1(t*) = 0.113783 f1(t*) = 0.082672 f1(t*) = 0.238883y* > f1 (t*) y* < f1 (t*) y* > f1 (t*) y* < f1 (t*)

Thus,reject t* Thus,Accept t* Thus,reject t* Thus,Accept t*


f1(t*) = 0.268675 f2 (t*) = 0.139026 f1(t*) = 0.288651 f1(t*) = 0.436015y* > f1 (t*) y* > f2 (t*) y* < f1 (t*) y* < f1 (t*)

Thus,reject t* Thus,reject t* Thus,accept t* Thus,accept t*


f1(t*) = 0.487392 f1(t*) = 0.235093 f2(t*) = 0.142538 f2(t*) = 0.086908y* < f1 (t*) y* < f1 (t*) y* < f2 (t*) y* > f2 (t*)

Thus,accept t* Thus,accept t* Thus,accept t* Thus,reject t*

r1 = 0.769932r2 = 0.574832t* = 3.079728y* = 0.287416

f2(t*) = 0.460136y* < f2 (t*)

Thus,accept t*

Thus, 10 random numbers generated are

3.794140 , 2.486604 , 0.682696 , 1.433296 , 1.731904 ,

2.616092 , 2.924352 , 1.410560 , 3.714924 , 3.079728


f. Generating random numbers by the inverse transformation method. Using the random number table again,starting with the first column, use the inverse transformation method to generate ten random numbers having thissame distribution.

Solution: Recall that the inverse CDF is given byF−1(R ) = T = 2 3R , if 0≤R≤0.75

= R124T −−= if 0.75≤R≤1We then do the computations:

i Ri 2√(3Ri) 4 - 2√(3−Ri) Ti

1 0.142582 1 . 3 0 8 0 4 5 8 71 . 3 0 8 0 4 5 8 711

2.148062636 1.308045871

2 0.097086 1 . 0 7 9 3 6 6 4 81 . 0 7 9 3 6 6 4 811

2.099564260 1.079366481

3 0.358324 2 . 0 7 3 6 1 7 1 32 . 0 7 3 6 1 7 1 3 2.397906370 2.073617134 0.257091 1 . 7 5 6 4 4 2 9 91 . 7 5 6 4 4 2 9 9

772.276156620 1.756442997

5 0.928731 3.338378648 3.466074912 3 . 4 6 6 0 7 4 9 13 . 4 6 6 0 7 4 9 122

6 0.729816 2 . 9 5 9 3 5 6 6 82 . 9 5 9 3 5 6 6 877

2.960415468 2.959356687

7 0.781062 3.061493753 3.064183779 3 . 0 6 4 1 8 3 7 73 . 0 6 4 1 8 3 7 799

8 0.093549 1 . 0 5 9 5 2 2 5 31 . 0 5 9 5 2 2 5 344

2.095845594 1.059522534

9 0.829708 3.155391576 3.174670975 3 . 1 7 4 6 7 0 9 73 . 1 7 4 6 7 0 9 755

10 0.550416 2 . 5 7 0 0 1 7 8 92 . 5 7 0 0 1 7 8 999

2.658979493 2.570017899

g. The CDF of the normal distribution N(µ , σ) is

F(t) = 1σ 2π

1 21 2exp –

t – µ2

2σ2

– ∞

t

Which, if any, of the two methods above would you suggest using to generate a sequence of numbers having theNormal distribution? (Explain why you would or would not choose each of the 2 methods above.)

Solution: Neither method is appropriate for generating numbers having a normal distribution. Since it is impossibleto derive a closed-form expression for F-1( R), the inverse transformation fails. The density function cannot beenclosed in a rectangular box, because of the tails extending infinitely far both to the right and left, and so therejection method fails.

2. Manual simulation of Centerville bank drive-up window:a. Use column A of the right half of the page (starting with 234903), generate the arrival times of twenty customers

having exponential distribution with mean 5 minutes (i.e., λ = 0.2/minute). Use the inverse transformation methodas we did in class: T = − (ln R)/λ.

b. Using the 20 inter-arrival times which you generated in (a), and the 20 service times which you generated in (1e)and (1f) above, simulate the operation of the system with a single teller window. Consider the entry in the"Length of Queue" column to be the number of cars waiting (not including that being served) after the arriving carhas joined the queue.


Car Inter-arrival Service Arrival Time Departure Waiting Increment Length

# Random# time Random# time Time Service Time Time in Waiting of Queue

Begins Time

1 0.234903 1.338763 0.948535 3.794140 1.338763 1.338763 5.132903 0.00 0.00 0.0

2 0.827916 8.798863 0.621651 2.486604 10.137626 10.137626 12.624230 0.00 0.00 0.0

3 0.072531 0.376480 0.170674 0.682696 10.514106 12.624230 13.306926 2.11 2.11 1.0

4 0.632794 5.009161 0.358324 1.433296 15.523267 15.523267 16.956563 2.11 0.00 0.0

5 0.382872 2.413394 0.432976 1.731904 17.936661 17.936661 19.668565 2.11 0.00 0.0

6 0.808573 8.266244 0.654023 2.616092 26.202905 26.202905 28.818997 2.11 0.00 0.0

7 0.354309 2.187171 0.731088 2.924352 28.390076 28.818997 31.743349 2.54 0.43 1.0

8 0.906243 11.835245 0.352640 1.410560 40.225321 40.225321 41.635881 2.54 0.00 0.0

9 0.525936 3.732065 0.928731 3.714924 43.957385 43.957385 47.672309 2.54 0.00 0.0

10 0.728432 6.517714 0.769932 3.079728 50.475099 50.475099 53.554827 2.54 0.00 0.0

11 0.133451 0.716183 0.142582 1.308046 51.191282 53.554827 54.862873 4.90 2.36 1.0

12 0.689269 5.844138 0.097086 1.079366 57.035420 57.035420 58.114786 4.90 0.00 0.0

13 0.724364 6.443371 0.358324 2.073617 63.478791 63.478791 65.552408 4.90 0.00 0.0

14 0.765851 7.258988 0.257091 1.756443 70.737779 70.737779 72.494222 4.90 0.00 0.0

15 0.947243 14.710294 0.928731 3.466075 85.448073 85.448073 88.914148 4.90 0.00 0.0

16 0.465051 3.127919 0.729816 2.959357 88.575993 88.914148 91.873505 5.24 0.34 1.0

17 0.257697 1.489989 0.781062 3.064184 90.065981 91.873505 94.937689 7.05 1.81 2.0

18 0.160987 0.877645 0.093549 1.059523 90.943627 94.937689 95.997212 11.04 3.99 3.0

19 0.821728 8.622224 0.829708 3.174671 99.565851 99.565851 102.740522 11.04 0.00 0.0

20 0.133820 0.718313 0.550416 2.570018 100.284164 102.740522 105.310540 13.50 2.46 1.0

c. What is the maximum length of the waiting line (not including the car currently being served at the window)? 3

d. What is the maximum waiting time? 3.99

e. What is the average waiting time? 0.675 (=13.5/20)




Bectol, Inc. is building a dam. A total of 10,000,000 cu ft of dirt is needed to construct the dam. A loaderis used to collect dirt for the dam. Then the dirt is moved via dump trucks to the dam site. Only one loaderis available, and it rents for $100 per hour. Bectol can rent, at $40 per hour, as many dump trucks asdesired. Each dump truck can hold 1000 cu ft of dirt. Triangular distributions are assumed to describe thefollowing various random quantities (primarily because the parameters are easily understood and estimatedby the work crews):

Random variable Best case(minimum time)

Most Likely Worst case(maximum time)

Loading truck 8 minutes 12 minutes 18 minutesTravel to unloading area 2 minutes 3 minutes 5 minutesUnloading truck 1 minute 2 minutes 4 minutesReturn to loader 2 minutes 3 minutes 4 minutes

You are asked to recommend the best number of dump trucks and to estimate the total expected rental cost(loader plus trucks) of moving the dirt needed to build the dam.

a. How many loads are required to deliver all of the dirt? ________________

b. If the loader could be kept busy continually, how long would be required to move all of the dirt? ______

Use an ARENA model to simulate the process, varying the number of trucks. (You need only do onereplication for each truck fleet size.) Simulate an 8-hour day to estimate the number of loads which can bemoved per hour, so that you can estimate the total completion time.

This will be a closed system, i.e., the trucks will not enter or leave the system, but move between theloading area and unloading area. The loading area has one server, and the unloading area (assuming thatmore than one truck can unload simultaneously) may be assumed to have as many servers as trucks. TheSERVER module of ARENA allows you to specify the probability distribution of the travel time of anentity from that server to the next module.

loadingarea

trucks

unloadingarea

Your recommendation:

# trucks # hours req'd Cost/hour Total cost_____ _______ _______ _______

(You may work in pairs. Be sure to submit the results of the simulations which you have done to arrive atyour recommendation.)

57:022 Principles of Design II HW#4 Solution page 1 of 3


Homework #4 SolutionDue Wednesday, February 16, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

Bectol, Inc. is building a dam. A total of 1,000,000 cu ft of dirt is needed to construct the dam. A loaderis used to collect dirt for the dam. Then the dirt is moved via dump trucks to the dam site. Only one loaderis available, and it rents for $100 per hour. Bectol can rent, at $40 per hour, as many dump trucks asdesired. Each dump truck can hold 1000 cu ft of dirt. Triangular distributions are assumed to describe thefollowing various random quantities (primarily because the parameters are easily understood and estimatedby the work crews):

Random variable Best case(minimum time)

Most Likely Worst case(maximum time)

Loading truck 8 minutes 12 minutes 18 minutesTravel to unloading area 2 minutes 3 minutes 5 minutesUnloading truck 1 minute 2 minutes 4 minutesReturn to loader 2 minutes 3 minutes 4 minutes

You are asked to recommend the best number of dump trucks and to estimate the total expected cost ofmoving the dirt needed to build the dam.

a. How many loads are required to deliver all of the dirt? ___1,000________

b. If the loader could be kept busy continually, how long would be required to move all of the dirt? ______• Case 1 : # truck = 1

The truckloads which were unloaded is 22, i.e., 22/ 8 hours = 2.75 / hourAt that rate, it would require 1,000 / 2.75 = 363.6 hours (= 45.45 days)The utilization of the loader is about 61.683 %, which is almost 100%, which would mean thateven if it were kept busy continually, it would still take about 224 hours ( = 28days) to completethe job.

• Case 2 : # truck 2The truckloads which were unloaded is 39, i.e., 39/ 8 hours = 4.875 / hourAt that rate, it would require 1,000 / 4.875 = 205.13 hours (= 25.64 days)The utilization of the loader is about 100 %, it would take about 205 hours ( = 25days) tocomplete the job

Use an ARENA model to simulate the process, varying the number of trucks. (You need only do onereplication for each truck fleet size.) Simulate an 8-hour day to estimate the number of loads which can bemoved per hour, so that you can estimate the total completion time.

This will be a closed system, i.e., the trucks will not enter or leave the system, but move between theloading area and unloading area. The loading area has one server, and the unloading area (assuming thatmore than one truck can unload simultaneously) may be assumed to have as many servers as trucks. TheSERVER module of ARENA allows you to specify the probability distribution of the travel time of anentity from that server to the next module.

Your recommendation:

# trucks # hours req'd Cost/hour Total cost__ 2__ __205.13___ ___180__ _$36923.4_

For the case the number of dump trucks equal 2 ,The truckloads which were unloaded is 39, i.e., 39/ 8 hours = 4.875 / hourAt that rate, it would require 1,000 / 4.875 = 205.13 hours (= 25.64 days)


The corresponding data with 1 dump truck is:# trucks # hours req'd Cost/hour Total cost__1___ __363.6_____ ___140____ __$50,904_____

(Note that because this is a simulation with random numbers, your results may differ slightly from thoseshown here.)

loadingarea

trucks

unloadingarea

ARENA Simulation Results for the number of trucks = 1

Replication ended at time : 480.0

TALLY VARIABLES

Identifier Average Half Width Minimum Maximum ObservationsUnload Area_R_Q Queue .00000 (Insuf) .00000 .00000 22Load Area_R_Q Queue Ti .00000 (Insuf) .00000 .00000 23

DISCRETE-CHANGE VARIABLES

Identifier Average Half Width Minimum Maximum Final Value# in Unload Area_R_Q .00000 (Insuf) .00000 .00000 .00000Load Area_R Busy .61683 (Insuf) .00000 1.0000 1.0000Unload Area_R Availabl 1.0000 (Insuf) 1.0000 1.0000 1.0000# in Load Area_R_Q .00000 (Insuf) .00000 .00000 .00000Load Area_R Available 1.0000 (Insuf) 1.0000 1.0000 1.0000Unload Area_R Busy .09783 (Insuf) .00000 1.0000 .00000

ARENA Simulation Results for the number of trucks = 2


TALLY VARIABLES

Identifier Average Half Width Minimum Maximum ObservationsUnload Area_R_Q Queue .00000 (Insuf) .00000 .00000 39Load Area_R_Q Queue Ti 3.5958 (Insuf) .00000 11.191 40


Identifier Average Half Width Minimum Maximum Final Value# in Unload Area_R_Q .00000 (Insuf) .00000 .00000 .00000Load Area_R Busy .99670 (Insuf) .00000 1.0000 1.0000Unload Area_R Availabl 2.0000 (Insuf) 2.0000 2.0000 2.0000# in Load Area_R_Q .30106 (Insuf) .00000 1.0000 1.0000Load Area_R Available 1.0000 (Insuf) 1.0000 1.0000 1.0000Unload Area_R Busy .18913 (Insuf) .00000 1.0000 .00000


The ARENA model:

Arrive Module :Enter Data : Station - Arrive 1Arrival Data : Time Between – 0

Max Batches - 3 ( = # of trucks)Leave Data : Station – Load Area

Server(Load Area) Module :Enter Data : Station – Load AreaServer Data : Capacity – 1

Process Time - TRIA( 8, 12, 18)Leave Data : Station – Unload Area

Route Time - TRIA( 2, 3, 5)

Server(Unload Area) Module :Enter Data : Station – Unload AreaServer Data : Capacity : # of trucks

Process Time - TRIA( 1, 2, 4)Leave Data : Station – Load Area

Route Time - TRIA( 2, 3, 4)

Simulate Module : Length of Replication - 480

Note that the "time between" parameter in the ARRIVE module should be specified as zero-- it would bereasonable to expect that the default value is zero, but instead it seems to be +∞, so that only the firsttruck's arrival would be simulated.

Arrive 1Arrive

480

Simulate

Load Area

Server

Unload Area

Server


❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍


Due Wednesday, February 23, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

1. Curve-fittingBelow are 5 sets of measurements (Y1 through Y5), each of which is a function of X (=1,2,…15). Choose a data setaccording to the last digit of your ID#:

Y1 if 0 or 1, Y2 if 2 or 3, Y3 if 4 or 5, Y4 if 6 or 7, and Y5 if 8 or 9.

Find the best fit for each of the five sets using one of the functional forms below:

1. Y = aX+b 2. Y = aXb 3. Y = aebX 4. Y = aeb/X

5. Y = aXb ecX 6. Y = X/(aX-b) 7. Y = 1/(a+be-X) 8. Y = a + b ln X

X Y1 Y2 Y3 Y4 Y51 7.67277 5.7625 1.68127 0.588587 1.4682 4.90977 7.7625 0.937634 0.605459 2.404763 4.13656 9.7625 0.490913 0.612056 3.20454 3.8342 11.7625 0.365237 0.61454 3.926525 3.75937 14.2375 0.469387 0.615462 4.659756 3.64969 16.2375 0.426559 0.183802 5.289727 3.57333 17.7625 0.224895 0.183927 5.824798 3.51712 20.2375 0.380662 0.183973 6.462649 3.37803 22.2375 0.19477 0.61599 6.95131

10 3.43994 24.2375 0.361419 0.615996 7.5498111 3.4123 25.7625 0.357503 0.615999 8.0684912 3.29344 28.2375 0.356312 0.184 8.5731913 3.27422 29.7625 0.183378 0.616 9.0654114 3.35383 31.7625 0.186387 0.184 9.5463915 3.24369 33.7625 0.365133 0.184 10.0172


2. Goodness-of-Fit test

Below is a table containing 100 values for a random variable (sorted in ascending order):

0.0639 0.1672 0.1684 0.1775 0.32750.3397 0.426 0.4276 0.4881 0.54150.6653 0.7337 0.746 0.8514 0.88080.9362 1.1 1.147 1.158 1.186

1.197 1.207 1.281 1.325 1.351.473 1.529 1.586 1.735 1.7731.775 1.787 1.801 1.855 1.882.022 2.089 2.166 2.175 2.2612.275 2.294 2.347 2.401 2.4762.755 2.831 2.916 3.103 3.1393.241 3.286 3.313 3.551 3.6093.803 3.955 4.024 4.06 4.1574.36 4.667 4.736 4.908 4.973

5.008 5.036 5.148 5.219 5.3235.353 5.626 5.669 5.893 5.9366.232 6.614 6.775 6.915 7.0097.405 7.422 7.432 7.463 7.5057.603 7.909 7.922 8.732 9.1899.636 10.18 10.2 10.67 10.8111.67 12.85 15.73 16.25 20.29

The mean of these values is 4.286, and the standard deviation is 3.812.

It is suspected that this random variable has the exponential distribution.a. What is the relationship between the mean and standard deviation of the exponential distribution?b. Group the observations into cells and perform a goodness-of-fit test for the exponential distribution with mean4.286. What are your conclusions? (Combine cells below so that you have at least 6 observations in each cell.)

0.0-1.0

1.0-2.0

2.0-3.0

3.0-4.0

4.0-5.0

5.0-6.0

6.0-7.0

7.0-8.0

8.0-9.0

9.0-10.0

10.0-11.0

11.0-12.0

12.0-15.0

15.0-20.0

>20.0

16 19 13 9 8 10 4 9 1 2 4 1 1 2 1

3. ARENA simulation:

a. Build a simulation model of the following system:• One hundred parts arrive at a processing center according to a Poisson process, an average of one every 5

minutes.• The processing center can process only one part at a time, requiring at least 3 minutes, no more than 8

minutes, and most likely 5 minutes.• There is only sufficient queue space for 3 waiting parts.• Parts which arrive and cannot enter the queue because it is full will depart and return to try again after an

average time of 10 minutes (with exponential distribution). This is called "balking".Perform 5 replications of the simulation.b. What is the average length of time required to process all of the parts?c. What is the average number of times that a part is turned away from the processing area because of a full queue?d. What fraction of time is the server busy?Hints: Use a DELAY module to handle the delay of a balking part before it returns to the processing center. Use aCOUNT module to count the number of times that a balking part returns to the processing center.


❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

57:022 Principles of Design IIHomework #5 Solution

Wednesday, February 23, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

1. Curve-fittingBelow are 5 sets of measurements (Y1 through Y5), each of which is a function of X (=1,2,…15). Choose a data setaccording to the last digit of your ID#:

Y1 if 0 or 1, Y2 if 2 or 3, Y3 if 4 or 5, Y4 if 6 or 7, and Y5 if 8 or 9.

Find the best fit for each of the five sets using one of the functional forms below:

1. Y = aX+b 2. Y = aXb 3. Y = aebX 4. Y = aeb/X

5. Y = aXb ecX 6. Y = X/(aX-b) 7. Y = 1/(a+be-X) 8. Y = a + b ln X

X Y1 Y2 Y3 Y4 Y51 7.67277 5.7625 1.68127 0.588587 1.4682 4.90977 7.7625 0.937634 0.605459 2.404763 4.13656 9.7625 0.490913 0.612056 3.20454 3.8342 11.7625 0.365237 0.61454 3.926525 3.75937 14.2375 0.469387 0.615462 4.659756 3.64969 16.2375 0.426559 0.183802 5.289727 3.57333 17.7625 0.224895 0.183927 5.824798 3.51712 20.2375 0.380662 0.183973 6.462649 3.37803 22.2375 0.19477 0.61599 6.95131

10 3.43994 24.2375 0.361419 0.615996 7.5498111 3.4123 25.7625 0.357503 0.615999 8.0684912 3.29344 28.2375 0.356312 0.184 8.5731913 3.27422 29.7625 0.183378 0.616 9.0654114 3.35383 31.7625 0.186387 0.184 9.5463915 3.24369 33.7625 0.365133 0.184 10.0172

Solutions: The actual data above was generated by adding random "noise" to the functions below:Y1 = 3.1e0.9/x

Y2 = 4 + 2xY3 = 1.6x-

−1.2 e0.1x

Y4 = 0.6723x-0.194e-0.027x

Y5 = 1.5x0.7

Y1 vs. XMTB > Regress 'Y' 1 'X';SUBC> Constant.Regression AnalysisThe regression equation isY = 5.28 - 0.173 Xs = 0.8535 R-sq = 47.0% R-sq(adj) = 42.9%

MTB > Regress 'ln Y' 1 'ln X';SUBC> Constant.Regression Analysis


The regression equation isln Y = 1.82 - 0.264 ln Xs = 0.08795 R-sq = 85.6% R-sq(adj) = 84.5%

MTB > Regress C8 1 C9;SUBC> Constant.Regression AnalysisThe regression equation isC8 = 1.63 - 0.0377 C9s = 0.1516 R-sq = 57.1% R-sq(adj) = 53.8%

MTB > Regress 'lnY' 1 '1 / X';SUBC> Constant.Regression AnalysisThe regression equation islnY = 1.13 + 0.906 1 / Xs = 0.01384 R-sq = 99.6% R-sq(adj) = 99.6%

MTB > Regress C14 2 C15 C16;SUBC> Constant.Regression AnalysisThe regression equation isC14 = 1.92 - 0.523 C15 + 0.0482 C16s = 0.04502 R-sq = 96.5% R-sq(adj) = 95.9%

MTB > Regress '1 / Y' 1 C19;SUBC> Constant.Regression AnalysisThe regression equation is1 / Y = 0.311 - 0.191 C19s = 0.007210 R-sq = 97.8% R-sq(adj) = 97.7%

MTB > Regress C21 1 'e^-X';SUBC> Constant.Regression AnalysisThe regression equation isC21 = 0.287 - 0.458 e^-Xs = 0.01583 R-sq = 89.6% R-sq(adj) = 88.8%


The largest R-sq (99.6%) is obtained using the relationshiplnY = 1.13 + 0.906 1 / X ⇒ Y = exp(1.13) × exp(0.906/X) ⇒ Y = 3.19 exp[0.906/X]>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>


Y2 vs. X

MTB > Regress 'Y' 1 'X';SUBC> Constant.Regression AnalysisThe regression equation isY = 3.93 + 2.00 Xs = 0.2495 R-sq = 99.9% R-sq(adj) = 99.9%

MTB > Regress 'ln Y' 1 'ln X';SUBC> Constant.Regression AnalysisThe regression equation isln Y = 1.60 + 0.685 ln Xs = 0.06387 R-sq = 98.7% R-sq(adj) = 98.6%

MTB > Regress C7 1 C8;SUBC> Constant.Regression AnalysisThe regression equation isC7 = 1.94 + 0.117 C8s = 0.1387 R-sq = 93.9% R-sq(adj) = 93.4%

MTB > Regress 'lnY' 1 '1 / X';SUBC> Constant.Regression AnalysisThe regression equation islnY = 3.30 - 1.92 1 / Xs = 0.2706 R-sq = 76.6% R-sq(adj) = 74.8%

MTB > Regress C13 2 C14 C15;SUBC> Constant.Regression AnalysisThe regression equation isC13 = 1.67 + 0.489 C14 + 0.0364 C15s = 0.02797 R-sq = 99.8% R-sq(adj) = 99.7%

MTB > Regress '1 / Y' 1 C18;SUBC> Constant.Regression AnalysisThe regression equation is1 / Y = 0.0297 + 0.162 C18s = 0.01157 R-sq = 92.7% R-sq(adj) = 92.2%

MTB > Regress C20 1 'e^-X';SUBC> Constant.Regression AnalysisThe regression equation isC20 = 0.0509 + 0.377 e^-Xs = 0.01931 R-sq = 79.7% R-sq(adj) = 78.2%

MTB > Regress C23 1 C24;SUBC> Constant.Regression AnalysisThe regression equation isC23 = - 0.17 + 10.8 C24s = 3.072 R-sq = 89.1% R-sq(adj) = 88.3%

The largest value of R-sq (99.9%) is obtained by the simple linear model: Y = 3.93 + 2.00 X>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>


Y3 vs. X

MTB > Regress 'Y' 1 'X';SUBC> Constant.Regression AnalysisThe regression equation isY = 0.917 - 0.0564 Xs = 0.2991 R-sq = 43.4% R-sq(adj) = 39.0%

MTB > Regress 'ln Y' 1 'ln X';SUBC> Constant.Regression AnalysisThe regression equation isln Y = 0.238 - 0.645 ln Xs = 0.3233 R-sq = 72.4% R-sq(adj) = 70.3%

MTB > Regress C8 1 C9;SUBC> Constant.Regression AnalysisThe regression equation isC8 = - 0.199 - 0.0953 C9s = 0.4277 R-sq = 51.7% R-sq(adj) = 48.0%

MTB > Regress 'lnY' 1 '1 / X';SUBC> Constant.Regression AnalysisThe regression equation islnY = - 1.43 + 2.11 1 / Xs = 0.3002 R-sq = 76.2% R-sq(adj) = 74.4%

MTB > Regress C14 2 C15 C16;SUBC> Constant.Regression AnalysisThe regression equation isC14 = 0.420 - 1.13 C15 + 0.0909 C16s = 0.3010 R-sq = 77.9% R-sq(adj) = 74.2%




The best fit (R-sq= 74.4%) for the models above is found using


lnY = - 1.43 + 2.11 1 / X ⇒ Y = 0.239 exp[2.11/X] which is not the function which was used togenerate the data. The function which was used requires the use of two independent variables (Xand ln X) and the dependent variable ln Y.>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Y4 vs. X

MTB > Regress 'Y' 1 'X';SUBC> Constant.Regression AnalysisThe regression equation isY = 0.605 - 0.0206 Xs = 0.2036 R-sq = 18.1% R-sq(adj) = 11.7%

MTB > Regress 'ln Y' 1 'ln X';SUBC> Constant.Regression AnalysisThe regression equation isln Y = - 0.343 - 0.338 ln Xs = 0.5691 R-sq = 18.9% R-sq(adj) = 12.6%

MTB > Regress C8 1 C9;SUBC> Constant.Regression AnalysisThe regression equation isC8 = - 0.503 - 0.0587 C9s = 0.5700 R-sq = 18.6% R-sq(adj) = 12.4%

MTB > Regress 'lnY' 1 '1 / X';SUBC> Constant.Regression AnalysisThe regression equation islnY = - 1.17 + 0.910 1 / Xs = 0.5877 R-sq = 13.5% R-sq(adj) = 6.9%

MTB > Regress C14 2 C15 C16;SUBC> Constant.Regression AnalysisThe regression equation isC14 = - 0.397 - 0.194 C15 - 0.027 C16s = 0.5906 R-sq = 19.3% R-sq(adj) = 5.9%



MTB > Regress C24 1 C25;SUBC> Constant.Regression AnalysisThe regression equation isC24 = 0.659 - 0.118 C25


s = 0.2036 R-sq = 18.1% R-sq(adj) = 11.8%

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Y5 vs. X

MTB > Regress 'Y' 1 'X';SUBC> Constant.Regression AnalysisThe regression equation isY = 1.44 + 0.594 Xs = 0.2601 R-sq = 99.1% R-sq(adj) = 99.1%

MTB > Regress 'ln Y' 1 'ln X';SUBC> Constant.Regression AnalysisThe regression equation isln Y = 0.387 + 0.709 ln Xs = 0.004801 R-sq = 100.0% R-sq(adj) = 100.0%


MTB > Regress 'lnY' 1 '1 / X';SUBC> Constant.Regression AnalysisThe regression equation islnY = 2.17 - 2.09 1 / Xs = 0.2180 R-sq = 85.7% R-sq(adj) = 84.5%

MTB > Regress C14 2 C15 C16;SUBC> Constant.Regression AnalysisThe regression equation isC14 = 0.384 + 0.716 C15 - 0.00132 C16s = 0.004497 R-sq = 100.0% R-sq(adj) = 100.0%

MTB > Regress '1 / Y' 1 C19;SUBC> Constant.Regression AnalysisThe regression equation is1 / Y = 0.0754 + 0.631 C19s = 0.01683 R-sq = 98.9% R-sq(adj) = 98.8%

MTB > Regress C21 1 'e^-X';SUBC> Constant.Regression AnalysisThe regression equation isC21 = 0.156 + 1.52 e^-Xs = 0.04814 R-sq = 91.2% R-sq(adj) = 90.5%



Two models gave R-sq=100%, one of which isln Y = 0.387 + 0.709 ln X ⇒ Y = 1.47256 X 0.709


2. Goodness-of-Fit test

Below is a table containing 100 values for a random variable (sorted in ascending order):

0.0639 0.1672 0.1684 0.1775 0.32750.3397 0.426 0.4276 0.4881 0.54150.6653 0.7337 0.746 0.8514 0.88080.9362 1.1 1.147 1.158 1.186

1.197 1.207 1.281 1.325 1.351.473 1.529 1.586 1.735 1.7731.775 1.787 1.801 1.855 1.882.022 2.089 2.166 2.175 2.2612.275 2.294 2.347 2.401 2.4762.755 2.831 2.916 3.103 3.1393.241 3.286 3.313 3.551 3.6093.803 3.955 4.024 4.06 4.1574.36 4.667 4.736 4.908 4.973

5.008 5.036 5.148 5.219 5.3235.353 5.626 5.669 5.893 5.9366.232 6.614 6.775 6.915 7.0097.405 7.422 7.432 7.463 7.5057.603 7.909 7.922 8.732 9.1899.636 10.18 10.2 10.67 10.8111.67 12.85 15.73 16.25 20.29

The mean of these values is 4.286, and the standard deviation is 3.812.

It is suspected that this random variable has the exponential distribution.a. What is the relationship between the mean and standard deviation of the exponential distribution?Solution : The mean and the standard deviation of the exponential distribution are identical.

b. Group the observations into cells and perform a goodness-of-fit test for the exponential distribution with mean4.286. What are your conclusions? (Combine cells below so that you have at least 6 observations in each cell.)

0.0-1.0

1.0-2.0

2.0-3.0

3.0-4.0

4.0-5.0

5.0-6.0

6.0-7.0

7.0-8.0

8.0-9.0

9.0-10.0

10.0-11.0

11.0-12.0

12.0-15.0

15.0-20.0

>20.0

16 19 13 9 8 10 4 9 1 2 4 1 1 2 1

Solution:Suppose we combine cells as follows:

interval Observed CDF probability Expected (Oi-Ei)2/Ei0-1 16 0.208098 0.208098 20.8098 1.11171-2 19 0.372891 0.164793 16.4793 0.3855612-3 13 0.503392 0.1305 13.05 0.000191663-4 9 0.606735 0.103343 10.3343 0.1722834-5 8 0.688573 0.0818377 8.18377 0.004126835-6 10 0.75338 0.0648075 6.48075 1.911076-8 13 0.845342 0.0919625 9.19625 1.573318-21 12 0.992551 0.147208 14.7208 0.502894

sum: 5.66113Note: Assuming λ = 1/4.286 = 0.233318, we compute F(t) for t=each end of the cells. Then take differences toobtain the probability for each cell, and multiply by 100 to get the expected number of observations.The chi-square distribution will have 6 degrees of freedom (8 cells, minus 1 because the total number ofobservations is fixed at 100, and minus 1 because a parameter (λ) was estimated using the data.)


degrees of freedom = 8 – 1 – 1 = 62

6,αχ = 12.592( i.e., probability that the observed value of D exceeds 12.592, given the Exponential model with meanvalue 0.233318, is only 5% ) i.e., P{D≥12.592} = 5%,

D = 5.661138 < 26,αχ = 12.592

⇒ Thus, the assumed model is valid…(P-value is big (much bigger than 0.05).⇒ Do not reject the Null Hypothesis that this random variable has the exponential

distribution.

deg.of Chi-square Dist'n P{D≥χ2}freedom 99% 95% 90% 10% 5% 1%

2 0.0201 0.103 0.211 4.605 5.991 9.2103 0.115 0.352 0.584 6.251 7.815 11.3414 0.297 0.711 1.064 7.779 9.488 13.2775 0.554 1.145 1.610 9.236 11.070 15.0866 0.872 1.635 2.204 10.645 12.592 16.8127 1.239 2.167 2.833 12.017 14.067 18.475

3. ARENA simulation:

a. Build a simulation model of the following system:• One hundred parts arrive at a processing center according to a Poisson process, an average of one every 5

minutes.• The processing center can process only one part at a time, requiring at least 3 minutes, no more than 8

minutes, and most likely 5 minutes.• There is only sufficient queue space for 3 waiting parts.• Parts which arrive and cannot enter the queue because it is full will depart and return to try again after an

average time of 10 minutes (with exponential distribution). This is called "balking".Perform 5 replications of the simulation.

b. What is the average length of time required to process all of the parts?Solution : 33.637

c. What is the average number of times that a part is turned away from the processing area because of a full queue?Solution : 176.00

d. What fraction of time is the server busy?Solution : 93.92%

Hints: Use a DELAY module to handle the delay of a balking part before it returns to the processing center. Use aCOUNT module to count the number of times that a balking part returns to the processing center.

ARENA Simulation Results

Summary for Replication 1 of 5

Project: HW5 Run execution date : 2/18/2000Analyst: DLB Model revision date: 2/18/2000


TALLY VARIABLES

Identifier Average Half Width Minimum Maximum Observations_______________________________________________________________________________


Server 1_R_Q Queue Tim 8.6034 (Insuf) .00000 17.427 100


Identifier Average Half Width Minimum Maximum Final Value_______________________________________________________________________________

# in Server 1_R_Q 1.4170 (Insuf) .00000 3.0000 .00000Server 1_R Available 1.0000 (Insuf) 1.0000 1.0000 1.0000Server 1_R Busy .88393 (Insuf) .00000 1.0000 .00000

COUNTERS Identifier Count Limit _________________________________________ N 58 Infinit




TALLY VARIABLES






COUNTERS Identifier Count Limit _________________________________________ N 123 Infinite




TALLY VARIABLES





# in Server 1_R_Q 2.5696 (Insuf) .00000 3.0000 .00000


Server 1_R Available 1.0000 (Insuf) 1.0000 1.0000 1.0000Server 1_R Busy .98343 (Insuf) .00000 1.0000 .00000

COUNTERS Identifier Count Limit _________________________________________ N 380 Infinite




TALLY VARIABLES






COUNTERS

Identifier Count Limit _________________________________________

N 267 Infinite




TALLY VARIABLES






COUNTERS


Identifier Count Limit _________________________________________

N 52 Infinite


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1. Suppose that your company wishes to estimate the reliability of an electric motor. Two hundred units are testedsimultaneously, and the time(in days) of failures is recorded until 200 days have passed (during which 102failures have occurred):

59.91 66.13 75.41 75.98 80.67 86.08 87.94 89.80 90.47 91.85 101.43 101.65 104.03 105.22 106.63 107.80 113.75 114.36 114.68 119.70 119.77 120.99 122.09 122.66 125.26 126.61 127.89 127.90 130.89 132.12 132.96 135.84 137.93 138.35 139.83 140.06 140.26 142.07 143.79 144.01 145.42 146.06 147.16 148.24 148.52 148.95 150.37 151.68 152.45 152.76 152.80 153.29 155.06 155.19 155.57 156.37 158.64 161.89 162.28 162.89 163.21 163.82 165.07 165.81 166.51 166.80 167.30 168.80 168.97 169.74 169.95 171.39 171.41 171.97 172.26 172.48 175.91 176.64 177.92 178.37 179.55 179.95 180.22 181.08 182.36 183.23 184.55 186.10 189.45 189.66 190.00 190.24 190.89 191.45 192.56 193.19 195.11 195.49 195.53 196.10 198.52 199.15

To simplify the computations, the data was aggregated, giving the table below showing the failure times of the fifth,tenth, fifteenth, twentieth, etc. motor:

NF T R(T) Ln(T) LnLn 1/R(T)5 80.67 0.975 4.39 −3.676

10 91.85 0.95 4.52 −2.9715 106.6 0.925 4.669 −2.55220 119.7 0.9 4.785 −2.2525 125.3 0.875 4.83 −2.01330 132.1 0.85 4.884 −1.81735 139.8 0.825 4.94 −1.64840 144 0.8 4.97 −1.545 148.5 0.775 5.001 −1.36750 152.8 0.75 5.029 −1.24655 155.6 0.725 5.047 −1.13460 162.9 0.7 5.093 −1.03165 166.5 0.675 5.115 −0.933870 169.7 0.65 5.134 −0.842275 172.3 0.625 5.149 −0.75580 178.4 0.6 5.184 −0.671785 182.4 0.575 5.206 −0.591790 189.7 0.55 5.245 −0.514495 192.6 0.525 5.26 −0.4395

100 196.1 0.5 5.279 −0.3665

a. Plot the value of (ln ln 1/R) on the vertical axis and ln T on the horizontal axis of ordinary graph paper.

b. By "eyeballing it", draw a straight line which seems best to fit the data point.

c. What is the slope of this line?

d. What is the y-intercept of this line?

e. What is therefore your estimate of the parameters k and u of the Weibull distribution for the lifetimes of thesemotors?


f. What is the expected lifetime of the motors, according to your Weibull probability model? You may use the table

below for the gamma function in the computation of µ. Values of Γ(1+1/k) are given for k=0.1, 0.2, ... 4.9.

For example, Γ(1+1/2.5) = 0.8882.

Γ 1 + 1k

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0 ∞ 3628800 120.00 9.2610 3.3230 2.0000 1.5050 1.2660 1.1330 1.05201 1 0.9649 0.9407 0.9236 0.9114 0.9027 0.8966 0.8922 0.8893 0.88742 0.8862 0.8857 0.8856 0.8859 0.8865 0.8873 0.8882 0.8893 0.8905 0.89173 0.8930 0.8943 0.8957 0.8970 0.8984 0.8997 0.9011 0.9025 0.9038 0.90514 0.9064 0.9077 0.9089 0.9102 0.9114 0.9126 0.9137 0.9149 0.9160 0.9171

g. Perform a Chi-Square goodness of fit test to decide whether the Weibull probability distribution model whichyou have found is a "good" fit of the data. Complete the table:

Intervali

ObservationsOi

probabilitypi

ExpectationEi

Oi – E i2

E i

0−80.67 580.67−91.85 591.85−106.6 5106.6−119.7 5119.7−125.3 5125.3−132.1 5132.1−139.8 5139.8−144 5144−148.5 5

148.5−152.8 5152.8−155.6 5155.6−162.9 5162.9−166.5 5166.5−172.3 5172.3−178.4 5178.4−182.4 5182.4−189.7 5189.7−192.6 5192.6−196.1 5196.1−198.5 5198.5 −199.1 5

Total: D = ______

h. What is the number of "degrees of freedom"? ______ (Keep in mind that two parameters, u & k, were estimatedbased upon the data!)i. Using α = 5%, should the probability distribution be accepted or rejected?

j. According to the value of k, is the failure rate increasing or decreasing with time?

k. According to this Weibull distribution, when should 90% of the motors have failed?

2. ARENA simulation. Consider a process whereby items come off an assembly line at the constant rate of oneevery 30 second. Next the item must be inspected. One inspector (A) performs a "standard inspection" whichrequires an amount of time having triangular distribution with min=10 seconds, max=30 seconds, and mode=15seconds. The other inspector (B) performs a more detailed inspection, the time of which also is assumed to have atriangular distribution but with min = 40 seconds, max= 90 seconds, and mode=60 seconds. If inspector B isavailable, an arriving item is sent to him for inspection. Otherwise (if B is busy) the item is sent to inspector A.Any defective items are immediately set aside at the end of the inspection, to be reworked later. (Inspector A finds2% of the items to be defective, whereas inspector B finds 5% of the items to be defective.) After an item is


inspected, it is sent to a packing station in which one worker places items in a carton, and when it is full (capacity is6), he/she seals it and carries it to a storage area. The sealing process requires at least 60 seconds, no more than 120seconds, and most likely 90 seconds. It requires the worker at least 45 seconds, no more than 90 seconds, and mostlikely 60 seconds to store the carton and return to the packing station.

a. Build an ARENA simulation model and simulate the operation of the above process for one 8-hour day, with 5replications. For each replication:b. What are the utilizations of the two inspectors and the worker at the packing station?c. What fraction of items receive the more detailed inspection?d. What is the maximum length of a queue in front of inspector A?


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Wednesday, March 1, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

(1-a)

(1-b)

5.35.25.15.04.94.84.74.64.54.4

-0.5

-1.5

-2.5

-3.5

Ln(T )

Ln C

5

R-Squared = 0.996

Y = -19.6692 + 3.66053X

Regression Plot

(1-c) slope of this line = 3.66053

(1-d) y-intercept of this line = -19.6692

(1-e) k = slope of this line = 3.66053 ≈ 3.66-k ln u = y-intercept of this line = -19.6692 ⇒ ∴ln u = 19.6692 / 3.66053 ⇒ ∴ u = 215.5775

(1-f)

+Γ⋅=µ

k1

1uY = (215.5775)(0.9019) = 194.4294 (days)

5.35.25.15.04.94.84.74.64.54.4

0

-1

-2

-3

-4

Ln(T)

Ln C

5


(1-g)Interval Oi Pi Ei ( (Oi-Ei)^2 ) / Ei

0 80.67 5 0.027017 5.4035 0.030180.67 91.85 5 0.016070 3.2140 0.992491.85 106.6 5 0.030062 6.0124 0.1705106.6 119.7 5 0.036465 7.2930 0.7210119.7 125.3 5 0.018633 3.7266 0.4351125.3 132.1 5 0.025162 5.0325 0.0002132.1 139.8 5 0.031871 6.3742 0.2963139.8 144 5 0.018880 3.7761 0.3967144 148.5 5 0.021373 4.2746 0.1231

148.5 152.8 5 0.021495 4.2989 0.1143152.8 155.6 5 0.014539 2.9078 1.5054155.6 162.9 5 0.039797 7.9594 1.1003162.9 166.5 5 0.020560 4.1120 0.1918166.5 172.3 5 0.034276 6.8551 0.5020172.3 178.4 5 0.037371 7.4743 0.8191178.4 182.4 5 0.025108 5.0216 0.0001182.4 189.7 5 0.046720 9.3440 2.0195189.7 192.6 5 0.018778 3.7556 0.4123192.6 196.1 5 0.022749 4.5498 0.0445196.1 198.5 5 0.015620 3.1241 1.1264198.5 199.1 5 0.003905 0.7809 22.7942

D = 33.7954

(1-h) Degree of Freedom = 21 – 1 – 2 = 18

(1-i) D = 33.7954 > 2,18 αχ = 28.8693

⇒ Reject this probability distribution!

(1-j) slope k = 3.66 > 1⇒ the failure rate is increasing with time.

(1-k) Exact estimates require solving the following equation for t :F(t) = 1 – exp{ -(t/u)k } , where F(t) is 90%.

The inverse of F(t) was derived in the notes :F-1(p) = u ( -ln[1-p] )1/k , where k = 3.66 & u = 215.5775

⇒ ∴ F-1(0.9) = 270.75


(2) ARENA

Arrive Module : Inspect Module :Time between : 30 Capacity : 2

Process Time : TRIA( 10, 15, 30)Failure Probability : 0.02

BatchQuantity : 6 Server Module :

Process Time : TRIA( 60, 90, 120)+TRIA(45, 60, 90)

Simulate Module :Number of Replication : 5Length of Replication : 28800

Replication1st 2nd 3rd 4th 5th

Utilization of 2 inspectors 60.511% 60.153% 61.031% 61.436% 60.997%Utilization of 1 inspector 30.2555% 30.0765% 30.5155% 30.718% 30.4985%

Utilization of worker 2.222% 2.221% 1.038% 1.408% 1.122%Max length of a queue in front of

inspectors0 0 0 0 0

Note : Maximum length of a queue in front of inspectors is zero through the 5 replication. This isreasonable since the processing time of inspect area is at most 30 which is equal to the timebetween arrival.

Arrive 1Arrive

Inspect A

Inspect

6Batch

Depart 1Depart

28800

SimulatePacking

Server

Depart 2

Depart?


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A: Weibull, with mean 2000 days and standard deviation 1200 daysB: Normal, with mean 1200 days and standard deviation 200 daysC: Exponential, with mean 2000 daysD: Exponential, with mean 4000 days


Device Reliability A ________ B ________ C ________ D ________



D fails: Yes or No


Subsystem ReliabilityB1B2 ________C1C2 ________ B1B2+ C1C2 ________Total system: ________






1. Draw a diagram showing the hybrid series/parallel configuration of the components, as in the Hypercard Stack"System Reliability".

2. Suppose that the system is required to survive for 1000 days. What is the reliability of each component, i.e., theprobability that it survives 1000 days?

3. What is the reliability of the system, i.e., the probability that the system survives 1000 days?

4. Construct an ARENA model which can simulate the lifetime of this system.

5. Run the ARENA simulation model, using 500 runs, collecting statistics on the time of system failure. Requestthat a histogram be printed. Select about 15 cells, with HLOW and HWID parameters which will give you a"nice" histogram with the mean approximately in the center and with small tails.

6. Suppose you will offer a warranty on this system, such that 95% of the systems will survive past the length of thewarranty. According to the ARENA model, what should be the length of the warranty?



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A: Weibull, with mean 2000 days and standard deviation 1200 days (u=2243.03, k=1.71708)B: Normal, with mean 1200 days and standard deviation 200 daysC: Exponential, with mean 2000 daysD: Exponential, with mean 4000 days


Device Reliability A 0.778958 B 0.840882 C 0.606531 D 0.778801



D fails: Yes or No


Subsystem ReliabilityB1B2 RB×RB = 0.70708C1C2 1 − (1 −RC)( 1 −RC)= 1− 0.15482 = 0.84518 B1B2+ C1C2 1 − (1 −RBB)( 1 −RCC) = 1 − (0.15482)(0.292917) = 1 − 0.04535 = 0.95465Total system: RA×RBBCC×RD = (0.778958)(0.95465)(0.7788) = 0.57914






1. Draw a diagram showing the hybrid series/parallel configuration of the components, as in the Hypercard Stack"System Reliability".

2. Suppose that the system is required to survive for 1000 days. What is the reliability of each component, i.e., theprobability that it survives 1000 days?Solution: R(t) = e−λt where λ is the failure rate, i.e., 1/2000days, etc.

RA = 0.6065306597RB = 0.7165313106RC = RD = 0.2865047969RE = RF = RG = 0.1353352832

3. What is the reliability of the system, i.e., the probability that the system survives 1000 days?RCD = 1 − (1 − RC )( 1 − RD ) = 0.490925REFG = 1 − (1 − RE )( 1 − RF ) )( 1 − RG ) = 0.353538Reliability of system = RA × RB × RCD × REFG = 0.205037

4. Construct an ARENA model which can simulate the lifetime of this system.

5. Run the ARENA simulation model, using 500 runs, collecting statistics on the time of system failure. Requestthat a histogram be printed. Select about 15 cells, with HLOW and HWID parameters which will give you a"nice" histogram with the mean approximately in the center and with small tails.


Output Summary for 500 Replications

Project: Reliability Run execution date : 3/13/2000Analyst: DLB Model revision date: 3/13/2000

OUTPUTS

Identifier Average Half-width Minimum Maximum # Replications_______________________________________________________________________________

TNOW 507.51 29.594 1.3470 2041.4 500

Histogram Summary System Lifetime

Cell Limits Abs. Freq. (Time) Rel. Freq. Cell From To Cell Cumul. Cell Cumul. 1 -Infinity 0 0 0 0 0 2 0 25 4 4 0.008016 0.008016 3 25 50 3 7 0.006012 0.01403 4 50 75 11 18 0.02204 0.03607 5 75 100 12 30 0.02405 0.06012 6 100 125 12 42 0.02405 0.08417 7 125 150 12 54 0.02405 0.1082 8 150 175 14 68 0.02806 0.1363 9 175 200 16 84 0.03206 0.1683 10 200 225 14 98 0.02806 0.1964 11 225 250 21 119 0.04208 0.2385 12 250 275 23 142 0.04609 0.2846


13 275 300 22 164 0.04409 0.3287 14 300 325 12 176 0.02405 0.3527 15 325 350 13 189 0.02605 0.3788 16 350 375 14 203 0.02806 0.4068 17 375 400 15 218 0.03006 0.4369 18 400 425 15 233 0.03006 0.4669 19 425 450 15 248 0.03006 0.497 20 450 475 13 261 0.02605 0.523 21 475 500 9 270 0.01804 0.5411 22 500 +Infinity 229 499 0.4589 1

6. Suppose you will offer a warranty on this system, such that 95% of the systems will survive past the length of thewarranty. According to the ARENA model, what should be the length of the warranty?

Solution: After 50 days, about 3.6% of the systems have failed, while after 75 days, about 6% have failed.If we perform linear interpolation, we get about 25+14.5 = 39.5 days. That is, a system has about 95%probability of surviving a 39.5-day warranty period.


57:022 Principles of Design II - HW#8 Spring 2000 page 1 of 2


Homework #8Due Friday, April 7, 2000

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1. Project Scheduling

The following table is a plan for a major freeway renovation project:

ActivityExpected time

(months)Immediatepredecessors

A. Obtain federal funding 16 --B. Obtain state funding 28 --C. Design/subcontract freeway lanes 14 AD. Design/subcontract bridges/exits 22 AE. Build new freeway sound walls 12 B,CF. Rebuild southbound lanes 26 B,CG. rebuild transition roads/exits 18 D,EH. Build new bridges/overpasses 50 D,EI. Rebuild northbound lanes 44 F,G

a. Draw the A-O-N network for the project.

b. Draw the A-O-A network for the project.

c. Determine the project completion time and the critical path.

d. What is the effect of• a delay in federal funding of five months?• a delay in state funding of five months?

e. Suppose that each activity has a standard deviation of four months. Assuming that the critical pathremains that found in (c) , what is the probability that the project will be completed in…• six years?• seven years?• eight years?• nine years?

f. Build an ARENA model, assuming that the durations have normal distributions with the expectedvalue as shown, and standard deviations of four months for every activity. Again answer thequestions in (e), based upon your simulation results.

57:022 Principles of Design II - HW#8 Spring 2000 page 2 of 2

2. The following is an A-O-N network for a project for a building construction project. The times aregiven in weeks.a. Draw the corresponding A-O-A network.

b. What is the critical path and the expected completion time?

c. Activity H is the building of a fence around the building. Because of other commitments, thesubcontractor in charge of this activity will be delayed 10 weeks in starting the work. Each week ofdelay in the construction of the building costs the investors $5000 due to lack of occupancy. Whichof the following alternatives would you recommend to the investors?

I. Keep the current subcontractorII. Hire an alternate subcontractor, even though it will require four weeks to complete the

activity (instead of two) and will cost $14,000.III. Use their own landscaping crew (currently doing landscaping activity E) design and build

the fence. The design and purchase of the materials for the fencing will cost $6400 andwill add four weeks to activity E. The actual construction of the fence will take sixweeks and cost $9000.


Homework #8Due Friday, April 7, 2000

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1. Project Scheduling

The following table is a plan for a major freeway renovation project:

a. Draw the A-O-N network for the project.

b. Draw the A-O-A network for the project.

A

C

B

D

E

F

G

H

I

Begin End

2

3

4

5

6

A

B

C

D

E

F

G

I

H

1

c. Determine the project completion time and the critical path.Project completion time : 104 monthsCritical path : A – C – E – G – I

d. What is the effect of• a delay in federal funding of five months?

Project completion time : 109 monthsCritical path remains the same ( i.e., A – C – E – G – I )

• a delay in state funding of five months?Project completion time : 107 monthsCritical path changes to B – E – G – I

e. Suppose that each activity has a standard deviation of four months. Assuming that the critical pathremains that found in (c) , what is the probability that the project will be completed in…

• six years?• seven years?• eight years?• nine years?

σ = 80 = 8.944∴ The completion time for the project is N(104,8.944)

∴P{T ≤ 6 years} = P{T ≤ 72} =

−

≤−

944.810472

944.8104T

P

= P{X ≤ -3.5778} = 0.000173 likewise, P{T ≤ 7 years} = P{T ≤ 84} = P{X ≤ -2.2361} = 0.012673

P{T ≤ 8 years} = P{T ≤ 96} = P{X ≤ -0.8945} = 0.185527

P{T ≤ 9 years} = P{T ≤ 108} = P{X ≤ 0.4472} = 0.672635

Activity µ σ2

A 16*** critical 16***B 28 16C 14*** critical 16***D 22 16E 12*** critical 16***F 26 16G 18*** critical 16***H 50 16I 44*** critical 16***

SUM = 80

2

3

4

5

7

A

B

C

D

E

F

G

I

H

1

16 16 42 42

104 104

60 6030 30

0 0

f. Build an ARENA model, assuming that the durations have normal distributions with the expected valueas shown, and standard deviations of four months for every activity. Again answer the questions in (e),based upon your simulation results.

Project: HW8_P1 Run execution date : 4/ 6/2000Analyst: Hansuk Model revision date: 4/ 6/2000

OUTPUTS

Identifier Average Half-width Minimum Maximum # ReplicationsProject Completion 107.27 .46263 82.824 129.17 1000

Thee 95% confidence interval for the expected duration of the project is 107.27 months(plus or minus 0.46263 months), with the completion times ranging from less than 82months to over 129 months.

Arrive 1Arrive

1

Duplicate

NORM( 16, 4)Delay

NORM( 28, 4)

Delay

1

Duplicate NORM( 22, 4)Delay

NORM( 14, 4)

Delay1

Duplicate

NORM( 12, 4)

Delay

NORM( 26, 4)

Delay

1

Duplicate

NORM( 18, 4)Delay

NORM( 50, 4)Delay

NORM( 44, 4)

Delay

Depart 1

Depart

A

B

C

D

2Batch

2Batch

2Batch

2Batch

2Batch

E

F

1

Duplicate

I

H1

Duplicate

Begin Complete

Simulate Statistics

2Batch

0

According to the ARENA output, the probability that the project will be completed in• six years ( = 72 months ) ? less than 1%• seven years ( = 84 months ) ? less than 1%• eight years ( = 96 months ) ? about 18%• nine years ( = 108 months ) ? about 65%

Histogram SummaryProject Completion(Lumped)

Cell Limits Abs. Freq. (Time) Rel. Freq. Cell From To Cell Cumul. Cell Cumul. 1 -Infinity 88 3 3 0.003003 0.003003 2 88 90 4 7 0.004004 0.007007 3 90 92 5 12 0.005005 0.01201 4 92 94 16 28 0.01602 0.02803 5 94 96 35 63 0.03504 0.06306 6 96 98 41 104 0.04104 0.1041 7 98 100 50 154 0.05005 0.1542 8 100 102 81 235 0.08108 0.2352 9 102 104 88 323 0.08809 0.3233 10 104 106 115 438 0.1151 0.4384 11 106 108 103 541 0.1031 0.5415 12 108 110 109 650 0.1091 0.6507 13 110 112 102 752 0.1021 0.7528 14 112 114 64 816 0.06406 0.8168 15 114 116 69 885 0.06907 0.8859 16 116 118 43 928 0.04304 0.9289 17 118 120 31 959 0.03103 0.96 18 120 122 16 975 0.01602 0.976 19 122 124 10 985 0.01001 0.986 20 124 126 7 992 0.007007 0.993 21 126 128 4 996 0.004004 0.997 22 128 +Infinity 3 999 0.003003 1

1

2

3

5

4 6

7

8

A

B

C

D

EFG

H

I

J

1

2

3

5

4 6

7

8

A

B

C

D

EFG

H

I

J

2. The following is an A-O-N network for a project for a building construction project. The times are givenin weeks.

a. Draw the corresponding A-O-A network.

b. What is the critical path and the expected completion time?The critical path : B – IThe expected completion time : 36

0 0

11 12

24 26

24 26

16 16

24 32

26 34

36 36

c. Activity H is the building of a fence around the building. Because of other commitments, thesubcontractor in charge of this activity will be delayed 10 weeks in starting the work. Each weekof delay in the construction of the building costs the investors $5000 due to lack of occupancy.Which of the following alternatives would you recommend to the investors?I. Keep the current subcontractor

⇒ For activity H, 2 weeks → 12 weeks⇒ Completion time will be 38 & it will cost you $10,000.

II. Hire an alternate subcontractor, even though it will require four weeks to complete the activity(instead of two) and will cost $14,000.⇒ For activity H, 2 weeks → 4 weeks⇒ Completion time will remain the same ( = 36) & total cost will be $14,000

III. Use their own landscaping crew (currently doing landscaping activity E) design and build thefence. The design and purchase of the materials for the fencing will cost $6400 and will addfour weeks to activity E. The actual construction of the fence will take six weeks and cost$9000.⇒ For activity E, 6 weeks → 16 ( = 6 + 4 + 6) weeks⇒ Completion time will remain the same ( = 36) & total cost will be $15,400

( = 6400 + 9000)

⇒ Therefore, choose Alt. II unless original contractor for it has a price < $4000 for the job.

57:022 Principles of Design II Spring 2000 page 1 of 2


Homework #9Due Monday, April 17, 2000

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1. A business office is trying to decide whether to rent a slow or fast copy machine. It is believed that anemployee's time is worth $15/hour. The slow copier rents for $4 per hour, and it takes an employee anaverage of 10 minutes to complete a copy job (exponentially distributed). The fast copier rents for $15 perhour, and it takes an employee an average of 6 minutes to complete a copy job (also exponentiallydistributed). An average of 4 employees per hour need to use the copying machines (interarrival times areexponentially distributed).

a. For each choice of machine, show the "birth" and "death" rates:

Slow copier:

Fast copier:

b. Which standard queueing model applies to this situation? (e.g., M/M/1, M/M/2, M/M/1/N, etc.)

c. For each machine, what is:Slow Fast

Utilization of the machine ______% ______%Average number of employees at

the copy center: ______ ______Total cost (rental + employee time) ______ ______

d. Which machine would you recommend?

2. Queueing model. A neighborhood grocery store has only one check-out counter. Customers arrive atthe check-out at a rate of one per 2 minutes. The grocery store clerk requires an average of one minute and20 seconds to serve each customer. However, as soon as the waiting line exceeds 2 customers, includingthe customer being served, the manager helps by packing the groceries, which reduces the average servicetime to one minute. Assume a Poisson arrival process and exponentially-distributed service times, andassume for ease of computation that the queue never includes more than six customers (seven, counting theone being served).

a. Indicate the "birth" and "death" rates on the flow diagram for a birth-death model of this system.

57:022 Principles of Design II Spring 2000 page 2 of 2

(b.) Compute the steady-state distribution of the number of customers at the check-out.1/πo =

Statei

Probabilityπ i

01234567

(c.) What fraction of the time will the check-out clerk be idle? _________%

(d.) What is the expected number of customers in the check-out area? __________

(e.) What is the expected length of time that a customer spends in the check-out area? _____ minutes

(f.) What fraction of the time will the manager spend at the check-out area? _________%

Suppose that the store is being remodeled, and space is being planned so that the waiting line does notoverflow the space allocated to it more than 5 percent of the time, and that 4 feet must be allocated percustomer (with cart).

(g.) How much space should be allocated for the waiting line? ___________ feet

57:022 HW#9 Solution Spring 2000 page 1 of 2


Homework #9 SolutionDue Monday, April 17, 2000

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1. A business office is trying to decide whether to rent a slow or fast copy machine. It is believed that anemployee's time is worth $15/hour. The slow copier rents for $4 per hour, and it takes an employee anaverage of 10 minutes to complete a copy job (exponentially distributed). The fast copier rents for $15 perhour, and it takes an employee an average of 6 minutes to complete a copy job (also exponentiallydistributed). An average of 4 employees per hour need to use the copying machines (interarrival times areexponentially distributed).

a. For each choice of machine, show the "birth" and "death" rates:

Slow copier: 4 / hr 4 / hr 4 / hr 4 / hr 4 / hr 4 / hr 4 / hr

6 / hr 6 / hr 6 / hr 6 / hr 6 / hr 6 / hr 6 / hr

Fast copier: 4 / hr 4 / hr 4 / hr 4 / hr 4 / hr 4 / hr 4 / hr

10 / hr 10/ hr 10/ hr 10/ hr 10/ hr 10/ hr 10/ hr

b. Which standard queueing model applies to this situation? (e.g., M/M/1 , M/M/2, M/M/1/N, etc.)

c. For each machine, what is:Slow Fast

Utilization of the machine 66.7 % ( ρ=λ/µ ) 40 %Average number of employees at

the copy center: 2 ( L = ρ/ (1-ρ) ) 0.667Total cost (rental + employee time) 34 (= $15L + $4) 25 ( = $15L + $15)

d. Which machine would you recommend?Fast Machine

2. Queueing model. A neighborhood grocery store has only one check-out counter. Customers arrive atthe check-out at a rate of one per 2 minutes. The grocery store clerk requires an average of one minute and20 seconds to serve each customer. However, as soon as the waiting line exceeds 2 customers, includingthe customer being served, the manager helps by packing the groceries, which reduces the average servicetime to one minute. Assume a Poisson arrival process and exponentially-distributed service times, andassume for ease of computation that the queue never includes more than six customers (seven, counting theone being served).


(a.) Indicate the "birth" and "death" rates on the flow diagram for a birth-death model of this system.

(b.) Compute the steady-state distribution of the number of customers at the check-out.1π0

= 1 + 1 21 23 43 4

+ 1 21 23 43 4

× 1 21 23 43 4

+ 1 21 23 43 4

× 1 21 23 43 4

× 1 21 21 + 1 21 2

3 43 4× 1 21 2

3 43 4× 1 21 2

1× 1 21 21

+ ...+ 1 21 23 43 4

× 1 21 23 43 4

× 1 21 21× 1 21 2

1× 1 21 21× 1 21 2

1 × 1 21 21

Statei

Probabilityπ i

Cumulativeprobability

0 0.39344262 0.393442621 0.26229508 0.65573772 0.17486339 0.830601093 0.087431694 0.918032794 0.043715847 0.961748635 0.021857923 0.983606566 0.010928962 0.994535527 0.005464481 1.00000000

(c.) What fraction of the time will the check-out clerk be idle? π0 = 39.3 %

(d.) What is the expected number of customers in the check-out area? 1.262Solution:

L = nπnΣn = 0

7

= 1.262(e.) What is the expected length of time that a customer spends in the check-out area? _____ minutesSolution: Average arrival rate is

λ = nλnΣn = 0

7

= 0.5π0 + ... 0.5π6 + 0π7 = 0.49726776 / minAccording to Little's Law, then,

L =λW ⇒ W = L λL λ = 1.2620.4973 / min = 2.538 min

(f.) What fraction of the time will the manager spend at the check-out area? _16.94_%Solution:

∑=

π7

3nn = 0.1694

Suppose that the store is being remodeled, and space is being planned so that the waiting line does notoverflow the space allocated to it more than 5 percent of the time, and that 4 feet must be allocated percustomer (with cart).

(g.) How much space should be allocated for the waiting line? 16 feetSolution: According to the cumulative probabilities shown in the table above, the probability that 4

or fewer customers are in the queue, i.e., five at the checkout area, is more than 96%, i.e., theprobability that more than five customers are at the checkout area is less than 4%.

57:022 HW#10 Spring 2000 page 1 of 3

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Due Monday, April 24, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

1. The figure below represents a job shop. The rectangles identify machine centers, each with one type ofmachine.

There are three classes of products, which take different routes through the shop. The order rates for theclasses, along with the routing information, are shown in the table below:

Product Order rate Route1 20/month A→ B→ D→ F2 30/month A→ B→ E→ F3 10/month A→ C→ E→ F

The average production rate of each type of machine are:Machine µ (# jobs/month)

A 40B 25C 30D 10E 25F 20

a. For each machine, calculate the minimum number of machines required to process the jobs.

b. Using the RAQS software (Network Neighborhood→W-ie03→RAQS→raqs32.exe) find, for eachproduct,• the average time spent in the system,• the average time spent in queues, and• the average inventory.

c. Which two machine centers are "bottlenecks"? Add a machine to each of these two centers andrecompute the average inventory of each product.

2. A job shop has twelve numerically controlled machines that are capable of operating on their own (i.e.,without a human operator) once they are set up with the proper cutting tools and all adjustments are made.Each setup requires the skills of an experienced machinist, and the time need to complete a setup isexponentially distributed with a mean of 40 minutes. When the setup is complete, the machinist pushes abutton, and the machine requires no further attention until it has finished its job and is ready for another

57:022 HW#10 Spring 2000 page 2 of 3

setup. The job times are exponentially distributed with a mean of 75 minutes. The question is, "how manymachinists should there be to tend the machines?" At opposite extremes, there could be one machinisttending all four machines, or there could be one machinist for each machine. The optimal number to haveobviously depends on a trade-off between the cost of machinists and the cost of idle machines. Of course,machinists are paid the same regardless of how much work they do, but each machine incurs idle-time costsonly when it is idle.

Assume that the cost of a machinist (including fringe benefits, etc.) is $30 per hour, and that the cost of anidle machine (including lost revenues, etc.) is $60 per hour of idleness. (Assume also that the machinistsare each assigned to a specific set of machines, and do not assist one another, so that it may happen thatmachinist A might be busy and one of machinist A's machines waiting, while machinist B is idle.) Foreach alternative (i.e., 1, 2, 3, or 4 machinists) answer (a) through (f):

a. Sketch the birth/death diagram with the transition rates.

b. Compute the steady-state distribution.

c. What is the percent of the time that each machinist is busy?

d. What is the average number of machines in operation?

e. What is the percent of the time that each machine is busy (i.e., the utilization)?

f. What is the total cost of the alternative?

Example: suppose that a machinist is assigned six machines. Then the birth/death process in whichthe "population" is the number of machines that have shut down (and are either being tended bythe machines or waiting) is:

The steadystate distribution is found by first computing1π0

= 1 + λ0µ 1

+ λ0µ 1

×λ1µ2

+ λ0µ1

×λ1µ2

×λ1µ2

+ ....= 1 + 3.2 + 8.5333333 + 18.204444 + 29.127111 + 31.068919 + 16.57009 = 107.7039⇒ π0 = 0.009285

and then π1 = 3.2× π0 = 0.02971, π2 = 8.5333333 × π0 = 0.07923,π3 = 0.169, π4 = 0.2704, π5 = 0.2885, π6 = 0.1538

The machinist is idle 0.9285%, i.e., his utilization is 99.0715%.The average number of machines idle is

i π iΣi = 0

6

= 4.1424088 , i.e., 6-4.14 =1.86 of the six machines are in operation (about 30%utilization.)

at a cost of 4.1424 × $60/hr = $248.54/hr.Double this, since this is the idle cost of only half of the twelve machines, and add 2×$30/hr for the

machinists' salaries, to get a total cost of $557.08/hr.

57:022 HW#10 Spring 2000 page 3 of 3

g. What is the optimal number of machinists?



Homework #10 SolutionDue Friday, April 21, 2000

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1. The figure below represents a job shop. The rectangles identify machine centers, each with one type ofmachine.

There are three classes of products, which take different routes through the shop. The order rates for theclasses, along with the routing information, are shown in the table below:

Product Order rate Route1 20/month A→ B→ D→ F2 30/month A→ B→ E→ F3 10/month A→ C→ E→ F

The average production rate of each type of machine are:Machine µ (# jobs/month)

A 40B 25C 30D 10E 25F 20

a. For each machine, calculate the minimum number of machines required to process the jobs.

Machine min. # requiredA 2 (3)B 2C 1D 2 (2)E 2F 3 (4)


b. Using the RAQS software (Network Neighborhood→W-ie03→RAQS→raqs32.exe) find, for eachproduct,

Product 1 Product 2 Product 3Route AvTAN Route AvTAN Route AvTAN

A 0.057 A 0.057 A 0.057B 0.058 B 0.058 C 0.049D 0.144 E 0.111 E 0.111F 0.075 F 0.075 F 0.075S 0.334 S 0.301 S 0.292

Weight 20 Weight 30 Weight 10S*W 6.68 S*W 9.03 S*W 2.92

• the average time spent in the system : 0.311 ( = (6.68 + 9.03 + 2.92)/ 60 )

Product 1 Product 2 Product 3Route AvTiQ Route AvTiQ Route AvTiQ

A 0.032 A 0.032 A 0.032B 0.018 B 0.018 C 0.016D 0.044 E 0.071 E 0.071F 0.025 F 0.025 F 0.025S 0.119 S 0.146 S 0.144

Weight 20 Weight 30 Weight 10S*W 2.38 S*W 4.38 S*W 1.44

• the average time spent in queues : 0.137 ( = (2.38 + 4.38 + 1.44)/60 )


c. Which two machine centers are "bottlenecks"? Add a machine to each of these two centers andrecompute the average inventory of each product.

⇒ From the average waiting time (in queue at a node) point of view, i.e., AvTIQ, machine centers D andE are Bottlenecks. Adding a machine to each of these two centers gives the following result :

2. A job shop has twelve numerically controlled machines that are capable of operating on their own (i.e.,without a human operator) once they are set up with the proper cutting tools and all adjustments are made.Each setup requires the skills of an experienced machinist, and the time need to complete a setup isexponentially distributed with a mean of 40 minutes. When the setup is complete, the machinist pushes abutton, and the machine requires no further attention until it has finished its job and is ready for anothersetup. The job times are exponentially distributed with a mean of 75 minutes. The question is, "how manymachinists should there be to tend the machines?" At opposite extremes, there could be one machinisttending all four machines, or there could be one machinist for each machine. The optimal number to haveobviously depends on a trade-off between the cost of machinists and the cost of idle machines. Of course,machinists are paid the same regardless of how much work they do, but each machine incurs idle-time costsonly when it is idle.

Assume that the cost of a machinist (including fringe benefits, etc.) is $30 per hour, and that the cost of anidle machine (including lost revenues, etc.) is $60 per hour of idleness. (Assume also that the machinistsare each assigned to a specific set of machines, and do not assist one another, so that it may happen thatmachinist A might be busy and one of machinist A's machines waiting, while machinist B is idle.) Foreach alternative (i.e., 1, 2, 3, or 4 machinists) answer (a) through (f):

a. Sketch the birth/death diagram with the transition rates.

b. Compute the steady-state distribution.

c. What is the percent of the time that each machinist is busy?


d. What is the average number of machines in operation?

e. What is the percent of the time that each machine is busy (i.e., the utilization)?

f. What is the total cost of the alternative?

Example: suppose that a machinist is assigned six machines. Then the birth/death process in whichthe "population" is the number of machines that have shut down (and are either being tended bythe machines or waiting) is:

The steadystate distribution is found by first computing1π0

= 1 + λ0µ 1

+ λ0µ 1

×λ1µ2

+ λ0µ1

×λ1µ2

×λ1µ2

+ ....= 1 + 3.2 + 8.5333333 + 18.204444 + 29.127111 + 31.068919 + 16.57009 = 107.7039⇒ π0 = 0.009285

and then π1 = 3.2× π0 = 0.02971, π2 = 8.5333333 × π0 = 0.07923,π3 = 0.169, π4 = 0.2704, π5 = 0.2885, π6 = 0.1538

The machinist is idle 0.9285%, i.e., his utilization is 99.0715%.The average number of machines idle is

i π iΣi = 0

6

= 4.1424088 , i.e., 6-4.14 =1.86 of the six machines are in operation (about 30%

utilization.)at a cost of 4.1424 × $60/hr = $248.54/hr.Double this, since this is the idle cost of only half of the twelve machines, and add 2×$30/hr for the

machinists' salaries, to get a total cost of $557.08/hr.

g. What is the optimal number of machinists?

Solution:n=1 n=2 n=3 n=4 n=6

π0 0.652174 0.379427 0.191685 0.0824446 0.00928472π1 0.347826 0.404722 0.306696 0.175882 0.0297111π2 0 0.215852 0.327143 0.281411 0.0792296π3 0 0 0.174476 0.300171 0.169023π4 0 0 0 0.160091 0.270437π5 0 0 0 0 0.288466π6 0 0 0 0 0.153849

# machines idle 0.347826 0.836425 1.48441 2.27958 4.14241Machine Utilization 65.22% 58.18% 43.01% 43.01% 30.96%Machinist Utilization 34.78% 62.06% 80.83% 91.76% 99.07%

Total Salaries 360 180 120 90 60Machine idleness cost 250.435 301.113 356.258 410.325 497.089

Total Cost 610.435 481.113 476.258 500.325 557.089

The optimal number of machines per machinist is 3, i.e., 4 machinists should be employed.

57:022 HW#11 Spring 2000 page 1 of 1

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Due Friday, April 28, 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

1. A certain large shop doing light fabrication work uses a single central storage facility (dispatch station) formaterial in in-process storage. The typical procedure is that each employee personally delivers his finished work(by hand, tote box, or hand cart) and receives new work and materials at the facility. Although this procedureworked well in earlier years when the shop was smaller, it appears that it may now be advisable to divide theshop into two semi-independent parts, with a separate storage facility for each one. You have been assigned thejob of comparing the use of two facilities and of one facility from a cost standpoint.

The factory has the shape of a rectangle 150 by 100 yards. Thus, by letting 1 yard be the unit of distance, the(x,y) coordinates of the corners are (0,0), (150,0), (150,100), and (0,100). With this coordinate system, theexisting facility is located at (75,50) and the locations for two facilities would be (37.5, 50) & (112.5, 50).

q Each facility would be operated by a single clerk.q The time required by a clerk to service a caller has an Erlang-2 distribution, with a mean of 2 minutes.q Employees arrive at the present facility according to a Poisson input process at a mean rate of 24 per

hour.q The employees are rather uniformly distributed throughout the shop, and if the second facility were

installed, each employee would normally use the nearer of the two facilities.q Employees walk at an average speed of about 1 yard/second.q All aisles are parallel to the outer walls of the shop.q The net cost of providing each facility is estimated to be about $25/hour, plus $15/hour for the clerk.q The estimated total cost of an employee being idled by traveling or waiting at the facility is $20/hour.

Given the preceding information, build an ARENA model and simulate the system in order to determine whichalternative will minimize the average total cost per hour.

2. A grocery store has three check-out counters. The sign by the check-out area states that an additional counterwill be opened any time the number of customers in any lane exceeds three. This means that for fewer than fourcustomers, only one counter will be in operation. For four to six customers, two counters will be open, and formore than six customers, all three counters will be open. The customers arrive at the check-out area according toa Poisson process, with a mean of 10 customers per hour. The check-out time per customer has exponentialdistribution, with average 12 minutes.

For convenience of computation, assume that a negligible probability that there are more than 10 customers inthe check-out area. Also assume that a customer may freely switch from one check-out counter to another if thequeue is shorter.

1. Using a birth-death model, determine the steady-state probability distribution of the number of customers in thecheck-out area.

2. What is the average number of customers in the check-out area?

3. What is the average waiting time of a customer?

57:022 POD2 HW#11 Solutions Spring 2000 page 1 of 3

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Spring 2000❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍ ❍

1. A certain large shop doing light fabrication work uses a single central storage facility (dispatch station)for material in in-process storage. The typical procedure is that each employee personally delivers hisfinished work (by hand, tote box, or hand cart) and receives new work and materials at the facility.Although this procedure worked well in earlier years when the shop was smaller, it appears that it maynow be advisable to divide the shop into two semi-independent parts, with a separate storage facilityfor each one. You have been assigned the job of comparing the use of two facilities and of one facilityfrom a cost standpoint.

The factory has the shape of a rectangle 150 by 100 yards. Thus, by letting 1 yard be the unit ofdistance, the (x,y) coordinates of the corners are (0,0), (150,0), (150,100), and (0,100). With thiscoordinate system, the existing facility is located at (75,50) and the locations for two facilities would be(37.5, 50) & (112.5, 50).

q Each facility would be operated by a single clerk.q The time required by a clerk to service a caller has an Erlang-2 distribution, with a mean of 2 min.q Employees arrive at the present facility according to a Poisson input process at a mean rate of 24

per hour.q The employees are rather uniformly distributed throughout the shop, and if the second facility were

installed, each employee would normally use the nearer of the two facilities.q Employees walk at an average speed of about 1 yard/second.q All aisles are parallel to the outer walls of the shop.q The net cost of providing each facility is estimated to be about $25/hour, plus $15/hour for the

clerk.q The estimated total cost of an employee being idled by traveling or waiting at the facility is

$20/hour.

Given the preceding information, build an ARENA model and simulate the system in order todetermine which alternative will minimize the average total cost per hour.

Solution: We would divide the original 150x100 rectangle into two rectangles, each 75x100. We needonly simulate one half, of course, since they are symmetric, and then double the cost of the half to get thecost of the whole. For the two facility case the travel time can be "UNIF(0,37.5) + UNIF(0,50)"And the arrival rate should be doubled, i.e., EXPO(300) instead of EXPO(150). The correspondingsummary of result is :

only one facility Two facilityAverage Travel Time 570.57 262.28Average Waiting Time 324.72 54.291

One facility


Net cost of providing a facility : ($25/hr + $15/hr)*(8 hr) = $320Net cost of traveling & waiting : (570.57 sec + 324.72 sec )*(1 hr / 3600 sec)*($20/hr) = $4.97Total cost / 8 hr : $324.97 per 8 hours

Two facilitiesNet cost of providing a facility : 2*{($25/hr + $15/hr)*(8 hr)} = $640Net cost of traveling & waiting : 2*{(262.28 sec + 54.291 sec)*(1 hr / 3600sec)*($20/hr)} = $3.52Total cost / 8 hr : $643.52 per 8 hours

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2. A grocery store has three check-out counters. The sign by the check-out area states that an additionalcounter will be opened any time the number of customers in any lane exceeds three. This means that forfewer than four customers, only one counter will be in operation. For four to six customers, twocounters will be open, and for more than six customers, all three counters will be open. The customersarrive at the check-out area according to a Poisson process, with a mean of 10 customers per hour. Thecheck-out time per customer has exponential distribution, with average 12 minutes.

For convenience of computation, assume that a negligible probability that there are more than 10customers in the check-out area. Also assume that a customer may freely switch from one check-outcounter to another if the queue is shorter. (This phenomena is known as "jockeying", and means thatthere is essentially a single queue with the number of servers changing, depending upon the number ofcustomers in the system.)

1. Using a birth-death model, determine the steady-state probability distribution of the number ofcustomers in the check-out area.

2. Solution:Arrival ("birth") rates are λi = 10/hour for i=0,1,2,3,4,5,6,7,8,9;Departure ("death") rates are µi = 5/hr for i=1,2,3; µi=10/hr for i=4,5,6; µi=15/hr for i=7,8,9,10

term#

Ratioλ / µ

Product of firsti+1 ratios

Steady-stateProbability

0 1 1 0.01929031 2 2 0.03858062 2 4 0.07716123 2 8 0.1543224 1 8 0.1543225 1 8 0.1543226 1 8 0.1543227 0.6666666667 5.333333333 0.1028828 0.6666666667 3.555555556 0.06858789 0.6666666667 2.37037037 0.045725210 0.6666666667 1.580246914 0.0304834


Summing the third column, we get 1/ π0 = 51.83950617 → π0 = 0.01929. Then we obtain the otherprobabilities by multiplying each entry in the third column by π0, for example, π1 = 2× π0 = 0.0385806,etc. (In retrospect, the assumption that terms after the tenth would be negligible and could be ignoredseems to be invalid!)

π0 π1 π2 π3 π4 π5 π6 π7 π8 π9 π100.0193 0.0386 0.0772 0.1543 0.1543 0.1543 0.1543 0.1029 0.0686 0.0457 0.0305

3. What is the average number of customers in the check-out area?

Solution:

L = n πnΣn = 0

10

= 4.955944. What is the average waiting time of a customer?

Solution:The average arrival rate is

λ = λn πnΣn = 0

10

= 9.69517 / hourHence, to first obtain W (the average time in the system) by Little's Law,

L= λW or W = L/ λ = 0.511177 hr. = 30.67 minutes.Since the average time being served is 12 minutes, that means that the average waiting time (exclusiveof the time being served) is

Wq = 30.67 - 12 minutes = 18.67 minutes.

57:022 Principles of Design II -- HW#12 Spring 2000 page 1 of 2

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Homework #12Due May 12, 2000

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1. Barges arrive at the La Crosse lock on the Mississippi River at an average rate of one every two hours.It requires an average of 30 minutes to move a barge through the lock.Assuming that the arrival process is Poisson and that the time to move the barge through the lock hasexponential distribution, find:

a. The average number of barges in the system, i.e., either using or waiting to use the lock.

b. The average time spent by a barge at the lock (including waiting time).

c. The fraction of the time that the lock is busy.

d. The standard deviation of the service time distribution.

Suppose that the time to move the barge through the lock has a distribution which, instead of Exponential,is Erlang-4 (i.e. the time is the sum of 4 times, each with exponential distribution) but with the same mean(i.e., 30 minutes).

e. What is now the standard deviation of the service time distribution?

f. With this new service time distribution, what is the average number of barges waiting to use thelock?

2. Consider a single-server queueing system where some potential customers balk (refuse to enter thesystem) and some customers who do enter the system later get impatient and renege (leave without beingserved). Potential customers arrive according to a Poisson process with a mean rate of 4 per hour. Anarriving potential customer who finds n customers already there will balk with the following probabilities:

n P{Balk | n customers already insystem}

0 01 25%2 50%3 75%4 100%

Service times have an exponential distribution with a mean of one hour.

A customer never reneges while being served, but the customers in the queue may renege. In particular, theremaining time that the customer at the front of the queue is willing to wait before reneging has anexponential distribution with a mean of one hour. For a customer in the second position in the queue, thetime that she or he is willing to wait in this position before reneging has an exponential distribution with amean of 30 minutes. For a customer in the third position in the queue, the time that she or he is willing towait in this position before reneging has an exponential distribution with a mean of 20 minutes.

57:022 Principles of Design II -- HW#12 Spring 2000 page 2 of 2

a. Indicate the "birth" and "death" rates for the birth/death model:

b. Find the steady-state distribution of the number of customers in the system:

n πn01234

c. Find the average arrival rate of customers.

d. Find the expected fraction of arrivals who are lost as customers because of balking.

e. Find the average number of customers in the queueing system (including any being served).

f. Find the average time that a customer spends in the system before leaving (either because service iscompleted or because the customer reneges).

3. A single-stall automated car wash can wash a car in an average of 10 minutes. Cars arrive at a rate of 4per hour (forming a Poisson process).

a. If the time required to wash a car has exponential distribution, what is the average number of carswaiting in the queue to be washed?

b. What is the average length of time that a car waits in the queue?

c. If the wash time could be speeded up, what wash time would reduce the average wait time to fiveminutes?

d. Can the same five-minute waiting time be achieved by reducing, not the average time to wash thecar, but only the standard deviation of the wash time?

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HW7 Soln S2K - user.engineering.uiowa.edu