Day 3

1
Day 3

• Section 5 - Event Approach to Simulation
• Events
• Implementation of the Event Approach
• Canceling Events
• Exercise 5
• Section 6 - Simulation Output Analysis
• Random Numbers and Statistics
• Modernizing a Bank - with replications
• Presentation Graphics
• Exercise 6

F
2
Section 6 - Simulation Output Analysis

• Part 1 - Random Numbers and Statistics

3
Generating Random Numbers

• Multiplicative congruent technique
• Single stream of numbers between 0.0 and 1.0.
• Provide a number called a seed
• Do the arithmetic
• Results are the random value and the seed for the
  next draw.
• Period gt 2 billion numbers on a 32 bit machine

4
Multiple Random Number Streams

• Seed.v is a 1-dimensional, integer array of size
  10.
• It holds the current value of seeds for 10
  different streams.
• Let .PROBABILITY random.f(3)
• List seed.v(3)
• Seeds are 100,000 numbers apart
• Can access and change seed.v
• Release seed.v
• Reserve seed.v as 200
• For I 1 to 200
• Do
• Read seed.v(I)
• Loop ''I 1 to 200 do

5
Reproducibility of Random Numbers

• Let .SAVE.SEED.1 seed.v(1)
• ltstatementsgt
• Let seed.v(1) .SAVE.SEED.1

6
Replications

• No stop statements
• Let event set empty
• All resources relinquished
• Reset output variables

7

• 1 Main
• 2
• 3 Define I as an integer variable
• 4
• 5 Call INITIALIZE
• 6
• 7 Print 6 lines thus
• Pacific Port Problem with
  Replications
• Number of Average
  Maximum
• Ships Unloaded Delay Time
  Delay Time
• ------------ ----------
  ----------
• 14 For I 1 to 20 do
• 15
• 16 Use unit 6 for output
• 17 Activate a GENERATOR now
• 18 Start simulation
• 19

8

• Pacific Port Problem with
  Replications
• Number of Average
  Maximum
• Ships Unloaded Delay Time
  Delay Time
• ------------ ----------
  ----------
• 795 4.10
  26.15
• 805 4.53
  34.71
• 813 5.28
  41.34
• 819 5.25
  40.86
• 793 4.67
  51.97
• 792 4.08
  24.56
• 801 4.32
  23.63
• 807 5.72
  33.99
• 786 4.31
  34.84
• 800 4.91
  39.13
• 821 4.74
  26.23
• 795 4.13
  24.88
• 781 3.73
  22.56

9
Collecting Statistics

• Number of occurrences
• Sum
• Mean
• Sum of squares
• Mean square
• Variance
• Standard deviation
• Maximum
• Minimum
• Histogram

10
Collecting Statistics (continued)

• Select a global variable or attribute you want
  statistics on.
• Write one statement in the preamble to collect
  them.
• In Preamble
• Accumulate SHORTEST.LINE.LENGTH as the
  minimum,
• LONGEST.LINE.LENGTH as the
  maximum,
• AVERAGE.LINE.LENGTH as the
  mean,
• SPREAD as the
  std.dev and
• LINE.HISTOGRAM (0 to 20 by 1) as the
  histogram
• of N.Q.RUNWAY
• In another routine
• List SHORTEST.LINE.LENGTH,
• LONGEST.LINE.LENGTH,
• AVERAGE.LINE.LENGTH and
• SPREAD

11
Collecting Statistics (continued)

• SIMSCRIPT II.5 will set up the required counters
  to collect the required statistics
• SIMSCRIPT II.5 will monitor the global variable
  N.Q.RUNWAY
• Every time N.Q.RUNWAY is about to change,
  SIMSCRIPT II.5 will intercept control and pass it
  to the library routines that update the counters.
• Then it lets N.Q.RUNWAY change.
• When you want a value of a statistic, SIMSCRIPT
  II.5 calls a function that calculates it using
  the current values of the counters.

12
Collecting Statistics (continued)

• Suppose you want to collect daily and weekly
  statistics on the same variable.
• In Preamble
• Accumulate
• DAILY.AVERAGE.LINE.LENGTH as the DAILY mean
  and
• WEEKLY.AVERAGE.LINE.LENGTH as the WEEKLY mean
• of N.Q.RUNWAY
• In another routine
• For I 1 to 7 do ''days
• Start simulation
• List DAILY.AVERAGE.LINE.LENGTH
• Reset DAILY totals of N.Q.RUNWAY
• Loop ''I 1 to 7 do
• List WEEKLY.AVERAGE.LINE.LENGTH

13
Using Statistics to Make Decisions

• May need to look at all possibilities before
  making a decision
• Length of shortest line
• For each RUNWAY do
• Compute SHORTEST.LINE as the minimum
• of N.Q.RUNWAY(RUNWAY)
• Loop ''each RUNWAY
• Runway with shortest line
• For each RUNWAY do
• Compute BEST.RUNWAY as the minimum(RUNWAY)
• of N.Q.RUNWAY(RUNWAY)
• Loop ''each RUNWAY
• Request 1 RUNWAY (BEST.RUNWAY)

14
Accumulate and Tally

• Accumulate gives the time averaged statistic.
  Used when the length of time spent at a value is
  important, as for example, in queue lengths.
• Tally does not time average, as for example, in
  time waited in queue.

• Accumulate gives a value of almost 2.0
• Tally gives a value of 1.5

15
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16
Section 6 - Simulation Output Analysis

• Part 2 - Modernizing a Bank - with Replications

17
Bank Teller Problem

• Model bank with single line or multiple lines for
  tellers
• Look at range of tellers for each case
• Run replications for each case and number of
  tellers
• Customer inter-arrival times and service times
  are exponentially distributed

18
Bank Teller Problem (continued)

• In the multi-line system, an arriving customer
  will go to any available teller. If no teller is
  available, the customer will choose the shortest
  line and wait there until served. When the bank
  closes, no more customers are admitted, but all
  are served before the tellers leave.
• Results will be printed after each replication as
  well as overall.
• The results desired are
• 1. Average utilization of tellers
• 2. Average and maximum waiting line length
• 3. A histogram of waiting line length
• 4. Average waiting time of customers
• 5. A histogram of the waiting time in five minute
  increments

19

• 1 Preamble
• 2
• 3 '' Modernizing a Bank - CACI Products
  Company
• 4 '' files SIMBANK.SRC
• 5
• 6 Normally mode is undefined
• 7
• 8 Processes include
• 9 GENERATOR and
• 10 CUSTOMER
• 11
• 12 Resources include
• 13 TELLER
• 14
• 15 Define CASE as a text variable
• 16
• 17 Define MIN.TELLERS,
• 18 MAX.TELLERS and
• 19 NO.OF.REPLICATIONS

20

• 37
• 38 Accumulate AVG.QUEUE.LENGTH as the
  average,
• 39 MAX.QUEUE.LENGTH as the
  maximum and
• 40 QUEUE.HISTOGRAM (0 to 20 by
  1) as the histogram
• 41 of N.Q.TELLER
• 42
• 43 Tally MEAN.WAITING.TIME as the mean and
• 44 WAIT.HISTOGRAM (0 to 100 by 5) as
  the histogram
• 45 of WAITING.TIME
• 46
• 47 End

21

• 1 Main
• 2 Define .NO.OF.TELLERS,
• 3 .REPLICATION,
• 4 .SAVESEED1 and
• 5 .SAVESEED2
• 6 as integer variables
• 7
• 8 Define .START.TIME as a real variable
• 9
• 10 Let .SAVESEED1 seed.v(1)
• 11 Let .SAVESEED2 seed.v(2)
• 12
• 13 Call READ.DATA
• 14
• 15 For .NO.OF.TELLERS MIN.TELLERS to
  MAX.TELLERS
• 16 Do
• 17 Call INITIALIZE.TELLERS
• 18 Given
• 19 .NO.OF.TELLERS

22

• 27 Call DAILY.REPORT
• 28 Given
• 29 .REPLICATION,
• 30 .START.TIME and
• 31 .NO.OF.TELLERS
• 32
• 33 For each TELLER
• 34 Do
• 35 Reset DAILY totals of
  N.X.TELLER(TELLER),
• 36
  N.Q.TELLER(TELLER) and
• 37
  WAITING.TIME
• 38 Loop ''each TELLER
• 39 Loop ''.REPLICATION 1 to
  NO.OF.REPLICATIONS
• 40
• 41 Call FINAL.REPORT
• 42 Let time.v 0.0
• 43 Let seed.v(1) .SAVESEED1
• 44 Let seed.v(2) .SAVESEED2
• 45

23

• 1 Routine READ.DATA
• 2
• 3 Print 1 line thus
• Enter case to be run (Single-queue (S) or
  Multi-queue (M)
• 5 Read CASE
• 6 Let CASE upper.f(CASE)
• 7
• 8 Print 1 line thus
• Enter minimum number of tellers
• 10 Read MIN.TELLERS
• 11
• 12 Print 1 line thus
• Enter maximum number of tellers
• 14 Read MAX.TELLERS
• 15
• 16 Print 1 line thus
• Enter number of replications
• 18 Read NO.OF.REPLICATIONS
• 19

24

• 32 Skip 2 output lines
• 33
• 34 Print 10 lines with MIN.TELLERS,
• 35 MAX.TELLERS,
• 36 NO.OF.REPLICATIONS,
  
• 37 MEAN.INTERARRIVAL.T
  IME,
• 38 MEAN.SERVICE.TIME
  and
• 39 DAY.LENGTH thus
• COMPARISON OF SINGLE AND MULTI-QUEUE BANK
  OPERATIONS
• The number of tellers ranges from to .
• There are replications for each number
  of tellers
• Customers arrive according to an
  exponential distribution
• of inter-arrival times with a mean of
  . minutes.
• Service time is exponentially distributed
• with a mean of . minutes.
• The bank doors are closed after .
  hours each day
• But all customers inside are served.
• 50 Skip 2 output lines

25

• 1 Routine INITIALIZE.TELLERS
• 2 Given
• 3 .NO.OF.TELLERS
• 4
• 5 Define .NO.OF.TELLERS as an integer
  variable
• 6
• 7 If N.TELLER is not zero
• 8 Destroy every TELLER
• 9 Endif ''N.TELLER is not zero
• 10
• 11 If CASE "S" ''single queue case
• 12 Create every TELLER(1)
• 13 Let U.TELLER(1) .NO.OF.TELLERS
• 14 Else ''multi-queue case
• 15 Create every TELLER(.NO.OF.TELLERS)
• 16 For each TELLER
• 17 Do
• 18 Let U.TELLER(TELLER) 1
• 19 Loop ''each TELLER do

26

• 1 Process GENERATOR
• 2
• 3 Define .TIME.TO.CLOSE as a real
  variable
• 4
• 5 Let .TIME.TO.CLOSE time.v
  (DAY.LENGTH / hours.v)
• 6
• 7 Until time.v gt .TIME.TO.CLOSE
• 8 Do
• 9 Activate a CUSTOMER now
• 10 Wait exponential.f(MEAN.INTERARRIVAL.
  TIME,1) minutes
• 11 Loop ''time.v gt .TIME.TO.CLOSE do
• 12
• 13 End ''GENERATOR

27

• 1 Process CUSTOMER
• 2
• 3 Define .ARRIVAL.TIME as a real variable
• 4
• 5 Define .MY.CHOICE as an integer
  variable
• 6
• 7 Let .ARRIVAL.TIME time.v
• 8 If CASE "S" ''single queue case
• 9 Let .MY.CHOICE 1
• 10 Else ''multi-queue case
• 11 For each TELLER
• 12 with N.X.TELLER(TELLER) 0
• 13 Find the first case
• 14 If found
• 15 Let .MY.CHOICE TELLER
• 16 Else ''they're all busy
• 17 For each TELLER
• 18 Do
• 19 Compute .MY.CHOICE as the
  minimum(TELLER)

28

• 1 Routine DAILY.REPORT
• 2 Given
• 3 .REPLICATION.NUMBER,
• 4 .START.TIME,
• 5 .NO.OF.TELLERS
• 6
• 7 Define .REPLICATION.NUMBER and
• 8 .NO.OF.TELLERS
• 9 as integer variables
• 10
• 11 Define .START.TIME as a real variable
• 12
• 13 If .REPLICATION.NUMBER 1
• 14 Start new page
• 15 Print 5 lines with .NO.OF.TELLERS
  thus
• Number of tellers
• Finish Teller Queue Length
  Average Customer
• Time Utilization Average Maximum
  Waiting time
• (Hours)
  (Minutes)

29

• 1 Routine FINAL.REPORT
• 2
• 3 Define .I as an integer variable
• 4
• 5 Print 2 lines thus
• Average over all replications
• 8
• 9 For each TELLER
• 10 Do
• 11 Print 1 line with UTILIZATION(TELLER)
  / U.TELLER(TELLER),
• 12 AVG.QUEUE.LENGTH(TE
  LLER),
• 13 MAX.QUEUE.LENGTH(TE
  LLER) and
• 14 MEAN.WAITING.TIME
  thus
• . .
  
• 16 Loop ''each TELLER
• 17 Skip 3 lines
• 18
• 19 Print 3 lines with WAIT.HISTOGRAM(1)
  and

30
35 Print 1 line with WAIT.HISTOGRAM(21)
and 36
QUEUE.HISTOGRAM(1,21) 100 /time.v thus 100 lt
T 20
. 38 39 Read as / using unit 5
40 41 End ''FINAL.REPORT
31

• Enter case to be run (Single-queue (S) or
  Multi-queue (M) S
• Enter minimum number of tellers 1
• Enter maximum number of tellers 3
• Enter number of replications 5
• Enter mean time between customer arrivals
  (in minutes) 5
• Enter mean time for servicing a customer
  (in minutes) 10
• Enter length of time that bank is open for
  one day 8
• COMPARISON OF SINGLE AND MULTI-QUEUE BANK
  OPERATIONS
• The number of tellers ranges from 1 to 3.
• There are 5 replications for each number
  of tellers
• Customers arrive according to an
  exponential distribution
• of inter-arrival times with a mean of
  5.00 minutes.
• Service time is exponentially distributed
• with a mean of 10.00 minutes.
• The bank doors are closed after 8.00
  hours each day
• But all customers inside are served.

32

• Number of tellers 1
• Finish Teller Queue Length
  Average Customer
• Time Utilization Average Maximum
  Waiting time
• (Hours)
  (Minutes)
• 17.82 1.0 25.8 53
  253
• 16.44 1.0 21.7 44
  227
• 19.28 1.0 32.3 58
  373
• 14.56 1.0 22.8 44
  210
• 14.03 1.0 20.0 44
  185
• Average over all replications
• 1.0 25.0 58
  252

33

• Waiting Time Number Who Waited Queue Length
  Percentage
• (minutes) This Time
  of Time
• T lt 5 13 0
  1.7
• 5 lt T lt 10 2 1
  1.8
• 10 lt T lt 15 1 2
  1.4
• 15 lt T lt 20 4 3
  2.1
• 20 lt T lt 25 2 4
  1.7
• 25 lt T lt 30 2 5
  1.6
• 30 lt T lt 35 6 6
  2.2
• 35 lt T lt 40 5 7
  2.1
• 40 lt T lt 45 6 8
  2.0
• 45 lt T lt 50 4 9
  2.5
• 50 lt T lt 55 5 10
  1.6
• 55 lt T lt 60 10 11
  2.3
• 60 lt T lt 65 7 12
  3.4
• 65 lt T lt 70 6 13
  2.0
• 70 lt T lt 75 9 14
  1.1
• 75 lt T lt 80 2 15
  2.5
• 80 lt T lt 85 12 16
  2.0

34

• Number of tellers 2
• Finish Teller Queue Length
  Average Customer
• Time Utilization Average Maximum
  Waiting time
• (Hours)
  (Minutes)
• 9.26 1.0 3.6 13
  18
• 9.11 .9 2.3 10
  13
• 9.91 1.0 12.3 28
  73
• 8.05 .9 1.4 8
  7
• 8.31 .8 2.2 9
  12
• Average over all replications
• .9 4.6 28
  25

35

• Waiting Time Number Who Waited Queue Length
  Percentage
• (minutes) This Time
  of Time
• T lt 5 177 0
  28.1
• 5 lt T lt 10 56 1
  13.9
• 10 lt T lt 15 43 2
  9.3
• 15 lt T lt 20 36 3
  7.7
• 20 lt T lt 25 23 4
  7.4
• 25 lt T lt 30 27 5
  5.3
• 30 lt T lt 35 22 6
  4.9
• 35 lt T lt 40 8 7
  3.5
• 40 lt T lt 45 3 8
  2.2
• 45 lt T lt 50 11 9
  2.3
• 50 lt T lt 55 4 10
  1.3
• 55 lt T lt 60 6 11
  1.3
• 60 lt T lt 65 9 12
  1.9
• 65 lt T lt 70 6 13
  2.0
• 70 lt T lt 75 5 14
  1.3
• 75 lt T lt 80 1 15
  1.0
• 80 lt T lt 85 3 16
  1.0

36

• Number of tellers 3
• Finish Teller Queue Length
  Average Customer
• Time Utilization Average Maximum
  Waiting time
• (Hours)
  (Minutes)
• 8.27 .7 .6 5
  3
• 8.34 .7 .4 4
  2
• 8.17 .8 1.3 6
  6
• 8.05 .6 .3 5
  2
• 8.08 .6 .4 5
  2
• Average over all replications
• .7 .6 6
  3

37

• Waiting Time Number Who Waited Queue Length
  Percentage
• (minutes) This Time
  of Time
• T lt 5 384 0
  72.6
• 5 lt T lt 10 36 1
  10.2
• 10 lt T lt 15 45 2
  7.3
• 15 lt T lt 20 15 3
  6.2
• 20 lt T lt 25 7 4
  2.5
• 25 lt T lt 30 1 5
  1.2
• 30 lt T lt 35 1 6
  .1
• 35 lt T lt 40 0 7
  0.
• 40 lt T lt 45 0 8
  0.
• 45 lt T lt 50 0 9
  0.
• 50 lt T lt 55 0 10
  0.
• 55 lt T lt 60 0 11
  0.
• 60 lt T lt 65 0 12
  0.
• 65 lt T lt 70 0 13
  0.
• 70 lt T lt 75 0 14
  0.
• 75 lt T lt 80 0 15
  0.
• 80 lt T lt 85 0 16
  0.

38

• Enter case to be run (Single-queue (S) or
  Multi-queue (M) M
• Enter minimum number of tellers 1
• Enter maximum number of tellers 3
• Enter number of replications 5
• Enter mean time between customer arrivals
  (in minutes) 5
• Enter mean time for servicing a customer
  (in minutes) 10
• Enter length of time that bank is open for
  one day 8
• COMPARISON OF SINGLE AND MULTI-QUEUE BANK
  OPERATIONS
• The number of tellers ranges from 1 to 3.
• There are 5 replications for each number
  of tellers
• Customers arrive according to an
  exponential distribution
• of inter-arrival times with a mean of
  5.00 minutes.
• Service time is exponentially distributed
• with a mean of 10.00 minutes.
• The bank doors are closed after 8.00
  hours each day
• But all customers inside are served.

39

• Number of tellers 1
• Finish Teller Queue Length
  Average Customer
• Time Utilization Average Maximum
  Waiting time
• (Hours)
  (Minutes)
• 17.82 1.0 25.8 53
  253
• 16.44 1.0 21.7 44
  227
• 19.28 1.0 32.3 58
  373
• 14.56 1.0 22.8 44
  210
• 14.03 1.0 20.0 44
  185
• Average over all replications
• 1.0 25.0 58
  252

40

• Waiting Time Number Who Waited Queue Length
  Percentage
• (minutes) This Time
  of Time
• T lt 5 13 0
  1.7
• 5 lt T lt 10 2 1
  1.8
• 10 lt T lt 15 1 2
  1.4
• 15 lt T lt 20 4 3
  2.1
• 20 lt T lt 25 2 4
  1.7
• 25 lt T lt 30 2 5
  1.6
• 30 lt T lt 35 6 6
  2.2
• 35 lt T lt 40 5 7
  2.1
• 40 lt T lt 45 6 8
  2.0
• 45 lt T lt 50 4 9
  2.5
• 50 lt T lt 55 5 10
  1.6
• 55 lt T lt 60 10 11
  2.3
• 60 lt T lt 65 7 12
  3.4
• 65 lt T lt 70 6 13
  2.0
• 70 lt T lt 75 9 14
  1.1
• 75 lt T lt 80 2 15
  2.5
• 80 lt T lt 85 12 16
  2.0

41

• Number of tellers 2
• Finish Teller Queue Length
  Average Customer
• Time Utilization Average Maximum
  Waiting time
• (Hours)
  (Minutes)
• 9.49 .9 2.1 7
  20
• 9.49 1.0 1.7 6
  20
• 9.25 .9 1.7 5
  18
• 9.25 .8 1.2 5
  18
• 10.12 .9 6.3 15
  75
• 10.12 1.0 6.0 14
  75
• 8.16 .9 1.1 4
  12
• 8.16 .9 1.2 4
  12
• 8.38 .9 1.3 5
  13
• 8.38 .7 1.1 4
  13
• Average over all replications
• .9 2.6 15
  28
• .9 2.4 14
  28

42

• Waiting Time Number Who Waited Queue Length
  Percentage
• (minutes) This Time
  of Time
• T lt 5 160 0
  23.4
• 5 lt T lt 10 47 1
  23.7
• 10 lt T lt 15 48 2
  18.3
• 15 lt T lt 20 39 3
  10.1
• 20 lt T lt 25 23 4
  6.6
• 25 lt T lt 30 24 5
  3.2
• 30 lt T lt 35 17 6
  2.6
• 35 lt T lt 40 17 7
  4.3
• 40 lt T lt 45 20 8
  2.3
• 45 lt T lt 50 6 9
  .2
• 50 lt T lt 55 11 10
  1.0
• 55 lt T lt 60 7 11
  .5
• 60 lt T lt 65 1 12
  .9
• 65 lt T lt 70 2 13
  1.7
• 70 lt T lt 75 5 14
  .6
• 75 lt T lt 80 4 15
  .4
• 80 lt T lt 85 4 16
  0.

43

• Number of tellers 3
• Finish Teller Queue Length
  Average Customer
• Time Utilization Average Maximum
  Waiting time
• (Hours)
  (Minutes)
• 8.48 .8 .4 2
  4
• 8.48 .8 .3 2
  4
• 8.48 .5 .1 1
  4
• 8.46 .9 .5 2
  4
• 8.46 .6 .1 1
  4
• 8.46 .5 .1 1
  4
• 8.17 .9 .6 2
  7
• 8.17 .8 .5 2
  7
• 8.17 .7 .3 2
  7
• 8.05 .8 .2 2
  3
• 8.05 .7 .2 2
  3
• 8.05 .4 .1 1
  3
• 8.14 .7 .3 2
  3
• 8.14 .5 .1 2
  3

44

• Waiting Time Number Who Waited Queue
  Length Percentage
• (minutes) This Time
  of Time
• T lt 5 370 0
  65.6
• 5 lt T lt 10 34 1
  28.0
• 10 lt T lt 15 30 2
  6.4
• 15 lt T lt 20 24 3
  0.
• 20 lt T lt 25 12 4
  0.
• 25 lt T lt 30 6 5
  0.
• 30 lt T lt 35 7 6
  0.
• 35 lt T lt 40 3 7
  0.
• 40 lt T lt 45 1 8
  0.
• 45 lt T lt 50 1 9
  0.
• 50 lt T lt 55 0 10
  0.
• 55 lt T lt 60 1 11
  0.
• 60 lt T lt 65 0 12
  0.
• 65 lt T lt 70 0 13
  0.
• 70 lt T lt 75 0 14
  0.
• 75 lt T lt 80 0 15
  0.
• 80 lt T lt 85 0 16
  0.

45
Section 6 - Simulation Output Analysis

• Part 3 - Presentation Graphics

46
Presentation Graphics
47
Displaying a Single Variable

• Preamble
• Resources include PASSENGER.AGENT
• Display variables include N.Q.PASSENGER.AGENT
• End ''Preamble
• Main
• Show N.Q.PASSENGER.AGENT with "QUEUE.GRF"
• End ''Main

48
Displaying a Histogram

• Preamble
• Define WAITING.TIME as a real variable
• Define LOW, HIGH and DELTA as integer
  variables
• Tally WAITING.TIME.HISTOGRAM (LOW to HIGH by
  DELTA)
• as the dynamic histogram of WAITING.TIME
• End ''Preamble
• Main
• Show WAITING.TIME.HISTOGRAM with "WAIT.GRF"
• End ''Main

49
Displaying a Clock

• Preamble
• Define CLOCKTIME as a double variable
• Display variables include CLOCKTIME
• End ''Preamble
• Main
• Let timesync.v 'CLOCK.UPDATE'
• Show CLOCKTIME with "CLOCK.GRF"
• End ''Main
• Routine CLOCK.UPDATE
• given .TIME.ATR yielding .NEW.TIME.ATR
• Define .TIME.ATR, .NEW.TIME.ATR as double
  variables
• Let CLOCKTIME .TIME.ATR
• Let .NEW.TIME.ATR .TIME.ATR
• End ''CLOCK.UPDATE

50
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51
Exercise 6

• The Pacific Port Problem (revisited)
• C\Program Files\Simscript3\models\ProblemC

52
Pacific Port Problem Revisited

• Now that the prototype simulation model is
  working, the Harbor Master has requested a better
  estimate of the port performance. In order to
  satisfy this requirement, the following
  statistics have been requested
• 1. Average and maximum waiting line for the
  docks
• 2. Average and maximum waiting line for the tugs
• 3. Utilization of the docks and the tug
• 4. Average and maximum ship waiting time
  (excluding unloading time)
• 5. Average and maximum in-port time for all
  ships
• 6. The number of ships served

53
Pacific Port Problem Revisited

• In addition, consider a shipper who wishes to
  ship crude oil from Valdez, Alaska, to San Pedro.
  He estimates that shipments will require five
  ships of a particular type. The time to unload
  one of the new ships follows an exponential
  distribution with a mean of 21 hours. After
  unloading, the new ship will travel to Valdez,
  load more crude oil, and return to San Pedro.
  The round trip times follow a normal distribution
  with a mean of 10 days and a standard deviation
  of 1 day.
• Before the port authorities agree to accommodate
  the five new ships, they want to determine the
  effect of the additional port traffic, especially
  on in-port residency time.
• The Harbor Master also desires to see a graphical
  presentation of a histogram for number of ships
  versus WAITING.TIME in hours.
• SELECT PROBLEMC (Or ANSWERB)

54
NOTES

• To identify the new ships and the old ships, give
  the ships attributes such as
• EVERY SHIP has
• a SHIP.TYPE
• Define ..OLD.SHIP to mean 1
• Define ..NEW.SHIP to mean 2
• Activate a SHIP now
• Let SHIP.TYPE(SHIP) ..OLD.SHIP

55
NOTES Calling a Synchronous Routine from a
Process

• 1 Process SHIP
• 2
• 3 Define .ARRIVAL.TIME,
• 4 .UNLOAD.TIME and
• 5 .PROBABILITY
• 6 as real variables
• 7
• 8 Let .PROBABILITY random.f(2)
• 9 Select case .PROBABILITY
• 10 Case 0.0 to .25
• 11 Let .UNLOAD.TIME
  uniform.f(16.0, 20.0, 3)
• 12 Case .25 to .80
• 13 Let .UNLOAD.TIME
  uniform.f(21.0, 27.0, 3)
• 14 Default
• 15 Let .UNLOAD.TIME
  uniform.f(32.0, 40.0, 3)
• 16 Endselect ''.PROBABILITY
• 17
• Call HARBOR.OPERATIONS
• giving .UNLOAD.TIME

56

• 18 Let .ARRIVAL.TIME time.v
• 19
• 20 Request 1 DOCK(1)
• 21 Request 1 TUG(1)
• 22 Wait 1.0 hour
• 23 Relinquish 1 TUG(1)
• 24
• 25 Work .UNLOAD.TIME hours
• 26
• 27 Request 1 TUG(1)
• 28 Wait 1.0 hour
• 29 Relinquish 1 TUG(1)
• 30 Relinquish 1 DOCK(1)
• 31
• 32 Let WAITING.TIME (time.v -
  .ARRIVAL.TIME) hours.v - .UNLOAD.TIME
• 33 Add WAITING.TIME to SUM.SHIP.DELAY
• 34 Let MAX.SHIP.DELAY max.f(WAITING.TIME,
  MAX.SHIP.DELAY)
• 35 Add 1 to NO.OF.SHIPS.SERVED
• 36