Decay chain simulation - with significantly different time scales - simulation

I would like to simulate a decay chain with Python. Normally, (in a loop over all nuclides) one calculates the number of decays per time step and updates the number of mother and daughter nuclei.
My problem is that the decay chain contains half-lives on very different time scales, i.e.
0.0001643 seconds for Po-214 and 307106512477175.9 seconds (= 1600 years) for Ra-226.
Using the same time step for all nuclides seems useless.
Is there a simulation method, preferably in Python, that can be used to handle this case?

Don't use time steps for this. Use event scheduling.
Half lives can be expressed as exponential decay, and the conversion between half life and rate of decay is straightforward. Start with the number of both types of nuclei, and schedule exponential inter-event times to figure out when the next decay of each type will occur. Whichever type has the lower time, decrement the corresponding number of nuclei and schedule the next decay for that type (and if need be, increment the count of whatever it decays into).
This can easily be generalized to multiple distinct event types by using a priority queue ordered by time of occurrence to determine which event will be the next one performed. This is the underlying principle behind discrete event simulation.
Update
This approach works with individual decay events, but we can leverage two important properties when we have exponential inter-event times.
The first is to note that exponentially distributed inter-event times means these are Poisson processes. The superposition property tells us that the union of two independent Poisson processes, each having rate λ, is a Poisson process with rate 2λ. Simple induction shows that if we have n independent Poisson properties with the same rate, their superposition is a Poisson process with rate nλ.
The second property is that the exponential distribution is memoryless. This means that when a Poisson event occurs, we can generate the time to the next event by generating a new exponentially distributed time at the current rate and adding it to the current time.
You haven't provided any information about what you want in the way of output, so I arbitrarily decided to print a report showing the time and the current numbers of nuclides whenever that number was halved. I also printed a report every 10 years, given the long half-life of Po-214.
I converted half-lifes to rates using the link provided at the top of the post, and then to means since that's what
Python numpy's exponential generator is parameterized to use. That's an easy conversion, since means and rates are inverses of each other.
Here's a Python implementation with comments:
from numpy.random import default_rng
from math import log
rng = default_rng()
# This creates a list of entries of quantities that will trigger a report.
# I've chosen to go with successive halvings of the original quantity.
def generate_report_qtys(n0):
report_qty = []
divisor = 2
while divisor < n0:
report_qty.append(n0 // divisor) # append next half-life qty to array
divisor *= 2
return report_qty
seconds_per_year = 365.25 * 24 * 60 * 60
po_214_half_life = 0.0001643 # seconds
ra_226_half_life = 1590 * seconds_per_year
log_2 = log(2)
po_mean = po_214_half_life / log_2 # per-nuclide decay rate for po_214
ra_mean = ra_226_half_life / log_2 # ditto for ra_226
po_n = po_n0 = 1_000_000_000
ra_n = ra_n0 = 1_000_000_000
time = 0.0
# Generate a report when the following sets of half-lifes are reached
po_report_qtys = generate_report_qtys(po_n0)
ra_report_qtys = generate_report_qtys(ra_n0)
# Initialize first event times for each type of event:
# - first entry is polonium next event time
# - second entry is radium next event time
# - third entry is next ten year report time
next_event_time = [
rng.exponential(po_mean / po_n),
rng.exponential(ra_mean / ra_n),
10 * seconds_per_year
]
# Print column labels and initial values
print("time,po_214,ra_226,time_in_years")
print(f"{time},{po_n},{ra_n},{time / seconds_per_year}")
while time < ra_226_half_life:
# Find the index of the next event time. Index tells us the event type.
min_index = next_event_time.index(min(next_event_time))
if min_index == 0:
po_n -= 1 # decrement polonium count
time = next_event_time[0] # update clock to the event time
if po_n > 0:
next_event_time[0] += rng.exponential(po_mean / po_n) # determine next event time for po
else:
next_event_time[0] = float('Inf')
# print report if this is a half-life occurrence
if len(po_report_qtys) > 0 and po_n == po_report_qtys[0]:
po_report_qtys.pop(0) # remove this occurrence from the list
print(f"{time},{po_n},{ra_n},{time / seconds_per_year}")
elif min_index == 1:
# same as above, but for radium
ra_n -= 1
time = next_event_time[1]
if ra_n > 0:
next_event_time[1] += rng.exponential(ra_mean / ra_n)
else:
next_event_time[1] = float('Inf')
if len(ra_report_qtys) > 0 and ra_n == ra_report_qtys[0]:
ra_report_qtys.pop(0)
print(f"{time},{po_n},{ra_n},{time / seconds_per_year}")
else:
# update clock, print ten year report
time = next_event_time[2]
next_event_time[2] += 10 * seconds_per_year
print(f"{time},{po_n},{ra_n},{time / seconds_per_year}")
Run times are proportional to the number of nuclides. Running with a billion of each took 831.28s on my M1 MacBook Pro, versus 2.19s for a million of each. I also ported this to Crystal, a compiled Ruby-like language, which produced comparable results in 32 seconds for a billion of each nuclide. I would recommend using a compiled language if you intend to run larger sized problems, but I will also point out that if you use half-life reporting as I did the results are virtually identical for smaller population sizes but are obtained much more rapidly.
I would also suggest that if you want to use this approach for a more complex model, you should use a priority queue of tuples containing time and type of event to store the set of pending future events rather than a simple list.
Last but not least, here's some sample output:
time,po_214,ra_226,time_in_years
0.0,1000000000,1000000000,0.0
0.0001642985647308265,500000000,1000000000,5.20630734690935e-12
0.0003286071415481526,250000000,1000000000,1.0412931957694901e-11
0.0004929007624958987,125000000,1000000000,1.5619082645571865e-11
0.0006571750701843468,62500000,1000000000,2.082462133319222e-11
0.0008214861652253772,31250000,1000000000,2.6031325741671646e-11
0.0009858208114474198,15625000,1000000000,3.1238776442043114e-11
0.0011502417677631668,7812500,1000000000,3.6448962144243124e-11
0.0013145712145548718,3906250,1000000000,4.165624808460947e-11
0.0014788866075394896,1953125,1000000000,4.686308868670272e-11
0.0016432124609700412,976562,1000000000,5.2070260760325286e-11
0.001807832817519779,488281,1000000000,5.728676507465013e-11
0.001972981254301889,244140,1000000000,6.252000324175124e-11
0.0021372947080755688,122070,1000000000,6.772678239395799e-11
0.002301139510796509,61035,1000000000,7.29187108904514e-11
0.0024642826956509244,30517,1000000000,7.808840645837847e-11
0.0026302282280720344,15258,1000000000,8.33469030620844e-11
0.0027944471221414947,7629,1000000000,8.855068579808016e-11
0.002954014120737834,3814,1000000000,9.3607058861822e-11
0.0031188370035748177,1907,1000000000,9.882998084692174e-11
0.003282466175503322,953,1000000000,1.0401507641592902e-10
0.003457552492113242,476,1000000000,1.0956322699169905e-10
0.003601851131916978,238,1000000000,1.1413577496124477e-10
0.0037747824699194033,119,1000000000,1.1961563838566314e-10
0.0039512825256332275,59,1000000000,1.252085876503038e-10
0.004124330529803301,29,1000000000,1.3069214800248755e-10
0.004337121375518753,14,1000000000,1.3743508300754027e-10
0.004535068261934763,7,1000000000,1.437076413268044e-10
0.004890820999020369,3,1000000000,1.5498076529965425e-10
0.004909065046898487,1,1000000000,1.555588842908994e-10
315576000.0,0,995654793,10.0
631152000.0,0,991322602,20.0
946728000.0,0,987010839,30.0
1262304000.0,0,982711723,40.0
1577880000.0,0,978442651,50.0
1893456000.0,0,974185269,60.0
2209032000.0,0,969948418,70.0
2524608000.0,0,965726762,80.0
2840184000.0,0,961524848,90.0
3155760000.0,0,957342148,100.0
3471336000.0,0,953178898,110.0
3786912000.0,0,949029294,120.0
4102488000.0,0,944898063,130.0
4418064000.0,0,940790494,140.0
4733640000.0,0,936699123,150.0
5049216000.0,0,932622334,160.0
5364792000.0,0,928565676,170.0
5680368000.0,0,924523267,180.0
5995944000.0,0,920499586,190.0
6311520000.0,0,916497996,200.0
6627096000.0,0,912511030,210.0
6942672000.0,0,908543175,220.0
7258248000.0,0,904590364,230.0
7573824000.0,0,900656301,240.0
7889400000.0,0,896738632,250.0
8204976000.0,0,892838664,260.0
8520552000.0,0,888956681,270.0
8836128000.0,0,885084855,280.0
9151704000.0,0,881232862,290.0
9467280000.0,0,877401861,300.0
9782856000.0,0,873581425,310.0
10098432000.0,0,869785364,320.0
10414008000.0,0,866002042,330.0
10729584000.0,0,862234212,340.0
11045160000.0,0,858485627,350.0
11360736000.0,0,854749939,360.0
11676312000.0,0,851032010,370.0
11991888000.0,0,847329028,380.0
12307464000.0,0,843640016,390.0
12623040000.0,0,839968529,400.0
12938616000.0,0,836314000,410.0
13254192000.0,0,832673999,420.0
13569768000.0,0,829054753,430.0
13885344000.0,0,825450233,440.0
14200920000.0,0,821859757,450.0
14516496000.0,0,818284787,460.0
14832072000.0,0,814727148,470.0
15147648000.0,0,811184419,480.0
15463224000.0,0,807655470,490.0
15778800000.0,0,804139970,500.0
16094376000.0,0,800643280,510.0
16409952000.0,0,797159389,520.0
16725528000.0,0,793692735,530.0
17041104000.0,0,790239221,540.0
17356680000.0,0,786802135,550.0
17672256000.0,0,783380326,560.0
17987832000.0,0,779970864,570.0
18303408000.0,0,776576174,580.0
18618984000.0,0,773197955,590.0
18934560000.0,0,769836170,600.0
19250136000.0,0,766488931,610.0
19565712000.0,0,763154778,620.0
19881288000.0,0,759831742,630.0
20196864000.0,0,756528400,640.0
20512440000.0,0,753237814,650.0
20828016000.0,0,749961747,660.0
21143592000.0,0,746699940,670.0
21459168000.0,0,743450395,680.0
21774744000.0,0,740219531,690.0
22090320000.0,0,736999181,700.0
22405896000.0,0,733793266,710.0
22721472000.0,0,730602000,720.0
23037048000.0,0,727427544,730.0
23352624000.0,0,724260327,740.0
23668200000.0,0,721110260,750.0
23983776000.0,0,717973915,760.0
24299352000.0,0,714851218,770.0
24614928000.0,0,711740161,780.0
24930504000.0,0,708645945,790.0
25246080000.0,0,705559170,800.0
25561656000.0,0,702490991,810.0
25877232000.0,0,699436919,820.0
26192808000.0,0,696394898,830.0
26508384000.0,0,693364883,840.0
26823960000.0,0,690348242,850.0
27139536000.0,0,687345934,860.0
27455112000.0,0,684354989,870.0
27770688000.0,0,681379178,880.0
28086264000.0,0,678414567,890.0
28401840000.0,0,675461363,900.0
28717416000.0,0,672522494,910.0
29032992000.0,0,669598412,920.0
29348568000.0,0,666687807,930.0
29664144000.0,0,663787671,940.0
29979720000.0,0,660901676,950.0
30295296000.0,0,658027332,960.0
30610872000.0,0,655164886,970.0
30926448000.0,0,652315268,980.0
31242024000.0,0,649481821,990.0
31557600000.0,0,646656096,1000.0
31873176000.0,0,643841377,1010.0
32188752000.0,0,641041609,1020.0
32504328000.0,0,638253759,1030.0
32819904000.0,0,635479981,1040.0
33135480000.0,0,632713706,1050.0
33451056000.0,0,629962868,1060.0
33766632000.0,0,627223350,1070.0
34082208000.0,0,624494821,1080.0
34397784000.0,0,621778045,1090.0
34713360000.0,0,619076414,1100.0
35028936000.0,0,616384399,1110.0
35344512000.0,0,613702920,1120.0
35660088000.0,0,611035112,1130.0
35975664000.0,0,608376650,1140.0
36291240000.0,0,605729994,1150.0
36606816000.0,0,603093946,1160.0
36922392000.0,0,600469403,1170.0
37237968000.0,0,597854872,1180.0
37553544000.0,0,595254881,1190.0
37869120000.0,0,592663681,1200.0
38184696000.0,0,590085028,1210.0
38500272000.0,0,587517782,1220.0
38815848000.0,0,584961743,1230.0
39131424000.0,0,582420312,1240.0
39447000000.0,0,579886455,1250.0
39762576000.0,0,577362514,1260.0
40078152000.0,0,574849251,1270.0
40393728000.0,0,572346625,1280.0
40709304000.0,0,569856166,1290.0
41024880000.0,0,567377753,1300.0
41340456000.0,0,564908008,1310.0
41656032000.0,0,562450828,1320.0
41971608000.0,0,560005832,1330.0
42287184000.0,0,557570018,1340.0
42602760000.0,0,555143734,1350.0
42918336000.0,0,552729893,1360.0
43233912000.0,0,550326162,1370.0
43549488000.0,0,547932312,1380.0
43865064000.0,0,545550017,1390.0
44180640000.0,0,543178924,1400.0
44496216000.0,0,540814950,1410.0
44811792000.0,0,538462704,1420.0
45127368000.0,0,536123339,1430.0
45442944000.0,0,533792776,1440.0
45758520000.0,0,531469163,1450.0
46074096000.0,0,529157093,1460.0
46389672000.0,0,526854383,1470.0
46705248000.0,0,524564196,1480.0
47020824000.0,0,522282564,1490.0
47336400000.0,0,520011985,1500.0
47651976000.0,0,517751635,1510.0
47967552000.0,0,515499791,1520.0
48283128000.0,0,513257373,1530.0
48598704000.0,0,511022885,1540.0
48914280000.0,0,508798440,1550.0
49229856000.0,0,506582663,1560.0
49545432000.0,0,504379227,1570.0
49861008000.0,0,502186693,1580.0
50176584000.0,0,500000869,1590.0
Expanded for More than 2 Nuclides
I mentioned that for more than a couple of nuclides you'd want to use a priority queue to track which decays occur next. I reorganized the code around functions, but that allowed greater flexibility in expanding the scope of the problem. Here you go:
#!/usr/bin/env python3
from numpy.random import default_rng
from math import log
import heapq
SECONDS_PER_YEAR = 365.25 * 24 * 60 * 60
LOG_2 = log(2)
rng = default_rng()
def generate_report_qtys(n0):
report_qty = []
divisor = 2
while divisor < n0:
report_qty.append(n0 // divisor) # append next half-life qty to array
divisor *= 2
return report_qty
po_n0 = 10_000_000
ra_n0 = 10_000_000
mu_n0 = 10_000_000
# mean is half-life / LOG_2
properties = dict(
po_214 = dict(
mean = 0.0001643 / LOG_2,
qty = po_n0,
report_qtys = generate_report_qtys(po_n0)
),
ra_226 = dict(
mean = 1590 * SECONDS_PER_YEAR / LOG_2,
qty = ra_n0,
report_qtys = generate_report_qtys(ra_n0)
),
made_up = dict(
mean = 75 * SECONDS_PER_YEAR / LOG_2,
qty = mu_n0,
report_qtys = generate_report_qtys(mu_n0)
)
)
nuclide_names = [name for name in properties.keys()]
def population_mean(nuclide):
return properties[nuclide]['mean'] / properties[nuclide]['qty']
def report(): # isolate as single point of maintenance even though it's a one-liner
nuc_qtys = [str(properties[nuclide]['qty']) for nuclide in nuclide_names]
print(f"{time},{time / SECONDS_PER_YEAR}," + ','.join(nuc_qtys))
def decay_event(nuclide):
properties[nuclide]['qty'] -= 1
current_qty = properties[nuclide]['qty']
if current_qty > 0:
heapq.heappush(event_q, (time + rng.exponential(population_mean(nuclide)), nuclide))
rep_qty = properties[nuclide]['report_qtys']
if len(rep_qty) > 0 and current_qty == rep_qty[0]:
rep_qty.pop(0) # remove this occurrence from the list
report()
def report_event():
heapq.heappush(event_q, (time + 10 * SECONDS_PER_YEAR, 'report_event'))
report()
event_q = [(rng.exponential(population_mean(nuclide)), nuclide) for nuclide in nuclide_names]
event_q.append((0.0, "report_event"))
heapq.heapify(event_q)
time = 0.0 # simulated time
print("time(seconds),time(years)," + ','.join(nuclide_names)) # column labels
while time < 1600 * SECONDS_PER_YEAR:
time, event_id = heapq.heappop(event_q)
if event_id == 'report_event':
report_event()
else:
decay_event(event_id)
To add more nuclides, add more entries to the properties dictionary, following the template of the current entries.

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I have a code that uses ode15s to solve. This is my semester question and I am not really good on Matlab but tried to solve the problem like traditional programming way but couldn't achieve success.
The thing I want to achieve is I would like to change the value of Xxc depends on time
Initially Xxc value is 20, but when time will reach 10 then I want to update Xxc to 30 (1.5 times higher) and after that when time is 30 then I want to update Xxc to 40(2.0 times)
There are 2 files
adm1init.m (Initial values)
function ret=adm1init()
global Xxc......
.....
Xxc = 20;
......
time=0:0.2:70;
[t,dX]=ode15s('adm1sys', time, [Ssu Saa Sfa Sva Sbu Spro Sac Sh2 Sch4 SIC SIN SI Xxc Xch Xpr Xli Xsu Xaa Xfa Xc4 Xpro Xac Xh2 XI Scat San Shva Shbu Shpro...
Shac Shco3 Snh3 S_H_ion S_gas_h2 S_gas_ch4 S_gas_co2 q_gas Xhomo XCE Slac Sca]);
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These are the only lines that I guess corelated.
Then the calculation file is adm1sys.m
function dX=adm1sys(t,X)
...
rho(1) = k_dis * X(13);
...
rho(13) = k_dec_Xsu * X(17);
rho(14) = k_dec_Xaa * X(18);
rho(15) = k_dec_Xfa * X(19);
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rho(17) = k_dec_Xpro * X(21);
rho(18) = k_dec_Xac * X(22);
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dX(13) = (q_in/V_liq) * (Xxc - X(13)) - rho(1) + sum(rho(13:19));
...
X0(13) = Xxc;
...
I just want to update Xxc value in formula of dX(13) depends on time
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Changing formula in ode15s

I have a code that uses ode15s to solve. This is my semester question and I am not really good on Matlab but tried to solve the problem like traditional programming way but couldn't achieve success.
The thing I want to achieve is I would like to change the value of Xxc depends on time
Initially Xxc value is 20, but when time will reach 10 then I want to update Xxc to 30 (1.5 times higher) and after that when time is 30 then I want to update Xxc to 40(2.0 times)
There are 2 files
adm1init.m (Initial values)
function ret=adm1init()
global Xxc......
.....
Xxc = 20;
......
time=0:0.2:70;
[t,dX]=ode15s('adm1sys', time, [Ssu Saa Sfa Sva Sbu Spro Sac Sh2 Sch4 SIC SIN SI Xxc Xch Xpr Xli Xsu Xaa Xfa Xc4 Xpro Xac Xh2 XI Scat San Shva Shbu Shpro...
Shac Shco3 Snh3 S_H_ion S_gas_h2 S_gas_ch4 S_gas_co2 q_gas Xhomo XCE Slac Sca]);
ret = dX
These are the only lines that I guess corelated.
Then the calculation file is adm1sys.m
function dX=adm1sys(t,X)
...
rho(1) = k_dis * X(13);
...
rho(13) = k_dec_Xsu * X(17);
rho(14) = k_dec_Xaa * X(18);
rho(15) = k_dec_Xfa * X(19);
rho(16) = k_dec_Xc4 * X(20);
rho(17) = k_dec_Xpro * X(21);
rho(18) = k_dec_Xac * X(22);
rho(19) = k_dec_Xh2 * X(23);
dX(13) = (q_in/V_liq) * (Xxc - X(13)) - rho(1) + sum(rho(13:19));
...
X0(13) = Xxc;
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I just want to update Xxc value in formula of dX(13) depends on time
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if t is 30 just Xxc == 50
But instead of making t >= 10 I just want to apply this patch once.
I can provide other variables depends on formulas if needed.
Thanks

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and 41% for above
For example, if salary_var is 30000 then, taxes are:
(25710-10085) * 0.11 + (30000-25711) * 0.3 = 1718 + 1286 = 3005
My question: how can I efficiently code my « taxes » objective?
Thanks for your help
Seb
This task looks rather strange, there is not much context and some parts of the task might touch some not-so-nice areas of finite-domain based solvers (large domains or scaling / divisions during solving).
Therefore: consider this as an idea / template!
Code
from ortools.sat.python import cp_model
# Data
INPUT = 30000
INPUT_UB = 1000000
TAX_A = 11
TAX_B = 30
TAX_C = 41
# Helpers
# new variable which is constrained to be equal to: given input-var MINUS constant
# can get negative / wrap-around
def aux_var_offset(model, var, offset):
aux_var = model.NewIntVar(-INPUT_UB, INPUT_UB, "")
model.Add(aux_var == var - offset)
return aux_var
# new variable which is equal to the given input-var IFF >= 0; else 0
def aux_var_nonnegative(model, var):
aux_var = model.NewIntVar(0, INPUT_UB, "")
model.AddMaxEquality(aux_var, [var, model.NewConstant(0)])
return aux_var
# Model
model = cp_model.CpModel()
# vars
salary_var = model.NewIntVar(0, INPUT_UB, "salary")
tax_component_a = model.NewIntVar(0, INPUT_UB, "tax_11")
tax_component_b = model.NewIntVar(0, INPUT_UB, "tax_30")
tax_component_c = model.NewIntVar(0, INPUT_UB, "tax_41")
# constraints
model.AddMinEquality(tax_component_a, [
aux_var_nonnegative(model, aux_var_offset(model, salary_var, 10085)),
model.NewConstant(25710 - 10085)])
model.AddMinEquality(tax_component_b, [
aux_var_nonnegative(model, aux_var_offset(model, salary_var, 25711)),
model.NewConstant(73516 - 25711)])
model.Add(tax_component_c == aux_var_nonnegative(model,
aux_var_offset(model, salary_var, 73516)))
tax_full_scaled = tax_component_a * TAX_A + tax_component_b * TAX_B + tax_component_c * TAX_C
# Demo
model.Add(salary_var == INPUT)
solver = cp_model.CpSolver()
status = solver.Solve(model)
print(list(map(lambda x: solver.Value(x), [tax_component_a, tax_component_b, tax_component_c, tax_full_scaled])))
Output
[15625, 4289, 0, 300545]
Remarks
As implemented:
uses scaled solving
produces scaled solution (300545)
no fiddling with non-integral / ratio / rounding stuff BUT large domains
Alternative:
Maybe something around AddDivisionEquality
Edit in regards to Laurents comments
In some scenarios, solving the scaled problem but being able to reason about the real unscaled values easier might make sense.
If i interpret the comment correctly, the following would be a demo (which i was not aware of and it's cool!):
Updated Demo Code (partial)
# Demo -> Attempt of demonstrating the objective-scaling suggestion
model.Add(salary_var >= 30000)
model.Add(salary_var <= 40000)
model.Minimize(salary_var)
model.Proto().objective.scaling_factor = 0.001 # DEFINE INVERSE SCALING
solver = cp_model.CpSolver()
solver.parameters.log_search_progress = True # SCALED BACK OBJECTIVE PROGRESS
status = solver.Solve(model)
print(list(map(lambda x: solver.Value(x), [tax_component_a, tax_component_b, tax_component_c, tax_full_scaled])))
print(solver.ObjectiveValue()) # SCALED BACK OBJECTIVE
Output (excerpt)
...
...
#1 0.00s best:30 next:[30,29.999] fixed_bools:0/1
#Done 0.00s
CpSolverResponse summary:
status: OPTIMAL
objective: 30
best_bound: 30
booleans: 1
conflicts: 0
branches: 1
propagations: 0
integer_propagations: 2
restarts: 1
lp_iterations: 0
walltime: 0.0039022
usertime: 0.0039023
deterministic_time: 8e-08
primal_integral: 1.91832e-07
[15625, 4289, 0, 300545]
30.0

how to reduce wait time in perl

I have script that queries a database a variable number of times per second.
For example, to achieve 36,000 queries per hour we input 600 queries per minute into our script. 600 x 60 = 36,000
This is the output we get you can see the delay between each query
{1} [2019-11-06 21:38:01.313]
{1} [2019-11-06 21:38:01.413]
{1} [2019-11-06 21:38:01.513]
{1} [2019-11-06 21:38:01.613]
{1} [2019-11-06 21:38:01.713]
{1} [2019-11-06 21:38:01.813]
My problem is we are missing out on that 0.0100 because we have a wait time inplace.
rates per minute = varies we can change this to a max of 960 queries per min but we would want fourmla that is flexible for 0-960.
my $wait_time = (1 / $rpm) * 60 * 1(connection); (max of 4 connections) wait time increases based on number of connections
Does anyone know how to reduce the wait time in between queries ?
thanks
This is the code line
my $wait_time = (1 / $rpm) * 60 * 1;
So when i enter in 600 queries per min
This code line calcuates the wait time based on number of connection we have
my $wait_time = (1 / 600) * 60 * 1;
1/600 * 60 * 1 = WAIT: 0.1
Well, your processing of the query needs time. A fragile solution, if i interpret your problem right, is to measure the time the current processing takes and substract that from the next sleep time. Of course that would break if the processing time equals or exceeds the sleep time.
A clean solution would be to have a dedicated main loop that does nothing but sleeping and firing off queries in separate threads.
I'm not sure if this will help because I have a very hard time understanding your question. I think you are concerned that you aren't making queries at the rate you desire.
It could be because you think of the wait time as being static. It's not the wait time that's static —that's dependent on how long the previous query took— it's the interval that's static.
use Time::HiRes qw( time sleep ); # Add support for fractional times.
my $interval = (1 / $qpm) * 60 * 1; # In (fractional) seconds.
my $next_run = time;
while (1) {
my $wait = $next_run - time;
sleep($wait) if $wait > 0;
$next_run += $interval;
... do work ...
}

Clarification between Epoch and iteration

This answer points to the difference between an Epoch and an iteration while training a neural network. However, when I look at the source code for the solver API in the Stanford CS231n course (and I'm assuming this is the case for most libraries out there as well), during each iteration, batch_size number of examples are randomly selected with replacement. Thus, there is no guarantee that all examples would been seen during each epoch is there?
Does an epoch then mean that all examples would be seen in expectation? Or am I understanding this wrong?
Relevant Source Code:
def _step(self):
"""
Make a single gradient update. This is called by train() and should not
be called manually.
"""
# Make a minibatch of training data
num_train = self.X_train.shape[0]
batch_mask = np.random.choice(num_train, self.batch_size)
X_batch = self.X_train[batch_mask]
y_batch = self.y_train[batch_mask]
# Compute loss and gradient
loss, grads = self.model.loss(X_batch, y_batch)
self.loss_history.append(loss)
# Perform a parameter update
for p, w in self.model.params.iteritems():
dw = grads[p]
config = self.optim_configs[p]
next_w, next_config = self.update_rule(w, dw, config)
self.model.params[p] = next_w
self.optim_configs[p] = next_config
def train(self):
"""
Run optimization to train the model.
"""
num_train = self.X_train.shape[0]
iterations_per_epoch = max(num_train / self.batch_size, 1)
num_iterations = self.num_epochs * iterations_per_epoch
for t in xrange(num_iterations):
self._step()
# Maybe print training loss
if self.verbose and t % self.print_every == 0:
print '(Iteration %d / %d) loss: %f' % (
t + 1, num_iterations, self.loss_history[-1])
# At the end of every epoch, increment the epoch counter and decay the
# learning rate.
epoch_end = (t + 1) % iterations_per_epoch == 0
if epoch_end:
self.epoch += 1
for k in self.optim_configs:
self.optim_configs[k]['learning_rate'] *= self.lr_decay
# Check train and val accuracy on the first iteration, the last
# iteration, and at the end of each epoch.
first_it = (t == 0)
last_it = (t == num_iterations + 1)
if first_it or last_it or epoch_end:
train_acc = self.check_accuracy(self.X_train, self.y_train,
num_samples=1000)
val_acc = self.check_accuracy(self.X_val, self.y_val)
self.train_acc_history.append(train_acc)
self.val_acc_history.append(val_acc)
if self.verbose:
print '(Epoch %d / %d) train acc: %f; val_acc: %f' % (
self.epoch, self.num_epochs, train_acc, val_acc)
# Keep track of the best model
if val_acc > self.best_val_acc:
self.best_val_acc = val_acc
self.best_params = {}
for k, v in self.model.params.iteritems():
self.best_params[k] = v.copy()
# At the end of training swap the best params into the model
self.model.params = self.best_params
Thanks.
I believe, as you say, that in the Stanford course they are effectively using "epoch" with the less strict meaning of "expected number of times each example is seen during training". However, in my experience, most implementations consider an epoch as running through every example in the training set once, and I'd say they only chose the sampling with replacement for simplicity. If you have a good amount of data, chances are that you will not see a difference, but still, it is more correct to sample without replacement until there are no more examples.
You can check, for example, how Keras does the training in its source code; it's a bit complicated, but the important point is that make_batches is called to split the (possibly shuffled) examples into batches, which matches your initial idea of "epoch".