[R] fitting a hyperbola to data points

[Manoj Srivastava]

On Mon, Apr 08 2013, PIKAL Petr wrote:

        Thanks for responding.

> without data we can provide just basic help.
> fit<-lm(Time~I(1/Requests))
> shall give you hyperbolic fit.

> You can test if your data follow this assumption by
> plot(1/Requests, Time)
> which shall for straight line.
>
> anyway, when you want to provide data use
> dput(your.data) and copy console output directly to your mail.

        Pardon. I had put my data on pastebin, but here is the dput:
> dput(dat)
structure(list(Requests = c(0.045364248295124, 0.11341062073781, 
0.16633557708212, 0.20413911732806, 0.26462478172156, 0.31754973806587, 
0.37047469441018, 0.42339965075449, 0.47632460709881, 0.52168885539393, 
0.58217451978743, 0.62753876808255, 0.68802443247605, 0.73338868077118, 
0.79387434516468, 0.8392385934598, 0.8997242578533, 0.94508850614843, 
0.99801346249274, 1.050938418837, 1.1038633751814, 1.1567883315257, 
1.20971328787, 1.2626382442143, 1.3155632005586, 1.3684881569029, 
1.4214131132472, 1.4743380695915, 1.5272630259359, 1.5801879822802, 
1.6255522305753, 1.6860378949688, 1.7389628513131, 1.7843270996082, 
1.8448127640017, 1.897737720346, 1.9431019686412, 1.9960269249855, 
2.056512589379, 2.1018768376741, 2.1548017940184, 2.2152874584119, 
2.260651706707, 2.3135766630513, 2.3665016193957, 2.41942657574, 
2.4723515320843, 2.5252764884286, 2.5782014447729, 2.6311264011172, 
2.6764906494123, 2.7369763138058, 2.7899012701502, 2.8428262264945, 
2.8881904747896, 2.9486761391831, 2.9940403874782, 3.1603759645603, 
3.2057402128555, 3.266225877249, 3.3115901255441, 3.3645150818884, 
3.4174400382327, 3.4779257026262, 3.5232899509213, 3.5762149072656, 
3.62913986361, 3.6820648199543, 3.7349897762986, 3.7879147326429, 
3.8408396889872, 3.8937646453315, 3.9391288936266, 3.9996145580201, 
4.0525395143644, 4.0979037626596, 4.1583894270531, 4.2113143833974, 
4.2566786316925, 4.317164296086, 4.3625285443811, 4.4230142087746, 
4.4683784570698, 4.5213034134141, 4.5817890778076, 4.7330032387913, 
4.7934889031848, 4.8388531514799, 4.9976280205129, 5.0505529768572, 
5.1034779332015, 5.1564028895458, 5.2017671378409, 5.2622528022344
), Time = c(1289L, 1262L, 1272L, 1222L, 1243L, 1259L, 1266L, 
1242L, 1249L, 1232L, 1303L, 1236L, 1251L, 1263L, 1234L, 1226L, 
1232L, 1246L, 1249L, 1272L, 1263L, 1247L, 1253L, 1257L, 1267L, 
1262L, 1284L, 1266L, 1269L, 1268L, 1273L, 1261L, 1261L, 1264L, 
1276L, 1283L, 1275L, 1277L, 1293L, 1285L, 1284L, 1289L, 1290L, 
1289L, 1300L, 1292L, 1300L, 1291L, 1297L, 1310L, 1303L, 1311L, 
1317L, 1345L, 1315L, 1327L, 1322L, 1597L, 1623L, 1510L, 1372L, 
1429L, 1371L, 1365L, 1357L, 1366L, 1402L, 1370L, 1407L, 1373L, 
1399L, 1420L, 1405L, 1393L, 1428L, 1394L, 1422L, 1425L, 1457L, 
1501L, 1426L, 1539L, 1492L, 1476L, 1493L, 1580L, 1670L, 1556L, 
1661L, 1593L, 3618L, 1903L, 1941L, 2908L)), .Names = c("Requests", 
"Time"), class = "data.frame", row.names = c(NA, -94L))


        When plotting with  plot(1/Requests, Time), I still get a
 parabolic line, not a linear one. Perhaps there is a data transform I
 can do to get this into a linear mode?

        I have also plotted the data (Time vs Requests and 1/Requests)
 on a log and log-log scale, but the curve remains stubbornly
 curved. This seems to argue against an exponential relationship, does
 it not?


        thanks again.

        manoj


>>         I am new to R, and I suspect I am missing something simple.

>>         I have a data set that  performance data that correlates
>> request  rate to response times
>>         http://pastebin.com/Xhg0RaUp
>>  There is some jitter in the data, but mostly it looks like a hockey
>> puck curve. It does not get converted into a straight line when I tried
>> log conversions, so it does not seem to be a power series relationship.

>>         My expectation is that the data will fit a curve that is a
>> hyperbola, but I don't know how to formulate that regression. How does
>> one fit data to a general function
>>    AX^2 + Bxy + Cy^2 +D = 0

>>         I have tried polynomial functions and inverse functions
>>  lm2 = lm(Time ~ Requests + I(Requests^2) +  I(Requests^3))  but that
>> does not appear to be close.

>>         Any pointers appreciated.

-- 
Many of us spend half our life wishing for things we could have if we
didn't spend half our time wishing.  -- Alexander Woollcott
Manoj Srivastava <srivasta at acm.org> <http://www.golden-gryphon.com/>
4096R/C5779A1C E37E 5EC5 2A01 DA25 AD20  05B6 CF48 9438 C577 9A1C
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