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#1 Re: Help Me ! » What is the Next #?? » 2010-01-13 02:42:11
oops.....
therefore next number will be
n=7 .........7^(7-3) ---> 7^4 =2401
j
#2 Re: Help Me ! » What is the Next #?? » 2010-01-13 02:41:00
Hello!
n^(n-3)
So for n=1 it'll be ...... 1^(1-3) --> 1^(-2) = 1
n=2..................2^(2-3) --> 2^(-1) = 0.5
n=3..................3^(3-3) --> 3^0 = 1
n=4...................4^(4-3) --> 4^1 = 4
n=5..................5^(5-3) --> 5^2 = 25
n=6..................6^(6-3) --> 6^3 = 216
J
#3 Help Me ! » how do you solve this excel problem??? » 2010-01-13 02:34:10
- cadjeff
- Replies: 0
Hello mathematicians,
Does anyone know how to solve a problem like this on excel?
There are six gambling departments, each is to be housed in one of six locations
P, A, M, C, F, T.
The distances between each house is as follows:
P A M C F T
P 10.6 7.1 8.6 9.4 8.1
A 10.6 7 2.4 1.9 2.6
M 7.1 7 4.3 4.3 5.3
C 8.6 2.4 4.3 6.7 2.5
F 9.4 1.9 4.3 6.7 2.1
T 8.1 2.6 5.3 2.5 2.1
The volumes of paperwork flowing each week from each department to every other are as follows. The unit of measurement is the number of departmental vans the documents would fill.
HORSES DOGS SLOT MAC LOTTERY CASINO INT. BETT
HORSES 22 16 41 12 18
DOGS 3 8 7 1 5
SLOT MAC 4 7 5 5 20
LOTTERY 26 17 18 16 17
CASINO 18 15 21 8 15
INT. BETT 4 13 5 14 19
Find where each department should be allocated whilst minimising cost (i.e. minimising distance)
Would really appreciate any help. Thanks!
J
#4 Help Me ! » LINGO problem. Please help!!! » 2010-01-10 09:11:52
- cadjeff
- Replies: 0
Hello,
i have a question about a problem i'm trying to solve.
I'll explain the problem here and show you what i've come up with after....
Ok so a clay company is considering the mining arrangements at four sites (A, B, C and D)
where it has the option of mining over a three-year period.
here are the data...
A B C D
Reserve 650 950 625 840
(thousands
of tonnes)
Quality of clay 66 64.1 63 62.1
Cost of mining 22 21 17 18
open or not open not open open open not open
(£ per tonne)
opening cost (£) 350,000 - - 300,000
Reserve is the maximum quantity of clay that can be extracted during the three-year period.
Quality of clay is expressed as a single figure. Better quality clay have higher ratings on this scale. It may be assumed that all the clay from a given mine is of the same quality.
Cost of mining is the total cost of mining and processing a single tonne of clay. It does not include the cost of opening new mines.
The company wishes to know which mines to work in each of the next three years in order to minimize costs, subject to the following restrictions.
i) produce 640,000 tonnes of mixed clay in year 1,
produce 850,000 tonnes of mixed clay in year 2,
produce 750,000 tonnes of mixed clay in year 3.
ii) The average quality of the mixed clay mined in any year must not be less than 64.0.
iii) Only two mines can be worked in the first year.
iv) The cost of mining clay will increase by £2 per tonne in year 2 and another £1 per tonne in year 3.
v) Clay (either mixed or from a single mine) cannot be stored from one year to the next.
Remember, there's a difference between a mine being 'open' and a mine being 'worked'.
Ok, here's what i've got so far....i'm sure there are mistakes in there and there are probably constraints or something missing.
Two things that are not included (because my attempts have been wrong) in my code are....
the constraint for i) the yearly production requirement mentioned above.
and also the constraint for ii) the average quality clay being at least 64 for every year.
Please help, i've tried many times, and now this is about as far as i can get. I appreciate your time, thank you.
MODEL:
! clay company problem;
SETS:
! I;
MINES/A1,B1,C1,D1/:CAP,QUAL,OPEN,FIXCOST;
! J;
YEAR/Y1,Y2,Y3/:DEMAND;
! I,J;
ALLOC(MINES,YEAR):COST,VOLUME,USED;
ENDSETS
@FOR(ALLOC(I,J):@SUM(YEAR(J):VOLUME(I,J)-DUMMY*USED(I,J))<=0);
@FOR(YEAR(J):@SUM(ALLOC(I,J):USED(I,J))=1);
! ONLY 2 MINES CAN BE WORKED ON IN YEAR 1;
@FOR(ALLOC(I,1):@SUM(YEAR(J):USED(I,J))<=2);
! THE OBJECTIVE;
MIN=@SUM(ALLOC(I,J):COST(I,J)*VOLUME(I,J))+@SUM(MINES(I):OPEN(I)*FIXCOST(I));
! DEMAND CONSTRAINTS;
@FOR(YEAR(J):@SUM(MINES(I):VOLUME(I,J))=DEMAND(J));
! SUPPLY CONSTRAINT;
@FOR(MINES(I):@SUM(YEAR(J):VOLUME(I,J))<=(CAP(I)*OPEN(I)));
! TIGHTENING CONSTRAINTS;
@FOR(ALLOC(I,J):VOLUME(I,J)<=DEMAND(J)*OPEN(I));
! MAKE THE OPEN AND USED VARIABLES BINARY;
@FOR(MINES(I):@BIN(OPEN(I)));
@FOR(ALLOC(I,J):@BIN(USED(I,J)));
DATA:
! CAPACITY OF EACH MINE;
CAP=650, 950, 625, 840;
! QUALITY OF CLAY FOR EACH MINE;
QUAL=66, 64.1, 63, 62.1;
! DEMAND FOR EACH YEAR;
DEMAND=640, 850, 750;
! FIXED COST OF OPENING MINES;
FIXCOST=350, 0, 0, 300;
! COST OF MINING CLAY FOR EACH SITE AND FOR EACH YEAR;
COST=22,24,25,
21,23,24,
17,19,20,
18,20,21;
DUMMY=10000;
ENDDATA
END
#5 Help Me ! » Cardinality - DOMINATION! » 2009-12-07 04:54:22
- cadjeff
- Replies: 1
Hi everybody,
I've been working my way through some cardinality questions (and constructions of number systems), the peano axiom and cauchy sequence questions have been ok but i'm stuck on this domination one now.....
i) Let A and B be sets such that A is dominated by B. Show that P(A) is dominated by P(B).
ii) Hence (or otherwise) show that if A is equivalent to B then P(A) is equivalent to P(B).
Would appreciate any help,
thanks!!
cadular
#6 Help Me ! » Recurrence relations problem, please help! » 2009-10-15 11:29:30
- cadjeff
- Replies: 1
Hello all,
i have a problem which is proving difficult (for me anyway!).
This is the question....
suppose a country is divided into N parliamentary constituencies C1,C2,...,Cn
The size of the electorate in C1 is ei (i=1,2,...N)
The constituency Ci is to be represented in parliament by mi members, where mi>1 for all i and
the Sum of mi = M > N. (But i think that's supposed to be M=>N)
Let ri = Mei/E, where the Sum of ei = E
The error in the representation of the constituency Ci is defined as zi = ABS (mi - ri), and the allocation of seats to constituencies is to be done so as to minimise max (z1,z2,...zN)
Let Gn(m) be defined, for n=1,2,...N, as the minimum possible value of max (z1,z2,...,zn) is the resricted problem in which m members are allocated to C1,C2,...,Cn.
A recurrence relation is obtained for Gn(m) and dynamic programming is used to solve the problem.
Ok so that's the problem.
I literally did a couple of weeks on recurrence relations about two years ago and the lecture notes are awful. Not an inspiring lecturer as i remember.
Anyway i don't have a clue what i'm doing, i don't know what sort of form i should be writing it in.....
should it be written in this kind of form for example.....
Xn+1 = KXn+1 + c X0 = a
X1 = KX0 +c = Ka + c
X2 = KX1 + c =K(Ka + c)c = (K^2)a + Kc + c
etc......
Here's a little made up example that i was hoping would help make the recurrence relation:
There are 8 seats available.
C1 has an electorate of 8500
C2 " " 13400
C3 " " 4500
C4 " " 13600
Therefore E = 40,000
And so their Representation ri = Mei/E
C1 has ri = Mei/E = 8*8500/40000 = 1.7
C2 has ri = 2.68
C3 has ri = 0.9
C4 has ri = 2.72
Obviously you must have whole numbers of people representing as you can't have half a person so i chose to round the numbers to their nearest integer so i got:
C1 = 2 C2 = 3 C3 = 1 C4 = 3 but this gives a total of 9 seats and there are only 8 available so i've taken the Ci with the least error and rounded it down, in this case it's C3 with the smallest error of 0.68.
BUT.....how on earth do you start putting something like this into a recurrence relation???
Any help would be much appreciated,
thanks for reading.
#7 Help Me ! » Could someone clarify this please? (on alternative hypothesis') » 2009-02-05 05:00:13
- cadjeff
- Replies: 0
Good day all,
- firstly, i have TWO INDEPENDENT samples (174 and 348) of peoples ages,
- i've checked for normality, and as they're not normal i'm carrying out a Mann-whitney test,
- When carrying out this test on the computer, it asks for the Alternative hypothesis and it lists three options: less than, not equal, or greater than.
RIGHT! Here's what i need to know.....
Which Alternative hypothesis do i use?? I always just thought it was 'equal' for the Null Hyp. and 'not equal' for the Alternative Hyp.
When i looked in my textbook (Devore - Probability and Statistics), there was an example on this and so i tried it on my computer testing each possible alternative and it turned out that they used the 'greater than' hypothesis!!!!
Can someone please tell me why!!
Thank you thank you thank you,
Cee
#8 Re: Help Me ! » Simple statistcs problem that's annoying me immensely. » 2009-02-04 01:38:09
I've realised my test stat (0.9939) is twice your test stat (0.496..).
Mine assumes a 2 tailed test. Is this then infact a one tailed test then? ...and why?
thank you
#9 Re: Help Me ! » Simple statistcs problem that's annoying me immensely. » 2009-02-04 01:28:25
Sorry for the delayed reply.
I'll show you the output i got from inputting those two columns into the mann-whitney test on two stats packages, MINITAB and SPSS. Can you possibly explain what these results mean because i have to say half the numbers they come out with mean nothing to me! oh dear, i know.
From the MINITAB package:
Mann-Whitney Test and CI: age_case, age_Controls
N Median
age_case 174 57.615
age_Controls 348 57.619
Point estimate for ETA1-ETA2 is 0.003
95.0 Percent CI for ETA1-ETA2 is (-0.772,0.775)
W = 45514.0
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.9939
The test is significant at 0.9939 (adjusted for ties)
From the SPSS package:
Descriptive Statistics
N Mean Std. Deviation Minimum Maximum
Age of respondant 522 58.0220 4.24643 46.29 67.44
Case or Control 522 .33 .472 0 1
MANN WHITNEY TEST
Ranks
Case or Control N Mean Rank Sum of Ranks
Age of respondant 0 348 261.46 90989.00
1 174 261.57 45514.00
Total 522
Test Statisticsa
Age of respondant
Mann-Whitney U 30263.000
Wilcoxon W 90989.000
Z -.008
Asymp. Sig. (2-tailed) .994
a. Grouping Variable: Case or Control
#10 Help Me ! » Simple statistcs problem that's annoying me immensely. » 2009-01-15 02:43:43
- cadjeff
- Replies: 3
Hello there,
Can someone please help because this is reeeeeally annoying me! (thankyou)
All Im wanting to do is carry out a Mann-Whitney test on some data. I have a sample of 522. 348 dont have the disease, 174 do have the disease. I have ages for all of them. Now Ive made a probability plot, and got p<0.005 so from concluding that its not normal Im going to carry out a Mann Whitney test (does that sound right to you??) anyway I have a bit of paper on which my lecturer had scribbled down their result getting p=0.197 therefore cannot reject Ho because p>0.05 BUT Im getting p=0.994!!!!
Can somebody please tell me what their results give them? and where Im maybe going wrong. This is so simple and its really frustrating because I cant proceed until I know whats wrong.
As you can see i've rather irritatingly copied and pasted all the data because i don't know how to upload a (non image) attachment.
SORRY!!!
the first column of numbers are ages, the column beneath represents the people with (1) and without (0) the disease. Just copy and paste them alongside eachother in whatever stats programme you use.
thank you!!!!!!!!!
J
57.72
56.5613
57.9138
53.4045
60.4627
56.9774
53.4155
64.9637
46.6749
52.7173
46.2861
55.6824
52.9309
62.9678
60.3121
51.9507
56.2382
51.6632
59.0062
57.3251
56.3422
54.976
62.6174
62.8665
48.783
49.9165
49.9028
54.0753
54.5243
65.2622
57.3333
65.1718
65.1417
61.5524
51.4908
51.7536
56.835
65.9932
59.1841
51.6961
55.3867
58.5736
52.3039
56.5339
61.8617
57.1088
55.3402
50.3929
61.1225
64.4928
62.7461
56.3915
53.5305
56.2793
67.436
55.2142
61.9247
51.8686
61.5633
56.3395
65.0404
58.0397
60.4271
55.6413
60.6899
66.0835
56.627
59.4524
57.0486
65.1362
55.7235
58.883
52.3231
64.2847
65.3744
55.384
50.7543
59.7153
54.987
54.6092
61.013
55.3292
61.7057
59.1239
65.3196
55.0363
58.601
66.7707
62.36
59.6687
61.1006
57.6673
63.1485
58.3381
66.6448
58.3546
59.1184
51.4853
58.2094
60.5914
55.2197
56.7912
61.859
63.2197
62.3354
62.5654
63.1266
53.5825
51.4935
57.3498
64.4983
62.5352
57.8179
57.4565
57.4648
62.36
61.4209
55.8604
59.4004
55.1485
58.642
53.8015
55.7865
56.3504
55.4141
58.1574
58.0999
62.3901
61.4155
65.5743
58.141
54.6229
64.6872
64.6927
57.6016
56.0055
51.3922
58.7379
63.3977
56.8597
66.0862
54.0835
66.2998
62.2642
63.5264
60.0794
60.282
55.1431
57.7878
60.8433
61.1964
48.0301
65.0787
57.0732
61.629
56.3231
60.2218
56.9993
59.1485
54.5517
57.0732
54.872
57.4429
58.5572
60.4928
55.9042
56.6023
56.2738
61.9192
57.6865
61.7933
54.7515
60.961
56.1807
54.7844
60.7228
64.4545
52.8405
61.0732
51.8385
53.1253
57.8097
57.2786
57.3005
60.7502
62.5161
56.3039
50.8036
58.204
64.115
57.1307
61.5551
61.1088
63.9535
62.1684
56.8624
67.1622
63.6413
64.3094
60.3614
60.3696
61.2375
57.5852
51.8357
58.5024
58.5708
52.0657
58.4066
59.447
53.7741
64.5503
65.5934
54.3025
54.1793
51.8111
53.41
55.5455
55.5455
57.4127
51.5756
65.588
60.742
61.4921
66.1793
61.9028
57.076
59.5154
52.9281
60.0575
65.473
51.6249
56.4928
61.6263
61.629
51.5647
54.2971
55.9288
49.4866
65.2211
61.9028
63.1595
60.7091
55.7043
61.1472
61.0568
58.976
52.0411
55.488
56.7885
61.1335
57.6181
56.7146
57.9384
56.0411
64.7228
51.9617
65.4949
54.7242
66.7378
59.7098
57.9849
61.7906
56.6516
61.8481
60.5886
55.8576
51.4689
56.4873
52.4873
54.3409
59.0171
62.8528
57.0267
63.5099
60.8871
64.4189
55.2006
59.7262
61.0568
52.2245
62.0726
54.9213
52.1396
60.282
51.9069
53.8727
58.0342
57.1253
55.6468
59.5921
56.6051
52.2026
56.0274
52.2108
62.6092
62.9295
52.7995
50.6201
63.2772
62.3518
56.282
60.2327
61.7933
60.4956
58.1164
60.9665
63.9726
54.8747
54.4969
54.7406
62.5298
60.7693
53.3251
64.7584
61.6509
54.6886
62.7296
61.5414
54.1684
53.3908
60.6188
60.7858
56.3203
57.859
51.8795
56.1013
54.8145
54.2724
54.1574
56.8296
52.3943
56.6078
66.3518
61.065
56.553
58.5982
54.6256
54.3628
60.8871
54.4175
58.1437
54.2067
62.4476
54.3628
55.9151
59.8604
54.245
63.6085
58.7077
56.8186
63.4114
57.9001
63.9179
52.8871
61.1307
54.204
52.3559
57.9083
59.3046
55.6879
53.1417
57.6208
55.9617
52.1068
52.9144
60.679
56.9774
56.7447
52.4271
54.7817
51.9836
56.5941
57.2156
57.744
54.7351
51.6085
56.3943
51.54
62.0068
63.7153
60.0876
56.7748
55.0308
54.2368
53.2594
54.9651
52.575
55.0089
51.9644
54.7926
57.0185
57.8946
59.5291
58.7899
56.2163
66.0096
60.7721
55.2662
65.3415
56.8186
52.2437
62.9295
65.3142
54.1273
56.6188
54.3956
57.1061
59.332
63.9808
60.8843
60.7173
64.9418
52.5613
52.4244
65.6427
66.3737
51.7536
57.4237
51.1978
55.4634
57.1827
51.6167
57.4374
56.0192
61.4675
58.2177
61.3799
52.4298
61.2101
64.6407
52.4107
54.486
56.2929
64.3477
51.7153
61.4565
60.1506
54.2368
61.2868
55.5893
57.2704
65.4045
53.8179
56.9719
63.4606
59.1869
62.5927
63.1212
58.4422
54.2313
64.6626
54.6913
57.8809
58.1218
54.256
57.7522
58.0288
54.6667
58.9897
54.6858
63.0226
55.0609
58.0835
58.935
62.3518
60.4709
65.1389
50.0178
59.655
51.7646
58.987
66.6283
61.577
58.5079
54.1164
61.7577
53.9439
61.8645
58.36
57.5442
63.436
55.833
58.59
58.59
56.9774
60.8159
46.475
66.4367
53.5934
57.9028
61.2621
57.0897
61.7029
55.5674
63.9179
55.4031
56.0465
54.0698
61.7084
61.5195
63.3374
51.9452
54.1711
61.0294
53.5633
57.065
56.7721
52.1615
60.3422
58.204
51.8549
55.9233
62.4394
59.7837
55.8877
52.3532
56.6927
54.6913
57.2621
54.7461
55.1513
56.4572
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#11 Help Me ! » number systems, natural, rationals, reals etc... & cardinality » 2008-12-02 10:59:23
- cadjeff
- Replies: 3
cool ok, i'll do what i can first and then get back to you.
j
#12 Help Me ! » expansions and approximations » 2008-04-30 22:28:22
- cadjeff
- Replies: 1
my knowledge on this subject is limited, for anyone who is into this, i'd very much appreciate any pointers/explanation on this.
thank you,
cad
#13 Help Me ! » Argh! can anybody help?! Problem solving ... VBA visual basic, macros » 2008-03-29 13:54:39
- cadjeff
- Replies: 2
Hello all,
before you read on, if you can't help or you don't have time...if you know of any sources where i may be able to find help that would be greatly appreciated.
i'm supposed to be doing this problem using Visual Basic, macros etc... except we were given two pathetic lectures on using it. I've never used it before and we've been taught virtually no suitable language for it. I've been looking on the internet and at the library for information, but it's just not enough and i'm at a complete stand still.
I know what i want visual basic to do roughly (i think) and i;ve written it after the question....anyway here's the question itself....
A brewery has 100 barrels of beer. It's been said that a small number of the barrels have been purposely contaminated with a deadly bacteria.
Unfortunately the owner is going to trick their workers into sampling the beer (from one or more barrels). For every worker who takes a sample/s the owner will pay £100 to the family of that worker.
A testing regime needs to be put in place that will minimise the cost whilst taking in account the collateral damage. An estimate of the cost (in money and lives) is required. Confidence intervals are needed for the estimates. Include the VBA programme you develop.
Because of the hugh lack of time that's been devoted to teaching us Visual basic and how it works.....i'm basically clueless.
But from what i've seen on the internet and in books, it looks to me that i'll porbably need to use IF statements and LOOPS most of the time.
This is my own untuitive outlook on the problem.
i = number of barrels
p = number of poisoned barrels (i don't whether or not to leave this as p or make up a value, or does it not matter?)
n = number of workers used
Probability that a worker will sample a poisoned barrel would be p/i
So i thinking i need to make some sort of command (but in visual basic language) like...
Sub barrel()
i = 100
p = 10 (for example)
n = 1
First sample taken
If poisonous Then
i = i - 1
p = p - 1
n = n + 1
If not poisonous Then
i = i - 1
LOOP
.....i don't know if i'm anyway near going in the right direction, the commands obviously need translating to VBA language. I'm not altogether sure how to work between excel and VBA either.
I just really need help if anyone could be of assistance i would be very appreciative.....THANK YOU!!!
cadd
#14 Help Me ! » HEEEELP!!!....please! - Laplace transforms » 2008-02-25 03:47:37
- cadjeff
- Replies: 0
I'm working through through a sheet, but i'm finding it very difficult to get started.
I'd appreciate any help to get the ball rolling for me.
Denote the Laplace transforms of y(t) and z(t) by Y(s) and Z(s) respectively. By taking the Laplace transform of Equation 1, determine the transfer function H(s) such that
Y(s) = H(s) Z(s)
(Equation 1 : Mÿ = - k (y - z) - c (ý - ź) The dots above the letters indicates differentiation with
respect to time t.
Thanks for looking,
cad
#15 Re: Help Me ! » eigenvalues etc... » 2008-01-31 01:37:11
also....
Consider the transformation R^5 --> R^5 defined by
8x1 + 6x2 + 4x3 + 6x4 + 2x5
8x1 + 11x2 +5x3 + 7x4 + 2x5
T(x)= -16x1 - 17x2 - 9x3 - 13x4 - 4x5
-18x1 - 19x2 - 11x3 - 17x4 - 6x5
34x1 + 36x2 + 20x3 + 30x4 +10x5
a) Find a matrix A such T(x) = Ax for all x in R^5.
b) Use an appropriate computer package to find all the eigenvalues and eigenvectors of A.
(Mathcad???)
c) Using the answer to part b), find the range, kernel, rank and nullity of T.
Thanks,
cads
#16 Help Me ! » eigenvalues etc... » 2008-01-31 01:09:14
- cadjeff
- Replies: 1
?????????
1.a) Let A be any non-singular real square matrix. Show that A^T A (A transform * A) is symmetric and positive definite.
b) Let A= the matrix 14 7 7 . Find the eigenvalues of A. (*show working. *no use of computer packages)
7 6 1
7 1 6
c) Using the answer to part b), or otherwise, show that the matrix B=A + pI where p is a positive constant and I the usual identity matrix, is both symmetric and positive definite.
Any help would be greatly appreciated.
caddy
#17 Help Me ! » The group S4 - group theory » 2007-12-13 11:08:30
- cadjeff
- Replies: 1
hello,
even after attending all my lessons in this subject, i'm still useless at it.
Some help on the following would be much appreciated.
1. Let F be a non square rectangle with the four corners labelled 1-4.
Identify the subgroup H1 of S4 which corresponds to S(F).
2.By finding appropriate plane figures, show that each of the following is a subgroup of S4.
a) H3=( e, (1234), (13)(24), (1432) )
b) H4=( e, (123), (132), (12), (13), (23) )
3. What are possible orders of subgroups of S4? ...explain.
4. Says to list each possible cycle structure of an element of S4. For each cycle structure, determine how many elements of S4 have the structure and state the order of each such element.
5. Write down all the cyclic subgroups of S4.
6. List all the subgroups of S4 which are isomorphic to K4.
Explain why your list is complete; that is, explain why there are no other subgroups of S4 isomorphic to K4.
7. Explain why S4 does not have cyclic subgroup of order 6.
Find all four subgroups of S4 of order 6.
Hint: Question 2 may help here.
You may assume that S4 has four subgroups of order 6.
There's more but i think this is more than enough to put up for now.
Any pokes at some of these questions would be a great help.
Thanks for your time,
caddy
#18 Re: Help Me ! » Population growth and effects of hunting » 2007-05-17 07:58:07
How do you go from: Co/e = 2k1P + Co
To: Co = 2ek1 p / 1 e ?
I cant seem to rearrange so as to get that answer.
Co/e = Coe^-2a , a = ½ , how do you know what a is?
Therefore Co = Ce1/2t How do get to this from the previous equation?
Hopefully if thats cleared the combined (1) and (2) equations will be clear to me.
Next post:
How do you get 1/K2 .dC/dt = K1/K2.P ?
Sorry for the mass of questions.
As you may have already noticed (haha!) but Ive been finding differentials difficult.
I covered 1st order/ 2nd order / Homogeneous differentials a month or so ago and I've struggled a lot with them.
Cheers,
caddy
#19 Re: Help Me ! » Population growth and effects of hunting » 2007-05-16 11:37:38
Ok, i see. It is supposed to be a sin wave according to my teacher. I'm trawling through 'Advanced engineering mathematics' by Kreyszig to see if any new equations/methods come to light...not going to well at the moment, haha!
I'm looking at this other part of the task now but i'm not sure where to start.
This is the basics of it anyway...Waste is being dumped by a factory into a river and it's contaminating the lake that the buffalo drink from.
1) The rate of change in the concentration, C, of the pollutant in the lake is directly proportional to the amount P of pollutant present at any time. The proportionality constant is K1.
2) When the amount of pollutant in the lake is constant the concentration appears to decay exponentially with time, from its starting value to 1/e of its starting value in 2 days.
However, if there is a change in the amount of pollutant in the lake, the concentration seems to become directly proportional to the rate of change in the amount of pollutant. The proportionality constant is K2. In other words, the concentration of pollutant in the lake seems to be affected by two phenomena independently ( a natural decay and any new waste arriving through the river).
I'm to provide a formula for the amount of pollutant in the lake as a function of time and the various proportionality constants above.
Bed time me tinks!
cheers,
caddy
#20 Re: Help Me ! » Population growth and effects of hunting » 2007-05-16 10:23:26
By assumption do you mean you made this equation up from nothing? If it doesn't come from some known formula or rule how would you know if it was right?
cheers,
caddy
#21 Re: Help Me ! » Population growth and effects of hunting » 2007-05-10 23:16:13
Sorry, i've been looking at what you started with,
H = Hmax - (T - 6)^2
but i'm still a little puzzled, it looks right but where does this equation come from? or what stages did you go through to get it?
thank you,
caddy
#22 Re: Help Me ! » Population growth and effects of hunting » 2007-05-09 07:25:06
Thank you so much, that was a great help.
Could you briefly explain how you got the second to last part equation that says:
'If we express T in terms of H:' just above it?
thank you very very much!
caddy
#23 Help Me ! » Population growth and effects of hunting » 2007-05-08 11:48:00
- cadjeff
- Replies: 12
Hello,
i'm having trouble with the following task.
Basically there's an island inhabited with buffalo.
For the last 12 months they've been the target of hunters.
The present population is 3920 heads.
The population 12 months ago was 3790.
The population 10 years ago was 580.
The number of hunters during the last 12 months is expected to have varied continuously but smoothly with the time of year, being the minimum at around the new year and peaking towards the end of june. At the beginning of january/end of december, six 'hard core' hunters were known to be operating on the island, but it is not known how many were around at the height of the season.
1) Is there any evidence to suggest that hunters managed to kill significant numbers (>10) of buffalo during the last twelve months?
2) Roughly how many hunters were active in mid-summer?
THIS DATA is also supplied:
The rate of change in the size of a herd of bison constrained to the area of a particular nature reserve, during the 3 month hunting seasons every year in the period 1990-2005, was invariably found to be directly proportional to the number of hunters. During the 2001 season, for example, the population decreased from 4734 to 4531 in the space of 3 months with 87 hunters constantly in attendance.
Ok, i got the population growth formula from dP/dt= KP
i ended up with N = No e^rt
(where N = present pop. No = initial pop. r= growth rate t=length of time)
So by substituting the 10 year and 12 month old populations and t=9 (time between the different populations) i obtained the 'r' (rate of growth without hunters)
I then used 'r' to find what the present population would be had there been no hunters.
I got the approx population of 4670 heads.
Clearly this suggested that there had been significant numbers of buffalo killed by hunters.
So thats part 1 done really.
Part 2: I've been told the following:
- i need to model the number of hunters around at any time (does not require differential equations at all)
- I need to consider how the population might vary as a function of the number of hunters present. (the extra data could be of use here)
I've managed to work out that the varying number of hunters can be represented by a sin wave (or cos) and that the minimum point represents the hunter no. at New year and the maximum point represents the no. of hunters at the end of june.
ASSUMPTION: The minimum and maximum no. of hunters are 6 months apart.
This is as far as i've got - i'm not sure how to do the rest of part 2.
Some help would be really appreciated!
Thank you
Caddy
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