**Origin**:jiaob@cas.mcmaster.ca

In this problem we have a polynomial of degree 18 in
and 11 in and we have to find at least one such
that one of the real root s of verify

We solve this problem as a non 0-dimensional problem: we will find ranges for s, such that is very small and s satisfies the above constraint.

Alternatively we may use our continuation module that will allow to display all the roots s as a function of and visually select the value of such that the constraint is satisfied.

The polynomial is too large to be displayed here but may be obtained with the following Maple program:

lambda:=0.8; mu:=1.5; muack:=0.1; G:= proc(x,muunc) (1/muack)^x*(muunc/(muunc-lambda))*sum('(muack/(mu-muunc))^k*(1/(x-k)!)* (1/(1-lambda/muunc)^k)','k'=0..x) end: Pis:=proc(M,k,muunc) (1/G(M,muunc))*(1/(mu-muunc))^(M-k)*(1/k!)*(1/muack)^k* (1/(1-lambda/muunc))^(M-k+1) end: findsig:=proc(s,M,muunc) sum('Pis(M,k,muunc)*(lambda+k*muack)/(mu-mu*s+lambda+k*muack)','k'=0..M) end: ew:=findsig(s,9,muunc)-s: ew:=numer(ew): P:=op(5,ew):

**Solving method**: `GeneralSolve`
**Solutions:**: 1 (approximate)
**Computation time** (April 2003):

Sun Blade | 6.03s |

**Origin**: COCONUT, taken from SpaceSolver, LIA/EPFL,
http://liawww.epfl.ch/lottaz/SpaceSolver/

We have 6 inequalities for 5 unknowns:

.1666666667e-2*u-(.318e-4*H_s+.54e-2)*S+.1e-11 <= 0 137.7-.8633e-1*S+.5511e-4*S^2-.8358e-8*S^3-H_s <= 0 .77e-1*((.1666666667e-2*u+.7783292378e-6*W^1.5161)*W^2)^.3976+.1e-11-H_b <= 0 .168e-1*(S*W^3)^.2839+.1e-11-H_b <= 0 .1666666667e-2*u+.7783292378e-6*W^1.5161-50 <= 0 15-.1666666667e-2*u-.7783292378e-6*W^1.5161 <= 0The unknowns are:

[H_b, H_s, S, W, u]

The original problem is to find one value for each of the unknowns so that the inequalities are satisfied. But with our interval analysis solver we are able to calculate not one value but a set of sets of possible values. Each set is provided as a range for each unknown and any value for the unknowns in this range will satisfy the inequalities. For example we get the set:

[306.9,400],[132.14344,137.53982],[2112.5,2200], [5625,6937.5],[10397.0999992237,12105.3999984474]Hence examples of possible value of the unknowns is (307, 133, 2113, 5700, 10398) or (399,136, 2199,600,12000).

Hence we obtain an approximation of the solution variety and furthermore we obtain an information of the quality of this approximation: we may compute the total volume of the approximation and we have also an upper bound of the total volume of the variety. For example we get here that the total volume of the variety volume is 1.20477e+12+ [0,2.66039e+14] or for a better approximation 2.04418e+12+ [0,2.51944e+14]. Furthermore we can adjust the accuracy of the approximation incrementally.

**Solving method**: `GradientSolve` +`HullIConsistency`+ 3B
**Computation time** (May 2004):

DELL D400 (1.7Ghz) | 1.74s (Vol=1.20477e+12), 5.41s (Vol=2.04418e+12) |

**Origin**: COCONUT, taken from the PhD Thesis by Claudio Lottaz,
LIA/EPFL.
**Physical meaning**: a civil engineering problem

We have 18 inequalities for 9 unknowns:

P-4.999999999999 <= 0 5.500000000001-1.*H_s1 <= 0 5.500000000001-1.*H_s2 <= 0 2.500000000001-1.*E <= 0 4.5*t-1.*c <= 0 c-85.*t <= 0 40.*d-1.*h_a <= 0 h_a-100.*d <= 0 252661.8726678*E^1.6*L^4-.220000*h_a^2*(3.37*h_a*d+42.84*t*c) <= 0 3300.00*L*E^(8/5)-67.8386566297810273*h_a*d <= 0 825.00*L*E^(8/5)-.47*t*c*h_a <= 0 H_s1+H_s2+.1e-2*h_a-10.999999999999-1.*P <= 0 100.*E^1.6*L-.1e7 <= 0 -100.*E^1.6*L+1. <= 0 200.*E^.6-2000. <= 0 -200.*E^.6+200. <= 0 .220000*h_a^2*(3.37*h_a*d+42.84*t*c)-.1e9*L^3 <= 0 -.220000*h_a^2*(3.37*h_a*d+42.84*t*c)+L^3 <= 0The unknowns are:

[E, H_s1, H_s2, L, c, d, h_a, P, t]and lie in the following ranges:

[2, 10], [5, 8], [5, 8], [10, 50], [200, 500], [20, 40], [800, 4000], [0, 10], [20, 100]As in the

E=[2.500000000001,2.50000058157186] H_s1=[5.500000000001,5.50000078503804] H_s2=[5.500000000001,5.50000078503804] L=[42.5,43.75] c=[445.1214375,472.56071875] d=[36.7682900289095,40] h_a=[3353.6580057819,3569.10533718793] P=[4.17682900289195,4.58841450144547] t=[90,95]

**Origin**: COCONUT, taken from the PhD Thesis by Claudio Lottaz,
LIA/EPFL.
**Physical meaning**: a civil engineering problem

Constants:

ZeroPlus= 1e-12 E= 0.21e12 F_r= 0.4 G_m= 1500 G_q= 1.5 G_r= 1.1 L_sc= 4000 M_FH= 3.5 Q_r= 5000 d_o= 0.5e-1 f_y= 2.35e8 t_fs= 0.03 t_s = 0.12 t_w = 0.012Equalities

L = 2*x + d + (n_h - 1)*e F_H = M_FH + h + t_s + t_fs B_V = F_H*A_b h = 2*a + d h = b + t N_1*a_N = M M = (q*L^2)/8 q = (1.3*G_m + G_q*Q_r)*b_s a_N = S/A_1 S = 2*c*t*b + t_w*(a - t)*(b - a) A_1 = 2*c*t + t_w*(a - t) N_1pl = f_y*A_1 T_d = V_1/A_1w A_1w = (a - t/2)*t_w V_1 = q/2 * (L/2 - y_1) y_1 = x + d/2 y_2 = y_1 + e V_1pl = f_y*A_1w/sqrt(3) V_2pl = f_y*A_2/sqrt(3) A_2 = (e - d)*t_w w_l = 5/384*q_sc*L^4/(E*I_y) q_sc = L_sc*b_s I_y = 4*c*t*(b/2)^2 + t_w*(a - t) * (b - a)^2/2 c_n = A_1/(2*t_w) Z_1 = c_n^2*t_w/2 + (a - c_n - t_w)^2/ 2*t_w + 2*(a - c_n - t/2)*c*t V_f = n_d*F_r*Pi*d_v^2/4Inequalities

n_h >= n_d # Constraints by contractor x >= 1.5*d + ZeroPlus e >= 2.5*d + ZeroPlus d <= 0.75*h - ZeroPlus # Constraints by civil engineer n_h >= 3 N_1 <= N_1pl/G_r - ZeroPlus (N_1/A_1 + V_1*d/4/Z_1)^2+ZeroPlus <= (f_y^2 - 3*T_d^2)/G_r V_1 <= V_1pl/G_r - ZeroPlus V_2 <= V_2pl/G_r - ZeroPlus (a - t/2)/t_w <= 11 - ZeroPlus 1.2*w_l <= L/350 - ZeroPlus t_w*(a - t) >= 2*c*t + ZeroPlus # Constraints by Ventilation expert d >= d_v + d_o + ZeroPlus V_f >= c_V*B_V + ZeroPlusThe range for the unknowns are defined by

# 42 variables A_1 >=1e-5 ,<=0.16 #8.36e-3 chord area A_1w >=0 ,<=0.08 #1.6e-3 web area of chord A_2 >=0 ,<=1.2 #6.24e-3 vertical x-sectional area A_b >=85 ,<=700 #360 floor area B_V >=0 ,<=7000 #72.8 Building volume F_H >=0 ,<=5 #3.64 Total floor height I_y >=1e-5 ,<=0.14 #8.28e-4 Moment of inertia L >=12 ,<=35 #15 beam span M >=0 ,<=3.125e6 #7.79e4 applied moment N_1 >=0 ,<=3e5 #1.76e5 applied flange force N_1pl >=0 ,<=7e7 #1.97e6 plastic flange strength S >=0 ,<=0.2 #3.69e-3 static moment T_d >=0 ,<=2e8 #1.13e7 tangent forces V_1 >=0 ,<=6e5 #1.81e4 shear force V_1pl >=0 ,<=2e7 #2.17e5 streer strength V_2 >=0 ,<=3.125e8 #1.4e5 shear force V_2pl >=0 ,<=3e8 #8.47e5 strear strength V_f >=0 ,<=3 #2.36e-2 ventilation flow Z_1 >=1e-7 ,<=0.06 #1.44e-4 plastic modulus a >=0 ,<=0.75 #0.145 distacne between outer fibre and hole edge a_N >=0 ,<=1.5 #0.442 distance between flange forces b >=0 ,<=1.5 #0.467 beam depth b_s >=1e-5 ,<=20 #1.0 beam spacing c >=0 ,<=0.4 #0.15 half flange width c_V >=8.5e-4,<=15e-4 #0.0015 constant depending on ventilation system chosen c_n >=1e-5 ,<=0.75 #0.133 Neutral axis of plastification d >=0 ,<=1 #0.2 hole diameter d_v >=0 ,<=1.2 #0.1 diameter needed for ventilation e >=0.7 ,<=1.2 #0.72 hole spacing h >=0.1 ,<=1 #0.55 beam depth n_d >=1 ,<=5 #2 number of ducts n_h >=0 ,<=20 #7 number of holes q >=0 ,<=65000 #16001 linear loading q_sc >=0 ,<=40000 #8400 short term live linear load t >=0 ,<=0.016 #0.012 flange thickness t_fs >=0.03 ,<=0.5 #0.03 floor surface thickness t_s >=0.12 ,<=0.5 #0.12 slab thickness t_w >=0 ,<=0.014 #0.012 web thickness w_l >=0 ,<=0.5 #0.001 beam deflection due to loads x >=0.8 ,<=1.2 #0.85 distance to first hole y_1 >=0 ,<=18 #0.86 distance from support to center of 1st hole y_2 >=0 ,<=30 #1.58 distance from support to center of 2nd hole

For this problem we have equality constraints U=F(X) and each occurrence of U into the inequalities are substituted by F(X). To respect the interval constraint on U, U in [a,b] we add the inequalities F(X)-b<= and a-F(X)<=0. To avoid the problem with denominator in the inequality we transform the any inequality A/B<=0 into AB<=0. We end up with a system of 69 inequalities with 14 unknowns.

**Solving method**: `GradientSolve` + 3B
**Solutions:**: 0 (exact)
**Computation time** (May 2004):

DELL D400 (1.7Ghz) | 0.71s |

However by enlarging the search space it is possible to find solutions for this problem. We have the following variables:

[A_b, N_1, a, b, b_s, c, c_V, d, d_v, e, n_d, n_h, t, x]and a possible solution is:

A_b=[ -0.0000000086,0.0000000414] N_1=[0,1483.1248046876] a=[0.0021900000,0.0022406250] b=[0.8641000000,0.8732737500] b_s=[-0.0000000000,0.0000000000] c=[0,0.0037500000] c_V=[0.0007200000,0.0016300000] d=[0.4017381641,0.4017383789] d_v=[0,0.0001147461] e=[ 1.2999999512, 1.3000000000] n_d=[1,20] n_h=[8,8] t=[0,0.0000691797] x=[1.2491306512,1.2491308203]