True global optimality of the pressure vessel design problem: a benchmark for bio-inspired optimisation algorithms

Int. J. Bio-Inspired Computation, Vol. 5, No. 6, 2013 329

True global optimality of the pressure vessel designproblem: a benchmark for bio-inspired optimisationalgorithms

Xin-She Yang*, Christian Huyck, Mehmet Karamanogluand Nawaz KhanSchool of Science and Technology,Middlesex University,The Burroughs, London NW4 4BT, UKE-mail: [email protected]: [email protected]: [email protected]: [email protected]*Corresponding author

Abstract: The pressure vessel design problem is a well-known design benchmark for validatingbio-inspired optimisation algorithms. However, its global optimality is not clear and there hasbeen no mathematical proof put forward. In this paper, a detailed mathematical analysis of thisproblem is provided that proves that 6,059.714335048436 is the global minimum. The Lagrangemultiplier method is also used as an alternative proof and this method is extended to find theglobal optimum of a cantilever beam design problem.

Keywords: algorithm; benchmarks; pressure vessel; global optimality; bio-inspired algorithm;metaheuristic; optimisation.

Reference to this paper should be made as follows: Yang, X-S., Huyck, C., Karamanoglu, M.and Khan, N. (2013) ‘True global optimality of the pressure vessel design problem: a benchmarkfor bio-inspired optimisation algorithms’, Int. J. Bio-Inspired Computation, Vol. 5, No. 6,pp.329–335.

Biographical notes: Xin-She Yang received his DPhil in Applied Mathematics from OxfordUniversity. After working as a Senior Research Scientist at National Physical Laboratory, he iscurrently a Reader in Modelling and Simulation at Middlesex University. He is also the Chairof IEEE CIS Task Force on Business Intelligence and Knowledge Management.

Christian Huyck is a Professor of Artificial Intelligence at Middlesex University having formedthe AI research group 15 years ago. His main areas of research are biological neural processingand natural language processing, and he explores questions more broadly in AI includingontologies and optimisation. During one of Dr Yang’s presentations, Yang stated that the optimalresult for the pressure vessel task was not known. It seemed that it was relatively simple to findand show the optimal result. We merely did a search using MATLAB to show that the resultwas optimal.

Mehmet Karamanoglu is a Professor of Design Engineering at Middlesex University. Hegraduated with a BEng degree in Mechanical Engineering and followed onto to completehis PhD in numerical methods, supported by British Aerospace. He is currently heading thedepartment of Design Engineering and Mathematics. His expertise includes manufacturingautomation, CAD, design engineering, modelling and robotics. His research interests includenumerical analysis, process simulation, design strategies and robotic systems.

Nawaz Khan received his PhD in Computing Science from Middlesex University. He has beencarrying out research on the area of bioinformatics. He has published well reputed articles andchaired a number of international conference sessions. Currently, he is working as a SeniorLecturer at Middlesex University and Programme Leader for Business Information Technology.

1 Introduction

Engineering optimisation is often non-linear with complexconstraints, which can be very challenging to solve.Sometimes, seemingly simple design problems may

in fact be very difficult indeed. Even in very simplecases, analytical solutions are usually not available, andresearchers have struggled to find the best possiblesolutions. For example, the well-known design benchmarkof a pressure vessel has only four design variables

Copyright © 2013 Inderscience Enterprises Ltd.

330 X-S. Yang et al.

(Cagnina et al., 2008; Gandomi et al., 2013; Yang, 2010);however, the global optimum solution for pressure vesseldesign benchmark is still unknown to the researchcommunity, despite a large number of attempts and studies,i.e., Annaratone (2007) and Deb and Gene (1997). Thus,a mathematical analysis will help to gain some insightinto the problem and thus guide researchers to validate iftheir solutions are globally optimal. This paper attempts toprovide a detailed mathematical analysis of the pressurevessel problem and find its global optimum. To the bestof the author’s knowledge, this is a novel result in theliterature.

Therefore, the rest of the paper is organised as follows:Section 2, introduces the basic formulation of the pressurevessel design benchmark and then highlights the relevantnumerical results from the literature. Section 3 provides ananalysis of the global optimum for the problem, whereasSection 4 uses Lagrange multipliers as an alternativemethod to prove that the analysis in Section 3 indeedgives the global optimum. Section 5 extends the samemethodology to analyse the optimal solution of anotherdesign benchmark: cantilever beam. Finally, we draw ourconclusions in Section 6.

2 Pressure vessel design benchmark

Bio-inspired optimisation algorithms have become popular,and many new algorithms have emerged in recent years(Yang and Gandomi, 2012; Che and Cui, 2011; Cui et al.,2013; Gandomi et al., 2012; Yang and Deb, 2013; Yang,2013). In order to validate new algorithms, a diverse set oftest functions and benchmarks are often used (Gandomi andYang, 2011; Jamil and Yang, 2013). Among the structuraldesign benchmarks, the pressure vessel design problem isone of the most widely used.

In fact, the pressure vessel design problem isa well-known benchmark for validating optimisationalgorithms (Cagnina et al., 2008; Yang, 2010). It has fourdesign variables: thickness (d1), thickness of the heads (d2),the inner radius (r) and the length (L) of the cylindricalsection. The main objective is to minimise the overallcost, under the non-linear constraints of stresses and yieldcriteria. The thickness can only take integer multiples of0.0625 inches.

This optimisation problem can be written as

minimise f(x) = 0.6224d1rL+ 1.7781d2r2

+ 3.1661d21L+ 19.84d21r, (1)

subject to

g1(x) = −d1 + 0.0193r ≤ 0

g2(x) = −d2 + 0.00954r ≤ 0

g3(x) = −πr2L− 4π3 r3 + 1, 296, 000 ≤ 0

g4(x) = L− 240 ≤ 0.

(2)

The simple bounds are

0.0625 ≤ d1, d2 ≤ 99× 0.0625, 10.0 ≤ r, L ≤ 200. (3)

This is a mixed-integer problem, which is usuallychallenging to solve. However, there are extensive studiesin the literature, and details can be found in several goodsurvey papers (Thanedar and Vanderplaats, 1995; Gandomiand Yang, 2011). The main results are summarised inTable 1. It is worth pointing out that some results are notvalid and marked with * in the footnote. These seeminglylower results actually violated some constraints and/or useddifferent limits.

As it can be seen from this table, the results varysignificantly from the highest value of 8,129.104 bySandgren (1988) to the lowest value of 6,059.714 bya few researchers (Cagnina et al., 2008; Coelho, 2010;He et al., 2004; Gandomi et al., 2013, 2011; Yang andGandomi, 2012). However, nobody is sure that 6059.714 isthe globally optimal solution for this problem.

The best solution by Gandomi et al. (2013) and Yangand Gandomi (2012) is

f∗ = 6059.714, (4)

with

x∗ = (0.8125, 0.4375, 42.0984, 176.6366). (5)

The rest of the paper analyses this problem mathematicallyand proves that this solution is indeed near the globaloptimum and concludes that the true globally minimalsolution is fmin = 6059.714335048436 at

x∗ = (0.8125, 0.4375,

42.0984455958549, 176.6365958424394). (6)

3 Analysis of global optimality

As all the design variables must have positive values andf is monotonic in all variables, the minimisation of frequires the minimisation of all the variables if there isno constraint. As there are 4 constraints, some of theconstraints may become tight or equalities. As the range ofL is 10 ≤ L ≤ 200, the constraint g4 automatically satisfiesL ≤ 240 and thus becomes redundant, which means that theupper bound for L is

L ≤ 200. (7)

This is a mixed integer programming problem, whichoften requires special techniques to deal with the integerconstraints. However, as the number of combinations of d1and d2 is not huge (just 1002 = 10, 000), it is possible togo through all the cases for d1 and d2, and then focus onsolving the optimisation problems in terms of r and L.

The first two constraints are about stresses. In order tosatisfy these conditions, the hoop stresses d1/r and d2/rshould be as small as possible. This means that r should

True global optimality of the pressure vessel design problem 331

be reasonably large. For any given d1 and d2, the first twoconstraints become

r ≤ d10.0193

, r ≤ d20.00954

. (8)

So the upper bound or limit for r becomes

Ur = min{

d10.0193

,d2

0.00954

}. (9)

The above argument that r should be moderately high, mayimply that one of the first two constraints can become tight,or an equality.

The third constraint g3 can be rewritten

πr2L+4π

3r3 ≥ K, K = 1, 296, 000. (10)

In fact, this is essentially the requirement that the volumeof the pressure vessel must be greater than a fixed volume.This provides the lower boundary in the search domain of(r, L).

Since f(r, L) is monotonic in r and L, the globalsolution must be on the lower boundary for any givend1 and d2. In other words, the inequality g3 becomes anequality

πr2L+4π

3r3 = K. (11)

Using equation (10), with L ≤ 200, r can be derived usingNewton’s method (or an online polynomial root calculator).The only positive root implies that

r ≥ 40.31961872409872 = r1. (12)

Similarly, L ≥ 10 means that

r ≤ 65.22523261350128 = r2. (13)

So the true value of r must lie in the interval of [r1, r2].

From the first inequality with r = r1, we have

d1 ≥ 0.7782. (14)

The second inequality gives

d2 ≥ 0.3846. (15)

As both d1 and d2 must be integer multiples I and J ,respectively, of d = 0.0625, the above two inequalitiesmean

I =

⌈0.7782

0.0625

⌉= 13, J =

⌈0.3846

0.0625

⌉= 7. (16)

In other words, we have

d1 ≥ 13d = 0.8125, d2 ≥ 7d = 0.4375. (17)

From the objective function [equation (1)], both d1 and d2should be as small as possible, so as to get the minimumpossible f . This means that the global minimum will occurat d1 = 0.8125 and d2 = 0.4375.

Now the objective function with these d1 and d2 valuescan be written as

f(r, L) = 0.5057rL+ 0.77791875r2

+ 2.090120703125L+ 13.0975r. (18)

As d1 = 0.8125 and d2 = 0.4375, the first inequalities(g1 and g2) will give an upper bound of r

R∗ = min{

d10.0193

,d2

0.00954

}

= min{42.0984455958549, 45.859538784067}

= 42.0984455958549. (19)

Table 1 Summary of main results

Authors Results Authors ResultsCoello (2000a) 6,288.745 Lee and Geem (2005) 7,198.433Li and Chou (1994) 7,127.3 Cai and Thierauf (1997) 7,006.931Cagnina et al. (2008) 6,059.714 Li and Chang (1998) 7,127.3Cao and Wu (1999) 7,108.616 Hu et al. (2003) 6,059.131∗

He et al. (2004) 6,059.714 Coello and Montes (2001) 6,059.946Huang et al. (2007) 6,059.734 He and Wang (2006) 6,061.078Litinetski and Abramzon (1998) 7,197.7 Coello (2000b) 6,263.793Sandgren (1988) 7,980.894 Kannan and Kramer (1994) 7,198.042Akhtar et al. (2002) 6 171 Yun Yun (2005) 7,198.424Tsai et al. (2002) 7,079.037 Cao and Wu (1997) 7,108.616Deb and Gene (1997) 6,410.381 Coello (1999) 6,228.744Montes et al. (2007) 6,059.702∗ Parsopoulos and Vrahatis (2005) 6,544.27Shih and Lai (1995) 7,462.1 Kannan and Kramer (2010) 6,059.73Sandgren (1990) 8,129.104 Coelho (2010) 6,059.714Wu and Chow (1995) 7,207.494 Ray and Liew (2003) 6,171Zhang and Wang (1993) 7,197.7 Coello and Cortes (2004) 6,061.123Joines and Houck (1994) 6,273.28 Michalewicz and Attia (1994) 6,572.62Hadj-Alouane and Bean (1997) 6,303.5 Fu et al. (1991) 8,048.6Yang and Gandomi (2012) 6,059.714 Gandomi et al. (2013) 6,059.714

Note: Results marked with ∗ are not valid, see text.


Again from the objective function, which is monotonic interms of r and L, the optimal solution should occur at thetwo extreme ends of the boundary governed by Eq. (11).

The one end at r = R∗ gives

L∗ =K

πR2∗− 4R∗

3= 176.6365958424394. (20)

This is the point for the global optimum with

fmin = 6, 059.714335048436. (21)

The other extreme point is at r′ = 40.31961872409872 andL = 200, which leads to an objective value of

f ′ = 6, 288.67704565344, (22)

and clearly is not the global optimum.

4 Method of Lagrange multipliers

The optimal solution (21) can alternatively be provedby solving the following constrained problem with oneequality because all of the upper bounds or inequalitiesare automatical satisfied. By minimising d1, and d2, theobjective function becomes:

minimise

f(r, L) = 0.5057rL+ 0.77791875r2

+ 2.090120703125L+ 13.0975r. (23)

subject to

ge = πr2L+4π

3r3 −K = 0, (24)

with the simple bounds

40.31961872409872 ≤ r ≤ 42.098445595854919,

10 ≤ L ≤ 176.6365958424394. (25)

To avoid writing long numbers, let us define

a = 0.5057, b = 0.77791875,

c = 2.090120703125, d = 13.0975. (26)

We have

minimise f(r, L) = arL+ br2 + cL+ dr. (27)

This problem can be solved by the Lagrange multipliermethod, and we have

minimise ϕ = f + λge, (28)

where λ is the Lagrange multiplier.The optimum should occur when

∂ϕ

∂r=

∂f

∂r+ λ

∂ge∂r

= aL+ 2br + d+ λ(2πrL+ 4πr2) = 0, (29)

∂ϕ

∂L=

∂f

∂L+ λ

∂ge∂L

= ar + c+ λ(πr2) = 0, (30)

∂ϕ

∂λ= πr2L+

4π

3r3 −K = 0. (31)

Now we have three equations for three unknownsaL+ 2br + d+ λ(2πrL+ 4πr2) = 0,

ar + c+ λ(πr2) = 0,

πr2L+ 4πr3

3 −K = 0.

(32)

The equation in the middle gives

λ = −ar + c

πr2, (33)

Substituting this, together with L = K/(2r2)− 4r/3, intothe first equation, we have

a

(Kr − 4πr3

3

)+ 2bπr4 + πdr3

−(ar + c)

(2K +

4πr3

3

)= 0, (34)

which is a quartic equation with four roots ingeneral. The only feasible solution within [r1, r2]is 42.098445595854919, which corresponds toL = 176.6365958424394. This solution is indeed the globalbest solution as given in (21).

5 Cantilever beam design benchmark

Another widely used benchmark for validating bio-inspiredalgorithms is the design optimisation of a cantilever beam,which is to minimise the overall weight of a cantileverbeam with square cross sections (Fleury and Braibant, 1986;Gandomi et al., 2013, 2011). It can be formulated as

minimise f(x) = 0.0624(x1 + x2 + x3 + x4 + x5), (35)

subject to the inequality

g(x) =61

x31

+37

x32

+19

x33

+7

x34

+1

x35

− 1 ≤ 0. (36)

The simple bounds/limits for the five design variables are

0.01 ≤ xi ≤ 100, i = 1, 2, ..., 5. (37)


Since the objective f(x) is linear in terms of all designvariables, and g(x) encloses a hypervolume, it can bethus expected that the global optimum occurs when theinequality becomes tight. That is, the inequality becomes anequality

g(x) =61

x31

+37

x32

+19

x33

+7

x34

+1

x35

− 1 = 0. (38)

For ease of analysis, we rewrite the above equation as

g(x) =5∑

i=1

aix3i

− 1 = 0, (39)

where

a = (a1, a2, a3, a4, a5) = (61, 37, 19, 7, 1). (40)

Hence, the cantilever beam problem becomes

minimise f(x) = k

5∑i=1

xi, k = 0.0624, (41)

subject to

g(x) =5∑

i=1

aix3i

− 1 = 0. (42)

Using the method of Lagrange multipliers, we have

Minimise ϕ = f + λg

= k

5∑i=1

xi + λ( n∑

i=1

aix5i

− 1). (43)

Then, the optimality conditions give

∂ϕ

∂xi= k + λ(−3)

aix4i

= 0, i = 1, 2, ..., 5, (44)

∂ϕ

∂λ=

5∑i=1

aix3i

− 1 = 0. (45)

From equation (44), we have

x4i =

3λaik

, oraix3i

=kxi

3λ. (46)

Substituting it into equation (45), we have

5∑i=1

(kxi

3λ

)− 1 =

k

3λ

(5∑

i=1

xi

)− 1 = 0. (47)

After rearranging and using results from (46), now we have

3λ

k=

5∑i=1

(3λaik

)1/4, (48)

which is a non-linear equation for λ. However, it isstraightforward to find that

λ ≈ 0.4466521202, (49)

which leads to the optimal solution

x∗ = (6.0160159, 5.3091739,

4.4943296, 3.5014750, 2.15266533), (50)

with the minimum

fmin(x∗) = 1.339956367. (51)

This is the global optimum. However, the authors havenot seen any studies that have found this solution in theliterature. Slightly higher values have been found by cuckoosearch and other methods (Chickermane and Gea, 1996;Gandomi et al., 2013). The best solution found so far byGandomi et al. (2013) is

xbest = (6.0089, 5.3049, 4.5023, 3.5077, 2.1504), (52)

and

fbest = 1.33999, (53)

which is near this global optimum.The above mathematical analysis can be very useful to

guide future validation of new optimisation methods whenthe above design benchmarks are used.

6 Conclusions

Pressure vessel design problem is a well-tested benchmarkthat has been used for validating optimisation algorithmsand their performance. We have provided a detailedmathematical analysis and obtained its global optimality.We have also used the method of Lagrange multipliersto double-check that the obtained optimum is indeed theglobal optimum for the pressure vessel design problem. Byusing the same methodology, we also analysed the designoptimisation of a cantilever beam.

However, it is worth pointing out that the method ofLagrange multipliers is only valid for optimisation problemswith equalities or when an inequality becomes tight. Forgeneral non-linear optimisation problems, we have to usethe full Karush-Kuhn-Tucker (KKT) conditions to analysetheir optimality (Yang, 2010), though such KKT can beextremely challenging to analyse in practice.

Even for design problems with only a few designvariables, an analytical solution will provide greater insightinto the problem and thus can act as better benchmarksfor validating new optimisation algorithms. Further workcan focus on the analysis of other non-linear designbenchmarks.

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