Risk management for sulfur dioxide abatement under multiple uncertainties

doi:10.1007/s11707-015-0495-6

Front. Earth Sci.

2016, Vol. 10

Issue (1) : 87-107 https://doi.org/10.1007/s11707-015-0495-6

RESEARCH ARTICLE

Risk management for sulfur dioxide abatement under multiple uncertainties

C. DAI¹,W. SUN²,Q. TAN^3,^2,^*(

),Y. LIU¹,W.T. LU¹,H.C. GUO¹

1. College of Environmental Science and Engineering, Peking University, Beijing 100871, China
2. Institute for Energy, Environment and Sustainable Communities, University of Regina, Regina, Saskatchewan S4S 7H9, Canada
3. College of Water Resources & Civil Engineering, China Agricultural University, Beijing 100083, China

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Abstract

In this study, interval-parameter programming, two-stage stochastic programming (TSP), and conditional value-at-risk (CVaR) were incorporated into a general optimization framework, leading to an interval-parameter CVaR-based two-stage programming (ICTP) method. The ICTP method had several advantages: (i) its objective function simultaneously took expected cost and risk cost into consideration, and also used discrete random variables and discrete intervals to reflect uncertain properties; (ii) it quantitatively evaluated the right tail of distributions of random variables which could better calculate the risk of violated environmental standards; (iii) it was useful for helping decision makers to analyze the trade-offs between cost and risk; and (iv) it was effective to penalize the second-stage costs, as well as to capture the notion of risk in stochastic programming. The developed model was applied to sulfur dioxide abatement in an air quality management system. The results indicated that the ICTP method could be used for generating a series of air quality management schemes under different risk-aversion levels, for identifying desired air quality management strategies for decision makers, and for considering a proper balance between system economy and environmental quality.

Keywords risk management conditional value-at-risk interval optimization two-stage programming uncertainty air quality management

Corresponding Author(s): Q. TAN

Just Accepted Date: 05 March 2015 Online First Date: 28 April 2015 Issue Date: 25 December 2015

Cite this article:

C. DAI,W. SUN,Q. TAN, et al. Risk management for sulfur dioxide abatement under multiple uncertainties[J]. Front. Earth Sci., 2016, 10(1): 87-107.

URL:

https://academic.hep.com.cn/fesci/EN/10.1007/s11707-015-0495-6
https://academic.hep.com.cn/fesci/EN/Y2016/V10/I1/87

Fig.1 The study system.

Fig.2 The frequency of SO₂ generation rate for (a) power plant 1, (b) power plant 2, (c) chemical industry plant, (d) petroleum refinery and (e) steel mill.

Scenario	Power plant 1	Power plant 2	Chemical industry plant	Petroleum refinery	Steel mill
Probability/%	Emission amount /(tonne·day^?1)	Probability/%	Emission amount /(tonne·day^?1)	Probability/%)/	Emission amount /(tonne·day^?1)	Probability/%	Emission amount /(tonne·day^?1)	Probability/%	Emission amount /(tonne·day^?1)
S1	0.2	[64.8, 68.4]	0.2	[41.3, 43.7]	0.2	[4.5, 4.7]	0.3	[10.3, 10.9]	0.5	[3.1, 3.3]
S2	0.7	[68.4, 72]	0.4	[43.7, 46]	0.4	[4.7, 5]	0.6	[10.9, 11.5]	0.9	[3.3, 3.5]
S3	2.3	[72, 75.6]	0.8	[46, 48.4]	1.2	[5, 5.3]	2	[11.5, 12]	1.8	[3.5, 3.7]
S4	3.1	[75.6, 79.2]	1	[48.4, 50.7]	1.6	[5.3, 5.6]	2.3	[12, 12.6]	2.9	[3.7, 3.9]
S5	3.5	[79.2, 82.7]	3	[50.7, 53.1]	3.7	[5.6, 5.8]	4	[12.6, 13.2]	4.2	[3.9, 4.1]
S6	6.2	[82.7, 86.3]	4.5	[53.1, 55.4]	4.3	[5.8, 6.1]	6.6	[13.2, 13.8]	7.3	[4.1, 4.3]
S7	7.8	[86.3, 89.9]	7	[55.4, 57.8]	7.5	[6.1, 6.4]	9.2	[13.8, 14.3]	11.1	[4.3, 4.5]
S8	10.5	[89.9, 93.5]	8.1	[57.8, 60.1]	10.4	[6.4, 6.6]	11.1	[14.3, 14.9]	13	[4.5, 4.7]
S9	9.8	[93.5, 97]	10.4	[60.1, 62.5]	11.4	[6.6, 6.9]	12.7	[14.9, 15.5]	15.5	[4.7, 4.9]
S10	13.3	[97, 100.6]	10.4	[62.5, 64.8]	13.4	[6.9, 7.2]	11.3	[15.5, 16.1]	14	[4.9, 5.1]
S11	10.6	[100.6, 104.2]	11.3	[64.8, 67.2]	12.5	[7.2, 7.5]	11.9	[16.1, 16.6]	11	[5.1, 5.3]
S12	9.7	[104.2, 107.8]	11.6	[67.2, 69.5]	12.4	[7.5, 7.7]	10.6	[16.6, 17.2]	8.2	[5.3, 5.5]
S13	9.4	[107.8, 111.4]	10.9	[69.5, 71.9]	7.2	[7.7, 8]	6	[17.2, 17.8]	5.5	[5.5, 5.7]
S14	5.3	[111.4, 114.9]	7.7	[71.9, 74.2]	6	[8, 8.3]	4.8	[17.8, 18.4]	2.1	[5.7, 5.9]
S15	3.3	[114.9, 118.5]	5.2	[74.2, 76.6]	4.2	[8.3, 8.6]	3.7	[18.4, 18.9]	0.8	[5.9, 6.1]
S16	2.4	[118.5, 122.1]	3.5	[76.6, 79]	1.7	[8.6, 8.8]	1.9	[18.9, 19.5]	0.6	[6.1, 6.3]
S17	1.1	[122.1, 125.7]	1.7	[79, 81.3]	1	[8.8, 9.1]	0.5	[19.5, 20.1]	0.1	[6.3, 6.5]
S18	0.4	[125.7, 129.3]	1.5	[81.3, 83.7]	0.6	[9.1, 9.4]	0.1	[20.1, 20.7]	0.2	[6.5, 6.7]
S19	0.3	[129.3, 132.8]	0.5	[83.7, 86]	0.1	[9.4, 9.7]	0.2	[20.7, 21.2]	0.2	[6.7, 7]
S20	0.1	[132.8, 136.4]	0.3	[86, 88.4]	0.2	[9.7, 9.9]	0.2	[21.2, 21.8]	0.1	[7, 7.2]

Tab.1 SO₂ generation rates under different probability levels

Tab.2 Parameters of pollution-control technologies

Fig.3 The framework of the ICTP method.

Source	Scenario	OPA (j = 1)	AA (j = 2)	LWS (j = 3)
Treated amount of SO₂ emission, Xij,opt±/(tonne·day^?1)	Treated amount of excess SO₂ emission, Yijs,opt±/(tonne·day^?1)	Treated amount of SO₂ emission, Xij,opt±/(tonne·day^?1)	Treated amount of excess SO₂ emission, Yijs,opt±/(tonne·day^?1)	Treated amount of SO₂ emission, Xij,opt±/(tonne·day^?1)	Treated amount of excess SO₂ emission, Yijs,opt±/(tonne·day^?1)
Power plant 1(i = 1)	s = 1	30	0	30	0	24	0
	s = 2	30	0	30	0	24	0
	s = 3	30	0	30	0	24	0
	s = 4	30	0	30	0	24	0
	s = 5	30	0	30	0	24	0
	s = 6	30	0	30	0	24	[0, 2.3]
	s = 7	30	0	30	0	24	[2.3, 5.9]
	s = 8	30	0	30	0	24	[5.9, 9.5]
	s = 9	30	0	30	0	24	[9.5, 13]
	s = 10	30	0	30	0	24	[13, 16.6]
	s = 11	30	0	30	0	24	[16.6, 20.2]
	s = 12	30	0	30	0	24	[20.2, 23.8]
	s = 13	30	[0, 3.4]	30	0	24	[23.8, 24]
	s = 14	30	[5.8, 6.9]	30	0	24	[21.6, 24]
	s = 15	30	12.6	30	0	24	[18.3, 21.9]
	s = 16	30	[13.6, 14.1]	30	0	24	[20.9, 24]
	s = 17	30	[17.4, 17.7]	30	0	24	[20.7, 24]
	s = 18	30	16.1	30	5.2	24	[20.4, 24]
	s = 19	30	13.1	30	12.1	24	[20.1, 23.6]
	s = 20	30	10.3	30	18.7	24	[19.8, 23.4]
Power plant 2(i = 2)	s = 1	30	0	18	0	12	0
	s = 2	30	0	18	0	12	0
	s = 3	30	0	18	0	12	0
	s = 4	30	0	18	0	12	0
	s = 5	30	0	18	0	12	0
	s = 6	30	0	18	0	12	0
	s = 7	30	0	18	0	12	0
	s = 8	30	0	18	0	12	[0, 0.1]
	s = 9	30	0	18	0	12	[0.1, 2.5]
	s = 10	30	0	18	0	12	[2.5, 4.8]
	s = 11	30	0	18	0	12	[4.8, 7.2]
	s = 12	30	0	18	0	12	[7.2, 9.5]
	s = 13	30	0	18	0	12	[9.5, 11.9]
	s = 14	30	[0, 2.2]	18	0	12	[11.9, 12]
	s = 15	30	[2.2, 4.6]	18	0	12	12
	s = 16	30	[4.6, 7]	18	0	12	12
	s = 17	30	[7, 9.3]	18	0	12	12
	s = 18	30	[9.3, 11.7]	18	0	12	12
	s = 19	30	[11.7, 14]	18	0	12	12
	s = 20	30	[14, 16.4]	18	0	12	12
Chemical industry plant (i = 3)	s = 1	0	0	0	0	[7.5, 7.7]	0
	s = 2	0	0	0	0	[7.5, 7.7]	0
	s = 3	0	0	0	0	[7.5, 7.7]	0
	s = 4	0	0	0	0	[7.5, 7.7]	0
	s = 5	0	0	0	0	[7.5, 7.7]	0
	s = 6	0	0	0	0	[7.5, 7.7]	0
	s = 7	0	0	0	0	[7.5, 7.7]	0
	s = 8	0	0	0	0	[7.5, 7.7]	0
	s = 9	0	0	0	0	[7.5, 7.7]	0
	s = 10	0	0	0	0	[7.5, 7.7]	0
	s = 11	0	0	0	0	[7.5, 7.7]	0
	s = 12	0	0	0	0	[7.5, 7.7]	0
	s = 13	0	0	0	0	[7.5, 7.7]	[0.2, 0.3]
	s = 14	0	0	0	0	[7.5, 7.7]	[0.5, 0.6]
	s = 15	0	0	0	0	[7.5, 7.7]	[0.8, 0.9]
	s = 16	0	0	0	0	[7.5, 7.7]	1.1
	s = 17	0	0	0	0	[7.5, 7.7]	[1.3, 1.4]
	s = 18	0	0	0	0	[7.5, 7.7]	[1.6, 1.7]
	s = 19	0	0	0	0	[7.5, 7.7]	[1.9, 2]
	s = 20	0	0	0	0	[7.5, 7.7]	2.2
Petroleum refinery (i = 4)	s = 1	8	0	0	0	7.5	0
	s = 2	8	0	0	0	7.5	0
	s = 3	8	0	0	0	7.5	0
	s = 4	8	0	0	0	7.5	0
	s = 5	8	0	0	0	7.5	0
	s = 6	8	0	0	0	7.5	0
	s = 7	8	0	0	0	7.5	0
	s = 8	8	0	0	0	7.5	0
	s = 9	8	0	0	0	7.5	0
	s = 10	8	0	0	0	7.5	[0, 0.6]
	s = 11	8	0	0	0	7.5	[0.6, 1.1]
	s = 12	8	0	0	0	7.5	[1.1, 1.7]
	s = 13	8	1.2	0	0	7.5	[0.5, 1.1]
	s = 14	8	2.3	0	0	7.5	[0, 0.6]
	s = 15	8	0	0	0	7.5	[2.9, 3.4]
	s = 16	8	3.4	0	0	7.5	[0, 0.6]
	s = 17	8	4	0	0	7.5	[0, 0.6]
	s = 18	8	4.6	0	0	7.5	[0, 0.6]
	s = 19	8	5.2	0	0	7.5	[0, 0.5]
	s = 20	8	5.7	0	0	7.5	[0, 0.6]
Steel mill (i = 5)	s = 1	2	0	2.7	0	[0, 0.2]	0
	s = 2	2	0	2.7	0	[0, 0.2]	0
	s = 3	2	0	2.7	0	[0, 0.2]	0
	s = 4	2	0	2.7	0	[0, 0.2]	0
	s = 5	2	0	2.7	0	[0, 0.2]	0
	s = 6	2	0	2.7	0	[0, 0.2]	0
	s = 7	2	0	2.7	0	[0, 0.2]	0
	s = 8	2	0	2.7	0	[0, 0.2]	0
	s = 9	2	0	2.7	0	[0, 0.2]	0
	s = 10	2	0.2	2.7	0	[0, 0.2]	0
	s = 11	2	0.4	2.7	0	[0, 0.2]	0
	s = 12	2	0.6	2.7	0	[0, 0.2]	0
	s = 13	2	0.8	2.7	0	[0, 0.2]	0
	s = 14	2	1	2.7	0	[0, 0.2]	0
	s = 15	2	1.1	2.7	0.1	[0, 0.2]	0
	s = 16	2	0.9	2.7	0.5	[0, 0.2]	0
	s = 17	2	0.7	2.7	0.9	[0, 0.2]	0
	s = 18	2	0	2.7	1.8	[0, 0.2]	0
	s = 19	2	0	2.7	2	[0, 0.2]	[0, 0.1]
	s = 20	2	0	2.7	2.3	[0, 0.2]	0

Tab.3 Solution obtained through the ICTP method under λ = 1 and β = 0.99

Confidence level	Source	Technology	Risk parameters/(tonne·day^?1)
λ = 0.5	λ = 1	λ = 2	λ = 5	λ = 10
β = 0.6	i = 1	j = 1	[34.2, 34.7]	[32.3, 32.5]	[31.5, 31.7]	[31, 31.7]	[30.9, 31.7]
		j = 2	22	27.3	30.1	30.1	30
		j = 3	[40.5, 43.4]	[37.6, 40.6]	[35.9, 38.8]	[36.4, 38.9]	[36.5, 38.9]
	i = 2	j = 1	[31.4, 32.3]	[31.1, 31.9]	[30.6, 31.1]	[30.3, 30.6]	[30.3, 30.6]
		j = 2	12.3	13.4	15.8	18	18
		j = 3	[19.1, 20.3]	[18.5, 19.7]	[17.2, 18.4]	[16, 17.2]	[16, 17.2]
	i = 3	j = 1	0	0	0	3.3	3.3
		j = 2	0	0	0	0	0
		j = 3	[7.3, 7.5]	[7.6, 7.8]	[8, 8.2]	[5, 5.2]	[5, 5.2]
	i = 4	j = 1	7.9	8.5	8.5	8.3	8.1
		j = 2	0	0	0	0	2.2
		j = 3	[8, 8.4]	[7.8, 8.1]	[7.8, 8.1]	[7.9, 8.3]	[7.5, 7.6]
	i = 5	j = 1	2.4	2.4	2.3	2.1	2.1
		j = 2	2.6	2.6	2.8	3.2	3.3
		j = 3	[0, 0.2]	[0, 0.2]	[0, 0.2]	[0, 0.2]	[0, 0.2]
β = 0.8	i = 1	j = 1	[33, 33.3]	[31.8, 32.1]	[31.4, 31.7]	[31, 31.7]	[30.9, 31.7]
		j = 2	25.4	28.9	30.1	30.1	30
		j = 3	[38.5, 41.6]	[36.6, 39.6]	[36, 38.8]	[36.5, 38.9]	[36.5, 38.9]
	i = 2	j = 1	[31.1, 31.9]	[31.1, 31.9]	[30.5, 30.8]	[30.3, 30.6]	[30.3, 30.6]
		j = 2	13.4	13.4	17.1	18	18
		j = 3	[18.5, 19.7]	[18.5, 19.7]	[16.5, 17.7]	[16, 17.2]	[16, 17.2]
	i = 3	j = 1	0	0	0.2	3.3	3.3
		j = 2	0	0	0	0	0
		j = 3	[7.3, 7.5]	[7.6, 7.8]	8.1	[5.3, 5.5]	[5.3, 5.5]
	i = 4	j = 1	7.9	8.5	8.5	8.3	8.1
		j = 2	0	0	0	1	3.1
		j = 3	[8, 8.4]	[7.8, 8.1]	[7.8, 8.1]	[7.6, 7.8]	7.5
	i = 5	j = 1	1.7	2.3	2.1	2.1	2.1
		j = 2	2.8	2.7	3.1	3.3	3.3
		j = 3	[0.5, 0.7]	[0, 0.2]	[0, 0.2]	[0, 0.2]	[0, 0.2]
β = 0.99	i = 1	j = 1	[33, 33.3]	[31.4, 31.7]	[31.3, 31.7]	[30.9, 31.7]	[31, 31.7]
		j = 2	25.4	30.1	30.1	30	30
		j = 3	[38.5, 41.6]	[36.1, 38.8]	[36.1, 38.9]	[36.5, 38.9]	[36.5, 38.9]
	i = 2	j = 1	[31.1, 31.9]	[30.3, 30.6]	[30.3, 30.6]	[30.3, 30.6]	[30.3, 30.6]
		j = 2	13.4	18	18	18	18
		j = 3	[18.5, 19.7]	[16, 17.2]	[16, 17.2]	[16, 17.2]	[16, 17.2]
	i = 3	j = 1	0	0	0.7	3.3	0
		j = 2	0	0	0	0	3.3
		j = 3	[7.3, 7.5]	[7.6, 7.8]	8.1	[6.4, 6.6]	[6.4, 6.6]
	i = 4	j = 1	8	8.4	8.3	8.1	7.4
		j = 2	0	0	1	6.5	7.2
		j = 3	[8, 8.4]	[7.8, 8.1]	[7.6, 7.8]	[6.6, 7.2]	[6.6, 7.2]
	i = 5	j = 1	1.7	2.3	2.1	2.1	2.1
		j = 2	2.8	2.7	3.1	3.3	3.3
		j = 3	[0.5, 0.7]	[0, 0.2]	[0, 0.2]	[0, 0.2]	[0, 0.2]

Tab.4 Optimal SO₂ emission allocation schemes under different risk parameters (tonne/day)

Fig.4 Excess SO₂ emissions treated by (a) OPA; (b) AA and (c) LWS technologies at different risk levels.

Fig.5 (a) Mean-risk function value and (b) total expected cost under different risk levels.

Fig.6 Recourse cost under different risk levels.

Fig.7 (a) lower bound and (b) upper bound of the CVaR under different risk levels.

	From power plant 1(i = 1)			From power plant 2(i = 2)			From chemical industry plant(i = 3)			From petroleum refinery(i = 4)			From steel mill(i = 5)
	j = 1	j = 2	j = 3	j = 1	j = 2	j = 3	j = 1	j = 2	j = 3	j = 1	j = 2	j = 3	j = 1	j = 2	j = 3
Allowable SO₂ emission ( X i j , o p t ± )/(tonne·day^?1)		30	21.6	24	30	11.5	12	0	0	[6.6, 6.8]	6.3	0	7.5	1	2.75	[0.5, 0.68]
Excess SO₂ emission ( Y i j s , o p t ± )/(tonne·day^?1)	s = 1	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
	s = 2	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
	s = 3	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
	s = 4	0	0	[0, 3.6]	0	0	0	0	0	0	0	0	0	0	0	0
	s = 5	0	0	[3.6, 7.1]	0	0	0	0	0	0	0	0	0	0	0	0
	s = 6	0	0	[7.1, 10.7]	0	0	[0, 1.9]	0	0	0	0	0	0	0	0	0
	s = 7	0	0	[10.7, 14.3]	0	0	[1.9, 4.3]	0	0	0	0	0	[0, 0.5]	0	0	[0.05, 0.07]
	s = 8	0	0	[14.3, 17.9]	0	0	[4.3, 6.6]	0	0	0	0	0	[0.5, 1.1]	0	0	[0.25, 0.27]
	s = 9	0	0	[17.9, 21.4]	0	0	[6.6, 9]	0	0	[0, 0.1]	0	0	[1.1, 1.7]	0	0	[0.4, 0.5]
	s = 10	[0, 1]	0	[21.4, 24]	0	0	[9, 11.3]	0	0	[0.3, 0.4]	0	0	[1.7, 2.3]	0.15	0	[0.5, 0.52]
	s = 11	[5.3, 5.6]	0	[19.7, 23]	[0, 1.7]	0	[11.3, 12]	0	0	[0.6, 0.7]	0	0	[2.3, 2.8]	0.35	0	[0.5, 0.52]
	s = 12	9.95	0	[18.6, 22.3]	[1.7, 4]	0	12	0	0	0.9	0	0	[2.8, 3.4]	1	0	[0.05, 0.07]
	s = 13	[11.1, 11.8]	0	[21.05, 24]	[4, 6.4]	0	12	0	0	[1.1, 1.2]	3.4	0	[0, 0.6]	1	0	[0.25, 0.27]
	s = 14	[14.8, 15.3]	0	[21, 24]	[6.4, 8.7]	0	12	0	0	[1.4, 1.5]	4	0	[0, 0.6]	1	0.45	[0, 0.02]
	s = 15	23.2	0	[16.1, 19.7]	[8.7, 11.1]	0	12	0	0	[1.7, 1.8]	0	0	[4.6, 5.1]	0.8	0.85	[0, 0.02]
	s = 16	16.5	6	[20.4, 24]	[11.1, 13.5]	0	12	0	0	2	5.1	0	[0, 0.6]	0	1.85	[0, 0.02]
	s = 17	13.5	12.8	[20.2, 23.8]	[13.5, 15.8]	0	12	0	0	[2.2, 2.3]	5.7	0	[0, 0.6]	0	2.05	[0, 0.02]
	s = 18	10.6	19.6	[19.9, 23.5]	[15.8, 18.2]	0	12	0	0	[2.5, 2.6]	6.3	0	[0, 0.6]	0	2.25	[0, 0.02]
	s = 19	19.4	21.6	[12.7, 16.2]	[13.3, 15.6]	4.9	12	0	0	[2.8, 2.9]	0	0	[6.9, 7.4]	0	2.45	[0, 0.12]
	s = 20	17.4	21.6	[18.2, 21.8]	[9, 11.4]	11.5	12	0	0	3.1	6.3	0	[1.1, 1.7]	[0, 0.1]	2.75	0

Tab.5 Solution of the ITSP model

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