1. A Secure and Efficient Cloud Resource Allocation Scheme with Trust
Evaluation Mechanism based on
Combinatorial Double Auction
Source：KSII Transactions on Internet and Information Systems,
vol. 11, no. 9, pp , Sep
Authors: Yun-Hao Xia, Han-Shu Hong, Guo-Feng Lin, Zhi-Xin Sun
Speaker: Chit-Jie Chew
Date: 7/12/2018

2. Outline Introduction Related works Proposed scheme
Experimental results
Conclusions

3. Introduction (1/3) – Double auction
$30
$40
Provider 1
Consumer 1
Profit: $30
$35
$50
Provider 2
Consumer 2
Auctioneer
$30
$20
Consumer 3
Provider 3

4. Introduction (2/3) – Combinatorial double auction
Case 1: Provider 1, 2, 3 Profit: $220
Case 3: Provider 1, 3 Profit: $250
Case 3: Provider 1, 3 Profit: $250
Case 2: Provider 1, 2 Profit: $70
Case 4: Provider 2, 3 Profit: $200
Case 1: Provider 1, 2, 3
Case 4: Provider 2, 3
Case 2: Provider 1, 2
Case 3: Provider 1, 3 1 2
$100 1
$200
Auctioneer
Auctioneer
Auctioneer
Auctioneer
-$300
-$400
-$250
-$250 1 1 2 2 3 5 3 4 5 2 6 5
Consumer 2
Consumer 2
Consumer 2
+$200
+$300
+$300
+$300 -1 -2 -1 -2 -2 -1 -3 -3 -3
Provider 1
Consumer 1
Consumer 1
-$50
-$100 $0
$50 1 1 1 2 3 3 1 1 1 2 2 3 1 2
$150 2 3
$300
Consumer 3
+$120 -1 -2
Consumer 1
Consumer 1
Consumer 1
+$200
+$200
+$200 -1 -1 -1 -1 -1 -1 -1 -1 -1
Profit:
$70 1 1 1 2 1 1 2
Profit:
Profit:
$250
$100
$200
Provider 2
Consumer 2 -2
Consumer 3
+$120 -1
𝑚𝑎𝑥 𝑗=1 𝑁 𝑝 𝑗 𝑥 𝑗
Auctioneer 1 2
$120
Profit:
$220 2 1 4
$150
Consumer 3
Provider 3

5. Introduction (3/3) – Model based on combinatorial double auction
· · · · · ·
Cloud resource agent
Cloud resource consumer n
Bidding request
Virtual
resource
Cloud resource provider 1
· · · · · ·
Providing
bid price
Records managements
Auctioneer
加上整个流程说这篇论文的方法用在哪里
Service managements
Cloud resource provider n
Providing
bid price
Computing resources
Storage resources

6. Related works (1/2) – Genetic algorithm(GA)
∆𝑓=250−200=50
∆𝑓=200−200=0 P1 P2 P3 C1 C2 C3
∆𝑓=−280− −50 =−230
Population number
Chromosome
Fitness
Parents
-$180
-$180 1 5
$200 1
-$50
-$50 2 4
$520 -1 -3 -3
Crossover
-$200 2 4
-$100 2 4
-$130 1 5
$250 1
Mutation
$20 -1 4
$200
$200 1
-$280
-$280 1 3 6
$250
$250 1
New parents 2 4
$250
$250 1
-$50
$200 1
-$180 1 5

7. Related works (2/2) – Simulated annealing algorithm(SA)P1 P2 P3 C1 C2 C3
Chromosome
Fitness 4 2
-$50 4 1
$70 1
$50
圖片記得改
∆𝑓=70 −(−$50)= 120
∆𝑓=50 −$70=− 20
𝑒𝑥𝑝 ∆𝑓 𝑇 =𝑒𝑥𝑝 = 3.32
𝑒𝑥𝑝 ∆𝑓 𝑇 =𝑒𝑥𝑝 − = 0.8
𝐼𝑓 𝑒𝑥𝑝 ∆𝑓 𝑇 >𝑟𝑎𝑛𝑑𝑜𝑚 0~1 :
𝐼𝑓 𝑒𝑥𝑝 ∆𝑓 𝑇 >𝑟𝑎𝑛𝑑𝑜𝑚 0~1 :
accept to new generation
accept to new generation
𝑇=𝑇×0.9=100×0.9=90
𝑇=𝑇×0.9=90×0.9=81

8. Proposed scheme (1/9) – The pricing determination model
Notations
Sign
Definition
𝑙𝑜𝑎𝑑 𝑡
Current system load
𝑢𝑠𝑒𝑟_𝑎𝑚𝑜𝑢𝑛𝑡
The amount of demand by buyers
𝑝𝑟𝑜𝑣𝑖𝑑𝑒_𝑎𝑚𝑜𝑢𝑛𝑡
The amount of resources supply 𝑖
Resource consumer 𝑗
Resource
𝜔 𝑗
The weight of each kind of resources
𝑎 𝑖𝑗
The amount of consumer 𝑖 requires type 𝑗 resource ℎ
Resource provider
𝑎 ℎ𝑗
The amount of provider ℎ requires type 𝑗 resource 𝑡
Timestamp
𝑝 ℎ 𝑡
The price of provider at the timestamp 𝑡
𝑝 ℎ
Auction price of provider ℎ
𝑝 𝑖 𝑡
The price of consumer at the timestamp 𝑡
𝑝 𝑖
Bidding price of user 𝑖
𝑙𝑜𝑎𝑑 𝑡 = 𝑢𝑠𝑒𝑟_𝑎𝑚𝑜𝑢𝑛𝑡 𝑝𝑟𝑜𝑣𝑖𝑑𝑒_𝑎𝑚𝑜𝑢𝑛𝑡
𝑢𝑠𝑒𝑟_𝑎𝑚𝑜𝑢𝑛𝑡= 𝑖=1 𝑁 𝑗=1 𝐾 | 𝜔 𝑗 𝑎 𝑖𝑗 |
𝑝𝑟𝑜𝑣𝑖𝑑𝑒_𝑎𝑚𝑜𝑢𝑛𝑡= ℎ=1 𝑁 𝑗=1 𝐾 | 𝜔 𝑗 𝑎 ℎ𝑗 |
𝑝 ℎ 𝑡 = 𝑝 ℎ × 𝑙𝑜𝑎𝑑 𝑡
𝑝 𝑖 𝑡 = 𝑝 𝑖 × 1− 𝑙𝑜𝑎𝑑 𝑡
If 𝑙𝑜𝑎𝑑 𝑡 >1
If 𝑙𝑜𝑎𝑑 𝑡 <1

9. Proposed scheme (2/9) – The pricing determination model1 2
$100
provide_𝑎𝑚𝑜𝑢𝑛𝑡= ℎ=1 𝑁 𝑗=1 𝐾 | 𝜔 𝑗 𝑎 ℎ𝑗 | = ℎ=1 2 𝑗=1 3 | 𝜔 𝑗 𝑎 ℎ𝑗 |
Provider 1
= 0.3×1+0.3×2+0.4× ×1+0.4×4
=3.2 1 4
$150
user_𝑎𝑚𝑜𝑢𝑛𝑡= 𝑖=1 𝑁 𝑗=1 𝐾 | 𝜔 𝑗 𝑎 ℎ𝑗 | = ℎ=1 2 𝑗=1 3 | 𝜔 𝑗 𝑎 𝑖𝑗 |
= 0.3×1+0.3×1+0.4× ×2+0.4×3
=2.8
Provider 2 1
$240
𝑝 𝑖 𝑡 = 𝑝 𝑖 × 1− 𝑙𝑜𝑎𝑑 𝑡
𝑙𝑜𝑎𝑑 𝑡 = 𝑢𝑠𝑒𝑟_𝑎𝑚𝑜𝑢𝑛𝑡 𝑝𝑟𝑜𝑣𝑖𝑑𝑒_𝑎𝑚𝑜𝑢𝑛𝑡
If 𝑙𝑜𝑎𝑑 𝑡 <1,
Consumer 1
𝑝 1 𝑡 =240× 1−
=112
𝑝 2 𝑡 =300× 1−
=140 =
=0.875 2 3
$300
Consumer 2

10. Proposed scheme (3/9) – The pricing determination model1
$240 2 3
$300
Provider 2
Provider 1
Consumer 1
Consumer 2 4
$150
$100 1
$112 2 3
$140
Provider 2
Provider 1
Consumer 1
Consumer 2 4
$150
$100 1 3 5
-$250
+$240 2 4
-$10
+$300
+$290 1 3 5
-$250
+$112 2 4
-$138
+$140
+$2

11. Proposed scheme (4/9) – SAGA
∆𝑓=250−200=50
∆𝑓=200−200=0 P1 P2 P3 C1 C2 C3
∆𝑓=−280− −50 =−230
Chromosome
Fitness
Parents
-$180 1 5
$200 1
-$50
-$50 2 4
$520 -1 -3 -3
Mutation
$20 -1 4
$200
$200 1
-$280
-$280 1 3 6
$250
$250 1
𝑒𝑥𝑝 ∆𝑓 𝑇 =𝑒𝑥𝑝 − = 0.1
𝐼𝑓 𝑒𝑥𝑝 ∆𝑓 𝑇 >𝑟𝑎𝑛𝑑𝑜𝑚 0~1 : SA
0.1>0.05
New Parents
$250
$250 1
-$280 1 3 6
$200 1
-$50 2 4

12. Proposed scheme (5/9) – Trust value evaluation mechanism
Statistical trust value(STV)
Direct trust score
Indirect trust score
User k
Provider j
𝑆𝑇𝑉 𝑖 =𝛼 𝐷 𝑖 +𝛽 𝐼𝐷 𝑖 ,𝛼+𝛽=1
Provider l
User i
Provider/User m
𝑆𝑇𝑉= 𝑛=1 𝑁𝑜𝑑𝑒 𝑆𝑇𝑉 𝑛 𝑁𝑜𝑑𝑒

13. Proposed scheme (6/9) – Trust value evaluation mechanism
Direct trust score
0.6,0.7,0.6
−0.2
Provider j
User i
Providers:
Users:
= 3−1 3+1 ≈0.67
𝐷 𝑖𝑗 = 𝑆𝑢𝑚 𝑖,𝑗 𝑆𝑎𝑡 𝑖,𝑗 +𝑈𝑛𝑆𝑎𝑡 𝑖,𝑗
= ≈0.64
𝐷 𝑖𝑗 = 𝐹𝑎𝑖𝑟 𝑖 −𝑈𝑛𝐹𝑎𝑖𝑟 𝑖 𝐹𝑎𝑖𝑟 𝑖 +𝑈𝑛𝐹𝑎𝑖𝑟 𝑖
𝐷 𝑖𝑗 ′= 𝐷 𝑖𝑗 × 𝑆𝑎𝑡 𝑖,𝑗 +𝑈𝑛𝑆𝑎𝑡 𝑖,𝑗 +𝑐𝑢𝑟𝑟𝑒𝑛𝑡_𝑠𝑐𝑜𝑟𝑒 𝑆𝑎𝑡 𝑖,𝑗 +𝑈𝑛𝑆𝑎𝑡 𝑖,𝑗 +1
∆= 𝑠𝑐𝑜𝑟𝑒 𝑖,𝑗 − 𝑆𝑇𝑉 𝑗
If ∆≤0.75
𝐹𝑎𝑖𝑟 𝑖 +1
Otherwise
𝑈𝑛𝐹𝑎𝑖𝑟 𝑖 +1
∆ 1 = 0.6−0.64 =0.04
∆ 1 = 0.7−0.64 =0.06
∆ 3 = 0.6−0.64 =0.04
∆ 4 = −0.2−0.64 =0.84
= 0.64× − ≈0.43
If 𝑐𝑢𝑟𝑟𝑒𝑛 𝑡 𝑠𝑐𝑜𝑟𝑒 >0
Value of range −1,1
𝑆𝑎𝑡 𝑖,𝑗 +1
Otherwise
𝑈𝑛𝑆𝑎𝑡 𝑖,𝑗 +1

14. Proposed scheme (7/9) – Trust value evaluation mechanism
Indirect trust score
k-j
0.65
k-j
0.65
m-j
0.6
l-k
0.7
User k
Provider j
l-k
0.6
i-l
0.6
i-m
0.65
m-j
0.7
i-l
0.55
i-m
0.65
Provider l
User i
Provider/User m
𝐼𝐷 𝑖𝑗_1 = 𝑛=1 𝑁 1 𝐷 𝑖𝑚 × 𝐷 𝑚𝑗 𝑁 1
= 0.65×0.6 1
=0.39
𝐼𝐷 𝑖𝑗_2 = 𝑛=1 𝑁 2 𝐷 𝑖𝑙 × 𝐷 𝑙𝑘 × 𝐷 𝑘𝑗 𝑁 2
≈0.30
= 0.6×0.7×0.65 1
𝐼𝐷 𝑖𝑗 =𝛿 𝐼𝐷 𝑖𝑗_1 +𝜗 𝐼𝐷 𝑖𝑗_2 , 𝛿+𝜗=1
𝐼𝐷 𝑖𝑘 =𝛿 𝐼𝐷 𝑖𝑙_1 +𝜗 𝐼𝐷 𝑙𝑘_2 ,
=0.8× ×0.39
≈0.37
=0.8× ×0.25
≈0.36

15. Proposed scheme (8/9) – Trust value evaluation mechanism
Statistical trust value(STV)
k-j
0.65
k-j
0.65
m-j
0.6
l-k
0.7
𝑆𝑇𝑉 1 =𝛼 𝐷 1 +𝛽 𝐼𝐷 1 ,𝛼+𝛽=1
User 3
Provider 4
=0.8× ×0.37
≈0.58
l-k
0.6
i-l
0.6
i-m
0.65
m-j
0.7
𝑆𝑇𝑉= 𝑛=1 𝑁𝑜𝑑𝑒 𝑆𝑇𝑉 𝑛 𝑁𝑜𝑑𝑒
i-l
0.55
i-m
0.65
Provider 2
User 1
Provider/User 5
𝐷 1 =
𝐼𝐷 1 =
≈0.63
≈0.37

16. Trust level of providers Trust level of consumers
Proposed scheme (9/9) – Trust value evaluation mechanism
𝑞 𝑗 = 𝑇𝐶 𝑗 × 𝑝 𝑗 𝑇𝑃 𝑗 × 𝑝 𝑗
𝑚𝑎𝑥 𝑗=1 𝑁 𝑝 𝑗 𝑥 𝑗
𝑚𝑎𝑥 𝑗=1 𝑁 𝑞 𝑗 𝑥 𝑗
STV of providers
Trust level of providers TP
[-1,0)
Very unreliable 5
[0,0.25)
Unreliable
1.5
[0.25,0.5)
Medium reliable 1
[0.5,0.75)
Reliable
0.85
[0.75,1
Very reliable
0.7
STV of consumers
Trust level of consumers TC
[-1,0)
Very unreliable
0.7
[0,0.25)
Unreliable
0.85
[0.25,0.5)
Medium reliable 1
[0.5,0.75)
Reliable
1.5
[0.75,1)
Very reliable 5

17. Experimental results (1/5) – The simulation of the efficiency of SAGA
Parameter
Value
Population number
500
Chromosome number 16
Crossover probability
0.7
Mutation probability
0.08
Number of population genetics 20
Variation of temperature T
𝑇=𝑇×0.95

18. Experimental results (2/5) – The simulation of the efficiency of SAGA

19. Experimental results (3/5) – The simulation of the efficiency of trust model
Parameter
Value 𝛿
0.8 𝜗
0.2 𝛼 𝛽
Providers
Users

20. Experimental results (4/5) – The simulation of the efficiency of trust model
The percentage of malicious nodes
The average number of successful transactions
Malicious SP
Normal SP
Malicious users
Normal users
10%
6.8
619.9
5.3
620.1
20%
5.4
534.6
2.2
535.4
30%
5.6
492.7
2.8
493.9
40%
5.9
507.3
3.7
508.7
50%
2.6
267.5
2.3
268.1

21. Experimental results (5/5) – The simulation of the efficiency of trust model

22. Conclusions
Secure and efficient
SAGA
Trust evaluation model