I have server and many clients or bidders bidders who are bidding with their resources (datasize,bandwidth) and price they want to sell their resources .. I am using first price sealed bid auction here.. I need to select winners.. I need to apply Nash equilibrium strategy for clients to maximize the expected profit and expected utility theory to give guidance to the aggregator to obtain the expected resources from clients. Note: Following is the solved exampe. q1, q2 are resources of bidder and p is price they demand for providing these resources First they are calculating resources q1, q2 and price p We have 5 bidders Bidders = (q1, q2, p) for five bidders are Bidder 1 = (4000, 85, 0.20), S(q, p) = min{α1q1, α2q2} − p, where coefficients α1 and α2 are to balance different types of resources and both set to 0.5 (its assumption). For example for Bidder 1 = (q1, q2, p) = (4000, 85, 0.20), q1 normalization = value - min/max-min = 4000 - 1000/ 5000 - 1000 = 0.75 q2 normalization = value - min/max-min = 85-5/100-5=0.84 p= 0.20 S(q, p) = min{α1q1, α2q2} − p, S(q, p) = min{α1(0.75), α2(0.84)} – 0.20 S(q, p) = min{0.5 (0.75), 0.5 (0.84)} – 0.20 min{0.375,0.42}– 0.20 min[0.375]-0.20 =0.175
I have server and many clients or bidders bidders who are bidding with their resources (datasize,bandwidth) and
Note: Following is the solved exampe.
q1, q2 are resources of bidder and p is price they demand for providing these resources
First they are calculating resources q1, q2 and price p
We have 5 bidders
Bidders = (q1, q2, p) for five bidders are
Bidder 1 = (4000, 85, 0.20),
S(q, p) = min{α1q1, α2q2} − p,
where coefficients α1 and α2 are to balance different types of resources and both set to 0.5 (its assumption).
For example for Bidder 1 = (q1, q2, p) = (4000, 85, 0.20),
q1 normalization = value - min/max-min = 4000 - 1000/ 5000 - 1000 = 0.75
q2 normalization = value - min/max-min = 85-5/100-5=0.84
p= 0.20
S(q, p) = min{α1q1, α2q2} − p,
S(q, p) = min{α1(0.75), α2(0.84)} – 0.20
S(q, p) = min{0.5 (0.75), 0.5 (0.84)} – 0.20
min{0.375,0.42}– 0.20
min[0.375]-0.20
=0.175
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