Then, the objective function of (57) can be more succinctly represented as a function of µ The optimization problem then becomes: Note that constraint (A) in (60) involves the quadratic terms ∑ J j $_{=1}$x$_{j}$ ( A $^{T}$ξ ) $_{j}$. Therefore, in this section, we suggest a way to linearize this constraint. Intuitively, x$_{j}$ ( A $^{T}$ξ ) j is simply the market maker’s revenue from the j th order. Let R$_{j}$ denote the market maker’s revenue from the j th order. x$_{j}$ is the number of shares of the option traded. ( A $^{T}$ξ ) j is the market-clearing price of the j th order. To begin the transformation, we first consider the feasible set of x and ξ . We say that the pair Figure 5 is a three-dimensional graph. The graph shows the market maker’s revenue as a (E1) ∀ j ∈ { 1 , 2 , .., J } , x$_{j}$ = 0 if ( A $^{T}$ξ ) j > B$_{j}$b$_{j}$ (E2) ∀ j ∈ { 1 , 2 , .., J } , x$_{j}$ ∈ [0 , Q$_{j}$ ] if ( A $^{T}$ξ ) j = B$_{j}$b$_{j}$ (E3) ∀ j ∈ { 1 , 2 , .., J } , x$_{j}$ = Q$_{j}$ if ( A $^{T}$ξ ) j < B$_{j}$b$_{j}$ (61) $^{10}$I can simply ignore cases for which Q$_{j}$ = 0. I can also simply remove the j th order from my optimization problem. 39 8.4.4 Linearization of Constraint (A) ℓ ( µ ) = µ Ω + µ ln ( N ∑ i =1 q$_{i}$e − d i µ ) = µ ln ( N ∑ i =1 q$_{i}$e $^{Ω}$e − d i µ ) = µ ln ( N ∑ i =1 θ$_{i}$e − d i µ ) (59) min ξ , x ,µ, $_{d}$µ ln ( ∑ N i $_{=1}$θ$_{i}$ e − d i µ ) such that (A) d$_{i}$ = z$_{i}$ exp [ − α ∑ J j $_{=1}$x$_{j}$ ( A $^{T}$ξ ) j − A$_{ij}$ )] for ∀ i ∈ { 1 , .., N } (B) µ ≥ 0 (C) ξ ≥ 0 (D) ∑ N i $_{=1}$ξ$_{i}$ = 1 (E) ( x , ξ ) ∈ F (60)