4826 A 4-2-20 Solution: Using dominate strategy method, we get the simpler 2X2matrix 2 4/,and we solved this matrix model, got the optimal solution: x,=0,x24'5NI 0 y=4. 22=0, y34y4=0, the value of this game is V=5/2 6. Consider the following linearly constrained optimization problem: (15 points) Mxf(x)=15x1+30x2+4x1x2-2x12-4x2 +2x,≤30 x1≥0.x1≥0 Use the KKT conditions to derive an optimal solution Solution: Using the KKt conditions, we derive the following equations u x(15+4x2-4x1-a1)=0 30+4x,-8x,-2l,≤0 x2(30+4x1-8x2-21)=0 2 30≤0 u1(x1+2x2-30)=0 x1≥0,x2≥0,120 Solving this equations system, we get the optimal solution x =12, X2=9 7. You are given a two-server queueing system in a steady-state condition where the number of customers in the system varies between 0 and 4. For n=0, 1,.., 4, the probability Pn that exactly n customers are in the system is po (a) Determine L, the expected number of customers in the system (b) Determine Lg, the expected number of customers in the queue (c) Determine the expected number of customers being served (d)Given that the mean arrival rate is 2 customers per hour, determine the expected waiting time in the system, W, and the expected waiting time in the queue, Wa (e) Given that both serves have the same expected service time, use the results from part(d) to determine this expected service time. (15 points)4 ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − = 4 2 2 0 2 0 4 2 4 8 2 6 2 4 0 2 A Solution: Using dominate strategy method, we get the simpler 2×2matrix： ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − 2 4 4 2 ，and we solved this matrix model, got the optimal solution ： , 0 4 1 , 4 3 0, x1 = x2 = x3 = x4 = ， , 0 4 3 , 0, 4 1 y1 = y2 = y3 = y4 = ，the value of this game is V=5/2。 6. Consider the following linearly constrained optimization problem: (15 points) ⎩ ⎨ ⎧ ≥ ≥ + ≤ = + + − − 0, 0 2 30 . . ( ) 15 30 4 2 4 1 2 1 2 2 2 2 1 2 1 2 1 x x x x st Max f x x x x x x x Use the KKT conditions to derive an optimal solution. Solution: Using the KKT conditions, we derive the following equations: ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎧ ≥ ≥ ≥ + − = + − ≤ + − − = + − − ≤ + − − = + − − ≤ 0, 0, 0 ( 2 30) 0 2 30 0 (30 4 8 2 ) 0 30 4 8 2 0 (15 4 4 ) 0 15 4 4 0 1 2 1 1 1 2 1 2 2 1 2 1 1 2 1 1 2 1 1 2 1 1 x x u u x x x x x x x u x x u x x x u x x u Solving this equations system, we get the optimal solution x1=12, x2=9. 7. You are given a two-server queueing system in a steady-state condition where the number of customers in the system varies between 0 and 4. For n=0,1,….,4, the probability Pn that exactly n customers are in the system is 16 1 , 16 4 , 16 6 , 16 4 , 16 1 p0 = p1 = p2 = p3 = p4 = . (a) Determine L, the expected number of customers in the system. (b) Determine Lq, the expected number of customers in the queue. (c) Determine the expected number of customers being served. (d) Given that the mean arrival rate is 2 customers per hour, determine the expected waiting time in the system, W, and the expected waiting time in the queue, Wq. (e) Given that both serves have the same expected service time, use the results from part (d) to determine this expected service time. (15 points)