Please answer Question 10 based on the data provided.

An auto parts company is in the construction phase of a new plant. The centerpiece of the operation is a stamping machine that (in its basic configuration) can process 40 parts per hour (on average). Parts arrive at an average of 36 units per hour. Assume all processing (service) times and interarrival times are exponentially distributed.

Question 10:

Look at your answers for questions 8 and 9. In question 8 you considered upgrading the machine's service time (call this Option 1). In question 9, you considered a second machine, with no upgrades to the first machine's service time (call this Option 2). Management has provided us with the relevant cost structure for either option in the form: f(s) = ($675s x p) + $400E [Lq] Here, the variable p represents a premium associated with any upgrades. For question 8, the increased service time to achieve the desired E [Lq ~ 2] required the purchase of a special regulator, resulting in a 12% premium (that is, p=1.12) for Option 1. The second machine considered in question 9 required no such upgrades, so the premium is simply p= 1 (for each machine) under Option 2. Which option should the manager choose? - 8. The plant manager has a goal that on average) no more than two parts will be in the waiting arca at any given time. How much faster (service time in minutes) will the stamping machine have to perform in order to meet this goal? (Hint: to get an E[Lq) ~ 2 parts we can simply use the quadratic equation and solve for utilization, getting p = 0.72, before deriving our new value for p.) 9. Suppose the manager now has the option of purchasing a second machine (instead of trying to make the one machine faster as in the above prol Under the current conditions (a = 36 40), will adding the second machine (in the form of an M/M/2 queue) achieve the manager's goal of no more than two parts in the waiting area? and u = Pol 34 - = ( = 2 361 () () 2 3 ( 2 (3-4) * (3 - ) () - 0: 2- 11- - 2 3- 8 = ua ug per hour solf cate de 36 penhor Parts arrive at an average machine can proces on average of u = 40 peahr parts in the waiting area de Mud) = 36x36 40 (40-36) To = 8.1 parts Area We can able to accomodate - und 36 4 - 9 Parte After buying and machine Parts astering will remain sarre, wherear machine processing on average will change M = ? riven a parts in waiting area. 2= 2 ) 2 = 36 x 36 (M-d) (4-36.) - 202_724 - 12960 & an solving the quadaatic equation M = 49.1 (2) M = -13 > M=-13 is not valid so Mzug Pea hour