Question

Kolkmeyer Manufacturing Company is considering adding two machines to its manufacturing operation. This addition will bring the number of machines to nine. The president of Kolkmeyer asked for a study of the need to add a second employee to the repair operation. The arrival rate is 0.06 machines per hour for each machine, and the service rate for each individual assigned to the repair operation is 0.6 machines per hour.

a. Compute the operating characteristics if the company retains the single-employee repair operation. If required, round your answers to four decimal places.

P₀ =

Lq=

L=

W_{q= _______hours}

_{W=    ______hours}       

b. Compute the operating characteristics if a second employee is added to the machine repair operation. If required, round your answers to four decimal places.

P₀ =

Lq=

L=

W_{q= _______hours}

_{W=    ______hours}       

Each employee is paid $30 per hour. Machine downtime is valued at $70 per hour. From an economic point of view, should one or two employees handle the machine repair operation? Explain. If required, round your answers to two decimal places.

Cost of one employee system: $_____

Cost of two employees system: $_____

Answer 1

a) For when there is only one employee,

M/M/C model where c =1.

= 0.06/0.6

= 0.1

Using this relation, we get,

Po = 0.9

Using this relation, we get :

Lq= 0.9 x 0.1 x 0.1 / ( 1 - 0.1 )2

= 0.011 machines.

Wq= Lq /

=0.011 / 0.06 = 0.183 hours = 11.1 minutes = 0.185 hours

W = Wq + 1/

  = 11.1 + (1/0.6)

= 12.76 minutes = 0.212 hours

L = . W

= 0.06 x 0.212

= 0.1272 machines

b) For when there are two employees,

M/M/C model where c =2.

= 0.06/ ( 2 x 0.6 )

= 0.05

Using this relation, we get,

Po = 0.9048

Using this relation, we get :

Lq= 0.9048 x 0.052 x 0.05 / (2 x ( 1 - 0.05 )2)

= 0.0003 machines.

Wq= Lq /

=0.0003 / 0.06 = 0.005 hours

W = Wq + 1/

  = 0.005 + (1/0.6)

= 1.67 hours

L = . W

= 0.06 x 1.67

= 0.1002 machines

Cost of one employee system = 70 x W + W x x 30 (where is the server utilization)

=70 x 0.212 + 0.212 x 0.1 x 30 = $ 15.47

Cost of two employee system = 70 x W + W x x 30 (where is the server utilization)

=70 x 1.67 + 1.67 x 0.05 x 30 = $ 119.4

Here one employee system is better.