Question

. A chemical plant stores spare parts for maintenance in a large warehouse. Throughout the working day, maintenance personnel go to the warehouse to pick up supplies needed for their jobs. The warehouse receives a request for supplies, on average, every three minutes. The average request requires 2.75 minutes to fill a request. Maintenance employees are paid $21.50 per hour and warehouse employees are paid $16 per hour. The warehouse operates 8 hours per day.

a) Based on the number of maintenance employees in the system, an 8 hour work day, and the given arrival and service rates. What is the system cost per day (to the nearest $) if there is only 1 warehouse employees working?

b) Based on the number of maintenance employees in the system, an 8 hour work day, and the given arrival and service rates. What is the system cost per day (to the nearest $) if there are 2 warehouse employees working?

c) Based on the number of maintenance employees in the system, an 8 hour work day, and the given arrival and service rates. What is the system cost per day (to the nearest $) if there are 3 warehouse employees working?

d) What is the optimal number of warehouse employees to staff the warehouse?

Answer 1

a) The number of maintenance employees is not specified in the problem, so consider that it is an infinite number. So, it is a M/M/1 system with following parameters,

Arrival rate, \lambda = 1/interarrival time = 1/3 per minute = (1/3)*60 = 20 per hour

Service rate, \mu = 1/service time = (1/2.75)*60 = 21.82 per hour

Number of maintenance employees in the system, L = \lambda/(\mu-\lambda) = 20/(21.82-20) = 11

System cost per hour = Cw*L + s*Cs

= 21.5*11+1*16

= $ 252.5

System cost per day = 252.5*8

= $ 2,020

b) With more than 1 warehouse employees (servers) working, it is a multi-channel M/M/s system , where

Number of servers, s = 2

Operating characteristics are computed using spreadsheet as below

D11 Q fx =(VLOOKUP(D6+1,$GS2SH$26,2,0)-VLOOKUPfD6,SG$2:SH$26,2,0))"D9"D10/(1-D10)^2+D4/D5 1 A Mutliple-Server Waiting Line System n Summation 2 2 2 2 2 Input per hour rate 21.818182 per hour Arrival rate- Service No. of servers, s Output P00.3714 Rho0.4583 10 11 Average number in the system (L)1.1604 12 Average number in the queue (Lq)0.2438 13 Average time in the system (W)0.0580 hour 14 15 16 17 18 19 20 21 10 12 13 Average time in the queue (Wq)0.0122 hour 15 16 17 18 19 20 21 23 24 23 24 25 26

EXCEL FORMULAS:

Cell Formula H3 (SDS4/SDS5)AG2/FACT(G2)+H2 D9 Copy to H3:H26 1/((VLOOKUP(D6+1,SGS2:SHS26,2,0)-VLOOKUP(D6,SGS2:SH$26,2,0)/(1 D4/(D5* D6))+VLOOKUP(D6,SGS2:SHS26,2,0)) D10 D4/D5/D6 D11 D12 D11-D4/D5 D13D11/SDS4 D14D12/SDS4 D10)A2+D4/D5

L = 1.1604

Total system cost per hour = 1.1604*21.5+2*16 = 56.95

Total system cost per day = 56.95*8 = $ 455.6

b) With more than 1 warehouse employees (servers) working, it is a multi-channel M/M/s system , where

Number of servers, s = 3

Operating characteristics are computed using spreadsheet as below:

n Summation 1 A Mutliple-Server Waiting Line System 2 2 Input Arrival rate- per hour 2 2 2 Service rate21.818182 per hour No. of servers, s- Output PO0.3966 Rho0.3056 10 11 Average number in the system (L0.9489 12 Average number in the queue (Lq)0.0323 13 10 Average time in the system (W)0.0474hour Average time in the queue (Wq)0.0016 hour 12 13 15 16 17 18 19 68.401837 56.9486 455.5888 455.6 15 16 17 18 19 20 21 23 24 25 25

L = 0.9489

Total system cost per hour = 0.9489*21.5+3*16 = 68.40

Total system cost per day = 68.40*8 = $ 547.2

d) We see that total system cost per day of 2 warehouse employees working is the lowest.

Therefore, optimal number of warehouse employees to staff the warehouse = 2