Question

Problem 4. The Security& Trust Bank employs 4 tellers to serve its customers. Customers arrive ac- cording to a Poisson process at a mean rate of 4 per minute. However, business is growing and management projects that the mean arrival rate will be 6 per minute a year from now. The transaction time between the teller and customer has an exponential distribution with a mean of 0.5 minute. Management has established the following guidelines for a satis- factory level of service to customers: The average number of customers waiting in line to begin service should not exceed 1; at least 95 percent of the time, the number of customers waiting in line should not exceed 6. (a) Determine how well these guidelines are currently being satisfied. (b) Evaluate how well the guidelines will be satisfied a year from now if no change is made in the number of tellers.

Answer 1

The Security and Trust Bank employs 4 tellers to serve its customers. The customers arrive at a mean rate of 4 per min, that is, a = 4 per min = 4x 60 = 240 per hour. Business is growing and management projects that the mean arrival rate is, 2 = 6 per min = 6x 60 = 360 per hour. The transaction time between the teller and customer has a mean of 0.5 minutes, that is, 60 u= = 120 per hour. 0.5 The guidelines established by the management is given as follows: (i) The average number of customers waiting in line to begin service should not exceed 1, that is, I, 31. (ii) For at least 95% of the time, the number of customers waiting in line should not exceed 6, that is, P[N, S6)20.95= P(N, 50)20.95 (a) 1-1 n! 240 20 120 - 240 + n! 4! 2 4.120 1 = 1+2 22 +- 2 23 +- 3! + 24 - . 4! 1 * 480) = [5+1.33+1.33) = 7.66-1 = 0.1305

For 0 <n<4, P = 1 LP, the following values are calculated as follows: -P = 2 x 0.1305 = 0.261 ? P= 2 x 0.1305 = 0.261 A =x0.1305=0.174 For n For n24, p. = kar) po, the followin 4, P. Po, the following values are calculated as follows: 41.4"-41 p - 16 x0.1305=0.087 x 0.1305 = 0.0435 4x4 64 -X0.1305 =0.02175 24x16 128 —x0.1305 = 0.010875 24 x 64 256 - 0.1305 =0.0054 24x 256 P- 512 - 0.1305 = 0.0027 24x1,024 P = 20$1,027 *0.1505 P 1,024 Po = 47.40 Po = 24x4,096 - 0.1305 = 0.0013 The average number of customers waiting in line is calculated as follows: 4240 0.1305*(220) (4x120) L =- 0.1305x29x= _ 0.1305x32 24 = 0.174 Hence, the L, <1, (i) is satisfied.

P(N, 56)=P(N <10) SP = 0.1305 +0.261+0.261+0.174+0.087 +0.0435+0.02175 +0.010875 +0.0054 +0.0027 +0.0013 = 0.999025 Hence, P(N, 56)=0.999025 > 0.95, (ii) is satisfied. Hence, all the guidelines are satisfied.

(b) 2 = 360 per hour 11-4 2014 4! 1134 -4 360 = 1+ n! 4! 4.120 32 33 =1+3+3+31 34 4 1 360 80 = 927 81 1+3+=+ = + = .4 2 6 24 1 = [4+4.5+4.5+13.5] = 26.5-1 = 0.0377

For 0 <n<4, P.=\ - Po, the following values are calculated as follows: (360 ?P= 3x 0.0377 = 0.1131 i, comezos (360? 120 P,= P = 2x0.0377 = 0.16965 2! 360 0.0377 = 0.16965 2 P, the following values are calculated as follows: 0.0377 =0.1272 243 --x 0.0377 = 0.0435 Ps4!445 24x4*0. 729 Po = 41.42 40 = 24x16 - x0.0377 = 0.0715 2187 P, = 4141 = 24x 64 --x 0.0377 = 0.0536 6561 P, = 41.4+Po = 24x 256 —x0.0377 = 0.0402 3º 19683 x0.0377 = 0.0301 P, = 2145 Po = 24x1,024 D31059049 10 4!46 +0 24x4 006 0.0377 = 0.0226 The average number of customers waiting in line is calculated as follows: 0.0377x( 360 x(360 0.0377* 120 * 4x120 _ 360 L = 0.0377x3* x3 248) = 1.526 Hence, the I, >1,(i) is not satisfied.

P(N, 56)=P(N <10) 10 n=1 = 0.1131+0.16965+0.16965 +0.1272 +0.0435+0.0715+0.0536+0.0402 +0.0301+0.0226 = 0.8411 Hence, P(N, 36)=0.8411<0.95, (ii)is not satisfied. Hence, all the guidelines are not satisfied.