Recitation 2

Solution:
a) According to Little’s law the average Processing time would be 500/1000=0.5
month or 0.5*30=15 days.
b) Under Process 2 the average requests undergoing Processing is LIR+LA+LB=375. So
the average Processing time is 375/1000=11.25 days. Or, if we look at the different
“Processing stations” we have that
W (IR) = 200/1000 = 0.2= 6 days.
W (A) = 25/ (1000*0.25) =0.1= 3 days.
W (B) = 150/ (1000*0.25) =0.6 = 18 days.
And therefore the average Processing would be
0 . 5*
0.
2 + 0.
25*
( 0.
2 + 0.
1)
+ 0.
25*
( 0.
2 + 0.
6)
= 0.
375 = (* 30)
= 11.
25 days
c) Note that for approved requests we have λ = 200 and all requests spend 0.2 in IR.
25*
0.
7 + 150*
0.
1 17.
5 + 15
Hence W = 0 . 2 +
= 0.
2 + = 0.
3625 = (* 30)
= 10.
875 days .
200
200
Or, we can calculate
200*
0.
25*
0.
7 + 200*
0.
25*
0.
1+
25*
0.
7 + 150*
0.
1
W =
= 0.
3625 = 10.
875 days
200
So we see that the Processing time for approved requests did improve. However it is
important to note that the averages do not tell the whole story and still all the
approved requests going through B “enjoy” a 24 days processing time. That’s
because
15
λ B = 1000 * 0.
25*
0.
1 = 25 AndWB = 0 . 2 + = 0.
8 = (* 30)
= 24 days .
25
d) The input rate to A is 250 and the Processing capacity is 5*75 so the utilization level
is 250/375=2/3.
In the same way for subgroup B we have Utilization level = 250/ (8*40) = 250/320
= 0.78125.
e) Now we have the input rate for B is 250+50=300, so the utilization level of B would
be 300/320=0.9375
f) 250/320=78% of the time is dedicated to requests from Initial Review and
50/320=16% of the time is dedicated to requests coming from A. Finally,
20/320=6% of the time agents are idle.
4