Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 16 | Low Level
3 ns after a read begins. Uninteresting time segments are surrounded by [brackets]. Each
character represents 1 ns.
0 - 79 ns 80 - 99 ns 100 - 707 ns 708 - 723 ns
Writer AAAABBBBCCCCDDDDEEEE XXXXYYYYZZZZ____
Worker ____aaaaabbbbbcccccd opppppqqqqqrrrrr
>= 724
Queue
11112221222222223222 3333333343333322 *
Load
Y = Writing word 159 @ 712 ns
Z = Writing word 160 @ 716 ns
q = Processing word 127 @ 714 ns
r = Processing word 128
* = Between 708 and 723 ns, the queue load is shown as 30 plus the
digit shown at each ns.
Note that the queue load does in fact reach a maximum of 34 at time = 716 ns.
As an interesting note, if the problem had required only 2 ns of the 5 ns processing time to
complete a read, the optimal queue depth would decrease to 33.
The below graphs are unnecessary, but show empirically that adding writes from the 3rd
period does not change the queue depth required.
< 796 ns 797 - 807 ns 808 - 873 ns 874 - 885 ns
Writer ____AAAABBBB !!@@@@####$$
Worker ^^^&&&&&**** yyyyyzzzzzaa
Queue Load 877788778887 112111221122 *
Writer
Worker
A = Writing word 161
& = Processing word 144
# = Writing word 181
z = Processing word 160 @ 779 ns
* = Between 874 and 885 ns, the queue load is shown as 20 plus the
digit shown at each ns.
< 1112 ns 1112 - 1123 ns
YYYYZZZZ____
^^&&&&&%%%%%
Queue Load 333343333322 *
Z = Writing word 240 @ 1116 ns
& = Processing word 207 @ 1114 ns
* = Between 1112 and 1123 ns, the queue load is shown as 30 plus the
digit shown at each ns.
ns
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