Asset Labels, Bytes and the 64 KB Requisition
The district vocational centre has 60 working desktops and 8 printers, and the principal has sanctioned 40 more desktops next year. A +2 student volunteer is asked to prepare an asset register. He decides to give every device a one-byte asset code written as two hexadecimal digits on a sticker, starting at 00 and counting upward. Separately, the lab's old inventory sheet lists a machine as having '64 KB cache' and a pen drive as '32 GB', but the teacher says the pen drive shows only about 29.8 GB free in Windows. The student also has to order RAM for 12 machines and writes '64,000 bytes' on the requisition because that is what he thinks 64 KB means. The accountant questions both the label scheme and the quantity. The student must defend or correct his work using number-system reasoning, not guesswork.
1. Will two hexadecimal digits cover the present 68 devices and the planned 108 devices? Show the range and state the minimum bits and hex digits needed for 108 devices. Should the student keep two hex digits or switch to one?
Read the worked answer
Two hex digits give 16^2 = 256 codes, running from 00 to FF, i.e. decimal 0 to 255. Present demand is 60 desktops + 8 printers = 68 devices, and after expansion it is 68 + 40 = 108 devices. Both fit inside 256, so the scheme survives, with 148 codes unused.
For 108 devices the minimum bit width is found by powers of two: 2^6 = 64, which is too small; 2^7 = 128, which is enough. So 7 bits are sufficient in theory. However, one hex digit represents only 4 bits and 16 values, so one digit is nowhere near enough. Two hex digits represent 8 bits (one byte), which is the smallest practical byte-aligned label and is easy to write by hand.
Judgement: keep two hex digits, but do not use the space carelessly. Reserve ranges, for example 00 for a spare/test machine, and F0–FF for printers and future devices, and record the allocation in the register. The spare capacity is a reason to keep the scheme, not a reason to shrink it.
2. The requisition says 64,000 bytes for 64 KB. Is it correct? What exactly should be ordered, and how do you explain the 32 GB pen drive showing 29.8 GB?
Read the worked answer
In memory and addressing, 1 KB means 2^10 = 1,024 bytes, so 64 KB = 64 × 1,024 = 65,536 bytes. Writing 64,000 bytes is short by 1,536 bytes, about 2.3%, so the requisition is not accurate if the requirement is a binary 64 KB. The precise wording is 64 KiB (65,536 bytes), or simply the exact RAM module specification.
The pen drive is a different convention. Drive manufacturers use decimal units: 32 GB = 32,000,000,000 bytes. Windows reports binary units (GiB): 32,000,000,000 ÷ 1,073,741,824 ≈ 29.8 GiB. Some space is also used by file-system structures, so the displayed free space can be slightly lower still.
Judgement: both figures can be 'right' but in different units. The student should label units explicitly, order the exact capacity or module, and explain the pen-drive figure as a unit convention rather than a defective device. For RAM, the binary figure is the correct one.