Question 4 of 6: Memory Volatility, IEEE-754 Decoding, and Variable-Length Encoding Savings

Question 4: Memory Volatility, IEEE-754 Decoding, and Variable-Length Encoding Savings (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) The general categories of volatile and non-volatile memory. (b) Two IEEE-754 single-precision bit patterns, $S{+}E(8){+}M(23)$. (c) An ISA with 56 registers, 221 opcodes, 16-bit immediates, and a program whose instructions are 20%/30%/25%/25% split across four operand-format types, with all instructions rounded up to a multiple of 8 bits.

Find. (a) The volatile/non-volatile distinction with examples. (b) The decimal value each bit pattern represents. (c)(1) The bit width of each instruction type. (c)(2) The memory saved by variable-length vs. fixed-length encoding.

Approach. (a) define by whether stored state survives loss of power; (b) apply $(-1)^S\times1.M\times2^{E-127}$; (c) size the opcode field from $\lceil\log_2 221\rceil$, each register field from $\lceil\log_2 56\rceil$, add the fixed 16-bit immediate where present, round each type up to the next multiple of 8, then compare a weighted average (variable-length) against the single worst-case width (fixed-length).

1. Part (a) — volatile vs. non-volatile memory. Volatile memory requires continuous power to retain its stored data; the instant power is removed, the contents are lost. Examples: SRAM (used for CPU caches/register files — each bit is a cross-coupled latch that needs power to hold state) and DRAM (used for main memory — each bit is a capacitor charge that leaks and must also be periodically refreshed even while powered). Non-volatile memory retains its stored data indefinitely with no power applied. Examples: ROM/PROM (mask- or fuse-programmed, unchangeable after manufacture/programming), EPROM/EEPROM/flash (electrically or UV-erasable and reprogrammable, used for firmware/BIOS and, for flash, for SSDs and USB drives), and magnetic/optical media (hard disks, magnetic tape, CDs/DVDs, which store data as persistent magnetic or optical states). The defining test is simple: disconnect power — volatile memory forgets, non-volatile memory remembers.
2. Part (b) — IEEE-754 decoding. The value is $(-1)^S\times(1+M/2^{23})\times2^{E-127}$, with the bias 127 subtracted from the stored 8-bit exponent.

   Field	(1)	(2)
   $S$	1	0
   $E$ (binary)	01111010	10101010
   $E$ (decimal)	122	170
   $E-127$	−5	43
   $M$ (leading 1.M bits)	1000…0	1100…0
   $1.M$ (decimal)	1.5	1.75

   (1): $(-1)^1\times1.5\times2^{-5}=-1.5/32=\boxed{-0.046875}$.
   (2): $(-1)^0\times1.75\times2^{43}=1.75\times8{,}796{,}093{,}022{,}208=\boxed{15{,}393{,}162{,}788{,}864}$.
3. Part (c) — instruction bit widths and variable- vs. fixed-length memory savings. The opcode field must distinguish 221 instructions, needing $\lceil\log_2 221\rceil=8$ bits (since $2^7=128<221\le256=2^8$); each register field must address 56 registers, needing $\lceil\log_2 56\rceil=6$ bits (since $2^5=32<56\le64=2^6$); each immediate field is the given fixed 16 bits.

   Type	Fields	Raw bits	Rounded to mult. of 8	Weight
   1 (20%)	opcode + 1 in-reg + 1 out-reg	$8+6+6=20$	$\boxed{24\ \text{bits}}$	0.20
   2 (30%)	opcode + 2 in-reg + 1 out-reg	$8+6+6+6=26$	$\boxed{32\ \text{bits}}$	0.30
   3 (25%)	opcode + 1 in-reg + 1 out-reg + immediate	$8+6+6+16=36$	$\boxed{40\ \text{bits}}$	0.25
   4 (25%)	opcode + 1 out-reg + immediate	$8+6+16=30$	$\boxed{32\ \text{bits}}$	0.25

   (1) Each raw field count is rounded UP to the next multiple of 8 bits, since the ISA requires whole bytes per instruction: 20→24, 26→32, 36→40, 30→32 bits, as tabulated above.
   (2) A fixed-length encoding must make every instruction as wide as the WIDEST type actually needed, i.e. $\boxed{40\ \text{bits}}$ (5 bytes) per instruction regardless of type. A variable-length encoding instead uses each type's own rounded width, so the average bits per instruction over the given mix is $$\bar b_{\text{var}}=0.20(24)+0.30(32)+0.25(40)+0.25(32)=4.8+9.6+10.0+8.0=\boxed{32.4\ \text{bits}}$$ The saving per instruction is $40-32.4=7.6$ bits, i.e. $$\frac{40-32.4}{40}\times100\%=\boxed{19\%}$$ so the variable-length program occupies 19% less memory than the fixed-length version of the same program (equivalently, 7.6 bits saved per instruction on average, or $7.6N$ bits total for a program of $N$ instructions).