That’s an interesting question. Let’s see. According to the Univac 1108 Processor and Storage [PDF] manual UP-4051r1 (link is to a scanned copy in Fourmilab’s Univac Memories archive), timing for double-precision floating point operations (the closest analogue to 64 bit IEEE floating point which most machines use now, although the Univac had 72 bit precision) were:
- Add/subtract 2.625 μsec
- Multiply 4.25 μsec
- Divide 17.25 μsec
Single precision (36-bit) floating point instructions were about half this time. The basic instruction clock speed was 750 nanoseconds (assuming optimal memory interleaving), so the number of clocks per instruction would be:
- Add/subtract 3.5
- Multiply 5.666
- Divide 23
This is a very different timing mix than contemporary high-performance CPUs, most of which can do at least one of any of the 64-bit floating point instructions in one clock cycle and, using the single instruction multiple data (SIMD) extensions, the latest generation of CPUs can do as many as 64 per clock. With the 1108, you have to assume a mix of floating point instructions to estimate average clocks, so I am going to arbitrarily assume an equal number of add/subtracts and multiplies as you do in matrix multiplication, and one divide for every eight multiply-add pairs. That gives an average of four clocks per instruction in this mix, or 3 microseconds per average floating point instruction.
This equates to 333,333 floating point operations per second, or 1/3 megaFLOPS. So dividing 2 exaFLOPS by 1/3 megaFLOPS, we find it would take 6\times 10^{12} (six trillion) Univac 1108s to achieve 2 exaFLOPS double precision computation.
For reference, here are Univac 1108 prices as of 1968, in 1968 dollars. If we just count the price of the CPU with the basic half megabyte core memory, that’s US$(1968) 1,389,960, which is about US$(2022) 12.1 million in today’s BidenBucks. So, to buy six trillion 1108s in today’s dollars would cost US$ 7.26\times 10^{19}, 72.6 exabucks, or or 72.6 quadrillion FRNs.
Let’s look at the storage capacity (solid-state disc) of the Aurora supercomputer, which is specified as 230 petabytes. The main mass storage device of the Univac 1108 was the FASTRAND II, which had a capacity of about 90 megabytes (depending on how you convert 36 bit words to 8 bit bytes). To obtain 230 petabytes of storage with FASTRANDs, you’d need 2.56 billion of them which, at their 1968 cost of US$(1968) 134,400 a pop (US$(2022) 1.17 million) would set you back around 3 quadrillion dollars (petabucks).
A FASTRAND II weighed around 2.25 tons, so 2.56 billion of them would weigh around 5.2\times 10^{12} kg, or about 1000 times the weight of the Great Pyramid of Khufu.
