# PM Yield Variance

PM Yield Variance
A company manufactures a chemical using two components, A and B.
The standard information for one unit of the chemical are as follows:
Material A 10 kg at \$4 per kg \$40
Material B 20 kg at \$6 per kg \$120
In a particular period, 160 units of the chemical were produced, using
1,000 kgs of material A and 1,460 kgs of material B.
Based on above question could someone please explain to me how did we get the answer for material yield variance using individual method. Please explain each number if possible
November 30th 2022

Retagged November 30th 2022

Hello!

The total yield variance is the easiest to work out.

If we go to the cost card, we see that 30 total KG should yield one unit of output output. Then we see the actual inputs were a total of 2460, this should yield 82 units (2460/30). But 2460 actual yielded 160 units, so the process was incredibly efficient (suspicously so) and yielded 78 extra units.

But for variance analysis, we need to put a dollar amount on the extra units, which is the standard cost per unit, 640 (10*\$40 + 20*\$12). So the Yield variance is favourable as the inputs yielded more than expected, and the \$ amount is 49,920.

The rest of this is too complicated to write out so please see this spreadsheet with the breakdown.