# Explain what is Direct Material Mix And Yield Variances

Direct Material MIX Variance |

- Measures the cost of any variation from the standard mix.
- Formula:
[Actual Input Quantity-Budgeted material input quantity for the output produced] x [Standard weighted average cost per input unit-Standard cost per input unit] |

Direct Material YIELD Variance |

· A measure of the cost effect of the difference between the actual material used and the standard material usage for the actual output.· Formula:
[ Budgeted material input quantity for the output produced less actual material input quantity] x [Standard weighted average cost per unit of material input] |

Illustration:
Company A produces Product XYX using 2 materials: material A and material B. After mixing both materials A & B, the product is packaged into airtight cans. The standard mix is 0.6kg of A + 0.6kg of B per 1 kg of Product XYZ. Details: Standard price for material A=$5 per kg; material B=$10 per kg. The weighted average standard price per kg input is $7.50 and The standard material cost of Product XYZ is $9 per kg produced($7.50×1.2) Previous month 2,000 kg of material A and 3,000 kg of material B were used to produce 5,000 1 kg cans of Product XYZ, a saving 1,000 kg compared with the standard input of 3,000 kg of material A and 3,000 kg of material B. Required: Compute: (a) the direct material MIX variance; (b) the direct material YIELD variance. Solution: (a) Direct material MIX variance for each material using the formula: [Actual quantity of material used-standard quantity] x[Weighted average standard price per kg-Standard price per kg of the material used] Material A=[2,000-3000]x[$7.50-$5.00]=$2,500A. Material B=[3,000-3,000]x[$7.50-$5.00]=Nil An adverse variance is due to the actual proportion of the cheaper input, material A being less than standard (b) Direct material YIELD in total [Standard quantity for output produced-actual quantity]X[Weighted average standard price per kg] =[6,000-5000]x$7.50=$7,500F A favorable variance has occurred because the quantity of inputs used was less than the standard required for the level of output. |

Comments are closed now.