Techniques of Pricing

In this section, the techniques of computation of unit prices and their application to measured quantities of work are explained. 

Basic Technique 

There are five major components of construction cost, i.e. material, labour, equipment, overhead and profit. For any particular item of work or an element of building, the total cost of a selected or a representative proportion of the total amount of work is divided by number of units of such work. This is the basic technique adopted at the unit prices of different items of work. 

Materials 

The cost of purchase, transport (delivery), taxes and all other factors to acquire sufficient quantities of proper (specified) materials to adequately complete the required work must be determined. Other factors include such things as waste, absorption, losses and accessories such as adhesives, anchors, nails, screws, trim and the like. The total material cost is then divided by the number of units to establish the estimated cost. 

As an example, let us take the item of work "construction of brick masonry built in cement mortar using metric bricks". The materials that constitute this work are metric bricks of standard size, cement and sand (for mortar). The unit of this work is 10 cu m (say), then, in order to arrive at the cost of these materials, we should know how much quantity of these materials go into 10 cu m of finished item of work. 

To calculate this, let us assume a wall 30 cm thick (lf brick) of known dimensions. With the specified bonding arrangement and thickness of bedding and prepared mortar, which is normally 1.0 cm, it is possible to theoretically arrive at the volume of bricks and mortar, as described below : 

Bricks 

Actual size of brick  = 190x90x90mm 

The actual thickness of wall  = 29 cm 

Therefore, actual volume  = 20x0.29x5  = 29cum 

Number of standard bricks of size 20 cm x 10 cm x 10 cm (nominal size) 

                29

=  --------------------  =14500 

       (0.2x0.1x0.1)

                                                                14500

Thus, number of bricks (10 cu m) =      -------------------- x 10 = 4833 nos. 

                                                          ( 20 x 5 x 0.3) 

Allowing about 3% to 4% for breakage, wastage etc, this may be taken as 5000 nos. 

Mortar 

Wet mortar requirement = Total number of brickwork -net volume of bricks 

= 29 - (0.19 x 0.09 x 0.09 x 14500) = 6.685 cu m 

For frog filling, for use of cut bricks, for bonding, for uniform joints  wastage etc., about 15% extra mortar will be required. 

Total actual volume of mortar  = 6.685 x 1.15 = 7.688 cu m 

For dry volume, increase wet volume by 25%. 

Then, dry volume of mortar required = 7.688 x 1.25 = 9.61 cu m 

For 10 cu m of brickwork, dry volume mortar required = 9.61 x 10130 = 3.2 cu m 

In practice, about 3.0 cu m dry cement mortar is taken for pricing. Further, to calculate materials of mortar, it is generally assumed that for every 1.0 cu m of dry mortar, 1.0 cu m of sand is required as cement is assumed to fill the voids in sand. 

For example, for cement mortar of proportion of 1 : 5 (by volume), we require 1 cu m of sand and 115 cu m of cement. As the dry density of cement is taken as 1440 kg/m3, the quantity of cement required is '/5  x 1440 = 288 kg. Therefore, for 3 cu m, we require 3 cu m of sand, and 864 kg (3 x 288 kg) of cement. 

Costing 

Now, cost of materials required for 10 cu m of brick masonry can be computed as the total cost of 5000 bricks, 864 kg of cement and 3 cum of sand. 

Assuming cost of the above materials at work site as below : 

Bricks (1000 nos.) = Rs. 2500, cement, (kg) = Rs. 4, and sand (cu m) = Rs. 50. 

Total cost of materials required for 10 cu m of brick masonry is 

= Rs. 12500.00+ Rs. 3456.00 + Rs. 150.00 = Rs. 16106.00 

Total cost of materials required for 1 cu m of brick masonry = Rs. 1611.00 

Similar procedure can be followed for estimating cost of materials of other elements of building. Help may be taken from standard texts relating to usage of materials.