Tender Penalty Calculations For Delay in Work

yogendra borse

Generally, in Government organizations Penalty clause is included in Tender stating that, if at any stage of work, it is found that the execution of work is not as per the planning (As mentioned in Programme & Progress Schedule) a fine shall be imposed on the contractor as mentioned in clause No.xxxxx of conditions of contract.

Compensation amount is generally equal to one percent or such smaller amount as Superintendent Engineer (whose decision in writing shall be final) may decide of the said estimated cost of the whole work for everyday that the due quantity of work remains incomplete provided always that the total amount of compensation to be paid under the provisions of this clause shall not exceed 10% of the estimated cost of the work as shown in the tender. Chief Engineer should be the final authority

If there is delay in completion of work, as per work completion period in Tender according to conditions Then

Starting Penalty Amount = a

Increment in Penalty = d

Penalty Period = n

Then Total Penalty (C) can be calculated as

$$C = {{n\over 2} \times\left[2a+(n-1) d\right]} ............................(1) $$

Now if the condition for Compensation is that total amount of compensation to be paid under the provisions of this clause shall not exceed 10% of the estimated cost of the work.

if ' A ' is Total estimated amount of Work then

compensation C < = 10% A

Therefore, 

$${\text{10 % A}} ={n\over2} \times {\left [2 a + ( n - 1) d \right]}  ................(2) $$

Now 

The work Executed in Phases, let's assume 4 phases of time period T as below

¼ of the work in ¼T of the time. 

½ of the work in ½T of the time. 

¾ of the work in ¾T of the time. 

Full work in T time............................ months including monsoon.

so, each phase have its own estimated cost

A_T = A_¼T + A_{(½T - ¼T )} + A_(¾T - _½T) + A_(T - _¾T)

Hence,

10% A_T = 10% {A_¼T + A_{(½T - ¼T )} + A_(¾T - _½T) + A_(T - _¾T)}

So, penalty for Phase I, if t₁ days remaining for completion of phase I for starting amount a₁

10% {A_¼T } = (t₁) /2 [2a₁ + (t₁-1)d] 

So, penalty for Phase II, if t₂ days remaining for completion of phase II for starting amount a₂

10% { A_{(½T - ¼T )} } = (t₂) /2 [2a₂ + (t₂-1)d] 

So, penalty for Phase III, if t₃ days remaining for completion of phase III for starting amount a₃

10% { A_(¾T - _½T) } = (t₃) /2 [2a₃ + (t₃-1)d] 

So, penalty for Phase IV, if t4 days remaining for completion of phase IV for starting amount a₄

10% { A_(T - _¾T)} = (t₄) /2 [2a₄ + (t₄-1)d] 

Note:- for constant Value of Penalty difference 'd' can be kept 0.

Any corrections or Suggestions will be appreciated.